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Instructor’s Manual for

Conceptual Physics,Twelfth Edition

The purpose of this manual is to help you combat the all-too-common

notion that a course in physics has to be a course in applied mathematics.

Rather than seeing the equations of physics as lifeless recipes for

plugging in numerical data, your students can be taught to see physics

equations as statements about the connections and relationships in

nature. You can teach them to see that terms in equations are like notes

on a musical score—they say something. Encountering conceptual

physics should be a delightful surprise for your students. When the first

experience with physics is delightful, rigor of the next experience will be

welcomed!

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I n s t r u c t o r ’ s M a n u a l t o C o n c e p t u a l P h y s i c s

Preface ix

Some Teaching Tips xi

On Class Lectures xii

New Ancillary Package for the

12th Edition xv

Flexibility of Material for Various

Course Designs xvii

Chapter Discussions, Lecture

Presentations, Answers and Solutions to

chapters, and Appendix E 1-383

1 About Science 1

Answers 4

P A R T O N E

M e c h a n i c s

2 Newton’s First Law of Motion—

Inertia 6

Answers and Solutions 10

3 Linear Motion 15

Answers and Solutions 20

4 Newton’s Second Law of

Motion 26

Answers and Solutions 30

5 Newton’s Third Law of Motion 37

Answers and Solutions 41

6 Momentum 46

Answers and Solutions 50

7 Energy 58

Answers and Solutions 63

8 Rotational Motion 71

Answers and Solutions 80

9 Gravity 88

Answers and Solutions 96

10 Projectile and Satellite Motion 104

Answers and Solutions 112

P A R T T W O

P r o p e r t i e s o f M a t t e r

11 The Atomic Nature of Matter 120

Answers and Solutions 123

12 Solids 126

Answers and Solutions 129

13 Liquids 137

Answers and Solutions 140

14 Gases 148

Answers and Solutions 154

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P A R T T H R E E

H e a t

15 Temperature, Heat, and

Expansion 161

Answers and Solutions 167

16 Heat Transfer 174

Answers and Solutions 180

17 Change of Phase 185

Answers and Solutions 192

18 Thermodynamics 199

Answers and Solutions 203

P A R T F O U R

S o u n d

19 Waves and Vibrations 209

Answers and Solutions 212

20 Sound 217

Answers and Solutions 221

21 Musical Sounds 227

Chapter 21 Problem Solutions 279

P A R T F I V E

E l e c t r i c i t y a n d M a g n e t i s m

22 Electrostatics 233

Answers and Solutions 238

23 Electric Current 245

Answers and Solutions 251

24 Magnetism 248

Answers and Solutions 252

25 Electromagnetic Induction 257

Answers and Solutions 261

P A R T S I X

L i g h t

26 Properties of Light 267

Answers and Solutions 272

27 Color 279

Answers and Solutions 286

28 Reflection and Refraction 291

Answers and Solutions 297

29 Light Waves 306

Answers and Solutions 310

30 Light Emission 314

Answers and Solutions 318

31 Light Quanta 324

Answers and Solutions 326

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P A R T S E V E N

A t o m i c a n d N u c l e a r P h y s i c s

32 The Atom and the Quantum 332

Answers and Solutions 335

33 Atomic Nucleus and

Radioactivity 339

Answers and Solutions 344

34 Nuclear Fission and Fusion 350

Answers and Solutions 354

P A R T E I G H T

R e l a t i v i t y

35 Special Theory of Relativity 360

Answers and Solutions 368

36 General Relativity 375

Answers and Solutions 377

Appendix E

Exponetial Growth and

Doubling Time 381

Answers to Appendix E 383

ix

Preface

People feel good about themselves when they exceed self-expectations. School is sometimes a

place where we fall below expectations—not only self-expectations, but the expectations of teachers,

family, and the overall community. Physics courses have been notorious in this regard.

Too often, physics has the reputation of being the “killer course“—the course that diminishes

average-ability students, who may drop out or take an incomplete, and spread the word about how

unpleasant it is. Or they hear about it and simply avoid it in the first place. But we physics instructors

have a secret: We know that the concepts of physics for the most part are much more comprehensible

than the public expects. And when that secret is shared with students in a non-intimidating way, one

that prompts them to discover they are learning more than they thought they could, they feel

wonderful—about us, about physics, but more important, about themselves. Because they are not

bogged down with time-consuming mathematical exercises of “the most threatening kind—word

problems,“ they instead get a deeper and wider overview of physics that can be their most enlightened

and positive school experience.

This manual describes a conceptual way of teaching. It helps relate physics to the students’

personal experience in the everyday world, so they learn to see physics not only as a classroom or

laboratory activity, but as a part of everyday living. People with a conceptual understanding of

physics are more alive to the world, just as a botanist taking a stroll through a wooded park is more

alive than most of us to the trees, plants, flora, and the life that teems in them. The richness of life is

not only seeing the world with wide-open eyes, but knowing what to look for. This puts you in a very

nice role—being one who points out the relationships of things in the world about us. You are in an

excellent position to add meaning to your student’s lives.

The appeal of the conceptual approach for nonscience students is obvious. Because conceptual

physics has minimum “mathematical road blocks“ and little or no prerequisites, it is a rare chance for

the nonscience student to learn solid science in a hard-core science course. I say rare chance, because

nonscience students do not have the opportunity to study science as science students have to study the

humanities. Any student, science or humanities, can take an intermediate course in literature, poetry,

or history at any time and in any order. But in no way can a humanities student take an intermediate

physics or chemistry course without first having a foundation in elementary physics and mathematics.

Science has a vertical structure, as noted by the prerequisites. So it is much easier for a science

student to become well rounded in the humanities than for a humanities student to become well

rounded in science. Hence the importance of this conceptual course.

Too often a physics course begins with a study of measurement, units of measure, and vector

notation. I feel this contributes to the unfortunate impression that physics is a dull subject. If you were

being instructed on some computer activity, wouldn’t you object to being shown everything that

might appear much later in your development? Don’t we prefer to be shown something when it is

needed? The same is true with a physics course. Rather than discuss vectors, wait until you’re dealing

with how fast an airplane is blown off course by a crosswind. When the vectors help to learn a topic

your class is immersed in, they are valued. Likewise with so much else in physics.

It is important to distinguish between physics concepts and the tools of physics. Why spend

valuable time teaching a class of nonscience

,46.The inertia of a whole roll resists the large acceleration of a sharp jerk and only a single piece tears. If a

towel is pulled slowly, a small acceleration is demanded of the roll and it unwinds. This is similar to the

hanging ball and string shown in Figure 2.5.

47.Your body tends to remain at rest, in accord with Newton’s first law. The back of the seat pushes you

forward. Without support at the back of your head, your head is not pushed forward with your body,

which likely injures your neck. Hence, headrests are recommended.

48.In a bus at rest your head tends to stay at rest. When the bus is rear-ended, the car lurches forward and

you and your head also move forward. Without headrest your body tends to leave your head behind.

Hence a neck injury.

49.The law of inertia applies in both cases. When the bus slows, you tend to keep moving at the previous

speed and lurch forward. When the bus picks up speed, you tend to keep moving at the previous

(lower) speed and you lurch backward.

50.The maximum resultant occurs when the forces are parallel in the same direction—32 N. The minimum

occurs when they oppose each other—8 N.

51. The vector sum of the forces equals zero. That means the net force must be zero.

52.Vector quantities are force and acceleration. Age and temperature are scalars.

53.You can correctly say the vectors are equal in magnitude and opposite in direction.

54.A hammock stretched tightly has more tension in the supporting ropes than one that sags. The tightly

stretched ropes are more likely to break.

55.The tension will be greater for a small sag. That’s because large vectors in each side of the rope

supporting the bird are needed for a resultant that is equal and opposite to the bird’s weight.

56. By the parallelogram rule, the tension is less than 50 N.

57. The upward force is the tension in the vine. The downward force is that due

to gravity. Both are equal when the monkey hangs in equilibrium.

58.By the parallelogram rule, the tension is greater than 50 N.

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59.No. If only a single nonzero force acts on an object, its motion

will change and it will not be in mechanical equilibrium. There would have to be other forces to result in

a zero net force for equilibrium.

60.At the top of its path (and everywhere else along its path) the force of gravity acts to change the ball’s

motion. Even though it momentarily stops at the top, the net force on the ball is not zero and it therefore

is not in equilibrium.

61.Yes. If the puck moves in a straight line with unchanging speed, the forces of friction are negligible. Then

the net force is practically zero, and the puck can be considered to be in dynamic equilibrium.

62.You can say that no net force acts on your friend at rest, but there may be any number of forces that

act—that produce a zero net force. When the net force is zero, your friend is in static equilibrium.

63. The scale will read half her weight. In this way, the net force (upward pull of left rope + upward pull of

right rope weight) = 0.

64.In the left figure, Harry is supported by two strands of rope that share his weight (like the little girl in the

previous exercise). So each strand supports only 250 N, below the breaking point. Total force up

supplied by ropes equals weight acting downward, giving a net force of zero and no acceleration. In the

right figure, Harry is now supported by one strand, which for Harry's well-being requires that the tension

be 500 N. Since this is above the breaking point of the rope, it breaks. The net force on Harry is then

only his weight, giving him a downward acceleration of g. The sudden return to zero velocity changes

his vacation plans.

65.The upper limit he can lift is a load equal to his weight. Beyond that he leaves the ground!

66.800 N; The pulley simply changes the direction of the applied force.

67. The force that prevents downward acceleration is the support (normal) force—the table pushing up on

the book.

68.Two significant forces act on the book: the force due to gravity and the support force (normal force) of the

table.

69.If the upward force were the only force acting, the book indeed would rise. But another force, that due to

gravity, results in the net force being zero.

70.When standing on a floor, the floor pushes upward against your feet with a force equal to that of gravity,

your weight. This upward force (normal force) and your weight are oppositely directed, and since they

both act on the same body, you, they cancel to produce a net force on you of zero—hence, you are not

accelerated.

71.Only when you are in equilibrium will the support force on you correctly show your weight. Then it is

equal to the force of gravity on you.

72.Without water, the support force is W. With water, the support force is W + w.

73.The friction on the crate has to be 200 N, opposite to your 200-N pull.

74.The friction force is 600 N for constant speed. Only then will F = 0.

75.The support force on the crate decreases as the load against the floor decreases. When the crate is

entirely lifted from the floor, the support force by the floor is zero. The support force on the workmen’s

feet correspondingly increases as the load transfers from the floor to them. When the crate is off the

floor and at rest, its weight is transferred to the men, whose normal force is then increased.

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76. The net force on the rope is zero. The force exerted by the rope on each person is 300 N (in opposite

directions).

77.Two forces must be equal and opposite so that the net force = 0. Then the parachutist is in dynamical

equilibrium.

78.We aren’t swept off because we are traveling just as fast as the Earth, just as in a fast-moving vehicle

you move along with the vehicle. Also, there is no atmosphere through which the Earth moves, which

would do more than blow our hats off!

Think and Discuss

79. Your friend should learn that inertia is not some kind of force that keeps things like the Earth moving,

but is the name given to the property of things to keep on doing what they are doing in the absence of a

force. So your friend should say that nothing is necessary to keep the Earth moving. Interestingly, the

Sun keeps it from following the straight-line path it would take if no forces acted, but it doesn’t keep it

moving. Nothing does. That’s the concept of inertia.

80.You should disagree with your friend. In the absence of external forces, a body at rest tends to remain at

rest; if moving, it tends to remain moving. Inertia is a property of matter to behave this way, not some

kind of force.

81.The tendency of the ball is to remain at rest. From a point of view outside the wagon, the ball stays in

place as the back of the wagon moves toward it. (Because of friction, the ball may roll along the cart

surface—without friction the surface would slide beneath the ball.)

82.The car has no tendency to resume to its original twice-as-fast speed. Instead, in accord with Newton’s

first law, it tends to continue at half speed, decreasing in speed over time due to air resistance and road

friction.

83.No. If there were no friction acting on the cart, it would continue in motion when you stop pushing. But

friction does act, and the cart slows. This doesn’t violate the law of inertia because an external force

indeed acts.

84.An object in motion tends to stay in motion, hence the discs tend to compress upon each other just as the

hammer head is compressed onto the handle in Figure 2.5. This compression results in people being

slightly shorter at the end of the day than in the morning. The discs tend to separate while sleeping in a

prone position, so you regain your full height by morning. This is easily noticed if you find a point you

can almost reach up to in the evening, and then find it is easily reached in the morning.

,Try it and see!

85.No. If there were no force acting on the ball, it would continue in motion without slowing. But air drag

does act, along with slight friction with the lane, and the ball slows. This doesn’t violate the law of inertia

because external forces indeed act.

86.Normal force is greatest when the table surface is horizontal, and progressively decreases as the angle

of tilt increases. As the angle of tilt approaches 90°, the normal force approaches zero. When the table

surface is vertical, it no longer presses on the book, then freely falls.

87.No. The normal force would be the same whether the book was on slippery ice or sandpaper. Friction

plays no role unless the book slides or tends to slide along the table surface.

88.A stone will fall vertically if released from rest. If the stone is dropped from the top of the mast of a

moving ship, the horizontal motion is not changed when the stone is dropped—providing air resistance

on the stone is negligible and the ship’s motion is steady and straight. From the frame of reference of

the moving ship, the stone falls in a vertical straight-line path, landing at the base of the mast.

89.A body in motion tends to remain in motion, so you move with the moving Earth whether or not your feet

are in contact with it. When you jump, your horizontal motion matches that of the Earth and you travel

with it. Hence the wall does not slam into you.

90.The coin is moving along with you when you toss it. While in the air it maintains this forward motion, so

the coin lands in your hand. If the train slows while the coin is in the air, it will land in front of you.

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91. If the train rounds a corner while the coin is in the air, it will land off to the side of you. The coin continues

in its horizontal motion, in accord with the law of inertia.

92.This is similar to Question 88. If the ball is shot while the train is moving at constant velocity (constant

speed in a straight line), its horizontal motion before, during, and after being fired is the same as that of

the train; so the ball falls back into the smokestack as it would have if the train were at rest. If the train

increases its speed, the ball will hit the train behind the smokestack because the ball’s horizontal speed

continues unchanged after it is fired, but the speeding-up train pulls ahead of the ball. Similarly, on a

circular track the ball will also miss the smokestack because the ball will move along a tangent to the

track while the train turns away from this tangent. So the ball returns to the smokestack in the first case,

and misses in the second and third cases because of the change in motion.

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3 Linear Motion

Conceptual Physics Instructor’s Manual, 12th Edition

3.1 Motion Is Relative

3.2 Speed

Instantaneous Speed

Average Speed

3.3 Velocity

Constant Velocity

Changing Velocity

3.4 Acceleration

Acceleration on Galileo’s Inclined Planes

3.5 Free Fall

How Fast

How Far

Hang Time

How Quickly “How Fast” Changes

3.6 Velocity Vectors

The photo openers begin with my niece, Joan Lucas, riding her horse Ghost. The second photo is of my

cherished friend from school days, Sue Johnson (wife of Dan Johnson) who with her racing-shell team won

national honors in rowing. Shown also is Norwegian friend Carl Angell who rolls a ball through a photo

timer. Also shown is friend and colleague from City College of San Francisco, Chelcie Liu, showing the

tracks he made while teaching his daughter Cindy some physics. The tracks have been and are well used.

This chapter opens with a profile on Galileo.

TAKE CARE NOT TO SPEND OVERTIME ON THIS CHAPTER!! Doing so is the greatest pacing

mistake in teaching physics! Time spent on kinematics is time not spent on why satellites continually fall

without touching Earth, why high temperatures and high voltages (for the same reason) can be safe to

touch, why rainbows are round, why the sky is blue, and how nuclear reactions keep the Earth’s interior

molten. Too much time on this chapter is folly. I strongly suggest making the distinction between speed,

velocity, and acceleration, and move quickly to Chapter 4. (I typically spend only one class lecture on this

chapter.) By all means, avoid the temptation to do the classic motion problems that involve 90% math

and 10% physics! Too much treatment of motion analysis can be counterproductive to maintaining the

interest in physics starting with the previous chapter. I suggest you tell your class that you’re skimming the

chapter so you’ll have more time for more interesting topics in your course—let them know they shouldn’t

expect to master this material, and that mastery will be expected in later material (that doesn’t have the

stumbling blocks of kinematics). It’s okay not to fully understand this early part of your course. Just as

wisdom is knowing what to overlook, good teaching is knowing what to omit.

Perchance you are getting into more problem solving than is customary in a conceptual course, be sure to

look at the student ancillary, Problem Solving Book, 3rd Edition. It has ample problems for a lightweight

algebra-trigonometry physics course.

The box on Hang Time on page 50 may be especially intriguing to your students if they’re unaware of the

short time involved. Even basketball legend Michael Jordan’s hang time was less than 0.9 s. Height jumped

is less than 1.25 m (4 feet—those who insist a hang time of 2 s are way off, for 1 s up is 16 feet—clearly,

no way!). A neat rule of thumb is that height jumped in feet, where g = 32 ft/s2, is equal to four times hang

time squared [d = g2T22 = g2T24 = 328T2 = 4T2].

I feel compelled to interject here (as I mean to stress all through this manual) the importance of the “check

with your neighbor” technique of teaching. Please do not spend your lecture talking to yourself in front of

your class! The procedure of “check with your neighbor” is a routine that keeps you and your class

engaged. I can’t stress enough its importance!

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The distinction between velocity and acceleration is prerequisite to the following chapters on mechanics.

The Practicing Physics Book of worksheets treats the distinction between velocity acquired and distance

fallen for free fall via a freely-falling speedometer-odometer. Students do learn from these, in class or out

of class, so whether you have your students buy their own from your bookstore or you photocopy select

pages for class distribution, get these to your students. There are four Practice Pages for this chapter:

• Free Fall Speed •Hang Time

• Acceleration of Free Fall • Non-Accelerated Motion

Problem Solving Book: Chapter 3 has abundant and insightful kinematics problems requiring straight-

forward algebra, some with solutions.

Laboratory Manual:

• Go! Go! Go! The Fundamentals of Graphing Motion (Experiment)

• Sonic Ranger Graphing Motion in Real Time (Tech Lab)

• Motivating the Moving Man Motion Graphing Simulation (Tech Lab)

The textbook does not treat motion graphically, but leaves that to the laboratory manual. Labs are

enhanced with the sonic ranger device, which is conceptual graphing at its best. If not done as lab

experiments, demonstrate the sonic ranger as part of your lecture.

Next-Time Questions (in the Instructor Resource DVD):

• Relative Speeds

• Bikes and Bee

Hewitt-Drew-It! Screencasts: (All accessed via QR code in the text)

•Free Fall •Sideways Drop

•Ball Toss • Bikes and Bee

•Velocity Vectors

SUGGESTED LECTURE PRESENTATION

Your first question: What means of motion has done more to change the way cities are built than any other?

[Answer: The elevator!]

Explain the importance of simplifying. Motion is best understood if you first neglect the effects of air

resistance, the effects of buoyancy, spin, and the shape of moving objects—that

,beneath these are simple

relationships that might otherwise be masked by “covering all bases,” and that these relationships are what

Chapter 3 and your lecture are about. State that by completely neglecting the effects of air resistance not

only exposes the simple relationships, but is a reasonable assumption for heavy and compact (dense)

objects traveling at moderate speeds; e.g., one would notice no difference between the rates of fall of a

heavy rock dropped from the classroom ceiling to the floor below, when falling through either air or a

complete vacuum. For a feather and heavy objects moving at high speeds, air resistance does become

important, and will be treated in Chapter 4.

Mention that there are few pure examples in physics, for most real situations involve a combination of

effects. There is usually a “first order” effect that is basic to the situation, but then there are 2nd, 3rd, and

even 4th or more order effects that interact also. If we begin our study of some concept by considering all

effects together before we have studied their contributions separately, understanding is likely to be difficult.

To have a better understanding of what is going on, we strip a situation of all but the first order effect, and

then examine that. When that is well understood, then we proceed to investigate the other effects for a

fuller understanding.

DEMONSTRATION: Drop a sheet of paper and note how slowly it falls because of air resistance.

Crumple the paper and note it falls faster. Air resistance has been reduced. Then drop a sheet of paper

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and a book, side by side. Of course the book falls faster, due to its greater weight compared to air

resistance. (Interestingly, the air drag is greater for the faster-falling book—an idea you’ll return to in

the next chapter.) Now place the paper against the lower surface of the raised horizontally-held book

and when you drop them, nobody is surprised to see they fall together. The book has pushed the paper

with it. Now repeat with the paper on top of the book and ask for predictions and neighbor discussion.

Then surprise your class by refusing to show it! Tell them to try it out of class! (Good teaching isn’t

giving answers, but raising good questions—good enough to prompt wondering. Let students discover

that the book will “plow through the air” leaving an air-resistance free path for the paper to follow!)

Air resistance will be treated in later chapters, but not this one. Again, simplifying brings out the concepts

better. You can briefly acknowledge the important effects of air drag: In a bicycle race, for example, the

lead cyclist carries along a flow of air that creates a “sweet spot” of low air pressure for the cyclist riding

close behind. Air resistance on spinning balls changes their course, and so on. In keeping with the adage

“Wisdom is knowing what to overlook,” we neglect the effects of air in this chapter to more clearly reveal

the connections between distance, time, speed, velocity, and acceleration. Let your students know that the

effects of air drag are treated in future chapters.

Speed and Velocity

Define speed by writing its equation in longhand form on the board while giving examples—automobile

speedometers, etc. Similarly define velocity, citing how a race car driver is interested in his speed, whereas

an airplane pilot is interested in her velocity—speed and direction. Cite the difference between a scalar and

a vector quantity and identify speed as a scalar and velocity as a vector. Tell your class that you’re not

going to make a big deal about distinguishing between speed and velocity, but you are going to make a big

deal of distinguishing between velocity and another concept—acceleration.

Acceleration

Define acceleration by identifying it as a vector quantity, and cite the importance of CHANGE. That’s

change in speed, or change in direction. Hence both are acknowledged by defining acceleration as a rate of

change in velocity rather than speed. Ask your students to identify the three controls in an automobile that

make the auto change its state of motion—that produce acceleration. Ask for them (accelerator, brakes,

and steering wheel). State how one lurches in a vehicle that is undergoing acceleration, especially for

circular motion, and state why the definition of velocity includes direction to make the definition of

acceleration all-encompassing. Talk of how without lurching one cannot sense motion, giving examples of

coin flipping in a high-speed aircraft versus doing the same when the same aircraft is at rest.

Units for Acceleration: Give numerical examples of acceleration in units of kilometershour per second to

establish the idea of acceleration. Be sure that your students are working on the examples with you. For

example, ask them to find the acceleration of a car that goes from rest to 100 kmh in 10 seconds. It is

important that you not use examples involving seconds twice until they taste success with the easier

kilometershour per second examples. Have them check their work with their neighbors as you go along.

Only after they get the hang of it, introduce meterssecondsecond in your examples to develop a sense for

the units ms2. This is treated in the screencasts, Unit Conversion and Acceleration Units.

Falling Objects: If you round 9.8 ms2 to 10 ms2 in your lecture, you’ll more easily establish the

relationships between velocity and distance. In lab you can use the more precise 9.8 ms2.

CHECK QUESTION: If an object is dropped from an initial position of rest from the top of a cliff,

how fast will it be traveling at the end of one second? (You might add, “Write the answer on your

notepaper.” And then, “Look at your neighbor’s paper—if your neighbor doesn’t have the right

answer, reach over and help him or her—talk about it.” And then possibly, “If your neighbor isn’t very

cooperative, sit somewhere else next time!”)

After explaining the answer when class discussion dies down, repeat the process asking for the

speed at the end of 2 seconds, and then for 10 seconds. This leads you into stating the relationship

v = gt, which by now you can express in shorthand notation. After any questions, discussion, and

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examples, state that you are going to pose a different question—not asking of how fast, but for

how far. Ask how far the object falls in one second. Ask for a written response and then ask if the

students could explain to their neighbors why the distance is only 5 m rather than 10 m. After

they’ve discussed this for almost a minute or so, ask “If you maintain a speed of 60 kmh for one

hour, how far do you go?”—then, “If you maintain a speed of 10 ms for one second, how far do

you go?” Important point: You’ll appreciably improve your instruction if you allow some thinking

time after you ask a question. Not doing so is the folly of too many instructors. Then continue,

“Then why is the answer to the first question not 10 meters?” After a suitable time, stress the idea

of average velocity and the relation d = vavet.

Show the general case by deriving on the board d = 12gt2. (We tell our students that the derivation is a

sidelight to the course—something that will be the crux of a follow-up physics course. In any event, the

derivation is not something that we expect of them, but to show that d = 12gt2 is a reasoned statement that

doesn’t just pop up from nowhere.)

CHECK QUESTIONS: How far will a freely falling object that is released from rest, fall in 2 s?

In 10 s? (When your class is comfortable with this, then ask how far in 12 second.)

To avoid information overload, we restrict all numerical examples of free fall to cases that begin at rest.

Why? Because it’s simpler that way. (We prefer our students to understand simple physics than to be

confused about not-so-simple physics!) We do go this far with them.

CHECK QUESTION: Consider a rifle fired straight downward from a high-altitude

,balloon. If the

muzzle velocity is 100 ms and air resistance can be neglected, what is the acceleration of the bullet

after one second? (If most of your class say that it’s g, you’re on!)

I suggest not asking for the time of fall for a freely-falling object, given the distance. Why? Unless the

distance given is the familiar 5 meters, algebraic manipulation is called for. If one of our teaching

objectives were to teach algebra, this would be a nice place to do it. But we don’t have time to present this

stumbling block and then teach how to overcome it. We’d rather put our energy and theirs into straight

physics!

Kinematics can be rich with puzzles, graphical analysis, ticker timers, photogates, and algebraic problems.

My strong suggestion is to resist these and move quickly into the rest of mechanics, and then into other

interesting areas of physics. Getting bogged down with kinematics, with so much physics ahead, is a

widespread practice. Please do your class a favor and hurry on to the next chapters. If at the end of your

course you have time (ha-ha), then bring out the kinematics toys and have a go at them.

The Two-Track Demo

Be sure to fashion a pair of tracks like those shown by Chelcie Liu

in the chapter opener photo. Chelcie simply bent a pair of angle iron

used as bookcase supports. The tracks are of equal length and can

be bent easily with a vice. Think and Discuss 95 and 96 refer to this

demo. Be prepared for the majority of your class to say they reach

the end of the track at the same time. Aha, they figure they have the same speed at the end, which throws

them off base. Same speed does not mean same time. I like to quip “Which will win the race, the fast ball or

the slower ball?” You can return to this demo when you discuss energy in Chapter 7.

Hang Time

As strange as it first may seem, the longest time a jumper can remain in air is less than a second. It is a

common illusion that jumping times are more. Even Michael Jordan’s best hang time (the time the feet are

off the ground) was 0.9 second. Then d = 12gt2 predicts how high a jumper can go vertically. For a hang

time of a full second, that’s 12 s up and 12 s down. Substituting, d = 50.52 = 1.25 m (which is about 4

feet!). So the great athletes and ballet dancers jump vertically no more than 4 feet high! Of course one can

clear a higher fence or bar; but one’s center of gravity cannot be raised more than 4 feet in free jumping. In

19

fact very few people can jump 2 feet high! To test this, stand against a wall with arms upstretched. Mark

the wall at the highest point. Then jump, and at the top, again mark the wall. For a human being, the

distance between marks is at most 4 feet! We’ll return to hang time for running jumps when we discuss

projectile motion in Chapter 10.

NEXT-TIME QUESTION: For OHT or posting. Note the sample reduced pages from the Next-

Time Questions book, full 8-12 11, just right for OHTs. Next-Time Questions for each chapter

are on your Resource DVD, each with its answer. Consider displaying printed NTQs at some

general area outside the classroom—perhaps in a glass case. This display generates general student

interest, as students in your class and those not in your class are stimulated to think physics. After

a few days of posting then turn the sheets over to reveal the answers. That’s when new NTQs can

be displayed. How better to adorn your corridors! Because of space limitations, those for other

chapters are not shown in this manual. [I’m not showing the answer here, but it makes the point to

solving this problem is consideration of time t. Whether or not one thinks about time should not be

a matter of cleverness or good insight, but a matter of letting the equation for distance guide

thinking. The v is given, but the time t is not. The equation instructs you to consider time.

Equations are important in guiding our thinking about physics.]

Velocity Vectors

Consider Figure 3.12 of the airplane flying in a cross wind. The resulting speed can only be found with

vectors. The only vector tools the student needs is the parallelogram rule, and perhaps the Pythagorean

Theorem. Avoid sines and cosines unless your students are studying to be scientists or engineers. Here we

distinguish between physics and the tools of physics. Tools for pre-engineers and scientists only. Physics

for everybody!

This is a good time for the Hewitt-Drew-It Screencast Velocity Vectors, which treats an airplane flying at

different directions relative to the wind.

20

Answers and Solutions for Chapter 3

Reading Check Questions

1. Relative to the chair your speed is zero. Relative to the Sun it’s 30 km/s.

2. The two necessary units are distance traveled and time of travel.

3. A speedometer registers instantaneous speed.

4. Average speed is 30 km/min.

5. The horse travels 25 km/h x 0.5 h = 12.5 km.

6. Speed is a scalar and velocity is speed and direction, a vector.

7. Yes. A car moving at constant velocity moves at constant speed.

8. The car maintains a constant speed but not a constant velocity because it changes direction as it rounds

the corner.

9. The acceleration is 10 km/h/s.

10. Acceleration is zero, because velocity doesn’t change.

11. You are aware of changes in your speed, but not steady motion. Therefore you are aware of

acceleration, but not constant velocity.

12. When motion is in one direction along a straight line, either may be used.

13. Galileo found that the ball gained the same amount of speed each second, which says the acceleration

is constant.

14. Galileo discovered that the greater the angle of incline, the greater the acceleration. When the incline is

vertical, the acceleration is that of free fall.

15. A freely-falling object is one on which the only force acting is the force of gravity. This means falling with

no air resistance.

16. Gain in speed is 10 m/s each second.

17. The speed acquired in 5 seconds is 50 m/s; in 6 seconds, 60 m/s.

18. The unit ‘seconds’ occurs in velocity, and again in the time velocity is divided by to compute

acceleration. Hence the square of seconds in acceleration.

19. The moving object loses 10 m/s for each second moving upward.

20. Galileo found distance traveled is directly proportional to the square of the time of travel

(d = ½ gt2).

21. The distance of fall in 1 second is 5 m. For a 4-s drop, falling distance is 80 m.

22. Air resistance reduces falling acceleration.

23. 10 m/s is speed, 10 m is distance, and 10 m/s2 is acceleration.

24. The resultant is 141 km/h at an angle of 45° to the airplane’s intended direction of travel.

Think and Do

25. Tell Grandma that, in effect, velocity is how fast you’re traveling and acceleration is how quickly how-fast

changes. Please don’t tell Grandma that velocity is speed!

26. This can be a social activity, with good physics.

27. This is a follow-up to the previous activity, a good one!

28. Hang times will vary, but won’t exceed 1 second!

Plug and Chug

29.

Average speed =

distance traveled

time

d

t

30 m

2 s

15 m/s .

30.

Average speed =

d

t

1.0 m

0 .5 s

2 m/s

31.

Acceleration

change in velocity

time

v

t

100 km /h

10 s

10 km/h s

32.

Acceleration

change in velocity

time

v

t

10 m/s

2 s

5 m/s

2

.

33. Starting from rest, distance = ½ at2 = ½ (5 m/s2)(3 s)2 = 22.5 m.

34. Distance of fall = ½ gt2 = ½ (10 m/s2)(3 s)2 = 45 m.

21

Think and Solve

35. Since it starts going up at 30 ms and loses 10 ms each second, its time going up is 3 seconds.

Its time returning is also 3 seconds, so it’s in the air for a total of 6 seconds. Distance up (or down) is ½

gt2 = 5 32 = 45 m. Or from d = vt, where average velocity is (30 + 0)2 = 15 ms, and time is 3

seconds, we also get d = 15 ms 3 s = 45 m.

36. (a) The velocity of the ball at the top of its

,vertical trajectory is instantaneously zero.

(b) Once second before reaching its top, its velocity is 10 ms.

(c) The amount of change in velocity is 10 ms during this 1-second interval (or any other 1-second

interval).

(d) One second after reaching its top, its velocity is -10 ms—equal in magnitude but oppositely

directed to its value 1 second before reaching the top.

(e) The amount of change in velocity during this (or any) 1-second interval is 10 ms.

(f) In 2 seconds, the amount of change in velocity, from 10 ms up to 10 ms down, is 20 ms (not zero!).

(g) The acceleration of the ball is 10 ms2 before reaching the top, when reaching the top, and after

reaching the top. In all cases acceleration is downward, toward the Earth.

37.Using g = 10 m s2, we see that v = gt = (10 ms2)(10 s) = 100 ms;

vav =

(vbeginning + vfinal)

2 =

(0 + 100)

2 = 50 m/s, downward.

We can get “how far” from either d = vavt = (50 ms)(10 s) = 500 m, or equivalently,

d = ½ gt2 = 5(10)2 = 500 m. (How nice we get the same distance using either formula!)

38.

a

v

t

25 m/s - 0

10 m/s

2 .5 m/s

2

.

39.From d = ½ gt2 = 5t2, t = d5 = 0.65 = 0.35 s. Doubling for a hang time of 0.7 s.

40. a.

t ? From v =

d

t

t

d

v

L

(v f v0 ) / 2

2L

v

.

b.

t

2L

v

2(1.4 m)

15.0 m/s

0.19s.

Think and Rank

41. D, C, A, B

42. C, B=D, A

43. a. B, A=C

b. A, B, C

c. C, B, A

44. a. C, B, A

b. A=B=C (10 m/s2)

45. B, A, C

46. a. B, A, C

b. C, B, A

Think and Explain

47. The shorter the better, so Mo has the more favorable reaction time and can respond quicker to situations

than Jo can.

48. Jo travels 1.2 m during the time between seeing the emergency and applying the brakes.

d = vt = 6 m/s x 0.20 s = 1.2 m.

49. The impact speed will be the relative speed, 2 kmh (100 kmh 98 kmh = 2 kmh).

50.She’ll be unsuccessful. Her velocity relative to the shore is zero (8 kmh 8 kmh = 0).

51.Your fine for speeding is based on your instantaneous speed; the speed registered on a speedometer or

a radar gun.

22

52.The speeds of both are exactly the same, but the velocities are not. Velocity includes direction, and since

the directions of the airplanes are opposite, their velocities are opposite. The velocities would be equal

only if both speed and direction were the same.

53.Constant velocity means no acceleration, so the acceleration of light is zero.

54.The car approaches you at twice the speed limit.

55.(a) Yes, because of the change of direction. (b) Yes, because velocity changes.

56.Emily is correct. Jacob is describing speed. Acceleration is the time rate of change in speed—“how fast

you get fast,” as Emily asserts.

57.No. You cannot say which car underwent the greater acceleration unless you know the times involved.

58.The acceleration is zero, for no change in velocity occurs. Whenever the change in velocity is zero, the

acceleration is zero. If the velocity is “steady,” “constant,” or “uniform,” the change in velocity is zero.

Remember the definition of acceleration!

59.The greater change in speeds occurs for (30 kmh 25 kmh = 5 kmh), which is greater than (100 kmh

96 kmh = 4 kmh). So for the same time, the slower one has the greater acceleration.

60.At 0 the acceleration is zero. At 90 the acceleration is that of free fall, g. So the range of accelerations

is 0 to g, or 0 to 10 ms2.

61.Its speed readings would increase by 10 ms each second.

62.Distance readings would indicate greater distances fallen in successive seconds. During each successive

second the object falls faster and covers greater distance.

63.The acceleration of free fall at the end of the 5th, 10th, or any number of seconds is g. Its velocity has

different values at different times, but since it is free from the effects of air resistance, its acceleration

remains a constant g.

64.In the absence of air resistance, the acceleration will be g no matter how the ball is released. The

acceleration of a ball and its speed are entirely different.

65. Whether up or down, the rate of change of speed with respect to time is 10 ms2, so each second while

going up the speed decreases by 10 ms. Coming down, the speed increases by 10 ms each second.

So with no air resistance, the time rising equals the time falling.

66. With no air resistance, both will strike the ground below at the same speed. Note that the ball thrown

upward will pass its starting point on the way down with the same speed it had when starting up. So its

trip on downward, below the starting point, is the same as for a ball thrown down with that speed.

67.When air resistance affects motion, the ball thrown upward returns to its starting level with less speed

than its initial speed; and also less speed than the ball tossed downward. So the downward thrown ball

hits the ground below with a greater speed.

68.Counting to twenty means twice the time. In twice the time the ball will roll 4 times as far (distance moved

is proportional to the square of the time).

69.The acceleration due to gravity remains a constant g at all points along its path as long as no other forces

like air drag act on the projectile.

70.Time (in seconds) Velocity (in meters/second) Distance (in meters)

0 0 0

1 10 5

2 20 20

3 30 45

4 40 80

5 50 125

23

6 60 180

7 70 245

8 80 320

9 90 405

10 100 500

71. If it were not for the slowing effect of the air, raindrops would strike the ground with the speed of high-

speed bullets!

72.No, free-fall acceleration is constant, which accounts for the constant increase of falling speed.

73.Air drag decreases speed. So a tossed ball will return with less speed than it possessed initially.

74. On the Moon the acceleration due to gravity is considerably less, so hang time would be considerably

more (six times more for the same takeoff speed!).

75.As water falls it picks up speed. Since the same amount of water issues from the faucet each second, it

stretches out as distance increases. It becomes thinner just as taffy that is stretched gets thinner the

more it is stretched. When the water is stretched too far, it breaks up into droplets.

76.The speed of falling rain and the speed of the automobile are the same.

77. Open ended.

78. Open ended.

Think and Discuss

79. Yes. Velocity and acceleration need not be in the same direction. A car moving north that slows down,

for example, accelerates toward the south.

80.Yes, again, velocity and acceleration need not be in the same direction. A ball tossed upward, for

example, reverses its direction of travel at its highest point while its acceleration g, directed downward,

remains constant (this idea will be explained further in Chapter 4). Note that if a ball had zero

acceleration at a point where its speed is zero, its speed would remain zero. It would sit still at the top

of its trajectory!

81. Acceleration occurs when the speedometer reading changes. No change, no acceleration.

82.“The dragster rounded the curve at a constant speed of 100 km/h.” Constant velocity means not only

constant speed but constant direction. A car rounding a curve changes its direction of motion.

83.Any object moving in a circle or along a curve is changing its velocity (accelerating) even if its speed is

constant, because direction is changing. Something with constant velocity has both constant direction

and constant speed, so there is no example of motion with constant velocity and varying speed.

84.A vertically-thrown ball has zero speed at the top of its trajectory, but acceleration there is g.

85.An object moving in a circular path at constant speed is a simple example of acceleration at constant

speed because its velocity

,is changing direction. No example can be given for the second case,

because constant velocity means zero acceleration. You can’t have a nonzero acceleration while

having a constant velocity. There are no examples of things that accelerate while not accelerating.

86.(a) Yes. For example, an object sliding or rolling horizontally on a frictionless plane. (b) Yes. For

example, a vertically thrown ball at the top of its trajectory.

87.The acceleration of an object is in a direction opposite to its velocity when velocity is decreasing—for

example, a ball rising or a car braking to a stop.

88.Only on the middle hill does speed along the path decrease with time, for the hill becomes less steep as

motion progresses. When the hill levels off, acceleration will be zero. On the left hill, acceleration is

constant. On the right hill, acceleration increases as the hill becomes steeper. In all three cases, speed

increases.

24

89.The one in the middle. The ball gains speed more quickly at the beginning where the slope is steeper, so

its average speed is greater even though it has less acceleration in the last part of its trip.

90.Free fall is defined as falling only under the influence of gravity, with no air resistance or other non-

gravitational forces. So your friend should omit “free” and say something like, “Air resistance is more

effective in slowing a falling feather than a falling coin.”

91. If air resistance is not a factor, an object’s acceleration is the same 10 ms2 regardless of its initial

velocity. If it is thrown downward, its velocity will be greater, but not its acceleration.

92. Its acceleration would actually be less if the air resistance it encounters at high speed retards its

motion. (We will treat this concept in detail in Chapter 4.)

93. When acceleration of the car is in a direction opposite to its velocity, the car is “decelerating,” slowing

down.

94. In the absence of air resistance both accelerations are g, the same. Their velocities may be in opposite

directions, but g is the same for both.

95.The ball on B finishes first, for its average speed along the lower part as well as the down and up slopes

is greater than the average speed of the ball along track A.

96.(a) Average speed is greater for the ball on track B.

(b) The instantaneous speed at the ends of the tracks is the same because the speed gained on the

down-ramp for B is equal to the speed lost on the up-ramp side. (Many people get the wrong answer

for the previous question because they assume that because the balls end up with the same speed

that they roll for the same time. Not so.)

97. The resultant speed is indeed 5 m/s. The resultant of any pair of 3-unit and 4-unit vectors at right angles

to each other is 5 units. This is confirmed by the Pythagorean theorem;

a2 + b2 = c2 gives 32 + 42 = 52. (Or, √[32 + 42] = 5.)

98. Again, from the Pythagorean theorem; a2 + b2 = c2 gives 32 + 42 = 52. (Or, √[32 + 42] = 5.) So the boat

has a speed of 5 m/s.

99. Again, from the Pythagorean theorem; a2 + b2 = c2 gives 1202 + 902 = 1502.

(Or, √[1202 + 902] = 150.) So the groundspeed is 150 km/h.

100.How you respond may or may not agree with the author’s response: There are few pure examples in

physics, for most real situations involve a combination of effects. There is usually a “first order” effect

that is basic to the situation, but then there are 2nd, 3rd, and even 4th or more order effects that interact

also. If we begin our study of some concept by considering all effects together before we have studied

their contributions separately, understanding is likely to be difficult. To have a better understanding of

what is going on, we strip a situation of all but the first order effect, and then examine that. When we

have a good understanding of that, then we proceed to investigate the other effects for a fuller

understanding. Consider Kepler, for example, who made the stunning discovery that planets move in

elliptical paths. Now we know that they don’t quite move in perfect ellipses because each planet affects

the motion of every other one. But if Kepler had been stopped by these second-order effects, he would

not have made his groundbreaking discovery. Similarly, if Galileo hadn’t been able to free his thinking

from real-world friction he may not have made his great discoveries in mechanics.

4 Newton’s Second Law of Motion

Conceptual Physics Instructor’s Manual, 12th Edition

4.1 Force Causes Acceleration

4.2 Friction

4.3 Mass and Weight

Mass Resists Acceleration

4.4 Force, Mass, and Acceleration

4.5 Newton’s Second Law of Motion

4.6 When Acceleration Is g—Free Fall

4.7 When Acceleration Is Less Than g—Nonfree Fall

Jill Johnsen of CCSF demonstrates ball drops in the opening photo of this chapter. Efrain Lopez, formerly

from CCSF and now at California State University at Hayward, demonstrates equilibrium. Regarding the

wingsuit skydiver in the center opening photo, I’m puzzled at the late date of this version of human flight.

First we went to the Moon, then we discovered hang gliding, then bungee jumping, then maneuverable

parachuting, and now, lastly, humans are doing what flying squirrels have been doing for eons! The order

simply doesn’t make sense! The bottom photo is my granddaughter Emily at soccer practice.

The personal profile features Isaac Newton.

Inertia, acceleration, and falling objects as introduced in Chapters 2 and 3, and are further developed in this

chapter. Here we distinguish between mass and weight without making a big deal about their units of

measurement (because I think time is better spent on physics concepts). A brief treatment of units and

systems of measurement is provided in Appendix A.

Practicing Physics Book:

• Mass and Weight • Cart

• Converting Mass to Weight • Force and Acceleration

• A Day at the Races with a = F/m • Friction

• Dropping Masses and Accelerating • Falling and Air Resistance

• Force-Vector Diagrams

Problem Solving Book:

More than 100 problems complement this chapter!

Laboratory Manual:

• The Weight Mass and Weight (Activity)

• Putting the Force Before the Cart Force, Mass, and Acceleration (Activity)

• Reaction Time Free Fall (Activity)

• The Newtonian Shot Force and Motion Puzzle (Activity)

Next-Time Questions (in the Instructor Resource DVD):

• Skidding Truck • Book Push Against the Wall

• Spool Pull • Acceleration at the Top

• Falling Balls • Net Force Half-Way Up

• Skydiver • Acceleration on the Way Up

• Truck and Car Collision • Balanced Scale

• Block Pull • Galileo

• Direction of Friction

Hewitt-Drew-It! Screencasts:

•Mass/Weight •Acceleration Units

•Newton’s Second Law •Skydiver Problem

SUGGESTED LECTURE PRESENTATION

27

In Chapter 2 the concept of inertia was introduced—the notion that once an object is in motion, it will

continue in motion if no forces are exerted on it. Moving things tend to remain moving at constant velocity.

In the previous chapter we learned about acceleration—the change in velocity that objects experience when

a force is exerted. In this chapter we’ll treat the relationship between force and acceleration.

Friction

Drag a block at constant velocity across your lecture table. Acknowledge the force of friction, and how it

must exactly counter your pulling force. Show that pulling force with a spring balance. Now since the block

moves without accelerating, ask for the magnitude of the friction force. It must be equal and opposite to

your scale reading. Then the net force is zero. While sliding the block is in dynamic equilibrium. That is,

F = 0.

CHECK QUESTIONS: (similar to one in the text.) Suppose in a high-flying airplane the captain

announces over the cabin public address system that the plane is flying at a constant 900 kmh and

the thrust of the engines is a constant

,80,000 newtons. What is the acceleration of the airplane?

[Answer: Zero, because velocity is constant.] What is the combined force of air resistance that acts

all over the plane’s outside surface? [Answer: 80,000 N. If it were less, the plane would speed up;

if it were more, the plane would slow down.]

Continue your activity of pulling the block across the table with a spring balance. Show what happens

when you pull harder. Your students see that when the pulling force is greater than the friction force, there

is a net force greater than zero, as evidenced by the observed acceleration. Show different constant speeds

across the table with the same applied force, which shows that friction is not dependent on speed.

Distinguish between static and sliding friction, and show how a greater force is needed to get the block

moving from a rest position. Show all this as you discuss these ideas. Cite the example in the book about

skidding with locked brakes in a car [where the distance of skid for sliding friction is greater than static

friction, where lower braking application results in nonsliding tires and shorter sliding distance]. Discuss

the new automatic braking systems (ABS) of cars.

Friction in the Practicing Physics Book nicely treats details of friction.

After you have adequately discussed friction and net force, pose the following (Be careful that your class

may not be ready for this, in which case you may confuse rather than enlighten.):

Mass and Weight

To distinguish between mass and weight compare the efforts of pushing horizontally on a block of slippery

ice on a frozen pond versus lifting it. Or consider the weightlessness of a massive anvil in outer space and

how it would be difficult to shake, weight or no weight. And if moving toward you, it would be harmful to

be in its way because of its great tendency to remain in motion. The following demo (often used to

illustrate impulse and momentum) makes the distinction nicely:

DEMONSTRATION: Hang a massive ball by a string and show that the top string

breaks when the bottom is pulled with gradually more force, but the bottom string

breaks when the string is jerked. Ask which of these cases illustrates weight.

[Interestingly enough, it’s the weight of the ball that makes for the greater tension

in the top string.] Then ask which of these cases illustrates inertia. [When jerked,

the tendency of the ball to resist the sudden downward acceleration, its inertia, is

responsible for the lower string breaking.] This is the best demo I know of for

showing the different effects of weight and mass.

Mass Resists Acceleration: The property of massive objects to resist changes is nicely shown with this

follow-up demonstration.

DEMONSTRATION: Lie on your back and have an assistant place a blacksmith’s anvil on your

stomach. Have the assistant strike the anvil rather hard with a sledge hammer. The principles here

are the same as the ball and string demo. Both the inertia of the ball and the inertia of the anvil

28

resist the changes in motion they would otherwise undergo. So the string doesn’t break, and your

body is not squashed. (Be sure that your assistant is good with the hammer. When I began

teaching I used to trust students to the task. In my fourth year the student who volunteered was

extra nervous in front of the class and missed the anvil entirely—but not me. The hammer

smashed into my hand breaking two fingers. I was lucky I was not harmed more.)

Relate the ideas of tightening a hammer head by slamming the opposite end of the handle

on a firm surface, with the bones of the human spine after jogging or even walking

around. Interestingly, we are similarly a bit shorter at night. Ask your students to find a

place in their homes that they can’t quite reach before going to bed—a place that is one or

two centimeters higher than their reach.

Then tell them to try again when they awake the next morning. Unforgettable, for you are

likely instructing them to discover something about themselves they were not aware of!

Newton’s 2nd Law

Briefly review the idea of acceleration and its definition, and state that it is produced by an imposed force.

Write this as a F and give examples of doubling the force and the resulting doubling of the acceleration,

etc. Introduce the ideas of net force, with appropriate examples—like applying twice the force to a stalled

car gives it twice as much acceleration—three times the force, three times the acceleration.

CHECK QUESTION: If one were able to produce and maintain a constant net force of only 1

newton on the Queen Mary 2 ocean liner, what would be its maximum speed? Give multiple

choices for an answer: a) 0 ms; b) 1 ms; c) less than 1 ms; d) about 10 ms; e) close to the

speed of light! In the discussion that follows, the key concept is net force. Point out the enormous

applied forces necessary to overcome the enormous water resistance at high speeds, to yield a net

force of 1 newton; and the meaning of acceleration—that every succeeding second the ship moves

a bit faster than the second before. This would go on seemingly without limit, except for

relativistic effects which result in e) being the correct answer.

Falling Objects:

Point out that although Galileo introduced the idea of inertia, discussed the role of forces, and defined

acceleration, he never tied these ideas together as Newton did with his second law. Although Galileo is

credited as the first to demonstrate that in the absence of air resistance, falling objects fall with equal

accelerations, he was not able to say why this is so. The answer is given by Newton’s 2nd law.

SKIT: Hold a heavy object like a kilogram weight and a piece of chalk with outstretched hands,

ready to drop them. Ask your class which will strike the ground first if you drop them

simultaneously. They know. Ask them to imagine you ask the same of a bright youngster, who

responds by asking to handle the two objects before giving an answer. Pretend you are the kid

judging the lifting of the two objects. “The metal object is heavier than chalk, which means there

is more gravity force acting on it, which means it will accelerate to the ground before the chalk

does.” Write the kids argument in symbol notation on the board. a F. Then go through the

motions of asking the same of another child, who responds with a good argument that takes inertia

rather than weight into account. This kid says, after shaking the metal and chalk back-and-forth in

his or her hands, “The piece of metal is more massive than the chalk, which means it has more

inertia, than the chalk, which means it will be harder to get moving than the chalk. So the chalk

will race to the ground first, while the inertia of the metal causes it to lag behind.” Write this kid’s

argument with, a 1m. State that a beauty of science is that such speculations can be ascertained

by experiment. Drop the weight and the chalk to show that however sound each child’s argument

seemed to be, the results do not support either. Then bring both arguments together with a Fm,

Newton’s 2nd law.

Relate your skit to the case of falling bricks, Figure 4.12, and the falling boulder and feather, Figure 4.13.

Once these concepts are clear, ask how the bricks would slide on a frictionless inclined plane, then illustrate

with examples such as the time required for a fully loaded roller coaster and an empty roller coaster to

29

make a complete run. In the absence of friction effects, the times are the same. Cite the case of a Cadillac

limousine and Volkswagen moving down a hill in the absence of friction. By now you are fielding

questions having to do with air resistance and friction. (Avoid getting into the buoyancy of falling

objects—information overload.)

DEMONSTRATION: After you have made clear the cases with no friction, then make a transition

to practical examples that involve friction—leading off with

,the dropping of sheets of paper, one

crumpled and one flat. Point out that the masses and weights are the same, and the only variable is

air resistance. Bring in the idea of net force again, asking what the net force is when the paper falls

at constant speed. (If you left the Chapter 3 demo of the falling book and paper on top of it

unexplained, reintroduce it here.)

CHECK QUESTIONS: What is the acceleration of a feather that “floats” slowly to the ground?

The net force acting on the feather? If the feather weighs 0.0 N, how much air resistance acts

upward against it? [Acceleration is zero at terminal speed, and air resistance = weight of object.]

These questions lead into a discussion of the parachutists in Figure 4.15. When the decrease of acceleration

that builds up to terminal velocity is clear, return to the point earlier about the Cadillac and Volkswagen

moving down an incline, only this time in the presence of air resistance. Then ask whether or not it would

be advantageous to have a heavy cart or a light cart in a soap-box-derby race. Ask which would reach the

finish line first if they were dropped through the air from a high-flying balloon. Then consider the carts on

an inclined plane.

For your information, the terminal velocity of a falling baseball is about 150 kmh (95 mih), and for a

falling Ping-Pong ball about 32 kmh (20 mih).

Make the distinction between how fast something hits the ground and with what force it hits. Dropping a

pebble on one foot, and a boulder on the other makes this clear. Although they both hit at the same speed,

the heavier boulder elicits the ouch!

So far we have regarded a force as a push or a pull. We will consider a deeper definition of force in the next

chapter. Onward!

30

Answers and Solutions for Chapter 4

Reading Check Questions

1. Acceleration and net force are proportional to each other, not equal to each other.

2. Your push and the force of friction have the same magnitude.

3. Yes. As you increase your push, friction also increases just as much.

4. Once moving, your push has the same magnitude as the force of friction.

5. Static friction is greater than sliding friction for the same object.

6. Friction does not vary with speed.

7. Yes. Fluid friction does vary with speed.

8. Mass is more fundamental than weight.

9. mass; weight.

10. kilogram; newton.

11. A quarter-pound hamburger after it is cooked weighs about 1 newton.

12. The weight of a 1-kg brick is about 10 newtons.

13. Breaking of the top string is due mainly to the ball’s weight.

14. Breaking of the lower string is due mainly to the ball’s mass.

15. Acceleration is inversely proportional to mass.

16. The acceleration produced by a net force on an object is directly proportional to the net force, is in the

same direction as the net force, and is inversely proportional to the mass of the object.

17. No. Weight is proportional to mass, but not equal to mass.

18. The acceleration triples.

19. The acceleration decreases to one-third.

20. The acceleration will be unchanged.

21. The acceleration and net force are in the same direction.

22. In free fall, the only force acting on an object is the force of gravity.

23. The ratio of force to mass is g.

24. The ratio of force to mass for both is the same, g.

25. The net force is 10 N.

26. The net force is 6 N; zero.

27. Speed and frontal area affect the force of air resistance.

28. Acceleration is zero.

29. A heavier parachutist must fall faster for air resistance to balance weight.

30. The faster one encounters greater air resistance.

Think and Do

31. Relate how Newton followed Galileo, and so on.

32. The coin hits the ground first; when crumpled, both fall in nearly the same time; from an elevated

starting point, the coin hits first. That’s because it has less frontal area.

33. When the paper is on top of the dropped book, no air resistance acts on the paper because the book

shields it from the air. So the paper and book fall with the same acceleration!

34. In all three kinds of motion they move in unison, in accord with a = F/m.

35. The spool will roll to the right! There is an angle at which it will not roll but slide. Any angle larger will

roll the spool to the left. But pulled horizontally it rolls in the direction of the pull.

Plug and Chug

36. Weight = (50 kg)(10 N/kg) = 500 N.

37. Weight = (2000 kg)(10 N/kg) = 20,000 N.

38. Weight = (2.5 kg)(10 N/kg) = 25 N; (25 N)(2.2 lb/kg)(10 N/1 kg) = 550 N.

39. (1 N)(1 kg/10 N) = 0.1 kg; (0.1 kg)(2.2 lb/1 kg) = 0.22 lb.

40. N to kg; (300 N)(1 kg/10 N) = 30 kg.

41. a = Fnet/m = (500 N)/(2000 kg) = 0.25 N/kg = 0.25 m/s2.

31

42. a = Fnet/m = (120,000 N)/(300,000 kg) = 0.4 N/kg = 0.4 m/s2.

43. a = Fnet/m = 200 N/40 kg = 5 N/kg = 5 m/s2.

44. a = ∆v/∆t = (6.0 m/s)/(1.2 m/s2) = 5.0 m/s2. .

45. a = Fnet/m = (15 N)/(3.0 kg) = 5.0 N/kg = 5.0 m/s2.

46. a = Fnet/m = (10 N)/(1 kg) = 10 N/kg = 10 m/s2.

47. Fnet = ma = (12 kg)(7.0 m/s2) = 84 kg·m/s2 = 84 N.

Think and Solve

48. (1 N)(1 lb/4.45 N) = 0.225 lb.

49. Lillian’s mass is (500N)/(10N/kg) = 50 kg. Her weight in pounds, (50 kg)(2.2 lb/kg) = 110 lb.

50. The acceleration of each is the same: a = F/m = 2 N/2 kg = 1 N/1 kg = 1 m/s2. (Incidentally, from the

definition that 1 N = 1 kg.m/s2, you can see that 1 N/kg is the same as 1 m/s2.)

51. For the jet: a = F/m = 2(30,000 N)/(30,000 kg) = 2 m/s2.

52. (a) a = ∆v/∆t = (9.0 m/s)/(0.2 s) = 45 m/s2. (b) F = ma = (100 kg)(45 m/s2) = 4500 N.

53. (a) The force on the bus is Ma. New acceleration = same force/new mass = Ma/(M+M/5) = 5Ma/(5M+M)

= 5Ma/6M = (5/6)a.

(b) New acceleration = (5/6)a = (5/6)(1.2 m/s2) = 1.0 m/s2.

Think and Rank

54. a. D, A=B=C; b. A=C, B=D

55. C, B, A

56. a. A=B=C; b. C, A, B

57. a. C, A, B; b. B, A, C

Think and Explain

58. The force you exert on the ball ceases as soon as contact with your hand ceases.

59.Yes, if the ball slows down, a force opposite to its motion is acting—likely air resistance and friction

between the ball and alley.

60. Constant velocity means zero acceleration, so yes, no net force acts on the motorcycle. But when

moving at constant acceleration there is a net force acting on it.

61.No, inertia involves mass, not weight.

62.Items like apples weigh less on the Moon, so there are more apples in a 1-pound bag of apples there.

Mass is another matter, for the same quantity of apples are in 1-kg bag on the Earth as on the Moon.

63. Buy by weight in Denver because the acceleration of gravity is less in Denver than in Death Valley.

Buying by mass would be the same amount in both locations.

64.Shake the boxes. The box that offers the greater resistance to acceleration is the more massive box, the

one containing the sand.

65.When you carry a heavy load there is more mass involved and a greater tendency to remain moving. If a

load in your hand moves toward a wall, its tendency is to remain moving when contact is made. This

tends to squash your hand if it’s between the load and the wall—an unfortunate example of Newton’s

first law in action.

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66.Mass is a measure of the amount of material in something, not gravitational pull that depends on its

location. So although the weight of the astronaut may change with location, mass does not.

67.A massive cleaver is more effective in chopping vegetables because its greater mass contributes to

greater tendency to keep moving as the cleaver chops the food.

68.Neither the mass nor the weight of a junked car changes when it is crushed. What does change is its

volume, not to be confused with mass and weight.

69.Ten kilograms weighs about 100 N on the Earth (weight = mg = 10 kg 10 m/s2 = 100 N, or 98 N if g =

9.8 m/s2 is used). On the Moon the weight is 1/6 of 100 N = 16.7 N

(or 16 N if g = 9.8 m/s2 is used). The mass is 10 kg everywhere.

70. The scale

,reading will increase during the throw. Your upward force on the heavy object is transmitted

to the scale.

71.The change of weight is the change of mass times g, so when mass changes by 2 kg, weight changes

by about 20 N.

72.One kg of mass weighs 2.2 pounds at the Earth’s surface. If you weigh 100 pounds, for example, your

mass is (100 lb)/(2.2 kg/lb) = 45 kg. Your weight in newtons, using the relationship weight = mg, is then

(45 kg)(10 N/kg) = 450 N.

73.A 1-kg mass weighs 10 N, so 30 kg weigh 300 N. The bag can safely hold 30 kg of apples—if you don’t

pick it up too quickly.

74. Since the crate remains at rest, the net force on it is zero, which means the force of friction by the

floor on the crate will be equal and opposite to your applied force.

75.The second law states the relationship between force and acceleration. If there is no net force, there is

no acceleration—which is what Newton’s first law states. So Newton’s first law is consistent with the

second law, and can be considered to be a special case of the second law.

76.Acceleration (slowing the car) is opposite to velocity (direction car moves).

77.Agree. Acceleration (slowing the car) is opposite to velocity (the direction the car is moving).

78.Acceleration is the ratio force/mass (Newton’s second law), which in free fall is just weight/mass = mg/m

=g. Since weight is proportional to mass, the ratio weight/mass is the same whatever the weight of a

body.

79.Lifting the opponent decreases the force with which the ground supports him, and correspondingly

decreases the force of friction he can muster. The reduced friction limits the opponent’s effectiveness.

80.The forces acting horizontally are the driving force provided by friction between the tires and the road,

and resistive forces—mainly air resistance. These forces cancel and the car is in dynamic equilibrium

with a net force of zero.

81.(a) No. Air resistance is also acting. Free fall means free of all forces other than that due to gravity. A

falling object may experience air resistance; a freely falling object experiences only the force due to

gravity. (b) Yes. Although getting no closer to the Earth, the satellite is falling (more about this in

Chapter 10).

82.The velocity of the ascending coin decreases while its acceleration remains constant (in the absence of

air resistance).

83.The only force on a tossed coin, except for air resistance, is mg. So the same mg acts on the coin at all

points in its trajectory.

84. The acceleration at the top or anywhere else in free fall is g, 10 m/s2, downward. The velocity of the

rock is momentarily zero while the rate of change of velocity is still present. Or better, by Newton’s 2nd

33

law, the force of gravity acts at the top as elsewhere; divide this net force by the mass for the

acceleration of free fall. That is, a = Fnet/m = mg/m = g.

85.You explain the distinction between an applied force and a net force. It would be correct to say no net

force acts on a car at rest.

86.When driving at constant velocity, the zero net force on the car results from the driving force that your

engine supplies against the friction drag force. You continue to apply a driving force to offset the drag

force that otherwise would slow the car.

87. When the apple is held at rest the upward support force equals the gravitational force on the apple and

the net force is zero. When the apple is released, the upward support force is no longer there and the

net force is the gravitational force, 1 N. (If the apple falls fast enough for air resistance to be important,

then the net force will be less than 1 N, and eventually can reach zero if the air resistance builds up to

1 N.)

88.High-speed grains of sand grazing the Earth’s atmosphere burn up because of friction against the air.

89.Both forces have the same magnitude. This is easier to understand if you visualize the parachutist at rest

in a strong updraft—static equilibrium. Whether equilibrium is static or dynamic, the net force is zero.

90.When anything falls at constant velocity, air resistance and gravitational force are equal in magnitude.

Raindrops are merely one example.

91.When a parachutist opens her chute she slows down. That means she accelerates upward.

92. There are usually two terminal speeds, one before the parachute opens, which is faster, and one after,

which is slower. The difference has mainly to do with the different areas presented to the air in falling.

The large area presented by the open chute results in a slower terminal speed, slow enough for a safe

landing.

93.Just before a falling body attains terminal velocity, there is still a downward acceleration because

gravitational force is still greater than air resistance. When the air resistance builds up to equal the

gravitational force, terminal velocity is reached. Then air resistance is equal and opposite to

gravitational force.

94.The terminal speed attained by the falling cat is the same whether it falls from 50 stories or 20 stories.

Once terminal speed is reached, falling extra distance does not affect the speed. (The low terminal

velocities of small creatures enables them to fall without harm from heights that would kill larger

creatures.)

95.The sphere will be in equilibrium when it reaches terminal speed—which occurs when the gravitational

force on it is balanced by an equal and opposite force of fluid drag.

96. Air resistance is not really negligible for so high a drop, so the heavier ball does strike the ground first.

(This idea is shown in Figure 4.16.) But although a twice-as-heavy ball strikes first, it falls only a little

faster, and not twice as fast, which is what followers of Aristotle believed. Galileo recognized that the

small difference is due to air friction, and both would fall together when air friction is negligible.

97.The heavier tennis ball will strike the ground first for the same reason the heavier parachutist in Figure

4.15 strikes the ground first. Note that although the air resistance on the heavier ball is smaller relative

to the ball’s weight, it is actually greater than the air resistance that acts on the other ball. Why?

Because the heavier ball falls faster, and air resistance is greater at greater speed.

98.Air resistance decreases the speed of a moving object. Hence the ball has less than its initial speed

when it returns to the level from which it was thrown. The effect is easy to see for a feather projected

upward by a slingshot. No way will it return to its starting point with its initial speed!

99.The ball rises in less time than it falls. If we exaggerate the circ*mstance and considering the feather

example in the preceding answer, the time for the feather to flutter from its maximum altitude is clearly

longer than the time it took to attain that altitude. The same is true for the not-so-obvious case of the

ball.

34

100. Open-ended.

Think and Discuss

101. Yes, as illustrated by a ball thrown vertically into the air. Its velocity is initially upward, and finally

downward, all the while accelerating at a constant downward g.

102. Neither a stick of dynamite nor anything else “contains” force. We will see later that a stick of dynamite

contains energy, which is capable of producing forces when an interaction of some kind occurs.

103. No. An object can move in a curve only when a force acts. With no force its path would be a straight

line.

104. The only force that acts on a dropped rock on the Moon is the gravitational force between the rock and

the Moon because there is no air and therefore no air drag on the rock.

105. A dieting person seeks to lose mass. Interestingly, a person can lose weight by simply being farther

from the center of the Earth, at the top of a mountain, for example.

106. Friction between the crate and the truck-bed is

,the force that keeps the crate picking up the same

amount of speed as the truck. With no friction, the accelerating truck would leave the crate behind.

107. Note that 30 N pulls three blocks. To pull two blocks then requires a 20-N pull, which is the tension in

the rope between the second and third block. The tension in the rope that pulls only the third block is

therefore 10 N. (Note that the net force on the first block,

30 N – 20 N = 10 N, is the force needed to accelerate that block, having one-third of the total mass.)

108. The net force on the wagon, your pull plus friction, is zero. So F = 0.

109. When you stop suddenly, your velocity changes rapidly, which means a large acceleration of stopping.

By Newton’s second law, this means the force that acts on you is also large. Experiencing a large

force is what hurts you.

110. The force vector mg is the same at all locations. Acceleration g is therefore the same at all locations

also.

111. The force you exert on the ground is greater. The ground must push up on you with a force greater than the downward force o

112. At the top of your jump your acceleration is g. Let the equation for acceleration via Newton’s second

law guide your thinking: a = F/m = mg/m = g. If you said zero, you’re implying the force of gravity

ceases to act at the top of your jump—not so!

113. For a decreasing acceleration the increase in speed becomes smaller each second, but nevertheless,

there’s greater speed each second than in the preceding second.

114. The net force is mg downward, 10 N (or more precisely, 9.8 N).

115. The net force is 10 N – 2 N = 8 N (or more precisely 9.8 N – 2 N = 7.8 N).

116. Agree with your friend. Although acceleration decreases, the ball is nevertheless gaining speed. It will

do so until it reaches terminal speed. Only then will it not continue gaining speed.

117. A sheet of paper presents a larger surface area to the air in falling (unless it is falling edge on), and

therefore has a lower terminal speed. A wadded piece of paper presents a smaller area and therefore

falls faster before reaching terminal speed.

35

118. In each case the paper reaches terminal speed, which means air resistance equals the weight of the

paper. So air resistance will be the same on each! Of course the wadded paper falls faster for air

resistance to equal the weight of the paper.

119. For low speeds, accelerations are nearly the same because air drag is small relative to the weights of

the falling objects. From a greater height, there is time for air resistance to build up and more

noticeably show its effects.

120. Sliding down at constant velocity means acceleration is zero and the net force is zero. This can occur if

friction equals the bear’s weight, which is 4000 N. Friction = bear’s weight = mg = (400 kg)(10 m/ s2) =

4000 N.

121. Nowhere is her velocity upward. The upward net force on Nellie during the short time that air

resistance exceeds the force of gravity produces a momentary upward net force and upward

acceleration. This produces a decrease in her downward speed, which is nevertheless still

downward.

37

5 Newton’s Third Law of Motion

Conceptual Physics Instructor’s Manual 12th Edition

5.2 Forces and Interactions

5.2 Newton’s Third Law of Motion

Defining Your System

5.3 Action and Reaction on Different Masses

5.4 Vectors and the Third Law

5.5 Summary of Newton’s Three Laws

The opening photos begin with Darlene Librero and Paul Doherty, my two dear friends at the

Exploratorium, Darlene goes back to the earliest days at the Exploratorium when she worked with Frank

Oppenheimer. Paul has been the senior scientist there since the 80s. The caption of the tennis ball and

racquet makes an important point, often overlooked by many: The racquet cannot hit the ball unless the

ball simultaneously hits the racquet! Toby Jacobson, who pushes on the pair of scales with wife Bruna,

sat with his mom on my Exploratorium Conceptual Physics class when he was 13 years old. Today, both

with physics PhDs, Toby and Bruna continue in their love of physics. The touching photo is of my son

Paul and his daughter, my granddaughter Gracie. Hooray for Newton’s third law!

The personality profile is of the Exploratorium’s senior scientist and good friend Paul Doherty.

Up to here a force is seen as a push or a pull. Newton’s third law defines it better—as part of an

interaction between one body and another. As the tennis ball and racquet attests, you cannot exert a force

on something—unless, and I pause, that something exerts an equal and opposite force on you. So you

can’t hit a ball unless the ball hits back. You can’t exert a force on the floor when you walk, unless the

floor exerts the same amount of force back on you, etc. In discussing action and reaction emphasize the

word “between,” for example, the forces between Earth and the Moon.

This chapter continues with a treatment of vectors. Trigonometry, no. The parallelogram rule, yes!

Vector components are also treated, which will be needed when projectiles are covered in Chapter 10.

Treatment of vectors continues in the Practice Book.

Practicing Physics Book:

• Action and Reaction Pairs • Force and Velocity Vectors

• Interactions • Force Vectors and the Parallelogram Rule

• Vectors and the Parallelogram Rule • Force-Vector Diagrams

• Velocity Vectors and Components • More on Vectors

Problem Solving Book:

Sample Problems and more, also with optional section on trigonometry instruction

Laboratory Manual:

• The Force Mirror Quantitative Observations of Force Pairs (Tech Lab)

• Blowout Newton’s Three Laws (Demonstration)

Next-Time Questions (in the Instructors Resource DVD):

• Reaction Forces • Leaning Tower of Pisa Drop

• Apple on a Table • Apple on Table

• Scale Reading • Atwood Pulley

• Tug of War • Airplane in the Wind

• Tug of War 2 • No-Recoil Cannon

Hewitt-Drew-It! Screencasts:

•Newton’s Third Law •Newton’s Laws Problem

38

SUGGESTED LECTURE PRESENTATION

Forces and Interactions

Hold a piece of tissue paper at arms length and ask if the heavyweight champion of the world could hit

the paper with 50 pounds of force. Ask your class to check their answer with their neighbors. Then don’t

give your answer. Instead, continue with your lecture. Reach out to your class and state, “I can’t touch

you, without you touching me in return—I can’t nudge this chair without the chair in turn nudging me—I

can’t exert a force on a body without that body in turn exerting a force on me.” In all these cases of

contact there is a single interaction between two things—contact requires a pair of forces, whether they

be slight nudges or great impacts, between two things. This is Newton’s 3rd law of motion. Call attention

to the examples of Figure 5.7.

Newton’s Third Law of Motion

Extend your arm horizontally and show the class that you can bend your fingers upward

only very little. Show that if you push with your other hand, and thereby apply a force

to them, or have a student do the same, they will bend appreciably more. Then walk

over to the wall and show that the inanimate wall does the same (as you push against the

wall). State that everybody will acknowledge that you are pushing on the wall, but only

a few realize the fundamental fact that the wall is simultaneously pushing on you also—

as evidenced by your bent fingers.

Do as Linda E. Roach does and place a sheet of paper between the wall and your hand. When you push

on the paper, it doesn’t accelerate—evidence of a zero net force on the paper. You can explain that in

addition to your push, the wall must be pushing just as hard in the opposite direction on the paper to

produce the zero net force. Linda recommends doing the same with an inflated balloon, whereupon your

class can easily see that both sides of the balloon are

,majors the tools needed for physics majors? By

minimizing time spent on graphical analysis, units conversions, measurement techniques,

mathematical notation, and problem solving techniques, time is provided to teach a broad survey of

physics—from Newton’s laws to E & M to rainbows to nuclear processes. Too often physics courses

spend overtime in kinematics because of its appealing tools, with the result that modern physics is

given short thrift. Many people who took a physics course can tell you that the acceleration due to

gravity is 9.8 m/s2, but they have no idea that radioactivity contributes to the molten state of Earth’s

x

interior. They didn’t get that far in their course, or if they did, they were rushing through the end of

the course. Modern physics gets too little attention.

The rest of my remarks here concern science majors. I maintain that science students who use this

book in their first physics course are even greater benefactors than nonscience students. Not because

it is an “easy“ introduction or even because it gets them excited about physics, but because it nurtures

that gut-level conceptual understanding that is the missing essential for so many science and

engineering students—who like their would-be poet counterparts, have mistaken being able to recite

poetry for understanding it.

I feel strongly that the ideas of physics should be understood conceptually before they are used as

a base for applied mathematics. We are all acquainted with students who can crank out the answers to

many problems by virtue of little-understood formulas and a knack for algebraic manipulation—

students who even in graduate school are able to do well in written exams (which are most always

exercises in problem solving), but who do poorly in oral exams (which are most always conceptual).

Is this a surprising outcome for students who have never had a good exposure to the concepts and

ideas of physics that weren’t at the same time paired with the techniques of mathematical problem

solving? To many of these people, physics is applied mathematics—so much so, that a physics course

without mathematical problem solving seems a contradiction! Conceptual understanding in every

physics course they ever encountered took a back seat to problem-solving techniques. The name of

the game in every physics course has been PROBLEM SOLVING. Students are solving problems

involving the manipulation of twigs and branches when they lack a conceptual understanding of the

trunk and base of the tree from which the branches stem.

We all know that the beauty of physics is its elegant mathematical structure. If you want to teach

mathematics to your students, a physics course is the way to go. This is because the mathematics is

applied to actual things and events. But if you want to teach physics to your students, put the niceties

of mathematical problem solving in the back seat for a semester and teach physics conceptually.

You’ll provide your students, especially your mathematical whizzes, a look at physics they may

otherwise miss. First having an understanding of concepts on a conceptual level is an essential

foundation for any serious further study of physics. Provide your students with a good look at the

overall forest before they make measurements of any single tree—place comprehension comfortably

before calculation.

For an algebra-trig based course that goes beyond the conceptual course, problem solving is

central. A blend of conceptual and problem solving is now an option, for Phil Wolf and I have written

a supplementary student problems book that we think will be greeted as being as novel as Conceptual

Physics was nearly 40 years ago. The book, now in its third edition, is described on the pages that

follow. With this supplement, Conceptual Physics can be the textbook for courses with a light

algebra-trig component.

The challenge is yours. Let’s get to work!

xi

Teaching Tips

• ATTITUDE toward students and about science are of utmost importance. Consider yourself not the master in

your classroom, but the main resource person, the pacesetter, the guide, the bridge between your student’s

ignorance and information you’ve acquired in your study. Guide their study—steer them away from the dead

ends you encountered, and keep them on essentials and away from time-draining peripherals. You are there to

help them. If they see you so, they’ll appreciate your efforts. This is a matter of self-interest. An appreciated

teacher has an altogether richer teaching experience than an under-appreciated teacher.

• ENGAGE your students. Recall your own student days with teachers whose engagement was with the subject

matter, but seldom with you or your fellow students. Engagement, eye contact for starters, is crucial to your

success as a teacher. Be with your students.

• Make your course ENJOYABLE. We all enjoy the discovery of learning more than we expected of ourselves.

Guide that discovery. When a student’s first encounter with physics is delightful, the rigor that comes later will

be welcomed.

• Don’t be a “know-it-all.” When you don’t know your material, don’t pretend you do. You’ll lose more respect

faking knowledge, than not knowing it. If you’re new to teaching, students will understand you’re still pulling it

together, and will respect you nonetheless. But if you fake it, and they CAN tell, whatever respect you’ve

earned plummets.

• Be firm, and expect good work of your students. Be fair and get papers graded and returned quickly. Be sure

the bell curve of grades reflects a reasonable average. If you have excellent students, some should score 100%

or near 100% on exams. This way you to avoid the practice of fudging grades at the end of the term to

compensate for off-the-mark low exam scores. The least respected professor in my memory was one who made

exams so difficult that the class average was near the noise level, where the highest marks were some 50%.

• Be sure that what knowledge you want from your students is reflected by your test items. The student

question, “Will that be on the test?” is a good question. What is important—by definition—is what’s on the test.

If you consider a topic important, provide an opportunity for students to demonstrate their understanding of that

topic. An excellent student should be able to predict what will be on your test. Remember your own frustration

in your student days of preparing for a topic only to find it not part of the test. Don’t let your students

experience the same. Using short questions that fairly span course content is the way to go.

• Consider having students repeat work that you judge to be poor—before it gets a final grade. A note on a

paper saying you’d rather not grade it until they’ve given it another try is the mark of a concerned and caring

teacher.

• Do less professing and more questioning. Valuable information should to be the answer to a question. Having

frequent “check-your-neighbor“ intervals should be an important feature of your class. Beware of the pitfall of

too quickly answering your own questions. Use “wait-time,” where you allow ample time before giving the next

hint.

• Show RESPECT for your students. Although all your students are more ignorant of physics than you are,

some are likely more intelligent than you are. Underestimating their intelligence is likely overestimating your

own. Respect is a two-way street. Students who know you care, respect you in return.

xii

On Class Lectures

Profess less and question more! Engage your students in lecture by frequently posing questions.

Instead of answering your own question, direct your students to come up with an answer, and check

their thinking with their neighbor—right then and there. This technique has enlivened my classes for

more than 25 years. I call it CHECK YOUR NEIGHBOR. The procedure goes something like this;

before moving on to new material, you want to summarize

,squashed.

CHECK QUESTION: Identify the action and reaction forces for the case of a bat striking the

ball.

Action and Reaction on Different Masses

Discuss walking on the floor in terms of the single interaction between you and the floor, and the pair of

action and reaction forces that comprise this interaction. Contrast this to walking on frictionless ice,

where no interaction occurs. Ask how one could get off a pond of frictionless ice. Make the answer easy

by saying one has a massive brick in hand. By throwing the brick there is an interaction between the

thrower and the brick. The reaction to the force on the brick, the recoiling force, sends one to shore. Or

without such a convenient brick, one has clothing. Or if on clothing, one has air in the lungs. One could

blow air in jet fashion. Exhale with the mouth facing away from shore, but be sure to inhale with the

mouth facing toward shore.

CHECK QUESTION: Identify the force that pushes a car along the road. [Interestingly enough,

the force that pushes cars is provided by the road. Why? The tires push on the road, action and

the road pushes on the tires, reaction. So roads push cars along. A somewhat different

viewpoint!]

Most people say that the Moon is attracted to Earth by gravity. Ask most people if Earth is also attracted

to the Moon, and if so, which pulls harder, Earth or the Moon? You’ll get mixed answers. Physicists

think differently than most people on this topic. Rather than saying the Moon is attracted to Earth by

gravity, a physicist would say there is an attractive gravitational force between Earth and the Moon.

Asking if the Moon pulls as hard on Earth as Earth pulls on the Moon is similar to asking if the distance

between New York and Los Angeles is the same as the distance between Los Angeles and New York.

Rather than thinking in terms of two distances, we think of a single distance between New York and Los

Angeles. Likewise there is a single gravitational interaction between Earth and the Moon.

39

Support this point by showing your outstretched hand where you have a stretched rubber band between

your thumb and forefinger. Ask which is pulling with the greater force, the thumb or the finger. Or, as

you increase the stretch, which is being pulled with more force toward the other—the thumb toward the

finger or the finger toward the thumb. After neighbor discussion, stress the single interaction between

things that pull on each other. Earth and the Moon are pulling on each other. Their pulls on each other

comprise a single interaction. This point of view makes a moot point of deciding which exerts the greater

force, the Moon on Earth or Earth on the Moon, or the ball on the bat or the bat on the ball, et cetera.

Pass a box of rubber bands to your class and have them do it.

DEMONSTRATION: Tug-of-war in class. Have a team of women engage in a tug-of-war with

a team of men. If you do this on a smooth floor, with men wearing socks and women wearing

rubber-soled shoes, the women will win. This illustrates that the team who wins in this game is

the team who pushes harder on the floor. This is featured at the bottom of page 80.

Discuss the firing of a bullet from a rifle, as treated in the chapter. Illustrate Newton’s 3rd law with a skit

about a man who is given one last wish before being shot, who states that his crime demands more

punishment than being struck by a tiny bullet, who wishes instead that the mass of the bullet match the

magnitude of his crime (being rational in a rigid totalitarian society), that the mass of the bullet be much

much more massive than the gun from which it is fired—and that his antagonist pull the trigger!

Return to your question about whether a heavyweight boxer could hit a piece of tissue

paper with a force of 50 pounds or so. Now your class understands (hopefully) that the

fist can’t produce any more force on the paper than the paper exerts on the fist. The

paper doesn’t have enough mass to do this, so the answer is no. The fighter can’t hit the

paper any harder than the paper can hit back. Consider solving Think and Solve 27 in

the end matter here.

Philosophically we know that if you try to do one thing, something else happens as a result. So we say

you can never do only one thing. Every equation reminds us of that—change a term on one side of an

equation and a term on the other correspondingly changes. In this chapter we similarly see that you can

never have only one force.

Defining Your System

Discuss the different systems of orange and apple as in Figures 5.8 - 5.11. This is also treated in the

Hewitt-Drew-It! Screencast on Newton’s Third law. Ask students to identify action and reaction parts of

the systems of Figures 5.14–5.18. That’s wife Lil and me in Figure 5.19. And continuing with the same

important concept of “you can’t touch without being touched”, my brother Steve and his daughter

Gretchen do the same in Think and Explain 35 in the back matter. A prior photo of them, when Gretchen

was a child, occurred in previous editions. The pushed bricks in the road of Figure 5.20 can illicit class

discussion. The photo is clear evidence that the bricks have been pushed, as they push the tires of

automobiles!

Vectors and their Components

Section 5.4 illustrates vectors and their components. The physics can be clearly seen without the use of

trigonometry. My assumption is that most readers of Conceptual Physics are not trig literate. You can

take physics time to teach some trigonometry, but my advise is that you resist that impulse and use class

time for the exciting physics beyond this chapter. If your school is typical, there are many math classes,

and perhaps your class is the only one focused on physics. In that case, learning trig can occur in the

math classes. Your math teaching colleagues are unlikely to teach much physics in their math classes!

Since you’re the physics person, go physics!

You’ll note examples involving vectors are simple ones. Why? Before one gets deeply into any subject,

they are better off with an understanding of the simplest examples first. When challenge comes, it should

be welcomed. It won’t be welcomed if the basics are missing. So go basics!

40

Think and Discuss 80 in the back matter, of the strongman pulled in opposite directions, is treated in the

screencast on Newton’s Third law. The situation elicits class discussion.

Force and Velocity Vectors

Have your students have a go at the vector exercises in the Practicing Physics book. Take care to avoid

force and velocity vectors on the same diagram. Having both on a vector diagram is an invitation to

confusion—what you don’t need.

Components of Vectors

For components of vectors, again, the Practicing Physics worksheet on page 27 is instructive. The notion

of component vectors will be useful in following chapters, particularly Chapters 6 and 10.

DEMONSTRATION: To highlight the parallelogram rule for vectors, here’s a good one: Have

two students hold the ends of a heavy chain. Ask them to pull it horizontally to make it as

straight as possible. Then ask what happens if a bird comes along and sits in the middle (as you

place a 1-kg hook mass on the middle of the chain!). What happens if another bird comes to join

the first (as you suspend another 1-kg mass)? Ask the students to keep the chain level. Now

what happens if a flock of birds join the others (as you hang additional masses). This works

well!

Invoke the parallelogram rule to show that the chain must be directed slightly upward to provide the

needed vertical components to offset the weight.

Appendix D nicely extends vectors, and describes the interesting case of a sailboat sailing into the wind.

This and the crossed Polaroids later in Chapter 29 are to my mind, the most intriguing illustrations of

vectors and what they can do. An interesting demo is the model sailboat which you can easily build

yourself with a

,small block of wood and a piece of aluminum. Cut slots in the wood and mount it on a

car (or ideally, on an air track). A square-foot sheet of aluminum serves as a sail, and wind from a hand-

held fan is directed against the sail in various directions. Most impressive is holding the fan in front, but

off to the side a bit, so that the cart will sail into the wind. This is indeed an excellent vehicle for teaching

vectors and their components!

41

Answers and Solutions for Chapter 5

Reading Check Questions

1. The force is the wall pushing on your fingers.

2. He can’t exert much force on the tissue paper because the paper can’t react with the same

magnitude of force.

3. A pair of forces are required for an interaction.

4. Whenever one object exerts a force on a second object, the second object exerts and equal and

opposite force on the first.

5. Action: bat against ball. Reaction: ball against bat.

6. Yes, and that external net force accelerates the orange system.

7. No, for the pair of forces are internal to the apple-orange system.

8. Yes, an external net force is required to accelerate the system.

9. Yes, the net force is provided by contact with your foot. If two opposite and equal forces act on the

ball, the net force on it is zero and it will not accelerate.

10. Yes, you pull upward with the same amount of force on Earth.

11. The different accelerations are due to different masses.

12. The force that propels a rocket is the exhaust gases pushing on the rocket.

13. A helicopter gets its lifting force by pushing air downward, in which case the reaction is the air

pushing the helicopter upward.

14. You cannot touch without being touched! And with the same amount of force.

15. The process of determining the components of a vector.

16. The magnitude of the normal force decreases.

17. The friction force has the same magnitude, with the sum of all forces being zero.

18. Moving upward, the vertical component of velocity decreases. The horizontal component remains

constant, in accord with Newton’s first law.

19. Inertia; acceleration; action and reaction.

20. Newton’s third law deals with interactions.

Think and Do

21.Your hand will be pushed upward, a reaction to the air it deflects downward.

22. Each will experience the same amount of force.

Plug and Chug

23. 100 km/h – 75 km/h = 25 km/h north. 100 km/h + 75 km/h = 175 km/h north

24. R = √(1002 + 1002) = 141 km/h

25. R = √(42 + 32) = 5.

26. R = √(2002 + 802) = 215 km/h

Think and Solve

27. a. a = ∆v/∆t =(25 m/s)/(0.05 s) = 500 m/s2. b. F = ma = (0.003 kg)(500 m/s2) =

1.5 N, which is about 1/3 pound. c. By Newton’s third law, the same amount, 1.5 N.

28.The wall pushes on you with 40 N.

a = F/m = 40 N/80 kg = 0.5 m/s2.

29.a = F/m, where F = √[(3.0 N)2 + (4.0 N)2] = 5 N. So a = F/m = 5 N/2.0 kg = 2.5 m/s2.

30. (a)

From the 3rd law Fon 2 m puck = Fon m puck 2m(a2 m ) = m (am ) 2m

v2 m

t

m

vm

t

Since

the force acts for exactly the same t for each mass v2m

1

2

vm . Since both masses start out at

rest

v

2m

1

2 v

m

.

(b)

v

2m

1

2 v

m

1

2 0.4 m

s 0.2 m

s .

42

Think and Rank

31. A = B = C

32. A, B, C; (b) B, C

33. (a) A = B = C; (b) C, B, A

Think and Explain

34. Action; hammer hits nail. Reaction; nail hits hammer. (b) Action; Earth pulls down on a book.

Reaction; book pulls up on Earth. (c) Action; helicopter blade pushes air downward. Reaction; air

pushes helicopter blade upward. (In these examples, action and reaction may be reversed—which is

called which is unimportant.)

35.In accord with Newton’s third law, Steve and Gretchen are touching each other. One may initiate the

touch, but the physical interaction can’t occur without contact between both Steve and Gretchen.

Indeed, you cannot touch without being touched!

36.No, for each hand pushes equally on the other in accord with Newton’s third law—you cannot push

harder on one hand than the other.

37. (a) Two force pairs act; Earth’s pull on apple (action), and apple’s pull on Earth (reaction). Hand

pushes apple upward (action), and apple pushes hand downward (reaction). (b) With no air

resistance, one force pair acts; Earth’s pull on apple, and apple’s pull on Earth.

38. (a) Action; Earth pulls you downward. Reaction; you pull Earth upward. (b) Action; you touch tutor’s

back. Reaction; tutor’s back touches you. (c) Action; wave hits shore. Reaction; shore hits wave.

39. (a) While the bat is in contact with the ball there are two interactions, one with the bat, and even

then, with Earth’s gravity. Action; bat hits ball. Reaction; ball hits bat. And, action, Earth pulls down

on ball (weight). Reaction; ball pulls up on Earth. (b) While in flight the major interactions are with

Earth’s gravity and the air. Action; Earth pulls down on ball (weight). Reaction; ball pulls up on Earth.

And, action; air pushes ball, and reaction; ball pushes air.

40. In accord with Newton’s first law, your body tends to remain in uniform motion. When the airplane

accelerates, the seat pushes you forward. In accord with Newton’s third law, you simultaneously

push backward against the seat.

41. When the ball exerts a force on the floor, the floor exerts an equal and opposite force on the ball—

hence bouncing. The force of the floor on the ball provides the bounce.

42. The billions of force pairs are internal to the book, and exert no net force on the book. An external

net force is necessary to accelerate the book.

43. The friction on the crate is 200 N, which cancels your 200-N push on the crate to yield the zero net

force that accounts for the constant velocity (zero acceleration). No, although the friction force is

equal and oppositely directed to the applied force, the two do not make an action-reaction pair of

forces. That’s because both forces do act on the same object—the crate. The reaction to your push

on the crate is the crate’s push back on you. The reaction to the frictional force of the floor on the

crate is the opposite friction force of the crate on the floor.

44. When the barbell is accelerated upward, the force exerted by the athlete is greater than the weight

of the barbell (the barbell, simultaneously, pushes with greater force against the athlete). When

acceleration is downward, the force supplied by the athlete is less.

45. The forces must be equal and opposite because they are the only forces acting on the person, who

obviously is not accelerating. Note that the pair of forces do not comprise an action-reaction pair,

however, for they act on the same body. The downward force, the man’s weight, Earth pulls down

on man, has the reaction man pulls up on Earth, not the floor pushing up on him. And the upward

force of the floor on the man has the reaction of man against the floor, not the interaction between

the man and Earth. (If you find this confusing, you may take solace in the fact that Newton himself

had trouble applying his 3rd law to certain situations. Apply the rule, A on B reacts to B on A, as in

Figure 5.7.)

43

46. When you pull up on the handlebars, the handlebars simultaneously pull down on you. This

downward force is transmitted to the pedals.

47. When the climber pulls the rope downward, the rope simultaneously pulls the climber upward—the

direction desired by the climber.

48. When you push the car, you exert a force on the car. When the car simultaneously pushes back on

you, that force is on you—not the car. You don’t cancel a force on the car with a force on you. For

cancellation, the forces have to be equal and opposite and act on the same object.

49. The strong man can exert only equal forces on both cars, just as your push against a wall equals the

push of the wall on you. Likewise for two walls, or two freight cars. Since their masses are equal,

,they will undergo equal accelerations and move equally.

50. As in the preceding exercise, the force on each cart will be the same. But since the masses are

different, the accelerations will differ. The twice-as-massive cart will undergo only half the

acceleration of the less massive cart.

51. In accord with Newton’s 3rd law, the force on each will be of the same magnitude. But the effect of

the force (acceleration) will be different for each because of the different mass. The more massive

truck undergoes less change in motion than the Civic.

52. Both will move. Ken’s pull on the rope is transmitted to Joanne, causing her to accelerate toward

him. By Newton’s third law, the rope pulls back on Ken, causing him to accelerate toward Joanne.

53. The winning team pushes harder against the ground. The ground then pushes harder on them,

producing a net force in their favor.

54. The tension in the rope is 250 N. With no acceleration, each must experience a 250-N force of friction

via the ground. This is provided by pushing against the ground with 250 N.

55. No. The net force on the rope is zero, meaning tension is the same on both ends, in accord with Newton’s

third law.

56. The forces on each are the same in magnitude, and their masses are the same, so their

accelerations will be the same. They will slide equal distances of 6 meters to meet at the midpoint.

57. The writer apparently didn’t know that the reaction to exhaust gases does not depend on a medium

for the gases. A gun, for example, will kick if fired in a vacuum. In fact, in a vacuum there is no air

drag and a bullet or rocket operates even better.

58. The slanted streaks are composed of two components. One is the vertical velocity of the falling rain.

The other is the horizontal velocity of the car. At 45° these components are equal, meaning the

speed of falling drops equals the speed of the car. (We saw this question back in Chapter 3.)

59. To climb upward means pulling the rope downward, which moves the balloon downward as the

person climbs.

60. The other interaction is between the stone and the ground on which it rests. The stone pushes down

on the ground surface, say action, and the reaction is the ground pushing up on the stone. This

upward force on the stone is called the normal force.

61. (a) The other vector is upward as shown.

(b) It is called the normal force.

62. (a) As shown.

(b) Yes.

(c) Because the stone is in equilibrium.

44

63. (a) As shown.

(b) Upward tension force is greater resulting in an upward net force.

64. As shown.

65. The acceleration of the stone at the top of its path, or anywhere where the net force on the stone is

mg, is g, downward.

66. (a) Weight and normal force only.

(b) As shown.

67. (a) As shown.

(b) Note the resultant of the two normal forces is equal and opposite to the stone’s weight.

68. Vector f will have the same magnitude as the vector sum of mg and N. If f is less, then a net force acts

on the shoe and it accelerated down the incline.

69. The magnitudes of mg and N will be equal.

70. When the rope is vertical, S is zero. If the rope were vertical, S would be at an angle such that its

vertical component would be equal and opposite to mg.

71. No force acts horizontally on the ball so the initial horizontal velocity remains constant as the ball

moves through the air in accord with Newton’s first law of inertia.

72. Earth pulls downward on the ball, action: the ball pulls upward on Earth, reaction. So the reaction force

is the ball’s upward pull on Earth. Acceleration all along the path is g

(a = F/m = mg/m = g).

Think and Discuss

73. The answer is given in the equation a = F/m. As fuel is burned, the mass of the rocket becomes less.

As m decreases as F remains the same, a increases! There is less mass to be accelerated as fuel is

consumed.

74.Action: your foot against the ball. Reaction: the ball against your foot. Both forces have the same

magnitude, in accord with Newton’s third law.

75.Yes, it’s true. The Earth can’t pull you downward without you simultaneously pulling Earth upward. The

acceleration of Earth is negligibly small, and not noticed, due to its enormous mass.

76.The scale will read 100 N, the same as it would read if one of the ends were tied to a wall instead of

tied to the 100-N hanging weight. Although the net force on the system is zero, the tension in the

rope within the system is 100 N, as would show on the scale reading.

77.Yes, a baseball exerts an external force on the bat, opposite to the bat’s motion. This external force

decelerates the oncoming bat.

78. The rapid deceleration of the speeding ball on the player’s glove produces the force on the player’s

glove. In this sense, deceleration produces force (cause and effect can sometimes be a matter of

interpretation).

45

79.The forces do not cancel because they act on different things—one acts on the horse, and the other

acts on the wagon. It’s true that the wagon pulls back on the horse, and this prevents the horse from

running as fast as it could without the attached wagon. But the force acting on the wagon (the pull by

the horse minus friction) divided by the mass of the wagon, produces the acceleration of the wagon.

To accelerate, the horse must push against the ground with more force than it exerts on the wagon

and the wagon exerts on it. So tell the horse to push backward on the ground.

80. Tension would be the same if one end of the rope were tied to a tree. If two horses pull in the same

direction, tension in the rope (and in the strongman) is doubled.

46

6 Momentum

Conceptual Physics Instructor’s Manual, 12th Edition

6.1 Momentum

6.2 Impulse

6.3 Impulse Changes Momentum

Case 1: Increasing Momentum

Case 2: Decreasing Momentum Over a Long Time

Case 3: Decreasing Momentum Over a Short Time

6.4 Bouncing

6.5 Conservation of Momentum

6.6 Collisions

6.7 More Complicated Collisions

Rising physics star Derek Muller rises to the occasion in the first of the photos that open this chapter. And

the personal profile is of Derek also. The second photo is of friend from school days, physics teacher

Howie Brand. The physics he shows applies nicely to the Pelton wheel. The photo below is of grandson

Alex.

This chapter begins where Chapter 5 leaves off. Newton’s 2nd and 3rd laws lead directly to momentum and

its conservation. We emphasize the impulse-momentum relationship with applications to many examples

that have been selected to engage the students’ interest. In your classroom I suggest the exaggerated symbol

technique as shown in Figures 6.5, 6.6, and 6.8. Draw a comparison between momentum conservation and

Newton’s 3rd law in explaining examples such as rocket propulsion. You might point out that either of these

is fundamental—i.e., momentum conservation may be regarded as a consequence of Newton’s 3rd law, or

equally, Newton’s 3rd law may be regarded as a consequence of momentum conservation.

The increased impulse that occurs for bouncing collisions is treated very briefly and is expanded in the next

chapter. Angular momentum is postponed to Chapter 8.

Interesting fact: The time of contact for a tennis ball on a racquet is about 5 milliseconds, whether or not a

player “follows through.” The idea that follow-through in tennis, baseball, or golf appreciably increases the

duration of contact is useful pedagogy and gets the point of extended time across. But it is not supported by

recent studies. Follow-through is more important in guiding one’s behavior in applying maximum force to

supply the impulse.

The swinging ball apparatus (Newton’s cradle) shown in the sketch is popular for

demonstrating momentum conservation. But any thorough analysis of it ought to be

postponed to the next

,chapter when energy is treated. This is because the question is

often raised, “Why cannot two balls be raised and allowed to swing into the array, and

one ball emerge with twice the speed?” Be careful. Momentum would indeed be

conserved if this were the case. But the case with different numbers of balls emerging

never happens. Why? Because energy would not be conserved. For the two-balls-one-

ball case, the KE after would be twice as much as the KE before impact. KE is

proportional to the square of the speed, and the conservation of both momentum and

KE cannot occur unless the numbers of balls for collision and ejection are the same.

Consider postponing this demo until the next chapter.

A system is not only isolated in space, but in time also. When we say that momentum is conserved when

one pool ball strikes the other, we mean that momentum is conserved during the brief duration of

interaction when outside forces can be neglected. After the interaction, friction quite soon brings both balls

to a halt. So when we isolate a system for purposes of analysis, we isolate both in space and in time.

System identification is developed in Systems, in the Practicing Physics Book.

47

You may want to assign an “Egg Drop” experiment. Students design and construct a case to hold an egg

that can and will be dropped from a three-or-four story building without breaking. The design cannot

include means to increase air resistance, so all cases should strike the ground with about the same speed.

By requiring the masses of all cases to be the same, the impulses of all will be the same upon impact. The

force of impact, of course, should be minimized by maximizing the time of impact. Or do as Peter

Hopkinson does (Think and Explain 56), and simply have students toss eggs into cloth sheets, suspended so

the eggs don’t hit the floor after impact. Either of these projects stir considerable interest, both for your

students and others who are not (yet?) taking your class.

In 2009 40-year old Paul Lewis from the UK survived a 10,000-foot skydiving fall after his parachute

failed to open. Amazingly, he landed on the roof of an aircraft hanger that broke his fall and flexed

sufficiently to reduce impact. That’s a wonderful Ft = ∆mv in action!

An economy air track is available from Arbor Scientific (P4-2710).

Practicing Physics Book:

• Changing Momentum

• Systems

Problem Solving Book:

Many problems on impulse, momentum, and the impulse-momentum relationship

Laboratory Manual:

• Bouncy Board Impact Time and Impact Force (Activity)

Next-Time Questions include:

• Car Crash

• Ice Sail craft

• Ball Catch

Hewitt-Drew-It! Screencasts:

•Momentum

•Conservation of Momentum

•Fish-Lunch Problem

•Freddy-Frog Momentum Problem

This chapter is important in its own right, and serves as a foundation for the concept of energy in the next

chapter.

SUGGESTED LECTURE PRESENTATION

Momentum

Begin by stating that there is something different between a Mack truck and a roller skate—they each have

a different inertia. And that there is still something different about a moving Mack truck and a moving

roller skate—they have different momenta. Define and discuss momentum as inertia in motion.

CHECK QUESTION: After stating that a Mack truck will always have more inertia than an

ordinary roller skate, ask if a Mack truck will always have more momentum than a roller skate.

[Only when mv for the truck is greater than mv for the skate.]

Cite the case of the supertanker shown in Figure 6.2, and why such huge ships normally cut off their power

when they are 25 or so kilometers from port. Because of their huge momentum (due mostly to their huge

mass), about 25 kilometers of water resistance are needed to bring them to a halt.

48

Impulse and Momentum

Derive the impulse-momentum relationship. In Chapter 3 you defined acceleration as a = vt (really t,

but you likely used t as the “time interval”). Then later in Chapter 4 you defined acceleration in terms of

the force needed, a = Fm. Now simply equate; a = a, or Fm = vt, with simple rearrangement you

have, Ft = mv (as in the footnote in the textbook on page 92).

Then choose your examples in careful sequence: First, those where the objective is to increase

momentum—pulling a slingshot or arrow in a bow all the way back, the effect of a long cannon for

maximum range, driving a golf ball. Second, those examples where small forces are the objective when

decreasing momentum—pulling your hand backward when catching a ball, driving into a haystack versus a

concrete wall, falling on a surface with give versus a rigid surface. Then lastly, those examples where the

objective is to obtain large forces when deceasing momentum—karate. Karate is more properly called “tae

kwon do.”

Point of confusion: In boxing, one “follows-through” whereas in karate one “pulls back.” But this is not

so—a karate expert does not pull back upon striking his target. He or she strikes in such a way that the hand

is made to bounce back, yielding up to twice the impulse to the target (just as a ball bouncing off a wall

delivers nearly twice the impulse to the wall than if it stuck to the wall).

CHECK QUESTION: Why is falling on a wooden floor in a roller rink less dangerous than falling

on the concrete pavement? [Superficial answer: Because the wooden floor has more “give.”

Emphasize that this is the beginning of a fuller answer—one that is prompted if the question is

reworded as follows:] Why is falling on a floor with more give less dangerous than falling on a

floor with less give? [Answer: Because the floor with more give allows a greater time for the

impulse that reduces the momentum of fall to zero. The greater time occurs because momentum

means less force.]

The loose coupling between railroad cars (Think and Discuss 88) makes good lecture topic. Discuss the

importance of loose coupling in bringing a long train initially at rest up to speed, and its importance in

braking the train as well. In effect the time factor in impulse is extended. The force needed to produce

motion is therefore decreased.

(I compare this to taking course load in proper sequence, rather than all at once where for sure one’s wheels

would likely spin.)

Conservation of Momentum

Distinguish between external and internal forces and progress to the conservation of momentum. Show

from the impulse-momentum equation that no change in momentum can occur in the absence of an external

net force.

DEMONSTRATION: Show momentum conservation with an air-track performance. Doing so can

be the focus of your lecture presentation.

Defining Your System

Momentum is not conserved in a system that experiences an external net force. This is developed in

Systems in the Practicing Physics book (next page, which is credited to Cedric Linder, the instructor

profiled in Chapter 2). The momentum of a system is conserved only when no external impulse is exerted

on the system. As the example of the girl jumping from the Earth’s surface suggests, momentum is always

conserved if you make your system big enough. Likewise when you jump up and down.

The momentum of the universe is without change.

The numerical example of lunchtime for the fish in Figure 6.17 should clarify the vector nature of

momentum—particularly for the case of the fishes approaching each other. Going over this should be

helpful—Think and Solve 35, and Think and Rank 42 on pages 104 and 105, for example. Vehicles, rather

than fish, are treated similarly.

49

Bouncing

When discussing bouncing, tell how the inventor of the Pelton wheel, Lester Pelton, made a fortune from

applying some simple physics to the old paddle wheels. Fortunately for him, he patented his ideas and was

one of the greatest financial beneficiaries of the Gold Rush Era in San Francisco.

Bouncing does not necessarily increase impact force. That depends on impact

,time. Point out that bouncing

involves some reversing of momentum, which means greater momentum change, and hence greater

impulse. If the greater impulse is over an extended time (bouncing from a circus net), impact force is small.

If over a short time (plant pot bouncing from your head), impact force is large. Damage from an object

colliding with a person may depend more on energy transfer than on momentum change, so in some cases

damage can be greater in an inelastic collision without bouncing.

Consider the demo of swinging a dart against a wooden block, as Howie Brand does in the photo that opens

this chapter, showing the effect of bouncing. A weak point of this demonstration is the fact that if the dart

securely sticks to the block, then the center of gravity of the block is changed to favor non-tipping. This

flaw is neatly circumvented by the following demo by Rich Langer of Beaumont High School in St. Louis,

MO, which considers sliding rather than tipping.

DEMONSTRATION: Toy dart gun and block of wood. Tape

some toothpicks to only one side of the block, so a suction-cup

dart won’t stick to it. First fire the dart against the smooth side

of the block. The dart sticks and the block slides an observed

distance across the table. Then repeat, but with the block turned

around so the dart hits the toothpick side. When the dart doesn’t

stick but instead bounces, note the appreciably greater distance the

block slides!

Or do as Fred Bucheit does and fashion a pendulum using the “happy-unhappy” rubber balls and let them

swing into an upright board. When the less-elastic ball makes impact, with very little bounce, the board

remains upright. But when the more-elastic ball makes impact, it undergoes a greater change in momentum

as it bounces. This imparts more impulse to the board, and it topples.

Think and Discuss 94-96 may need your elaboration if you wish to go this deep in your lecture. Simply

removing the sail, as 94 suggests, is the option used by propeller-driven aircraft. Consider suddenly

producing a sail in the airstream produced by the propeller of an airplane. The result would be a loss of

thrust, and if bouncing of the air occurred, there would be a reverse thrust on the craft. This is precisely

what happens in the case of jet planes landing on the runway. Metal “sails” move into place behind the

engine in the path of the ejected exhaust, which cause the exhaust to reverse direction. The resulting reverse

thrust appreciably slows the aircraft.

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Answers and Solutions for Chapter 6

Reading Check Questions

1. The moving skateboard has more momentum since only it is moving.

2. Impulse is force x time, not merely force.

3. Impulse can be increased by increasing force or increasing time of application.

4. More speed is imparted because the force on the cannonball acts for a longer time.

5. The impulse-momentum relationship is derived from Newton’s second law.

6. For greatest increase in momentum, use both the largest force for the longest time.

7. Less force will occur if momentum is decreased over a long time.

8. When the momentum of impact is quick, less time means more force.

9. By rolling with the punch, more time of impact occurs, which means a less forceful punch.

10. Choice (c) represents the greatest change in momentum.

11. Choice (c) also represents the greatest impulse.

12. Only external forces produce changes in momentum, so sitting in a car and pushing on the dash is an

internal force, and no momentum change of the car occurs. Likewise with the internal forces within a

baseball.

13. Yes, the statement is correct.

14. To say a quantity is conserved is to say its magnitude before an event is the same as its magnitude after

the event. Momentum in a collision, for example is the same before and after providing no external

forces act.

15. Momentum would not be conserved if force, and therefore impulse, was not a vector quantity.

16. Momentum is conserved in both an elastic and an inelastic collision.

17. Car B will have the speed of Car A before the collision.

18. After collision, the cars will move at half the initial speed of Car A.

19. Since they are same-magnitude vectors at right angles to each other, the combined momentum is √2

kg.m/s.

20. The total momentum before and after collision is the same, √2 kg.m/s.

Think and Do

21. Open ended.

Plug and Chug

22. Momentum (p) = mv = (8 kg)(2 m/s) = 16 kg.m/s.

23. p = mv = (50 kg)(4 m/s) = 200 kg.m/s.

24. I = (10 N)(2.5 s) = 25 N.s.

25. I = (10 N)(5 s) = 50 N.s.

26. I = ∆mv = (8 kg)(2 m/s) = 16 kg.m/s = 16 N.s.

27. I = ∆mv = (50 kg)(4 m/s) = 200 kg.m/s = 200 N.s.

28. From mvbef + 0 = (m + m)vaft; vaft = mvbef/2m = vbef/2 = (3 m/s)/2 = 1.5 m/s.

Think and Solve

29. The bowling ball has a momentum of (10 kg)(6 m/s) = 60 kg.m/s, which has the magnitude of the impulse

to stop it. That’s 60 N.s. (Note that units N.s = kg.m/s.)

30. From Ft = ∆mv, F =

∆mv

t = [(1000 kg)(20 m/s)]/10 s = 2000 N.

31. From Ft = ∆mv, F =

∆mv

t = [(75 kg)(25 m/s)]/0.1 s = 18,750 N.

32. From the conservation of momentum,

Momentumdog = momentumJudy + dog

(15 kg)(3.0 m/s) = (40.0 kg + 15 kg)v

45 kg m/s = (55 kg) v , so v = 0.8 m/s.

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33. Momentum after collision is zero, which means the net momentum before collision must have been zero.

So the 1-kg ball must be moving twice as fast as the 2-kg ball so that the magnitudes of their momenta

are equal.

34. Let m be the mass of the freight car, and 4m the mass of the diesel engine, and v the speed after both

have coupled together. Before collision, the total momentum is due only to the diesel engine, 4m(5

km/h), because the momentum of the freight car is 0. After collision, the combined mass is (4m + m),

and combined momentum is (4m + m)v. By the conservation of momentum equation:

Momentumbefore = momentumafter

4m(5 km/h) + 0 = (4m + m)v

v =

(20m.km/h)

5m = 4 km/h

(Note that you don’t have to know m to solve the problem.)

35. Momentumbefore = momentumafter

(5 kg)(1 m/s) + (1 kg)v = 0

5 m/s + v = 0

v = -5 m/s

So if the little fish approaches the big fish at 5 m/s, the momentum after will be zero.

36.By momentum conservation,

asteroid mass 800 m/s = Superman’s mass v.

Since asteroid’s mass is 1000 times Superman’s,

(1000m)(800 m/s) = mv

v = 800,000 m/s. This is nearly 2 million miles per hour!

37.Momentum conservation can be applied in both cases.

(a) For head-on motion the total momentum is zero, so the wreckage after collision is motionless.

(b) As shown in Figure 6.18, the total momentum is directed to the northeast—the resultant of two

perpendicular vectors, each of magnitude 20,000 kgm/s. It has magnitude 28,200 kg.m/s. The

speed of the wreckage is this momentum divided by the total mass, v = (28,200 kg.m/s)/(2000 kg) = 14

m/s.

38. (a,b)

From Ft p mv F

mv

t

(1 kg )(2 m /s )

(0 .2s )

10 kg m /s

2 10 N .

Think and Rank

39. a. B, D, C, A

b. B, D, C, A

40. a. B=D, A=C

b. D, C, A=B

41. a. A, B, C

b. A, B, C

c. C, B, A

d. A, B, C

42. C, A, B

Think and Explain

43. The momentum of a supertanker is enormous, which means enormous impulses are needed for

changing motion—which are produced by applying modest forces over long periods of time. Due to the

force of water resistance, over time it coasts 25 kilometers to sufficiently reduce the momentum.

44.When you are brought to a halt in a moving car, an impulse, the product of force and time, reduces your

momentum. During a collision, padded dashboards increase the time of impact while reducing the

force of impact. The impulse equals your change in momentum.

45.Air bags lengthen the time of impact thereby reducing the force of impact.

46.The extra thickness extends the time during which momentum changes

,and reduces impact force.

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47.Stretching ropes extend the time during which momentum decreases, thereby decreasing the jolting force

of the rope. Note that bringing a person to a stop more gently does not reduce the impulse. It only

reduces the force.

48.The steel cord will stretch only a little, resulting in a short time of stop and a correspondingly large force.

Ouch!

49.Bent knees will allow more time for momentum to decrease, therefore reducing the force of landing.

50.The time during which the stopping force acts is different for the different situations. Stopping time is least

on concrete and most on water, hence the different impact speeds. So there are three concepts; speed

at impact, time of impact, and force of impact—which are all related by the impulse-momentum

relationship.

51.An extended hands allow more time for reducing the momentum of the ball to zero, resulting in a smaller

force of impact on your hand.

52. The time during which the ball stops is small, producing a greater force.

53.Crumpling allows more time for reducing the momentum of the car, resulting in a smaller force of impact

on the occupants.

54.The blades impart a downward impulse to the air and produce a downward change in the momentum of

the air. The air at the same time exerts an upward impulse on the blades, providing lift. (Newton’s third

law applies to impulses as well as forces.)

55.Its momentum is the same (its weight might change, but not its mass).

56.The egg hitting the sagging sheet has a longer impact time, which decreases the force that would

otherwise break it.

57.The large momentum of the spurting water is met by a recoil that makes the hose difficult to hold, just as

a shotgun is difficult to hold when it fires birdshot.

58.Not a good idea. The gun would recoil with a speed ten times the muzzle velocity. Firing such a gun in

the conventional way would not be a good idea!

59.Impulse is force time. The forces are equal and opposite, by Newton’s third law, and the times are the

same, so the impulses are equal and opposite.

60.The momentum of recoil of Earth is 10 kg m/s. Again, this is not apparent because the mass of the Earth

is so enormous that its recoil velocity is imperceptible. (If the masses of Earth and person were equal,

both would move at equal speeds in opposite directions.)

61. The momentum of the falling apple is transferred to the Earth. Interestingly, when the apple is released,

the Earth and the apple move toward each other with equal and oppositely directed momenta.

Because of the Earth’s enormous mass, its motion is imperceptible. When the apple and Earth hit each

other, their momenta are brought to a halt—zero, the same value as before.

62. There is usually greater speed and therefore impact on a catcher’s mitt than the mitts of other players.

That’s why extra padding is used to prolong the time of the impulse to stop the ball and lessen the

catching force.

63.The lighter gloves have less padding, and less ability to extend the time of impact, and therefore result in

greater forces of impact for a given punch.

64. In jumping, you impart the same momentum to both you and the canoe. This means you jump from a

canoe that is moving away from the dock, reducing your speed relative to the dock, so you don’t jump

as far as you expected to.

65.The swarm will have a net momentum of zero if the swarm stays in the same location; then the momenta

of the many insects cancel and there is no net momentum in any given direction.

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66.To get to shore, the person may throw keys, coins or an item of clothing. The momentum of what is

thrown will be accompanied by the thrower’s oppositely-directed momentum. In this way, one can

recoil towards shore. (One can also inhale facing the shore and exhale facing away from the shore.)

67.If no momentum is imparted to the ball, no oppositely directed momentum will be imparted to the thrower.

Going through the motions of throwing has no net effect. If at the beginning of the throw you begin

recoiling backward, at the end of the throw when you stop the motion of your arm and hold onto the

ball, you stop moving too. Your position may change a little, but you end up at rest. No momentum

given to the ball means no recoil momentum gained by you.

68.Regarding Question 66: If one throws clothing, the force on the clothes will be paired with an equal and

opposite force on the thrower. This force can provide recoil toward shore. Regarding Question 67:

According to Newton’s third law, whatever forces you exert on the ball, first in one direction, then in the

other, are balanced by equal forces that the ball exerts on you. Since the forces on the ball give it no

final momentum, the forces it exerts on you also give no final momentum.

69.Both recoiling carts have the same amount of momentum. So the cart with twice the mass will have half

the speed of the less massive cart. That is, 2m(v/2) = mv.

70.An impulse is responsible for the change in momentum, resulting from a component of gravitational force

parallel to the inclined plane.

71.Momentum is not conserved for the ball itself because an impulse is exerted on it (gravitational force

time). So the ball gains momentum. Only in the absence of an external force does momentum not

change. If the whole Earth and the rolling ball are taken together as a system, then the gravitational

interaction between Earth and the ball are internal forces and no external impulse acts. Then the

change of momentum of the ball is accompanied by an equal and opposite change of momentum of

Earth, which results in no change in momentum.

72.A system is any object or collection of objects. Whatever momentum such a system has, in the absence

of external forces, that momentum remains unchanged—what the conservation of momentum is about.

73.For the system comprised of only the ball, momentum changes, and is therefore not conserved. But for

the larger system of ball + Earth, momentum is conserved for the impulses acting are internal

impulses. The change of momentum of the ball is equal and opposite to the change of momentum of

the recoiling Earth.

74.For the system comprised of ball + Earth, momentum is conserved for the impulses acting are internal

impulses. The momentum of the falling apple is equal in magnitude to the momentum of the Earth

toward the apple.

75.If the system is the stone only, its momentum certainly changes as it falls. If the system is enlarged to

include the stone plus the Earth, then the downward momentum of the stone is cancelled by the equal

but opposite momentum of the Earth “racing” up to meet the stone.

76.Yes, because you push upward on the ball you toss, which means the ball pushes downward on you,

which is transmitted to the ground. So normal force increases as the ball is thrown (and goes back to

equal mg after the ball is released). Likewise, in catching the ball you exert an upward force while

stopping it, which is matched by a downward force by your feet on the ground, which increases the

normal force.

77.By Newton’s 3rd law, the force on the bug is equal in magnitude and opposite in direction to the force on

the car windshield. The rest is logic: Since the time of impact is the same for both, the amount of

impulse is the same for both, which means they both undergo the same change in momentum. The

change in momentum of the bug is evident because of its large change in speed. The same change in

momentum of the considerably more massive car is not evident, for the change in speed is

correspondingly very small. Nevertheless, the magnitude of m∆V for the bug is equal to M∆v for the

car!

78.In accord with Newton’s third law, the forces on each are equal in magnitude, which means the impulses

are likewise equal in magnitude, which means both undergo equal changes in momentum.

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79.The magnitude of force, impulse, and change in momentum will be the same for each. The MiniCooper

undergoes the greater deceleration because its mass is less.

80.Cars brought to a rapid halt experience a change in momentum, and a corresponding impulse. But

greater momentum change occurs if the cars bounce, with correspondingly greater impulse and

therefore greater damage. Less damage results if the cars stick upon impact than if they bounce apart.

81.The direction of momentum is to the left, for the momentum of the 0.8-kg car is greater. By magnitude,

net momentum = (0.5)(1) – (0.8)(1.2) = -0.46.

82. The combined momentum is √2 times the magnitude of that of each cart before collision.

83.Momentum conservation is being violated. The momentum of the boat before the hero lands on it will be

the same as the momentum of boat + hero after. The boat will slow down. If, for example, the masses

of the hero and boat were the same, the boat should be slowed to half speed; mvbefore = 2m(v/2)after.

From an impulse-momentum point of view, when the hero makes contact with the boat, he is moved

along with the boat by a friction force between his feet and the boat surface. The equal and opposite

friction force on the boat surface provides the impulse that slows the boat. (Here we consider only

horizontal forces and horizontal component of momentum.)

84. Yes, you exert an impulse on a ball that you throw. You also exert an impulse on the ball when you

catch it. Since you change its momentum by the same amount in both cases, the impulse you exert in

both cases is the same. To catch the ball and then throw it back again at the same speed requires

twice as much impulse. On a skateboard, you’d recoil and gain momentum when throwing the ball,

you’d also gain the same momentum by catching the ball, and you’d gain twice the momentum if you

did both—catch and then throw the ball at its initial speed in the opposite direction.

85.The impulse will be greater if the hand is made to bounce because there is a greater change in the

momentum of hand and arm, accompanied by a greater impulse. The force exerted on the bricks is

equal and opposite to the force of the bricks on the hand. Fortunately, the hand is resilient and

toughened by long practice.

Think and Discuss

86. The impulse required to stop the heavy truck is considerably more than the impulse required to stop a

skateboard moving with the same speed. The force required to stop either, however, depends on the

time during which it is applied. Stopping the skateboard in a split second results in a certain force.

Apply less than this amount of force on the moving truck and given enough time, the truck will come to

a halt.

87.When a boxer hits his opponent, the opponent contributes to the impulse that changes the momentum of

the punch. When punches miss, no impulse is supplied by the opponent—all effort that goes into

reducing the momentum of the punches is supplied by the boxer himself. This tires the boxer. This is

very evident to a boxer who can punch a heavy bag in the gym for hours and not tire, but who finds by

contrast that a few minutes in the ring with an opponent is a tiring experience.

88.Without this slack, a locomotive might simply sit still and spin its wheels. The loose coupling enables a

longer time for the entire train to gain momentum, requiring less force of the locomotive wheels against

the track. In this way, the overall required impulse is broken into a series of smaller impulses. (This

loose coupling can be very important for braking as well.)

89.The internal force of the brake brings the wheel to rest. But the wheel, after all, is attached to the tire

which makes contact with the road surface. It is the force of the road on the tires that stops the car.

90.If the rocket and its exhaust gases are treated as a single system, the forces between rocket and exhaust

gases are internal, and momentum in the rocket-gases system is conserved. So any momentum given

to the gases is equal and opposite to momentum given to the rocket. A rocket attains momentum by

giving momentum to the exhaust gases.

91.When two objects interact, the forces they exert on each other are equal and opposite and these forces

act simultaneously, so the impulses are equal and opposite. Therefore their changes of momenta are

equal and opposite, and the total change of momentum of both objects is zero.

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92.Let the system be the car and the Earth together. As the car gains downward momentum during its fall,

Earth gains equal upward momentum. When the car crashes and its momentum is reduced to zero,

Earth stops its upward motion, also reducing its momentum to zero.

93.This exercise is similar to the previous one. If we consider Bronco to be the system, then a net force acts

and momentum changes. In this case, momentum is not conserved. If, however we consider the

system to be Bronco and the world (including the air), then all the forces that act are internal forces

and momentum is conserved. Momentum is conserved only in systems not subject to external forces.

94. The craft moves to the right. This is because there are

two horizontal impulses that act on the craft: One is that of

the wind against the sail, and the other is that of the fan

recoiling from the wind it produces. These impulses are

oppositely directed, but are they equal in magnitude? No,

because of bouncing. The wind bounces from the sail and

produces a greater impulse than if it merely stopped. This

greater impulse on the sail produces a net impulse in the

forward direction, toward the right. We can see this in

terms of forces as well. Note in the sketch there are two

force pairs to consider: (1) the fan-air force pair, and (2)

the air-sail force pair. Because of bouncing, the air-sail

pair is greater. The net force on the craft is forward, to the

right. The principle described here is applied in thrust reversers used to slow jet planes after they land.

Also, you can see that after the fan is turned on, there is a net motion of air to the left, so the boat, to

conserve momentum, will move to the right.

95. If the air is brought to a halt by the sail, then the impulse against the sail will be equal and opposite to

the impulse on the fan. There will be no net impulse and no change in momentum. The boat will

remain motionless. Bouncing counts!

96. Removing the sail and turning the fan around is the best means of propelling the boat! Then maximum

impulse is exerted on the craft. If the fan is not turned around, the boat is propelled backward, to the

left. (Such propeller-driven boats are used where the water is very shallow, as in the Florida

Everglades.)

97. Bullets bouncing from the steel plate experience a greater impulse. The plate will be moved more by

bouncing bullets than by bullets that stick.

98. In terms of force: When Freddy lands on the skateboard he is brought up to the skateboard’s speed.

This means a horizontal force provided by the board acts on Freddy. By action-reaction, Freddy exerts

a force on the board in the opposite direction—which slows the skateboard. In terms of momentum

conservation: Since no external forces act in the horizontal direction, the momentum after the

skateboard catches Freddy is equal to the momentum before. Since mass is added, velocity must

decrease.

99. Agree with the first friend because after the collision the bowling ball will have a greater momentum than

the golf ball. Note that before collision the momentum of the system of two balls is all in the moving

golf ball. Call this +1 unit. Then after collision the momentum of the rebounding golf ball is nearly –1

unit. The momentum (not the speed!) of the bowling ball will have to be nearly +2 units. Why? Because

only then is momentum conserved. Momentum before is +1 unit: momentum after is (+2 – 1) = +1.

100.We assume the equal

,lowest, and where it is highest it doesn’t move at all.

The bob transforms energy of motion to energy of position in cyclic fashion. Allow the pendulum to swing

to-and-fro while you’re talking. Its motion decays. Why? Then point out the transformation of energy from

the moving bob to the molecules of air that are encountered, and to the molecules in the bending string or

wire at the pivot point. The energy of the pendulum will end up as heat energy. I quip that on a very hot

day, somebody, somewhere, is swinging a giant pendulum to-and-fro.

Work

Define work and compare it to impulse of the previous chapter. In both case, the effect of exerting a force

on something depends on how long the force acts. In the previous chapter, how long was meant as time,

and we spoke of impulse. In this chapter, how long is meant as distance, and we speak of work. Cite the

examples of the drawn slingshot and the long barreled cannon, where the added length produces greater

speed. We described this greater speed in terms of greater momentum. Now we describe this greater speed

in terms of greater energy—that is, greater KE.

CHECK QUESTION: Is work done when a weightlifter (Figure 7.3) holds a barbell stationary

above her head? [Yes and no. With each contraction of the weightlifter’s heart, a force is exerted

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through a distance on her blood and so does work on the blood. But this work is not done on the

barbell.]

Work-Energy Theorem

When discussing whether or not work is done, be sure to specify done on what. If you push a stationary

wall, you may be doing work on your muscles (that involve forces and distances in flexing), but you do no

work on the wall. Key point: If work is done on something, then the energy of that something changes.

Distinguish between the energy one expends in doing things, and the work that is actually done on

something.

CHECK QUESTION: When a car slows down due to air resistance, does its KE decrease? [Most

certainly!]

CHECK QUESTION: Which is greater, 1 joule or 1 newton? [Whoops! The comparison is silly,

for they’re units of completely different things—work and force.] An idea about the magnitude of

1 joule is that it’s the work done in vertically lifting a quarter-pound hamburger with cheese

(approximately 1 N) one meter.

Power

A watt of power is the work done in vertically lifting a quarter-pound hamburger with cheese

(approximately 1 N) one meter in one second.

Potential Energy

Return to your pendulum: With the pendulum at equilibrium show how the force necessary to pull it

sideways (which varies with the angle made by the string) is very small compared to the force necessary to

lift it vertically (its weight). Point out that for equal elevations, the arced path is correspondingly longer

than the vertical path—with the result that the product of the applied force and distance traveled—the work

done—is the same for both cases. (Without overdoing it, this is a good place to let your students know

about integral calculus—how calculus is required to add up the work segments that continuously increase

in a nonlinear way.) Then discuss the work needed to elevate the ball in Figure 7.6.

CHECK QUESTIONS: Does a car hoisted for lubrication in a service station have PE? How much

work will raise the car twice as high? Three times as high? How much more PE will it have in

these cases?

You can give the example of dropping a bowling ball on your toe—first from a distance of a couple of

centimeters above your toe, then to various distances up to 1 m. Each time, the bowling ball would do more

work on your toe because it would transfer more gravitational potential energy when released.

Kinetic Energy

Relate force distance = KE to examples of pushing a car, and then to braking a car as treated in the text.

You may do Problem 3 (about skidding distance as a function of speed) at this point.

To a close approximation, skidding force is independent of speed. Hence change in KE is approximately

equal to change in skidding distance. When the car’s brakes are applied, the car’s kinetic energy is changed

into internal energy in the brake pads, tire, and road as they become warmer.

You may or may not at this point preview future material by relating the idea of the KE of molecules and

the idea of temperature. State that molecules in a substance having the same temperature have the same

average KE. If the masses of the molecules are the same, then it follows that the speeds of the molecules

are the same. But what if the masses are different, for example in a sample of gas composed of light and

heavy molecules at the same temperature? Which molecules would move faster? (If you shook a container

of billiard balls mixed with Ping-Pong balls so that both kinds of balls had the same kinetic energy, which

would move faster in the container? If an elephant and a mouse run with the same kinetic energy, which

means that both will do the same amount of work if bumping into the door of a barn, can you say which of

the two is running faster?) You might consider the demonstration of inhaling helium and talking at this

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point—particularly if you are not including the chapters on sound in your course design. Relate the higher

temperature due to the faster moving helium molecules to the higher temperature in a bugle when faster

moving air is blown through it.

Energy Conservation

Discuss Figures 7.9 and 7.11 and then return to your pendulum. Explain how the kinetic energy and hence,

the speed of the bob at the bottom of its swing is equal to the speed it would have if dropped vertically

through the same height.

CHECK QUESTION: Refer to Figure 7.6 in “inclines” (a) and (b): How does the speed of the ball

compare at ground level when released from equal elevations? [It is impressive that the speeds

will be the same. The lesser acceleration down the sloped ramp is compensated by a longer time.

But return to the situation and ask how the times to reach the bottom compare and be prepared for

an incorrect response, “The same!” (NOT true!) Quip and ask if the colors and temperatures will

also be the same. Straight-forward physics can be confusing enough!]

DEMONSTRATION: Preview electricity and magnetism and bring out the horseshoe magnet

hand-cranked generator that lights up the lamp shown ahead in photo 5 that opens Chapter 25

(Sheron Snyder producing light). Have student volunteers attest to the fact that more work is

needed to turn the crank when the lamp is connected than when it is not. Then relate this to Think

and Discuss 101 (about the car burning more fuel with lights on).

When gasoline combines with oxygen in a car’s engine, the chemical potential energy stored in the fuel is

converted mainly into molecular KE (thermal energy). Some of this energy in effect is transferred to the

piston and some of this causes motion of the car.

We think of electric cars as something new. But they were more popular than gasoline-driven cars in the

late 19th and early 20th century. They could go all day on a single charge and move a driver around a city

with ease. They required no hand crank to start and had no gears to shift. But back then speed limits were

set below 20 mph to accommodate horse-drawn carriages. After World War I these limits were lifted and

gasoline powered cars began to dominate. Sooner of later when most cars go electric, we’ll be going full-

circle!

Go over the Check Yourself question about fuel economy on page 117—very important. (I pose the same

question on my exams, which to the student is the definition of what’s important!) This is a pre-hybrid

question about cars. As a side point, gas economy is increased when tires are inflated to maximum

pressures, where less flattening of the tire occurs as it turns. The very important point of this exercise is the

upper limit possible.

I extend this idea of an upper limit to the supposed notion that certain gadgets attached

,to automobile

engines will give phenomenal performance—so much in fact, (tongue in cheek) that the oil companies have

gobbled up the patents and are keeping them off the market. Charlatans stand ready to benefit from this

public perception, and offer the public a chance to invest in their energy producing machines. They prey on

people who are ignorant of or do not understand the message of the energy conservation law. You can’t get

something for nothing. You can’t even break-even, because of the inevitable transformation of available

energy to heat. For more on such charlatans, read Bob Park’s book, Voodoo Science.

Scams that sell energy-making machines rely on funding from deep pockets and shallow brains!

Solar Power

Government subsidies for solar power have made Europe the world’s solar capitol. Even the first large

solar plant in the U.S, Solar One in Nevada, belongs to Acciona, a Spanish company that generates

electricity that it sells to NV Energy, the regional utility. Nevada One uses solar thermal, where sunlight is

reflected onto long rows of pipes that make steam to run a 64-megawatt power plant. The mirrors were

made in Germany.

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Another method of getting electricity from sunshine is employed by SunCatchers, huge mirrors at Sandia

National Labs in New Mexico that power Stirling engines held at the focal points of the arrays. Electricity

is made by pistons in the engines. It is the most efficient system for converting photon energy to grid-ready

AC power.

Nearly all big solar plants lack a storage system, a means of storing some of the heat produced during

daylight hours for release when the Sun isn’t shining. Check the commercial solar plant near Granada in

Spain where sunlight from mirrors is used to heat molten salt. In the evening the salt cools and gives back

heat to make steam. In this way, molten salt is used for storage. As the book mentions, energy can be stored

in compressed air, which a plant in Alabama is using, and which has been used in Germany for decades.

Another way is with batteries. With a storage system of one kind or another, electricity can be generated

continuously on demand.

Solar photovoltaic panels are expensive to produce and normally provide efficiencies of 10 to 20%.

Parabolic troughs that turn heat to steam get about 24%. Researchers can produce PV panels somewhat

more than 40% efficient. Check the Internet for current information.

Efficiency

It should be enough that your students become acquainted with the idea of efficiency, so I don’t

recommend setting the plow too deep for this topic. The key idea to impart is that of useful energy. To say

that an incandescent lamp is 10% efficient is to say that only 10% of the energy input is converted to the

useful form of light. All the rest goes to heat. But even the light energy converts to heat upon absorption.

So all the energy input to an incandescent lamp is converted to heat. This means that it is a 100% efficient

heater (but not a 100% device for emitting light)! Much more efficient light sources are treated in Chapter

23 and 30 (CFLs and LEDs).

Dark Energy: Not discussed in the text is the current serious speculation of dark energy, which is

postulated to be speeding up the expanding universe. You may want to discuss this current finding, which

may be one of the most important discoveries in science in the past quarter century.

NEXT-TIME QUESTION: Think and Discuss 120, when you’ve shown the

swinging ball apparatus, Newton’s cradle, in class (available from Arbor

Scientific. P1-6001.)

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Answers and Solutions for Chapter 7

Reading Check Questions

1. Energy is most evident when it is changing.

2. Force multiplied by distance is work.

3. No work is done in pushing on a stationary wall, as in Figure 7.4.

4. It is the same, for the product of each is the same; (50 kg)(2 m) = (25 kg)(4 m).

5. Energy enables an object to do work.

6. The same power when both are raised in the same time; Twice the power for the lighter sack raised in

half the time.

7. It would have twice because distance raised is twice.

8. Twice-as-massive car has twice the PE.

9. PE is significant when it changes, does work or transforms to energy of another form.

10. Four times as much (as 22 = 4).

11. Four times as much work; 4 times as much stopping distance (as 22 = 4).

12. ∆KE = work done = (100 N – 70 N)(10 m) = (30 N)(10 m) = 300 Nm = 300 J.

13. Speed has little or no effect on friction.

14. Its gain in KE will equal its decrease in PE, 10 kJ.

15. Immediately before hitting the ground its initial PE becomes KE. When it hits the ground its energy

becomes thermal energy.

16. The source of the energy of sunshine is fusion power in the Sun.

17. Recycled energy is the reemployment of energy that otherwise would be wasted.

18. A machine can multiply input force or input distance, but NEVER input energy.

19. As force is increased, distance is decreased by the same factor.

20. The end moving 1/3 as far can exert 3 times the input force, 150 N.

21. Efficiency would be 100%.

22. Efficiency will be 60%.

23. The Sun is the source of these energies.

24. Radioactivity is the source of geothermal energy.

25. Like electricity, hydrogen is a carrier of energy, not a source. That’s because it takes energy to separate

hydrogen from molecules.

Think and Do

26. The temperature of the sand is more after shaking than before. You do work on the sand in shaking it,

which increases its temperature.

27. Some of the basketball’s energy is transferred to the tennis ball by compression. During

decompressing, the basketball pushes the tennis ball upward, while the tennis ball pushes the

basketball downward. So PE of the bounced basketball is less and PE of the tennis ball is more,

but both add to equal the original PEs of the balls before dropped.

Plug and Chug

28. W = Fd = (5 N)(1.2 m) = 6 N.m = 6 J.

29. W = Fd = (2.0 N)(1.2 m) = 2.4 N.m = 2.4 J.

30. W = Fd = (20 N)(3.5 m) = 70 N.m = 70 J.

31. W = Fd = (500 N)(2.2 m) = 1100 N.m = 1100 J, which is also the gain in PE.

32. P = W/t = (100 J)/(2 s) = 50 W.

33. P = W/t = Fd/t = (500 N)(2.2 m)/(1.4 s) = 786 W.

34. PE = mgh = (3.0 kg)(10 N/kg)(2.0 m) = 60 N.m = 60 J.

35. PE = mgh = (1000 kg)(10 N/kg)(5 m) = 50,000 N.m = 50,000 J.

36. KE = ½ mv2 = ½(1.0 kg)(3.0 m/s)2 = 4.5 kg(m/s)2 = 4.5 J.

37. KE = ½ mv2 = ½(84 kg)(10 m/s)2 = 4200 kg(m/s)2 = 4200 J.

38. W = ∆KE = ∆½ mv2 = ½(3.0 kg)(4.0 m/s)2 = 24 J.

39. From W = ∆KE, ∆KE = Fd = (5000 N)(500 m) = 2,500,000 J.

40. Efficiency = energy output/energy input x 100% = (40 J)/(100 J) = 0.40 or 40%

Think and Solve

41. Work = ∆E = ∆mgh = 300 kg 10 N/k 6 m = 18,000 J.

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42. (a) You do F d = 100 N 10 m = 1000 J of work.

(b) Because of friction, net work on the crate is less. ∆KE = Net work = net force distance = (100 N –

70 N)(10 m) = 300 J.

(c) So the rest, 700 J, goes into heating the crate and floor.

43. At three times the speed, it has 9 times (32) the KE and will skid 9 times as far—135 m. Since the

frictional force is about the same in both cases, the distance has to be 9 times as great for 9 times as

much work done by the pavement on the car.

44.PE + KE = Total E; KE = 10,000 J – 1000 J = 9000 J.

45. From F d = F’ d/4, we see F’ = 4F = 200 N.

46. Your input work is 50 J, so 200-N h = 50 J. h = 50/200 = 0.25 m.

47.(F d)in=( F d)out

F 2 m = 5000 N 0.2 m

F = [(5000 N)(0.2 m)]/2 m = 500 N.

48.(F d)in = (F d)out

(100 N 10 cm)in = (? 1 cm)out

So we see that the output force and weight held is 1000 N (less if efficiency < 100%).

49. Power = Fd/t = (50N)(8m)/(4s) = 100J/1s = 100 watts.

50. The initial PE of the banana is transformed to KE as it falls. When the banana is about to hit the water,

all of its initial PE becomes KE.

From PE0 KEf mgh 1 / 2 mv

2 v

2 2gh v 2gh.

Think and Rank

51. a. B, A,

,what you’ve already discussed. So you

pose the challenge, “If you understand this—if you really do—then you can answer the following

question.“ Then pose your question slowly and clearly, perhaps in multiple-choice form or one

requiring a short answer. Direct your class to make a response—usually written. Tell them to

“CHECK YOUR NEIGHBOR“; look at their neighbors’ papers, and briefly discuss the answer with

them. At the beginning of the course you can add that if their neighbors aren’t helpful, to sit

somewhere else next time! The check-your-neighbor practice changes teaching by telling to teaching

by questioning—perhaps first admonished by Socrates. Questioning brings your students into an

active role, no matter how large the class. It also clears misconceptions before they are carried along

into new material. In the suggested lectures of this manual, I call such questions, CHECK

QUESTIONS. The check-question procedure may also be used to introduce ideas. A discussion of the

question, the answer, and some of the misconceptions associated with it, will get more attention than

the same idea presented as a statement of fact. And one of the very nice features of asking for

neighbor participation is that it gives you pause to reflect on your delivery.

Harvard’s Eric Mazur, profiled at the outset of Chapter 9, pioneers the conceptual approach to

physics with science and engineering majors. Eric is a strong advocate of what he calls CONVINCE

YOUR NEIGHBOR, much akin to Next-Time Questions. Students answer questions with clickers, an

extension of whiteboards. This feature is also a central component of the Modeling Workshops that

are gaining in popularity. I regret that I didn’t employ whiteboards in my classes before I retired in

2000. And electronic clickers are now popular, giving the instructor immediate feedback on questions

posed. By whatever method, have your students check their neighbors!

On homework, a note of caution: Please, please, do not overwhelm your students with excessive

written homework! (Remember those courses you took as a student where you were so busy with the

chapter-end material that you didn’t get into the chapter material itself?) The chapter-end exercises

are significantly more numerous in this edition only to provide you a wide selection to consider.

Depending on your style of teaching, you may find that posing and answering exercises in class is an

effective way to develop physics concepts. A successful course may place either very much or very

little emphasis on the exercises. Likewise with the problems, which are meant to be assigned after

concepts are treated and tested. Please don’t let your course end up as a watered-down physics

major’s problem solving course!

I strongly recommend lecture notes. In all of my teaching years I brought a note or two to every

lecture. A list of topics gives you a checklist to glance at when students are going through a check-

your-neighbor routine. Such notes insure you don’t forget main points, and a mark or two will let you

know in your next lecture what you may have missed or where you stopped.

You may find that your students are an excellent source of new analogies and examples to

supplement those in the text. A productive class assignment is:

Choose one (or more) of the concepts presented in the reading assignment and cite any

illustrative analogies or examples that you can think of.

This exercise not only prompts your students to relate physics to their own experiences, but adds to

your future teaching material. I’ve relied on this procedure to provide me with credible wrong

answers for devising multiple-choice exams!

xiii

Equations are important in a conceptual course—not as a recipe for plugging in numerical values,

but more important, as a guide to thinking. The equation tells the student what variables to consider in

treating an idea. How much an object accelerates, for example, depends not only on the net force, but

on mass as well. The equation a = F/m reminds one to consider both quantities. Does gravitation

depend on an object’s speed? Consideration of F ~ mM/d2 shows that it doesn’t, and so forth. The

problem sets, Think and Solves, at the ends of most chapters involve computations that help to

illustrate concepts rather than challenge your students’ mathematical abilities. They are relatively few

in number to avoid overload. Again, for those who make problem solving a greater part of the course,

see the student supplement Problem Solving in Conceptual Physics that complements this 12th

Edition.

Getting students to come to class prepared is a perennial problem. An ineffective way to address

this is to preach about the importance of reading assigned material before coming to class. When you

do that, you might as well be whistling Dixie. What does work is rewarding the reading directly.

What a great idea: If we want students to behave a certain way, we reward them when they do! Start

your class with a short quiz on the reading assignment. Suk R. Hwang of the University of Hawaii at

Hilo begins each class by handing out a half sheet of paper with one or two questions that highlight

the reading assignment. Before lecturing on gravity, for example, the students will take one or two

minutes to respond to “State Newton’s law of gravity in both words and equation form.“ Suk collects

the sheets and then begins his lecture. The whole process takes less than five minutes. He assigns a

grade to the sheets, with brief comments, and returns them. But the grades do not count at all when

tallying the final course grade. He is out front with his class when he tells them that the only purpose

of the quizzes is to increase the probability of coming to class having first read the assigned reading

material. Suk finds that because students abhor returning blank sheets, or dislike not being able to

correctly answer the simple questions, they DO the reading assignment. Evidently a well-answered

paper, even though it doesn’t count to the final grade, is sufficient reward for the student.

More and more instructors are finding that giving daily short quizzes or assigning summary

reports gets students to class prepared. Importantly, the instructor needn’t be submerged in

paperwork. Spot grading or even no grading is sufficient. With a prepared class, instead of presenting

material, you can refine and polish, with students that can fully benefit by the questions you pose.

Less professing—more questioning!

Make use of the NEXT-TIME QUESTIONS located on your Instructor Resource DVD, which

poses intriguing questions accompanied by my cartoons, with and accompanying answer page. These

can be photocopied and posted on display boards to capture attention and create discussion. They are

also available on the Arbor Scientific website: www.arborsci.com. What I passionately ask is that you

heed this advice: Employ “wait time“ before displaying the answers. If you put them in printed form,

you can display them in a designated space, perhaps a glass case near your office. My policy was to

display four or six of them weekly—answers to the ones posted the previous week, with new ones.

These can also be projected in your classroom, perhaps via PowerPoint. They will certainly prompt

out-of-class discussion. When impatient students want to check their answers with me before posting

time I advise them to consult with their friends. When they tell me they have done so and that their

friends are also perplexed, I suggest they seek new friends! So post them in a hallway for all to

ponder or conclude your lessons with them in class as ties to the next class meeting—hence their

name, Next-Time Questions. Most all of these first appeared as Figuring Physics in The Physics

Teacher, the must-read magazine of the American Association of Physics Teachers (AAPT).

New with this edition are my Hewitt-Drew-it! PHYSICS screencasts, which are QR coded

throughout the textbook.

,C

b. C, B, A

c. C, B, A

52. a. C, B=D, A

b. C, B=D, A

c. A, B=D, C

53. a. D, B, C, E, A

b. D, B, C, E, A

c. A, E, C, B, D

54. B=C, A (same as two supporting ropes)

Think and Explain

55. Stopping a lightly loaded truck of the same speed is easier because it has less KE and will therefore

require less work to stop. (An answer in terms of impulse and momentum is also acceptable.)

56. You do no work because you haven’t exerted more than a negligible force on the backpack in the

direction of motion. Also, the energy of the backpack hasn’t changed. No change in energy means no

work done.

57. Your friend does twice as much work (4 1/2 > 1 1).

58. Although no work is done on the wall, work is nevertheless done on internal parts of your body (which

generate heat).

59. More force is required to stretch the strong spring, so more work is done in stretching it the same

distance as a weaker spring.

60. Work done by each is the same, for they reach the same height. The one who climbs in 30 s uses

more power because work is done in a shorter time.

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61. The PE of the drawn bow as calculated would be an overestimate (in fact, about twice its actual value)

because the force applied in drawing the bow begins at zero and increases to its maximum value when

fully drawn. It’s easy to see that less force and therefore less work is required to draw the bow halfway

than to draw it the second half of the way to its fully-drawn position. So the work done is not maximum

force distance drawn, but average force distance drawn. In this case where force varies almost

directly with distance (and not as the square or some other complicated factor) the average force is

simply equal to the initial force + final force, divided by 2. So the PE is equal to the average force

applied (which would be approximately half the force at its full-drawn position) multiplied by the

distance through which the arrow is drawn.

62. When a rifle with a long barrel is fired, more work is done as the bullet is pushed through the longer

distance. A greater KE is the result of the greater work, so of course, the bullet emerges with a greater

velocity. (Note that the force acting on the bullet is not constant, but decreases with increasing

distance inside the barrel.)

63. Agree, because speed itself is relative to the frame of reference (Chapter 3). Hence ½ mv2 is also

relative to a frame of reference.

64. The KE of the tossed ball relative to occupants in the airplane does not depend on the speed of the

airplane. The KE of the ball relative to observers on the ground below, however, is a different matter.

KE, like velocity, is relative.

65. You’re both correct, with respect to the frames of reference you’re inferring. KE is relative. From your

frame of reference she has considerable KE for she has a great speed. But from her frame of

reference her speed is zero and KE also zero.

66. The energy goes mostly into frictional heating of the air.

67. Without the use of a pole, the KE of running horizontally cannot easily be transformed to gravitational

PE. But bending a pole stores elastic PE in the pole, which can be transformed to gravitational PE.

Hence the greater heights reached by vaulters with very elastic poles.

68. The KE of a pendulum bob is maximum where it moves fastest, at the lowest point; PE is maximum at

the uppermost points. When the pendulum bob swings by the point that marks half its maximum

height, it has half its maximum KE, and its PE is halfway between its minimum and maximum values. If

we define PE = 0 at the bottom of the swing, the place where KE is half its maximum value is also the

place where PE is half its maximum value, and KE = PE at this point. (By energy conservation: Total

energy = KE + PE.)

69. If the ball is given an initial KE, it will return to its starting position with that KE (moving in the other

direction!) and hit the instructor. (The usual classroom procedure is to release the ball from the nose at

rest. Then when it returns it will have no KE and will stop short of bumping the nose.)

70. Yes to both, relative to Earth, because work was done to lift it in Earth’s gravitational field and to impart

speed to it.

71. In accord with the theorem, once moving, no work is done on the satellite (because the gravitational

force has no component parallel to motion), so no change in energy occurs. Hence the satellite cruises

at a constant speed.

72. According to the work-energy theorem, twice the speed corresponds to 4 times the energy, and

therefore 4 times the driving distance. At 3 times the speed, driving distance is 9 times as much.

73. The answers to both (a) and (b) are the same: When the direction of the force is perpendicular to the

direction of motion, as is the force of gravity on both the bowling ball on the alley and the satellite in

circular orbit, there is no force component in, or parallel to, the direction of motion and no work is done

by the force.

74. On the hill there is a component of gravitational force parallel to the car’s motion. This component of

force does work on the car. But on the level, there is no component of gravitational force parallel to the

direction of the car’s motion, so the force of gravity does no work in this case.

66

75. The string tension is everywhere perpendicular to the bob’s direction of motion, which means there is

no component of tension parallel to the bob’s path, and therefore no work done by the tension. The

force of gravity, on the other hand, has a component parallel to the direction of motion everywhere

except at the bottom of the swing, and does work, which changes the bob’s KE.

76. The fact that the crate pulls back on the rope in action-reaction fashion is irrelevant. The work done on

the crate by the rope is the horizontal component of rope force that acts on the crate multiplied by the

distance the crate is moved by that force—period. How much of this work produces KE or thermal

energy depends on the amount of friction acting.

77. The 100 J of potential energy that doesn’t go into increasing her kinetic energy goes into thermal

energy—heating her bottom and the slide.

78. A Superball will bounce higher than its original height if thrown downward, but if simply dropped, no

way. Such would violate the conservation of energy.

79. When a Superball hits the floor some of its energy is transformed to heat. This means it will have less

kinetic energy after the bounce than just before and will not reach its original level.

80. Kinetic energy is a maximum as soon as the ball leaves the hand. Potential energy is a maximum

when the ball has reached its highest point.

81. The design is impractical. Note that the summit of each hill on the roller coaster is the same height, so

the PE of the car at the top of each hill would be the same. If no energy were spent in overcoming

friction, the car would get to the second summit with as much energy as it starts with. But in practice

there is considerable friction, and the car would not roll to its initial height and have the same energy.

So the maximum height of succeeding summits should be lower to compensate for friction.

82. You agree with your second classmate. The coaster could just as well encounter a low summit before

or after a higher one, so long as the higher one is enough lower than the initial summit to compensate

for energy dissipation by friction.

83. Sufficient work occurs because with each pump of the jack handle, the force she exerts acts over a

much greater distance than the car is raised. A small force acting over a long distance can do

significant work.

84. Einstein’s E = mc2. (More on this in Chapters 34 and 35).

85. When the mass is doubled with no change in speed, both momentum and KE are doubled.

86. When the velocity is doubled, the momentum

,is doubled and the KE is increased by a factor of four.

Momentum is proportional to speed, KE to speed squared.

87. Both have the same momentum, but the faster 1-kg one has the greater KE.

88. The momentum of the car is equal in magnitude but opposite in direction in the two cases—not the

same since momentum is a vector quantity.

89. Zero KE means zero speed, so momentum is also zero.

90. Yes, if we’re talking about only you, which would mean your speed is zero. But a system of two or

more objects can have zero net momentum, yet have substantial total KE.

91. Not at all. For two objects of the same KE, the one of greater mass has greater momentum. (The

mathematical relationship is p2 = 2m KE.)

92. Net momentum before the lumps collide is zero and after collision is zero. Momentum is indeed

conserved. Kinetic energy after is zero, but was greater than zero before collision. The lumps are

warmer after colliding because the initial kinetic energy of the lumps transforms into thermal energy.

Momentum has only one form. There is no way to “transform” momentum from one form to another, so

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it is conserved. But energy comes in various forms and can easily be transformed. No single form of

energy such as KE need be conserved.

93. Scissors and shears are levers. The applied force is normally exerted over a short distance for scissors

so that the output force is exerted over a relatively long distance (except when you want a large cutting

force like cutting a piece of tough rope, and you place the rope close to the “fulcrum” so you can

multiply force). With metal-cutting shears, the handles are long so that a relatively small input force is

exerted over a long distance to produce a large output force over a short distance.

94. Energy is transformed into nonuseful forms in an inefficient machine, and is “lost” only in the loose

sense of the word. In the strict sense, it can be accounted for and is therefore not lost.

95. An engine that is 100% efficient would not be warm to the touch, nor would its exhaust heat the air, nor

would it make any noise, nor would it vibrate. This is because all these are transfers of energy, which

cannot happen if all the energy given to the engine is transformed to useful work. (Actually, an engine

of 100% efficiency is not even possible in principle. We discuss this in Chapter 18.)

96. Your friend is correct, for changing KE requires work, which means more fuel consumption and

decreased air quality.

97. In accord with energy conservation, a person who takes in more energy than is expended stores

what’s left over as added chemical energy in the body—which in practice means more fat. One who

expends more energy than is taken in gets extra energy by “burning” body fat. An undernourished

person who performs extra work does so by consuming stored chemical energy in the body—

something that cannot long occur without losing health—and life.

Think and Discuss

98. Once used, energy cannot be regenerated, for it dissipates into less useful forms in the environment—

inconsistent with the term “renewable energy.” Renewable energy refers to energy derived from

renewable resources—trees, for example.

99. As world population continues to increase, energy production must also increase to provide decent

standards of living. Without peace, cooperation, and security, global-scale energy production likely

decreases rather than increases.

100. Both will have the same speed. This is easier to see here because both balls convert the same PE to

KE. (Think energy when solving motion problems!)

101. Yes, a car burns more gasoline when its lights are on. The overall consumption of gasoline does not

depend on whether or not the engine is running. Lights and other devices are run off the battery, which

“runs down” the battery. The energy used to recharge the battery ultimately comes from the gasoline.

102.Except for the very center of the plane, the force of gravity acts at an angle to the plane, with a

component of gravitational force along the plane—along the block’s path. Hence the block goes

somewhat against gravity when moving away from the central position, and moves somewhat with

gravity when coming back. As the object slides farther out on the plane, it is effectively traveling

“upward” against Earth’s gravity, and slows down. It finally comes to rest and then slides back and the

process repeats itself. The block slides back and forth along the plane. From a flat-Earth point of view

the situation is equivalent to that shown in the sketch.

103. Solar energy is merely energy from the Sun. Solar power, like power in general, is the rate at which

energy is transferred. Solar power is therefore the same from hour to hour, whereas the amount of

solar energy depends on the amount of time energy is transferred.

104. If KEs are the same but masses differ, then the ball with smaller mass has the greater speed. That is,

1/2 Mv2 = 1/2 mV2. Likewise with molecules, where lighter ones move faster on the average than more

massive ones. (We will see in Chapter 15 that temperature is a measure of average molecular KE—

lighter molecules in a gas move faster than same-temperature heavier molecules.)

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105. A car with windows open experiences more air drag, which causes more fuel to be burned in

maintaining motion. This may more than offset the saving from turning off the air conditioner.

106.A machine can multiply force or multiply distance, both of which can be of value.

107.Your friend may not realize that mass itself is congealed energy, so you tell your friend that much more

energy in its congealed form is put into the reactor than is taken out from the reactor. About 1% of the

mass that undergoes fission is converted to energy of other forms.

108. The work that the rock does on the ground is equal to its PE before being dropped, mgh = 100 joules.

The force of impact, however, depends on the distance that the rock penetrates into the ground. If we

do not know this distance we cannot calculate the force. (If we knew the time during which the impulse

occurs we could calculate the force from the impulse-momentum relationship—but not knowing the

distance or time of the rock’s penetration into the ground, we cannot calculate the force.)

109. When we speak of work done, we must understand work done on what, by what. Work is done on the

car by applied forces that originate in the engine. The work done by the road in reacting to the

backward push of the tires is equal to the product of the applied force and the distance moved, not the

net force that involves air resistance and other friction forces. When doing work, we think of applied

force; when considering acceleration, we think of net force. Actually, the frictional forces of the internal

mechanisms in the car, and to some extent the road itself are doing negative work on the car. The zero

total work explains why the car’s speed doesn’t change.

110. When air resistance is a factor, the ball will return with less speed (as discussed in Chapter 4). It

therefore will have less KE. You can see this directly from the fact that the ball loses mechanical

energy to the air molecules it encounters, so when it returns to its starting point and to its original PE, it

will have less KE. This does not contradict energy conservation, for energy is transformed, not

destroyed.

111. The ball strikes the ground with the same speed, whether thrown upward or downward. The ball starts

with the same energy at the same place, so they will have the same energy when they reach the

ground. This means they will strike with the same speed. This is assuming negligible air resistance, for

if air resistance is a factor, then the ball thrown upward will lose more energy to the air in its longer

path and strike with somewhat less speed. Another way to look at this is to consider Figure

,3.8 back

on page 50; in the absence of air resistance, the ball thrown upward will return to its starting level with

the same speed as the ball thrown downward. Both hit the ground at the same speed (but at different

times).

112. Tension in the string supporting the 10-kg block is 100 N (which is the same all along the string). So

Block B is supported by two strands of string, each 100 N, which means the mass of Block B is twice

that of Block A. So Block B has a mass of 20 kg.

113.The other 15 horsepower is supplied by electric energy from the batteries (which are ultimately recharged

using energy from gasoline).

114. In a conventional car, braking converts KE to heat. In a hybrid car, braking charges up the batteries. In

this way, braking energy can soon be transformed to KE.

115. The question can be restated; Is (302 - 202) greater or less than (202 - 102)? We see that (302 - 202) =

(900 - 400) = 500, which is considerably greater than (202 - 102) =

(400 - 100) = 300. So KE changes more for a given ∆v at the higher speed.

116.If an object has KE, then it must have momentum—for it is moving. But it can have potential energy

without being in motion, and therefore without having momentum. And every object has “energy of

being”—stated in the celebrated equation E = mc2. So whether an object moves or not, it has some

form of energy. If it has KE, then with respect to the frame of reference in which its KE is measured, it

also has momentum.

117. (a) In accord with Newton’s second law, the component of gravitational force that is parallel to the incline

in B produces an acceleration parallel to the incline. (b) In accord with the work-energy theorem, that

parallel force component multiplied by the distance the ball travels is equal to the change in the ball’s

KE.

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118. The physics here is similar to that of the ball on the horizontal alley in the previous problem. (a) Tension

in the string is everywhere perpendicular to the arc of the pendulum, with no component of tension

force parallel to its motion. (b) In the case of gravity, a component of gravitational force on the

pendulum exists parallel to the arc, which does work and changes the KE of the pendulum. (c) When

the pendulum is at its lowest point, however, there is no component of gravitational force parallel to

motion. At that instant of motion, gravity does no work (as it doesn’t when the pendulum hangs at rest

when the sting is vertical).

119. This is very similar to the previous two problems. In circular orbit, the force of gravity is everywhere

perpendicular to the satellite’s motion. With no component of force parallel to its motion, no work is

done and its KE remains constant.

120.There is more to the “swinging balls” problem than momentum conservation, which is why the problem

wasn’t posed in the previous chapter. Momentum is certainly conserved if two balls strike with

momentum 2mv and one ball pops out with momentum m(2v). That is, 2mv = m2v. We must also

consider KE. Two balls would strike with 2(1/2 mv2) = mv2. The single ball popping out with twice the

speed would carry away twice as much energy as was put in:

1/2 m(2v)2= 1/2 m(4 v2) = 2mv2. So popping out with twice its initial energy is clearly a conservation of

energy no-no!

121.In the popular sense, conserving energy means not wasting energy. In the physics sense energy

conservation refers to a law of nature that underlies natural processes. Although energy can be wasted

(which really means transforming it from a more useful to a less useful form), it cannot be destroyed.

Nor can it be created. Energy is transferred or transformed, without gain or loss. That’s what a

physicist means in saying energy is conserved.

122. The rate at which energy can be supplied is more central to consumers than the amount of energy

that may be available, so “power crisis” more accurately describes a short-term situation where

demand exceeds supply. (In the long term, the world may be facing an energy crisis when

supplies of fuel are insufficient to meet demand.)

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8 Rotational Motion

Conceptual Physics Instructor’s Manual, 12th Edition

8.1 Circular Motion

Wheels on Railroad Trains

8.2 Rotational Inertia

8.3 Torque

8.4 Center of Mass and Center of Gravity

Locating the Center of Gravity

Stability

8.5 Centripetal Force

8.6 Centrifugal Force

Centrifugal Force in a Rotating Reference Frame

Simulated Gravity

8.7 Angular Momentum

8.9 Conservation of Angular Momentum

The photo openers credit four influential educators: Paul Stokstad, the founder of Pasco, a supplier of high-

quality physics apparatus, Jacque Fresco, my original inspiration guiding me to physics (who is featured in

this chapter’s personality profile), late friend Mary Beth Monroe, who was very active in the organization

AAPT (American Association of Physics Teachers), and CCSF physics instructor Diana Lininger Markham.

It is often said that rotational motion is analogous to linear motion and therefore should not be difficult to

learn. Really? Consider the numerous distinctions between motions that are (1) linear, (2)

rotational, (3) revolutional, (4) radial, (5) tangential, and (6) angular. I remember as a student being

told that rotational motion would be easy to learn since it is an extension of linear motion. But alas, at

that time my grasp of linear motion was anything but a secure foundation. Many students are still

grappling with speed, velocity, and acceleration. And we have centripetal and centrifugal forces, real and

fictitious, not to mention torques. So there is a myriad of ideas and material to understand in this chapter. Is it

any wonder why students find rotational motion a steep hill to climb? Since a study of rotational motion is

considerably more complex than a study of linear motion, caution your students to be patient with themselves

if they don’t immediately comprehend what has taken centuries to master. To keep coverage manageable, the

chapter does not treat rotational kinetic energy.

One of the most intriguing examples of v = r is the beveled shape of railroad wheels. A train is able to

round a corner in the same way a tapered glass rolls in a circle along a tabletop. Or the way a person with one

leg shorter than the other tends to walk in a circle when lost in the woods. This is the feature of the box

Wheels on Railroad Trains. Fascinating, especially when demonstrated with a pair of tapered cups taped at

their wide ends that roll along a pair of metersticks. At CCSF, Will Maynez built a beautiful “roller coaster”

along which a set of tapered wheels faithfully follow the curved track. Most impressive!

Martha Lietz at Niles West High School does a nice activity with torques. She places the ends of a board on

two bathroom scales and sets the scales to zero. Then she challenges her students to calculate where a person

of a given weight should stand on the board so one scale would read 75 pounds. The scale displays are

covered when students do this, until they think they are in the proper position. Whole-body physics!

Although torques is a vector quantity, I don’t emphasize it in this chapter. For example I omit entirely the

“right hand rule,” where fingers of the right hand represent the motion of a rotating body and the thumb

represents the positive vector of motion. I have always felt that the reason for this and other hand rules in

introductory physics courses has been to provide some instructors the opportunity to write tricky exam

questions. Wisdom in general, and in physics teaching, is knowing what can be overlooked. I suggest you

overlook the vector nature of torque, which continuing students can get into in a follow-up course.

A nice activity for demonstrating centripetal force was introduced to me by physics teacher Howie Brand:

Have a small group of students around a small

,table (ideally circular) blow air through straws at a Ping-Pong

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ball so that it will move in a circular path. They will experience the fact that the ball must be blown radially

inward.

Rotation often involves what is called the Coriolis effect. As the name implies, it is an effect and not a force.

It occurs only in situations involving rotation. Our Earth rotates. A cannon fired northward from the equator

has a horizontal component of velocity equal to the tangential speed of the rotating Earth at that point. But it

lands at a location far enough north where Earth’s tangential speed is less. Hence it misses the true-north

target. Likewise firing from any latitude to another. It “seems” as if the shell were deflected by some force.

Toss a ball from a rotating carousel and you’ll see the ball deflect from a straight-line path. But a non-rotating

observer sees the path as straight. The effect is dramatic with winds that tend to flow around regions of high

and low pressure, running parallel to the lines of constant pressure on a weather map (isobars), instead of

flowing in a direct path. In the Northern Hemisphere, air flowing radially inward across the isobars toward

the low pressure deflects to the right. In the Southern Hemisphere, the deflection is to the left.

A common misconception is that water flowing down a drain turns in one direction in the Northern

hemisphere and in the opposite direction in the Southern Hemisphere. This is not so in something as small as

a kitchen sink. But yes for larger parcels of air. The Coriolis force that is strong enough to direct winds of

hurricanes when acting over hundreds of miles, is far too weak to stir a small bowl of water as it runs down a

drain. To say it does is to say that one side of the bowl is moving at a different speed relative to Earth’s axis

than the other side. It does. But how much? That’s the amount of your Coriolis force.

The classic oldie but goodie PSSC film, “Frames of Reference” goes well with this chapter.

This chapter can be skipped or skimmed if a short treatment of mechanics is desired. Note that this chapter

has more figures than any in the book—54 of them.

Practicing Physics Book:

• Torques • Banked Airplanes

• Torques and Rotation • Banked Track

• Acceleration and Circular Motion • Leaning On

• The Flying Pig • Simulated Gravity and Frames of Reference

Problem Solving Book:

Many problems with a bit of trigonometry are employed.

Laboratory Manual:

• Twin-Baton Paradox A Puzzle, With a Twist (Activity)

• It’s All in the Wrist Experiencing Torque “Firsthand” (Activity)

• Will it Go Round in Circles? Accelerating at Constant Speed (Demonstration)

• Sit On It and Rotate Take Physics for a Spin (Activity)

Next-Time Questions (in the Instructor Resource DVD):

• Falling Metersticks • Woman on the Plank • Trucks on a Hill • Tether Ball • Rotating Disk

• Kagan Roll • Wrench Pull • Two Spheres • Rolling Cans • Can Spurt • Berry Shake • Normal Forces

• Broom Balance • Centrifugal Force • Skateboard Lift • Post Wrap

Hewitt-Drew-It! Screencasts:

•Circular Motion •RR Wheels •Centripetal Force •Centrifugal Force •Torque •Balanced Torques

•Torques on a Plank •Skateboard Torques •Angular Momentum

The suggested lecture should take two or three class periods.

SUGGESTED LECTURE PRESENTATION

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Cite the difference between rotational and linear speed—examples of riding at various radial positions on a

merry-go-round, or the various speeds of different parts of a rotating turntable. A couple of coins on a

turntable, one close to the axis and the other near the edge, dramatically show the greater speed of the outer

one. Cite the motion of “tail-end Charlie” at the skating rink.

Circular Motion

Only for a rigid rotating system such as a solid turntable or a stiff spoke does the equation v = r apply—the

greater the distance from the axis of rotation, the greater the linear speed. Don’t be surprised to find students

applying this relationship to a nonrigid system, such as a system of planets. They are confused about

Mercury, which orbits relatively fast about the Sun, and Neptune, which orbits very slow. Horses running

around a circular track obey v = r only if they are constrained, like joined by a giant nonflexible spoke.

Railroad Train Wheels (RR Wheels)

A fascinating application of v = r is presented in the box on railroad train wheels. Fasten a pair of cups with

wide ends connected, and with small ends connected, and roll them along a pair of metersticks. Very

impressive! That’s my niece, Professor Cathy Candler, in Figure 8.8. The screencast on RR Wheels nicely

ties together the taper of cups with the taper of the rims of RR wheels.

Side point: Toilet tissue rolls are smaller in diameter than rolls of toilet tissue years ago. Since more tissue

makes a complete circle on the outer part of the roll, decreasing the diameter only slightly means appreciably

less tissue per roll.

While we’re on the subject of circles, you might ask why manhole covers are round (asked in the Check Point

on page 138 of the textbook). The answer is so that some moron type doesn’t drop them accidentally into the

manhole. If they were square, they could be tipped up on edge and dropped through the hole on the diagonal.

Similarly with ovals. But a circular hole will defy the most determined efforts. Of course there is a lip around

the inside of the manhole that cover rests on, making the diameter of the hole somewhat less than the

diameter of the cover.

Rotational Inertia

Compare the idea of inertia and its role in linear motion to rotational inertia (moment of inertia) in rotational

motion. The difference between the two involves the role of distance from a rotational axis. The greater the

distance of mass concentration, the greater the resistance to rotation. Discuss the role of the pole for the

tightrope walker in Figure 8.10. A novice tightrope walker might begin with the ends of the pole in

supporting slots, similar to the training wheels on a beginner’s bicycle. If the pole has adequate rotational

inertia, the slots mainly provide psychological comfort as well as actual safety. Just as the training wheels

could be safely removed without the rider’s knowledge, the slots could be safely removed without the

walker’s knowledge.

Show how a longer pendulum has a greater period and relate this to the different strides of long and short-

legged people. Imitate these strides yourself—or at least with your fingers walking across the desk.

DEMONSTRATION: This is a good one. Have two 1 meter pipes, one with two lead plugs in the

center, the other with plugs in each end. They appear identical. Weigh both to show the same weight.

Give one to a student (with plugs in ends) and ask her to rotate it about its center (like in Figure 8.9).

Have another student do the same with the pipe that has the plugs in the middle. Then have them

switch. Good fun. Then ask for speculations as to why one was noticeably more difficult to rotate

than the other.

DEMONSTRATION: As in Check Point 1 on page 138, have students try to balance on a

finger a long stick with a massive lead weight at one end. Try it first with the weight at

the fingertip, then with the weight at the top. Or you can use a broom, or long-handled

hammer. Relate this to the ease with which a circus performer balances a pole full of

people doing acrobatics, and cite how much more difficult it would be for the performer

to balance an empty pole!

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Also relate this demonstration and the continued adjustments you have to execute to keep the object balanced

to the similar adjustments that must be made in keeping a rocket vertical when it is first fired. Amazing! As

Tenny Lim and Mark Clark demonstrate in Figure 8.35, the Segway Transporter employs the same physics. The

Segway behaves as we do—when it leans forward

,it increases its speed to keep its CG above a point of

support.

DEMONSTRATION (As in Check Point 2 on page 138): Fasten a mass to

the end of a meterstick. A blob of clay works fine. Set it on end on your

lecture table, along with another meterstick with no attached mass. When you

let go of the sticks, they’ll topple to the tabletop. Ask which stick will reach

the tabletop first. [The plain stick wins due to the greater rotational inertia of

the clay-top stick. There is more to this than simply greater rotational inertia,

for torque is increased as well. If the clay is located at the middle of the stick,

the effects of greater torque and greater rotational inertia balance each other

and both sticks fall together.]

Discuss the variety of rotational inertias shown in Figure 8.15. Stress the formulas are for comparison, and

point out why the same formula applies to the pendulum and the hoop (all the mass of each is at the same

distance from the rotational axis). State how reasonable the smaller value is for a solid disk, given that much

of its mass is close to the rotational axis.

The rotational inertia of a thin-walled hollow sphere, missing from the drawings in Figure 8.15, is given by

Sanjay Rebello in Figure 8.16. Sanjay was an enormous help in developing the PowerPoint presentations of

Conceptual Physics. Thanx Sanjay!

DEMONSTRATION: Place a hoop and disk at the top of an incline and

ask which will have the greater acceleration down the incline. Do not

release the hoop and disk until students have discussed this with their

neighbors. Try other shapes after your class makes reasoned estimates.

Center of Mass and Torque

Depart a bit from the order of the chapter and begin a discussion of center of mass before treating torque. Do

this by tossing a small metal ball across the room, stating it follows a smooth curved path—a parabola. Then

pick up an irregularly shaped piece of wood, perhaps an L-shape, and state that if this were thrown across the

room it would not follow a smooth path, but would wobble all over the place—a special place, the place

presently being discussed—the center of mass, or center of gravity. Illustrate your definition with figures of

different shapes, first those where the center of mass lies within the object and then to shapes where the center

of mass lies outside the objects.

CHECK QUESTION: Where is the center of mass of a donut?

Consider the motion of a basketball tossed across the room when a heavy weight is attached to one side. The

wobble is evident. Likewise for suns with planet, a welcome feature to astronomy types.

Ask your students if they “have” a CG. Acknowledge that the CG in men is generally higher than in women

(1%-2%), mainly because women tend to be proportionally smaller in the upper body, and heavier in the

pelvis. On the average it lies about 6 inches above the crotch, a bit below the bellybutton. Interestingly

enough, the reason for the bellybutton being where it is relates to CG. A fetus turning in its mother’s womb

would rotate about its CG, the likely place for its umbilical cord. Standing erect with heavy side down

simulates an average woman. Standing with heavy side up, simulates an average man. A baseball bat likewise

makes this point. Interestingly, When we bend over, of course, the CG extends beyond the physical body.

Pass around a meterstick with a weight that can be suspended at different

places. This is “Torque Feeler,” an important activity that can be done in

your classroom as Mary Beth Monroe shows in the chapter photo opener.

Students hold the meterstick horizontally and note that different torques

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when the weight’s distance is varied. The difference between force and torque is felt! How nice when

students can feel physics!

Place an L-shaped body on the table and show how it topples—because its center of mass lies outside a point

of support. Sketch this on the board. Then stand against a wall and ask if it is possible for one to bend over

and touch their toes without toppling forward. Attempt to do so. Sketch this next to the L-shape as shown. By

now your board looks like the following:

Discuss a remedy for such toppling, like longer shoes or the wearing of snowshoes or skis. Sketch a pair of

skis on the feet of the person in your drawing. Seem to change the subject and ask why a pregnant woman

often gets back pains. Sketch a woman before and after getting pregnant, showing how the CG shifts

forward—beyond a point of support for the same posture. (This whole idea goes over much better in lecture

than as reading material, so is not found in this edition. So now you can introduce it as a fresh idea in class.)

Make a third sketch showing how a woman can adjust her posture so that her CG is above the support base

bounded by her feet, sketching lastly, the “marks of pain.” Ask the class how she could prevent these pains,

and if someone in class doesn’t volunteer the idea of wearing skis, do so yourself and sketch skis on her feet

in the second drawing.

Lead your class into an alternate solution, that of carrying a pole on her shoulder, near the end of which is a

load. Erase the skis and sketch in the pole and load as shown. Acknowledge the objection that she would have

to increase the mass of the load as the months go by, and ask what else can be done. Someone should

volunteer that she need only move the load closer to the end, which in effect shifts the overall CG in a

favorable direction. This routine is effective and sparks much class interest. However, you must be very

careful that you don’t offend your students, particularly your female students. Whenever you single out any

“minority”(?) you run the risk of offending members of that minority group or those sensitive to the feelings

of members of that group. We instructors, whether male or female ourselves, are for the most part conscious

of this and therefore make our examples as general as possible—mixing “shes” and “hes” whenever these

pronouns come up. But in the case of a person becoming pregnant, it’s a definite “she.” Any classroom

laughter that your presentation elicits should be, after all, directed to the situation and not particularly toward

the woman. In any event, we are in sad shape when we cannot laugh at ourselves occasionally.

CHECK QUESTION: Why does a hiker with a heavy backpack lean forward when standing or

walking?

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Return to your chalkboard sketches of L-shaped objects and relate their

tipping to the torques that exist. Point out the lever arms in the sketches.

CHECK QUESTION: An L-shaped object with CG marked by the X

rests on an incline as shown. Draw this on your paper and mark it

appropriately to determine whether the object will topple or not.

Comment: Be prepared for some students to sketch

in the “vertical” line through the CG perpendicular

to the slope as shown.

A simple example of this is to balance a pipe

(smoking kind) on your hand when held at an angle.

Cite examples involving the CG in animals and people—how the long tails of monkeys enable them to lean

forward without losing balance—and how people lean backwards when carrying a heavy load at their chests,

and how the coolie method with the load distributed in two parts suspended at the ends of a pole supported in

the middle is a better way.

Ask why a ball rolls down a hill. State that “because of gravity” is an

incomplete answer. Gravity would have it slide down the hill. The fact it rolls,

or rotates, is evidence of an unbalanced torque. Sketch this on your chalkboard.

DEMONSTRATION: Show how a “loaded disk” rolls up

an inclined plane. After class speculation, show how the

disk remains at rest on the incline. Modify your chalkboard

sketch to show how both the CG with respect to the

support point is altered, and the absence of a lever arm and

therefore the absence of a torque.

On rolling: Cliff birds

,lay eggs that are somewhat pear-shaped. This shape assures that the eggs roll in circles,

and don’t easily roll off precarious nesting places.

Discuss wrenches and clarify lever arm distances (Figure 8.20). Cite how a steering wheel is simply a

modified wrench, and why trucks and heavy vehicles before the advent of power steering used large-diameter

steering wheels.

DEMONSTRATION: Attempt to stand from a seated position without

putting your feet under the chair. Explain with center of gravity and

torques.

DEMONSTRATION: Do as Michael Bimmerle does and stick a piece of masking tape on an easy-

to-move door. Place the tape near the middle and when you pull the door, the tape becomes unstuck.

Progressively move the tape closer toward the edge away from the hinges and the tape sticks better.

Near the edge the tape will stick and open the door without pulling off. More torque for less force.

Seesaws

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Extend rotation to seesaws, as in Figures 8.18 and 8.19. Explain how participants on a seesaw can vary the

net torque by not only sliding back and forth, but by leaning. In this way the location of their CGs and hence

the lever arm distance is changed. Discuss the boy playing by himself in the park (Think and Discuss 111),

and how he is able to rotate up and down by leaning toward and away from the fulcrum.

DEMONSTRATION: Make a candle seesaw by trimming the wick so

both ends are exposed, and balance the candle by a needle through the

center. Rest the ends of the needle on a pair of drinking glasses. Light

both ends of the candle. As the wax drips, the CG shifts, causing the

candle to oscillate.

CHECK QUESTION: To balance a horizontal meterstick on one finger,

you’d place your finger at the 50-cm mark. Suppose you suspend an

identical meterstick vertically from one end, say the 0-cm end. Where

would you place your finger to balance the horizontal stick? [At the 25-cm

mark, where equal weights would each be 25 cm distant.]

DEMONSTRATION: Place a heavy plank on your lecture table so that it overhangs. Walk out on

the overhanging part and ask why you don’t topple. Relate this to a solitary seesaw example.

(Note the version of this in the NTQs.) This is also treated in the screencast ‘More on Torques.’

Here’s a neat application of CG that is not in the text, but is another NTQ. If you gently shake a basket of

berries, the larger berries will make their way to the top. In so doing the CG is lowered by the more compact

smaller berries settling to the bottom. You can demonstrate this with a Ping-Pong ball at the bottom of a

container of dried beans, peas, or smaller objects. When the container is shaken, the Ping-Pong ball surfaces,

lowering the CG of the system. This idea can be extended to the Ping-Pong ball in a glass of water. The CG

of the system is lowest when the Ping-Pong ball floats. Push it under the surface and the CG is raised. If you

do the same with something more dense than water, the CG is lowest when it is at the bottom.

Centripetal Force

Whirl an object tied to the end of a string overhead and ask if there is an outward or an inward force exerted

on the whirling object. Explain how no outward or centrifugal force acts on the whirling object (the only

outward directed force is the reaction force on the string, but not on the object). Emphasize also that

centripetal force is not a force in its own right, like gravity, but is the name for any force that pulls an object

into a curved path.

DEMONSTRATION: Swing a bucket of water in a vertical circle and show that

the water doesn’t spill (when centripetal force is at least equal to the weight of

the water). All your students have heard of this demonstration, but only a few

have actually seen it done. Why doesn’t the water fall at the top of the path? [The

answer is intriguing—it does! You have to pull the bucket down as fast as the

water falls. Similarly, a space shuttle above falls—just as much as the round

Earth curves! Nothing holds the water up; nothing holds the satellite up.] Both

are falling—nice physics!

The “trick” of this demonstration is to pull the bucket down as fast as the water falls so both fall the same

vertical distance in the same time. Too slow a swing produces a wet teacher. As said, the water in the

swinging bucket is analogous to the orbiting of a satellite. Both the swinging water and a satellite such as the

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orbiting space shuttle are falling. Because of their tangential velocities, they fall in a curve; just the right

speed for the water in the bucket, and just the right greater speed for the space shuttle. Tying these related

ideas together is good teaching!

CHECK QUESTION: A motorcycle runs on the inside of a bowl-shaped track (sketched in Think

and Explain 89). Is the force that holds the motorcycle in a circular path an inward- or outward-

directed force? [It is an inward-directed force—a centripetal force. An outward-directed force acts

on the inner wall, which may bulge as a result, but no outward-directed force acts on the

motorcycle.]

Centrifugal Force in a Rotating Frame

The concept of centrifugal force is useful when viewed from a rotating frame of reference. Then it seems as

real as gravity to an occupant—like inside a rotating space habitat. State how it differs from a real force in

that there is no agent such as mass. The magnetic force on a magnet, for example, is caused by the presence

of another magnet; the force on a charge is caused by the presence of another charge. Whereas a real force is

an interaction between one body and another, there is no reaction counterpart to centrifugal force. Distinguish

centrifugal force from the action-reaction pairs of forces at

the feet of an astronaut in a rotating habitat.

Discuss rotating space habitats. Show how g varies with both

the radial distance from the hub and the rotational rate of the

structure. The Earth has been the cradle of humankind; but

humans do not live in the cradle forever. We will likely leave

our cradle and inhabit structures of our own building;

structures that will serve as lifeboats for the planet Earth.

Their prospect is exciting.

DEMONSTRATION: Do as Diana Lininger Markham does in the chapter opening photos and swing

a drink in an overhead circle without spilling a drop. The surface of the liquid remains parallel to the

dish when freely swinging. (I witnessed this method of carrying cups of tea or other beverages

though crowded areas without spilling while visiting Turkey—very impressive.) A smaller gadget

that does the same, SpillNot, (P4-2500) is available from Arbor Scientific.

Angular momentum

Just as inertia and rotational inertia differ by a radial distance, and just as force and torque also differ by a

radial distance, so momentum and angular momentum also differ by a radial distance. Relate linear

momentum to angular momentum for the case of a small mass at a relatively large radial distance—an object

you swing overhead.

For the more general case, angular momentum is simply the product of rotational inertia I and angular

velocity . This is indicated in Figure 8.52.

DEMONSTRATION: With weights in your hand, rotate on a platform as shown in Figure 8.52. Simulate

the slowing down of the Earth when ice caps melt and spread out.

DEMONSTRATION: Show the operation of a gyroscope—either a model or a rotating bicycle

wheel as my late son James demonstrates in Figure 8.50.

Regarding the falling cat of Figure 8.54, J. Ronald Galli of Weber State University in Utah cautions that a

falling cat bends its spine to swing about and twist in an opposite direction to land feet first—all the while

maintaining a total angular momentum of zero.

Regarding Think and Explains 93 through 97: When answering these, demonstrate again on the rotating

platform, holding the weights over your head to simulate Earth washing toward the equator, melting ice caps

spreading toward the

,equator by lowering your hands in an outstretched position to simulate Earth and water

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flowing toward the equator. To simulate the effects of skyscraper construction, hold the weights short of fully

stretched, then extend your arms full-length.

Going Further with Rolling

Rolling things have two kinds of kinetic energy: That due to linear motion, and that due to rotational motion.

So an object rolling down an incline will lag behind a freely sliding object because part of a rolling object’s

kinetic energy is in rotation. If this is clear, then the following question is in order for your better students.

NEXT-TIME QUESTION: Which will roll with the greater acceleration down an incline, a can of

water or a frozen can of ice? Double credit for a good explanation of what is seen. [The can of liquid

will undergo appreciably more acceleration because the liquid is not made to rotate with the rotating

can. It in effect “slides” rather than rolls down the incline, so practically all the KE at the bottom is

in linear speed with next-to none in rotation. Fine, one might say, then if the liquid doesn’t rotate, the

can ought to behave as an empty can, with the larger rotational inertia of a “hoop” and lag behind.

This brings up an interesting point: The issue is not which can has the greater rotational inertia, but

which has the greater rotational inertia compared to its mass (note the qualifier in the legend of

Figure 8.14). The liquid content has appreciably more mass than the can that contains it; hence the

non-rolling liquid serves to increase the mass of the can without contributing to its rotational inertia.

It gives the can of liquid a relatively small rotational inertia compared to its mass.]

You can follow through by asking which can will be first in rolling to a stop once they meet a horizontal

surface. The can hardest to “get going” is also the can hardest to stop—so given enough horizontal distance,

the slowest can down the incline rolls farther and wins the race!

CHALLENGE: At the bottom of an incline are two balls of equal mass—a solid one and a thin-

walled hollow one. Each is given the same initial speed. Which rolls higher up the incline before

coming to a stop? [The answer is the hollow ball.] In terms of rotational inertia, whether a ball is

hollow or solid makes a big difference. A thin-walled hollow ball, having much of its mass along

its radius, has a relatively large rotational inertia (2/3MR2). A solid ball, having much of its mass

near its center, has less rotational inertia (2/5MR2). The ball with the greater rotational inertia out-

rolls a lower-inertia ball. Hence the hollow ball rolls farther up the incline before it comes to a stop.

Another way to look at it is in terms of energy. The balls begin their upward travel with kinetic energy of two

kinds—translational and rotational. Although their initial translational KEs are the same, the hollow ball

begins with more rotational KE due to its greater rotational inertia. So the hollow ball has more total KE at

the base of the incline, which means it must have more PE at the top. The hollow ball indeed goes higher.

Does mass make a difference? No. As with the mass of a pendulum bob, or the mass of a freely-falling object,

mass makes no difference. Sent rolling up an incline with equal speeds, any hollow ball will out-roll any solid

ball. That’s right—a tennis ball will roll higher than a bowling ball or a marble.

If you instead release both balls from a rest position at the top of an incline, the hollow ball “out-rests” the

solid ball, is slower to gain speed, and lags behind the solid ball. The solid ball reaches the bottom first.

Inertia is a resistance to change.

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Answers and Solutions for Chapter 8

Reading Check Questions

1. Tangential speed is measured in meters per second; rotational speed in RPM (revolutions per minute) or

rotations per second.

2. Only tangential speed varies with distance from the center.

3. The wide part has a greater tangential speed than the narrow part.

4. The wide part of the wheel has a greater radius than the narrow part, and hence a greater tangential speed

when the wheel rolls.

5. Rotational inertia is the resistance to a change in rotational motion, which is similar to plane inertia which is

a resistance to a change in velocity.

6. Rotational inertia also depends on the distribution of mass about an objects axis of rotation.

7. Rotational inertia increases with increasing distance.

8. Smallest when rotation is about the lead; next when at a right angle about the middle, and most when

about a right angle at the end.

9. Easier to get swinging when held closer to the massive end.

10. Bent legs have mass closer to the axis of rotation and therefore have less rotational inertia.

11.A solid disk has less rotational inertial and will accelerate more.

12. A torque tends to change the rotational motion of an object.

13. The lever arm is the shortest distance between the applied force and rotational axis.

14. For a balanced system, both clockwise and counterclockwise torques have equal magnitudes.

15. The stick ‘wobbles,’ spins really, about its CM (or CG).

16. A baseball’s CM and CG are at its center. Both are closer to the massive end of a baseball bat.

17. Your CG is beneath the rope.

18. The CM of a soccer ball is at its center.

19. For stable equilibrium the CG must be above a support base, and not extend beyond it.

20. The CG of the tower lies above and within the support base of the tower.

21. In attempting so, your CG extends beyond your support base, so you topple.

22. The direction of the force is inward, toward the center of rotation.

23. The force on the clothes is inward.

24. When the string breaks, no inward force acts and via with the law of inertia, the can moves in a straight

line.

25. No force is responsible, for you tend to move forward in a straight line and the car curves into you.

26. It’s called fictitious because there is no reaction counterpart to centrifugal force.

27. Rotational motion results in a centrifugal force that behaves like the force of gravity.

28. Linear momentum involves straight-line motion; angular momentum involves rotational motion.

29. The angular momentum of a system remains constant when no net torque acts.

30. Angular momentum remains the same, while her rate of spin doubles.

Think and Do

31. Open ended.

32. You’ll note the cups roll off the track!

33. Yes, this happens because the CG hangs below the point of support.

34. Women have lower CGs than men. Their feet are also smaller. So women have the advantage in the

toppling contest because their CG is more likely to be above a support base.

35. Facing the wall is more difficult! For both sexes the CG extends beyond the support base defined by the

balls of the feet to the wall.

36. Your fingers will meet in the center. When a finger is farther from the center than the other, it presses with

less force on the stick and slides. The process alternates until both fingers are at the center.

37. If the coin is on the line to the center of rotation, the ‘normal’ force on the coin provides a centripetal force

to keep it steadily rotating.

Plug and Chug

38. = 0.2 m 50 N = 10 m.N.

39. = 0.5 m 50 N = 25 m.N.

40. F = (2 kg)(3 m/s)2/2.5 m = 7.2 N.

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41. F = (80 kg)(3 m/s)2/2 m = 360 N.

42. Ang momentum = mvr = (80 kg)(3 m/s)(2 m) = 480 kg.m2/s.

43. Twice: Ang momentum = mvr = (80 kg)(6 m/s)(2 m) = 960 kg.m2/s.

Think and Solve

44. In accord with v = r, the greater the radius (or diameter), the greater the tangential speed. So the wide

part rolls faster. It rolls 9/6 = 3/2 = 1.5 times faster.

45.(a) Torque = force lever arm = (0.25 m) (80 N) = 20 N.m.

(b) Force = 200 N. Then (200 N)(0.10 m) = 20 N.m.

(c) Yes. These answers assume that you are pushing perpendicular to the wrench handle. Otherwise,

you would need to exert more

,force to get the same torque.

46.The mass of the rock is 1 kg. (A reverse of the Check Point on page 142.)

47. The 1-kg mass weighs 10 N. At the 50-cm mark, torque = 10 N 0.5 m = 5 N.m.

At the 75-cm mark, torque = 10 N 0.75 m = 7.5 N.m, and at the 100-cm mark,

torque = 10 N 1.0 m = 10 N.m. So at the 75-cm mark the torque is 7.5/5 = 1.5 times as much, and at

the 100-cm mark the torque is twice what it is at the 50-cm mark.

48. From F = mv2/r, substituting, T = mv2/L. (a) Rearranging, m = TL/v2.

(b) Substituting numerical values, m = (10N)(2m)/(2m/s)2 = 5 kg.

49.The artist will rotate 3 times per second. By the conservation of angular momentum, the artist will increase

rotation rate by 3. That is

Ibefore = Iafter

Ibefore = [(I/3)I(3)]after

50. (a) In the absence of an unbalanced external torque the angular momentum of the system is conserved.

So (angular momentum)initial = (angular momentum)final., where angular momentum is mvL.

From mv0L = mvnew(0.33L) we get vnew = v0(L/0.33L) = v0/0.33 = 3.0v0.

(b) vnew = v0/(0.33) = (1.0 m/s)/(0.33) = 3.0 m/s.

Think and Rank

51. B, C, A

52. C, A, B

53. B, A, C

54. B, C, A

55. C, A, B

Think and Explain

56. Sam’s rotational speed , RPMs, remains the same, assuming the Ferris wheel is powered and not “free

wheeling.” Sam’s tangential speed, v = r is half because the radial distance r is half. Answers are

different because tangential speed v depends on distance from the spin axis, while rotational speed

does not.

57.Sue’s tires have a greater rotational speed for they have to turn more times to cover the same distance.

58.For the same twisting speed , the greater distance r means a much greater speed v.

59.The amount of taper is related to the amount of curve the railroad tracks take. On a curve where the

outermost track is say 10% longer than the inner track, the wide part of the wheel will also have to be at

least 10% wider than the narrow part. If it’s less than this, the outer wheel will rely on the rim to stay on

the track, and scraping will occur as the train makes the curve. The “sharper” the curve, the more the

taper needs to be on the wheels.

60.Yes, rotational inertia is enhanced with long legs. The bird’s foot is directly below the bird’s CM.

61.Rotational inertia and torque are most predominantly illustrated with this vehicle, and the conservation of

angular momentum also plays a role. The long distance to the front wheels means greater rotational

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inertia of the vehicle relative to the back wheels, and also increases the lever arm of the front wheels

without appreciably adding to the vehicle’s weight. As the back wheels are driven clockwise, the chassis

tends to rotate counterclockwise (conservation of angular momentum) and thereby lift the front wheels off

the ground. The greater rotational inertia and the increased clockwise torque of the more distant front

wheels counter this effect.

62.The bowling ball wins. A solid sphere of any mass and size beats both a solid cylinder and a hollow ball of

any mass and size. That’s because a solid sphere has less rotational inertia per mass than the other

shapes. A solid sphere has the bulk of its mass nearer the rotational axis that extends through its center

of mass, whereas a cylinder or hollow ball has more of its mass farther from the axis. The object with the

least rotational inertia per mass is the “least lazy” and will win races.

63.The ball to reach the bottom first is the one with the least rotational inertia compared with its mass—that’s

the softball (as in the answer to the previous question).

64.The lever arm is the same whether a person stands, sits, or hangs from the end of the seesaw, and

certainly the person’s weight is the same. So the net torque is the same also.

65.No, for by definition, a torque requires both force and a lever arm.

66.In the horizontal position the lever arm equals the length of the sprocket arm, but in the vertical position, the

lever arm is zero because the line of action of forces passes right through the axis of rotation. (With

cycling cleats, a cyclist pedals in a circle, which means they push their feet over the top of the spoke and

pull around the bottom and even pull up on the recovery. This allows torque to be applied over a greater

portion of the revolution.)

67. No, because there is zero lever arm about the CM. Zero lever arm means zero torque.

68. Friction between the ball and the lane provides a torque, which spins the ball.

69.A rocking bus partially rotates about its CM, which is near its middle. The farther one sits from the CM, the

greater is the up and down motion—as on a seesaw. Likewise for motion of a ship in choppy water or an

airplane in turbulent air.

70.With your legs straight out, your CG is farther away and you exert more torque sitting up. So sit-ups are

more difficult with legs straight out, a longer lever arm.

71.The long drooping pole lowers the CG of the balanced system—the tightrope walker and the pole. The

rotational inertia of the pole contributes to the stability of the system also.

72. You bend forward when carrying a heavy load on your back to shift the CG of you and your load above

the area bounded by your feet—otherwise you topple backward.

73.The wobbly motion of a star is an indication that it is revolving about a center of mass that is not at its

geometric center, implying that there is some other mass nearby to pull the center of mass away from the

star’s center. This is one of the ways in which astronomers have discovered planets existing around stars

other than our own.

74. Two buckets are easier because you may stand upright while carrying a bucket in each hand. With two

buckets, the CG will be in the center of the support base provided by your feet, so there is no need to

lean. (The same can be accomplished by carrying a single bucket on your head.)

75.The Earth’s atmosphere is a nearly spherical shell, which like a basketball, has its center of mass at its

center, i.e., at the center of the Earth.

76. The CG of a ball is not above a point of support when the ball is on an

incline. The weight of the ball therefore acts at some distance from the point

of support which behaves like a fulcrum. A torque is produced and the ball

rotates. This is why a ball rolls down a hill.

77. It is dangerous to pull open the upper drawers of a fully-loaded file cabinet

that is not secured to the floor because the CG of the cabinet can easily be shifted beyond the support

base of the cabinet. When this happens, the torque that is produced causes the cabinet to topple over.

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78.An object is stable when its PE must be raised in order to tip it over, or equivalently, when its PE must be

increased before it can topple. By inspection, because of its narrow base the first cylinder undergoes the

least change in PE compared to its weight in tipping. So it is the least stable. The third truncated pyramid

requires the most work, so it is the most stable.

79.The CG of truck at the left on the lower part of the incline, is not above its support base, and will tip. The

CGs of the two other trucks are above their support bases and won’t tip. So only the first of the three

trucks will tip.

80.In accord with the equation for centripetal force, twice the speed corresponds to four times the force.

81.No—in accord with Newton’s first law, in the absence of force a moving object follows a straight-line path.

82.Yes. Letting the equation for centripetal force guide our thinking, increased speed at the same radial

distance means greater centripetal force. If this greater centripetal force isn’t provided, the car will skid.

83.Newton’s first and third laws provide a straight-forward explanation. You tend to move in a straight line

(Newton’s first law) but are intercepted by the door. You press

,against the door because the door is

pressing against you (Newton’s third law). The push by the door provides the centripetal force that keeps

you moving in a curved path. Without the door’s push, you wouldn’t turn with the car—you’d move along

a straight line and be “thrown out.” Explanation doesn’t require invoking centrifugal force.

84.On a banked road the normal force, at right angles to the road surface, has a horizontal component that

provides the centripetal force. Even on a perfectly slippery surface, this component of the normal force

can provide sufficient centripetal force to keep the car on the track.

85.A car can remain on a perfectly slippery banked track if the horizontal component of its normal force is

sufficient to provide the required centripetal force.

86.There is no component of force parallel to the direction of motion, which work requires.

87.In accord with Newton’s first law, at every moment her tendency is to move in a straight-line path. But the

floor intercepts this path and a pair of forces occur; the floor pressing against her feet and her feet

pressing against the floor—Newton’s third law. The push by the floor on her feet provides the centripetal

force that keeps her moving in a circle with the habitat. She senses this as an artificial gravity.

88.

89. (a) Except for the vertical force of friction, no other vertical force except the weight of the motorcycle +

rider exists. Since there is no change of motion in the vertical direction, the force of friction must be equal

and opposite to the weight of motorcycle + rider. (b) The horizontal vector indeed represents the normal

force. Since it is the only force acting in the radial direction, horizontally, it is also the centripetal force. So

it’s both.

90. The resultant is a centripetal force.

91. As you crawl outward, the rotational inertia of the system increases (like the

masses held outward in Figure 8.52). In accord with the conservation of angular

momentum, crawling toward the outer rim increases the rotational inertia of the

spinning system and decreases the angular speed.

92.

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93. Soil that washed down the river is being deposited at a greater distance from the Earth’s rotational axis.

Just as the man on the turntable slows down when one of the masses is extended, the Earth slows in its

rotational motion, extending the length of the day. The amount of slowing, of course, is exceedingly

small.

94. Rotational inertia would increase. By angular momentum conservation, the rotation of the Earth would

slow (just as a skater spins slower with arms outstretched), tending to make a longer day.

95. In accord with the conservation of angular momentum, as the radial distance of mass increases, the

angular speed decreases. The mass of material used to construct skyscrapers is lifted, slightly increasing

the radial distance from the Earth’s spin axis, which tends to slightly decrease the Earth’s rate of rotation,

making the days a bit longer. The opposite effect occurs for falling leaves as their radial distance from

the Earth’s axis decreases. As a practical matter, these effects are entirely negligible!

96. In accord with the conservation of angular momentum, if mass moves closer to the axis of rotation,

rotational speed increases. So the day would be ever so slightly shorter.

97. In accord with the conservation of angular momentum, if mass moves farther from the axis of rotation, as

occurs with ice caps melting, rotational speed decreases. So the Earth would slow in its daily rotation.

98. Without the small rotor on its tail, the helicopter and the main rotor would rotate in opposite directions.

The small rotor on the tail provides a torque to offset the rotational motion that the helicopter would

otherwise have.

99. Gravitational force acting on every particle by every other particle causes the cloud to condense. The

decreased radius of the cloud is then accompanied by an increased angular speed because of angular

momentum conservation. The increased speed results in many stars being thrown out into a dish-like

shape.

100. In accord with Newton’s first law, moving things tend to travel in straight lines. Surface regions of a

rotating planet tend to fly off tangentially, especially at the equator where tangential speed is greatest.

More predominantly, the surface is also pulled by gravity toward the center of the planet. Gravity wins,

but bulging occurs at the equator because the tendency to fly off is greater there. Hence a rotating planet

has a greater diameter at the equator than along the polar axis.

Think and Discuss

101. Large diameter tires mean you travel farther with each revolution of the tire. So you’ll be moving faster

than your speedometer indicates. (A speedometer actually measures the RPM of the wheels and

displays this as mi/h or km/h. The conversion from RPM to the mi/h or km/h reading assumes the wheels

are a certain size.) Oversize wheels give too low a reading because they really travel farther per

revolution than the speedometer indicates, and undersize wheels give too high a reading because the

wheels do not go as far per revolution.

102. The tangential speeds are equal, for they have the same speed as the belt. The smaller wheel rotates

twice as fast because for the same tangential speed, and r half, must be twice. v(big wheel) = r;

v(small wheel) = (r/2 2).

103. Two conditions are necessary for mechanical equilibrium, F = 0 and Torque = 0.

104. Before leaving the cliff, front and back wheels provide the support base to support the car’s weight. The

car’s CM is well within this support base. But when the car drives off the cliff, the front wheels are the first

to leave the surface. This shifts the support base to the region between the rear wheels, so the car tips

forward. In terms of torques, before driving off the cliff, the torques are balanced about the CM between

the front and back wheels. But when the support force of the front wheels is absent, torque due to the

support force of the rear wheels rotates the car forward about its CM making it nose forward as shown.

At high speed, the time that this torque acts is less, so less rotation occurs as it falls.

105. Friction by the road on the tires produces a torque about the car’s CM. When the car accelerates

forward, the friction force points forward and rotates the front of the car upward. When braking, the

direction of friction is rearward, and the torque rotates the car in the opposite direction so the rear end

rotates upward (and the nose downward).

85

106. If you roll them down an incline, the solid ball will roll faster. (The hollow ball has more rotational inertia

compared with its weight.)

107. Don’t say the same, for the water slides inside the can while the ice is made to roll along with the can.

When the water inside slides, it contributes weight rather than rotational inertia to the can. So the can of

water will roll faster. (It will even beat a hollow can.)

108. Lightweight tires have less rotational inertia, and are easier to get up to speed.

109. Advise the youngster to use wheels with the least rotational inertia—lightweight solid ones without

spokes (more like a disk than hooplike).

110. In all three cases the spool moves to the right. In (a) there is a torque about the point of contact with the

table that rotates the spool clockwise, so the spool rolls to the right. In (b) the pull’s line of action

extends through (not about) the point of table contact, yielding no lever arm and therefore no torque; but

with a force component to the right; hence the spool slides to the right without rolling. In (c) the torque

produces clockwise rotation so the spool rolls to the right.

111. The weight of the boy is counterbalanced by the weight of the board, which can

,be considered to be

concentrated at its CG on the opposite side of the fulcrum. He is in balance when his weight multiplied by

his distance from the fulcrum is equal to the weight of the entire board multiplied by the distance between

the fulcrum and the midpoint (CG) of the board. (How do the relative weight of boy and board relate to

the relative lever arms?)

112. The top brick would overhang 3/4 of a brick length as shown. This is best

explained by considering the top brick and moving downward; i.e., the CG of

the top brick is at its midpoint; the CG of the top two bricks is midway

between their combined length. Inspection will show that this is 1/4 of a brick

length, the overhang of the middle brick. (Interestingly, with a few more

bricks, the overhang can be greater than a brick length, and with a limitless

number of bricks, the overhang can be made as large as you like.)

113. The track will remain in equilibrium as the balls roll outward and until the ball rolls off the track. This is

because the CG of the system remains over the fulcrum. For example, suppose the billiard ball has twice

the mass of the golf ball. By conservation of momentum, the twice-as-massive ball will roll outward at half

the speed of the lighter ball, and at any time be half as far from the starting point as the lighter ball. So

there is no CG change in the system of the two balls. So the torques produced by the weights of the balls

multiplied by their relative distances from the fulcrum are equal at all points—because at any time the

less massive ball has a correspondingly larger lever arm.

114. The center of mass of the bird is slightly below its beak, the point at which it rests on Diana’s finger. So

the bird is “hanging” on Diana’s finger. This is accomplished by lead or some very dense metal

embedded in the wing tips of the bird.

115. The equator has a greater tangential speed than latitudes north or south. When a projectile is launched

from any latitude, the tangential speed of the Earth is imparted to the projectile, and unless corrections

are made, the projectile will miss a target that travels with the Earth at a different tangential speed. For

example, if the rocket is fired south from the Canadian border toward the Mexican border, its Canadian

component of speed due to the Earth’s turning is smaller than Earth’s tangential speed further south. The

Mexican border is moving faster and the rocket falls behind. Since the Earth turns toward the east, the

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rocket lands west of its intended longitude. (On a merry-go-round, try tossing a ball back and forth with

your friends. The name for this alteration due to rotation is the Coriolis effect.)

116. Acceleration caused this force. His body was accelerated by support at his head, but his brain was not so

supported. In effect, the back of his head exerted a force on his head, with the cause being too-great an

acceleration.

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9 Gravity

Conceptual Physics Instructor’s Manual, 12th Edition

9.1 The Universal Law of Gravity

9.2 The Universal Gravitation Constant, G

9.3 Gravity and Distance: The Inverse-Square Law

9.4 Weight and Weightlessness

9.5 Ocean Tides

Tides in Earth and Atmosphere

Tidal Bulges on the Moon

9.6 Gravitational Fields

Gravitational Field Inside a Planet

Einstein’s Theory of Gravitation

9.7 Black Holes

9.8 Universal Gravitation

Dutch friend Ed Van den Berg uses balls to pose questions about the inverse-square law in the opening

photo to this chapter. I am still moved by photos of astronauts performing space walks! Photo 3 shows Eric

Mazur engaging students in class. When engagement occurs between professor and student, learning can

occur. Without this engagement, likely less learning occurs. Hats off to Eric! Tomas Brage, physics

department head at Lund University in Sweden shows a version of the Cavendish apparatus to measure G.

The personality profile is of Eric Mazur.

This chapter begins with a historical approach and ends on an astronomical theme. It offers a good place to

reiterate the idea of a scientific theory, and comment on the all-too common and mistaken idea that because

something has the status of scientific theory, it is somehow short of being valid. This view is evident in

those who say, “But it’s only a theory.” Bring the essence of the first and last footnotes in the chapter into

your discussion (about scientific homework and being unable to see radically new ways of viewing the

world). The last chapter on Cargo Cult Science of Feynman’s book, Surely You’re Joking Mr. Feynman

(Norton, 1985), expands nicely on this. (When I first read this delightful book I allowed myself only one

chapter per day—to extend the pleasure. It’s THAT good!)

Kepler’s 3rd law follows logically from Newton’s law of gravitation. Equate the force of gravity between

planet m and the Sun M to the centripetal force on m. Then,

GmM

r2

mv 2

r

m 2r /T

r

2

where the speed of the planet is 2 per period T. Cancel and collect terms,

GM

4 2

r3

T 2

This is Kepler’s 3d law, for GM/42 is a constant.

The idea of the force field is introduced in this chapter and is a good background for the electric field

treated later in Chapter 22. The gravitational field here is applied to regions outside as well as inside the

Earth.

In the text I say without explanation that the gravitational field increases linearly with radial distance inside

a planet of uniform density. Figure 9.24 shows that the field increases linearly from zero at its center to

maximum at the surface. This is also without explanation. The text states that “perhaps your instructor will

provide the explanation.” Here it is: We know that the gravitational force F between a particle m and a

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spherical mass M, when m is outside M is simply F = GmMd2. But when m is inside a

uniform density solid sphere of mass M, the force on m is due only to the mass M´

contained within the sphere of radius r < R, represented by the dotted line in the figure.

Contributions from the shell > r cancel out (Figure 9.25, and again for the analogous case

of the electric field in Figure 22.20, later in the book). So, F = GmM´r2. From the ratio of

M´M, you can show that M´ = Mr3R3. [That is, M´M = V´V = (43 πr3)(43 πR3) = r3R3.]

Substitute M´ in Newton’s equation for gravitation and we get F = GmMrR3. All terms on

the right are constant except r. So F = kr; force is linearly proportional to radial distance when r < R.

Interestingly enough, the condition for simple harmonic motion is that the restoring force be proportional to

displacement, F = kr. Hence the simple harmonic motion of one who falls in the tunnel through the Earth.

(Hence also the simple harmonic motion of one who slides without friction to-and-fro along any straight

line tunnel through any part of the Earth. The displacement is then the component r sin .) The period of

simple harmonic motion, T = 2π√(R3GM) is the same as that of a satellite in close circular orbit about the

Earth. Note that it is independent of the length of the tunnel. I treat falling through a vertical tunnel in the

screencast Tunnel Through Earth.

You can compare the pull of the Moon that is exerted on you with the pull exerted by more local masses,

via the gravitational equation. Consider the ratio of the mass of the Moon to its distance squared:

7.4 1022 kg(4 105 km)2 = 5 1012 kgkm2.

This is a sizeable ratio, one that buildings in your vicinity cannot match. (City buildings of greatest mass

are typically on the order of 106 or 107 kilograms.) However, if you stand 1 kilometer away from

the foot of a mountain of mass 5 x 1012 kilograms (about the mass of Mount Kilimanjaro), then the pull of

the mountain and the pull of the Moon are about the same on you. Simply put, with no friction you would

tend to gravitate from your spot

,These are quickie tutorials that complement the textbook, often with visual

explanations of concepts. Creating these has been my passion for the past two years.

The student ancillary PRACTICING PHYSICS BOOK can serve as a tutor on the side. At CCSF

it is carried in the student bookstore as “recommended but not required“ and used by about one-third

of the students taking the course. Answered practice pages are in the back of the book, reduced in

xiv

size. I consider the Practicing Physics Book my best pedagogical creation, along with my new

screencasts.

The Conceptual Physics package of text and ancillaries lend themselves to teaching by way of the

3-stage LEARNING CYCLE, developed by Robert Karplus some 40 years ago.

EXPLORATION—giving all students a common set of experiences that provide opportunities for

student discussion. Activities are both in the Laboratory Manual and the chapter-end Think-and-

Do’s in the textbook.

CONCEPT DEVELOPMENT—lectures, textbook reading, doing practice pages from Practicing

Physics, viewing Hewitt-Drew-it! screencasts, and class discussions.

APPLICATION—doing end-of-chapter exercises and problems, Next-Time Questions,

experiments from the Laboratory Manual, and for your math-savvy students, Problem Solving in

Conceptual Physics.

The first step of the learning cycle increases the effectiveness of instruction by insuring students

have first-hand experience with much of the phenomena to be discussed. For example, before hearing

a lecture on torques, have your students pass around a meterstick with a weight dangling from a string

(as nicely shown by Mary Beth Monroe in the third chapter-opening photo for Chapter 8 on page

132). Holding the meterstick horizontally with the weight near the end, students feel the greater effort

needed to rotate the stick. When the weight is positioned closer to their hand, rotational effort is much

less. Aha, now you’re ready to discuss the concept of torque, and to distinguish it from weight.

Many of the suggested lectures in this manual will require more than one class period, depending

on your pace of instruction and what you choose to add or omit. The lectures of each instructor, of

course, must be developed to fit his or her style of teaching. My suggested lectures may or may not be

useful to you. If you’re new to teaching conceptual physics and your lecture tendency is to lean on

chalkboard derivations, you may find them quite useful, and a means of jumping off and developing

your own non-computational way of teaching.

DVDs of my classroom lectures are described on page xv.

Please bring to my attention any errors you find in this manual, in the text, in the test bank, or in

any of the ancillaries. I welcome correspondence suggesting improvements. E-mail,

Pghewitt@aol.com. Good luck in your course!

xv

ANCILLARY PACKAGE FOR THE 12th EDITION

For Instructors

INSTRUCTOR RESOURCE DVD (IR-DVD):

Contains a wealth of goodies, including all textbook illustrations, tables from the text, interactive

presentation applets and animations, parts of Hewitt’s videoed lectures and demos, in-class clicker

questions for use with PRS and HiTT Classroom Response Systems, most of which can be edited and

customized for classroom presentations. The main ancillaries are:

NEXT-TIME QUESTIONS:

These are insightful questions, with answers, to central ideas in physics. Each was formerly

published as Figuring Physics in the American Association of Physics Teachers (AAPT)

magazine, The Physics Teacher. Aside from projecting these via PowerPoint or otherwise,

consider printing copies and posting for student display. Allow a sufficient ‘wait time’

before posting solutions. There are Next-Time-Questions for every chapter.

TEST BANK:

Contains more than 2000 multiple-choice questions, categorized by level of difficulty and

skill type. The friendly graphical interface enables you to easily view, edit, and add

questions, transfer questions to tests, and print tests in a variety of fonts and forms. Search

and sort features let you quickly locate questions and arrange them in a preferred order. A

built-in question editor gives you power to create graphs, import graphics, insert

mathematical symbols and templates, and insert variable numbers or text.

VIDEOS Conceptual Physics Alive!:

Features my classroom lectures while teaching Conceptual Physics at the University of Hawaii in

1989-1990. These are available in DVD or rental of individual lessons streamed from Arbor

Scientific, (www.arborsci.com) P.O. Box 2750, Ann Arbor, MI 48106-2750.

Additionally, the 12-lecture set of videos taken at CCSF in 1982 have been resurrected by Marshall

Ellenstein, and with other “goodies,“ comprise a 3-disc DVD set “Conceptual Physics Alive! The San

Francisco Years.” The goodies include the 60-minute Teaching Conceptual Physics, which

documents how I teach physics conceptually, and the 55-minute Lecture Demonstrations in

Conceptual Physics, which is more classroom footage with emphasis on demonstrations (most of

which are in the “Suggested Lectures“ in this manual). Another goodie is a 45-minute general-interest

opening lecture, The Fusion Torch and Ripe Tomatoes. At one-quarter the price of the Hawaii tapes,

they are available from Media Solutions, 1128 Irving St., San Francisco, CA 94122

(www.mediasolutions-sf.com/hewitt/sfyorders.pdf).

For Students

PRACTICING PHYSICS BOOK:

This booklet of more than 100 practice pages helps students to learn concepts. This is very different

from traditional workbooks that are seen as drudgery by students. These are insightful and interesting

activities that prompt your students to engage their minds and DO physics. They play the role of a

tutor when you post solutions at appropriate times (like the posting of Next-Time Questions). Practice

Book solutions, reduced to one-half size, are included at the end of the book. Practicing Physics is

xvi

low priced and can be offered as a suggested supplement to the textbook in your

student bookstore. ISBN: 0805391983.

PROBLEM SOLVING BOOK:

Now in its Third Edition, this is my latest effort to boost the teaching of physics, with co-author Phil

Wolf. It is meant for those who wish a stronger problem-solving component in their teaching, and

particularly for those wishing to extend Conceptual Physics to an algebra-trig course. I feel the

novelty of the problems and the simple method employed for solving them will be as important to the

way physics is taught as was Conceptual Physics when it was introduced some 40 years ago. No

longer does the instructor have to plead with students to complete the problem before plugging in

numerical values. Instructors no longer have to plead with students to show their work. Why?

Because the phrasing of the problems makes these concerns mandatory. Variables are given in letters,

not numbers (mass is m, velocity is v, and so forth). Not until a second part of a problem are

numerical values given and a numerical solution asked for. Each chapter set of problems is followed

by a second set of Show-That Problems, which give the numerical answer and ask the student to show

how it comes about. I’ve been using this method for decades when teaching the algebra-trig and

calculus based physics courses. Now it is available to users of Conceptual Physics. Solutions to the

problems are given on the website in the Instructor’s Resource area. At your discret ion you can

post solutions for your students. ISBN: 0805393773.

LABORATORY MANUAL:

This manual, written by myself and mostly by Dean Baird, is rich with simple activities to precede the

coverage of course material, as well as experiments that are a follow through to course material. It

also employs the computer in tech labs. New to this 12th edition, the instructor manual for the

laboratory manual is separate from this manual.

xvii

Flexibility of Material for

,toward the mountain—but you experience no tendency at all to gravitate

from your spot toward the Moon! That’s because the spot you stand on undergoes the same gravitational

acceleration toward the Moon as you do. Both you and the whole Earth are accelerating toward the Moon.

Whatever the lunar force on you, it has no tendency to pull you off a weighing scale—which is the essence

of Think and Discuss 97 and 98. This is not an easy notion to grasp—at first.

Not covered in this edition is the inverse-cube nature of tidal forces. This follows from subtracting the tidal

force on the far side of a body from the tidal force on the near side. Consider a kilogram of water on the

side of the Earth nearest the Moon that is gravitationally attracted to the Moon with a greater force than a

kilogram of water on the side of the Earth farthest from the Moon. The difference in force per kilogram of

mass, ∆Fm, which we’ll call the tidal force TF is

TF = Fd+R - Fd-R

GM

1

d R 2

1

d R 2

4GMdR

d2 R2 2

where M is the Moon’s mass, (d+R) is the distance to the far side of Earth, (d-R) is the distance to the near

side.

When d is very much greater than R, the (d2-R2)2 is very nearly equal to d4. Then the inverse-cube nature of

tidal force is evident, for

TF ~ 4GMR

d3

.

90

Some interesting results occur when calculating the tidal force of the Moon on planet

Earth. TF is 2.2 10-6 N/kg. In contrast TF of an overhead Moon on a person on Earth

is 3 10-13 N/kg, a hundred million times weaker because of the tiny differences in

pulls across the body. The tidal force of the Earth on the same person is 6 10-6 N/kg,

more than the Moon’s influence. And as the text reports, the tidal force due to a 1-kg

mass held 1 m above your head is about 200 times as much effective as the Moon!

Have those who believe the tidal effects of planets influence people make the

calculations themselves.

A brief treatment of black holes is included in this chapter. The idea that light is influenced by a

gravitational field isn’t treated until Chapter 36, so may merit further explanation. You’ll probably want to

acknowledge that light bends in a gravitational field as does a thrown baseball. We say light travels in

straight lines much for the same reason that some people say that a high-speed bullet doesn’t curve

downward in the first part of its trajectory. Over short distances the bullet doesn’t appear to drop only

because of its high speed and the short time involved. Likewise for light’s speed, which we don’t notice

because of the vast distance it travels in the brief time it’s in the strong part of Earth’s gravitational field.

Look ahead to the treatment of this idea in Figure 36.6.

Black holes at the center of galaxies are bigger than those found in binary star systems. The biggest

recently reported galactic black holes have equivalent masses of some 10 to 40 billion Suns.

Dark matter is briefly mentioned in this chapter and is discussed in Chapter 11. Present consensus among

astrophysicists is that dark energy is working against the force of gravity to accelerate the expansion of the

universe. These findings downplay the oscillating universe scenario speculated about in the earlier editions

of this text (although there remains speculation that the present outward acceleration may change to rapid

deceleration and lead to a Big Crunch). The concepts of dark matter and dark energy are at the forefront of

physics at this point, and are quite mysterious. Dark matter is out of sight, but not out of mind.

This chapter is prerequisite to the following chapter on satellite motion. It also provides useful background

information for Chapter 22 (the inverse-square law, and the analogy between a gravitational and electric

field) and Chapter 36 (general relativity). This chapter may be skipped without complicating the treatment

of material in Chapters other than 22 and 36. It’s an especially interesting chapter because the material is

high interest, historical, quite understandable, and closely related to areas of space science that are currently

in the public eye.

Practicing Physics Book:

• Inverse-Square Law • Our Ocean Tides

Problem Solving Book:

Some 30 problems

Laboratory Manual:

No labs for this chapter

Next-Time Questions:

• Earth-Moon Cable • Giant Plane

• Moon Tides • Body Tide

• Solar Black Hole • Gravity Force on Shuttle

• Earth Rise • Weight

• Normal Force and Weight

Hewitt-Drew-It! Screencasts:

• Weight/Weightlessness • Tunnel Through Earth

• Gravity • Ocean Tides

• Gravity Inside Earth

91

SUGGESTED LECTURE PRESENTATION

Begin by briefly discussing the simple codes and patterns that underlie the complex things around us,

whether musical compositions or DNA molecules, and then briefly describe the harmonious motion of the

solar system, the Milky Way and other galaxies in the universe—stating that the shapes of the planets,

stars, and galaxies, and their motions are all governed by an extremely simple code, or if you will, a

pattern. Then write the gravitational equation on the board. Give examples of bodies pulling on each other

to convey a clear idea of what the symbols in the equation mean and how they relate. (Acknowledge that

many other texts and references use the symbol r instead of the d in this text. The r is used to indicate the

radial distance from a body’s CG, and to emphasize the center-to-center rather than surface-to-surface

nature for distance, and to prepare for r as a displacement vector. We don’t set our plow that deep,

however, and use d for distance.)

Inverse-Square Law

Discuss the inverse-square law and go over Figures 9.5 and 9.6 or their equivalents with candlelight or

radioactivity.

Plot to scale an inverse-square curve on the board, showing the steepness of the curve—14, 19, and 116, for

twice, three times, and four times the separation distance. This is indicated in Figures 9.5 and 9.6. (You

may return to the curve of Figure 9.6 when you explain tides.)

CHECK QUESTIONS: A photosensitive surface is exposed to a point source of light that is a

certain distance away. If the surface were instead exposed to the same light four times as far away,

how would the intensity upon it compare? A radioactive detector registers a certain amount of

radioactivity when it is a certain distance away from a small piece of uranium. If the detector is

four times as far from the uranium, how will the radioactivity reading compare?

CHECK QUESTIONS: How is the gravitational force between a pair of planets altered when one

of the planets is twice as massive? When both are twice as massive? When they are twice as far

apart? When they are three times as far apart? Ten times as far apart? [The screencast on Gravity

explains this.]

CHECK QUESTION: What do you say to a furniture mover who claims that gravity increases

with increased distance from the Earth, as evident to him when he’s carrying heavy loads up

flights of stairs?

Weight and Weightlessness

Note that we define weight as a support force. Even in a gravity-free region inside a rotating toroid, you’d

experience weight. So weight needn’t always be related to gravity. Discuss weightlessness and relate it to

the queasy feeling your students experience when in a car that goes too fast over the top of a hill. State that

this feeling is what an astronaut is confronted with all the time in orbit! Ask how many of your class would

still welcome the opportunity to take a field trip to Cape Canaveral and take a ride aboard an orbiting

vehicle. What an exciting prospect!

A marvelous space station called Skylab was in orbit in the 1970s. When it underwent unavoidable orbital

decay the space shuttle was not yet in operation to give it the boost it needed to keep it in orbit. Quite

unfortunate. But fortunately, there is movie footage of antics of astronauts

,aboard Skylab. The NASA film

is “Zero g,” which I showed every semester in my classes. It not only is fascinating in its shots of astronaut

acrobatics in the orbiting lab, but illustrates Newton’s laws as they apply to intriguing situations. The film

shows the good sense of humor of the astronauts. A must! Also check out the screencast on Weight/

Weightlessness.

Discuss the differences in a baseball game on the Moon, and your favorite gravity-related topics.

Tides: Begin your treatment of tides by asking the class to consider the consequences of someone pulling

your coat. If they pulled only on the sleeve, for example, it would tear. But if every part of your coat were

92

pulled equally, it and you would accelerate—but it wouldn’t tear. It tears when one part is pulled harder

than another—or it tears because of a difference in forces acting on the coat. In a similar way, the spherical

Earth is “torn” into an elliptical shape by differences in gravitational forces by the Moon and Sun.

CHECK QUESTION: Why do the tides not occur at the same time each day? [As the Earth takes

24 hours to rotate, the Moon advances in its orbit one hour ahead of the Earth. If the Moon didn’t

move in its orbit, the high-tide bulge would be at the same time each day as the Earth spins

beneath the water.]

Misconceptions About the Moon: This is an appropriate place for you to dispel two popular misconceptions

about the Moon. One is that since one side of the Moon’s face is “frozen” to the Earth it doesn’t spin like a

top about its polar axis; and two, that the crescent shape commonly seen is not the Earth’s shadow. To

convince your class that the Moon spins about its polar axis, simulate the situation by holding your eraser at

arms length in front of your face. Tell your class that the eraser represents the Moon and your head

represents the Earth. Rotate slowly keeping one face of the eraser in your view. Call attention to the class

that from your frame of reference, the eraser doesn’t spin

as it rotates about you—as evidenced by your observation

of only one face, with the backside hidden. But your

students occupy the frame of reference of the stars. (Each

of them is a star.) From their point of view they can see all

sides of the eraser as it rotates because it spins about its

own axis as often as it rotates about your head. Show them

how the eraser, if not slowly spinning and rotationally

frozen with one face always facing the same stars, would show all of its sides to you as it circles around

you. See one face, then wait 14 days later and the backside is in your view. The Moon’s spin rate is the

same as its rotational rate .

Misconception 2: Draw a half moon on the board. The shadow is along the diameter and is perfectly

straight. If that were the shadow of the Earth, then the Earth would have to be flat, or be a big block shape!

Discuss playing “flashlight tag” with a suspended basketball in a dark room that is illuminated by a

flashlight in various locations. Ask your class if they could estimate the location of the flashlight by only

looking at the illumination of the ball. Likewise with the Moon illuminated by the Sun!

Sketch the picture on the right on the board and

ask what is wrong with it.

[Answer: The Moon is in a daytime position as

evidenced by the upper part of the Moon being

illuminated. This means the Sun must be above.

Dispel notions that the crescent shape of the

Moon is a partial eclipse by considering a half

moon and the shape of the Earth to cast such a

shadow.]

Back to Tides

Explain tides via the accelerating ball of Jell-O as in Figure 9.14. Equal

pulls result in an undistorted ball as it accelerates, but unequal pulls

cause a stretching. This stretching is evident in the Earth’s oceans,

where the side nearest the Moon is appreciably closer to the Moon than

the side farthest away. Carefully draw Figure 9.16 on the board, which

explains why closeness is so important for tides. The figure shows that

the magnitude of F rather than F itself is responsible for tidal effects.

Hence the greater attraction of the distant Sun produces only a small

difference in pulls on the Earth, and compared to the Moon makes a small

contribution to the tides on Earth.

93

Explain why the highest high tides occur when the Earth, Moon, and Sun are aligned—at the time of a new

and a full Moon.

Discuss tides in the molten Earth and in the atmosphere.

Amplify Figure 9.16 with a comparison of Fs for both the Sun and the Moon as sketched at the upper

right.

Clearly F is smaller for the larger but farther Sun.

The text treats tides in terms of forces rather than fields. In terms of the latter, tidal forces are related to

differences in gravitational field strengths across a body, and occur only for bodies in a nonuniform

gravitational field. The gravitational fields of the Earth, Moon, and Sun, for example, are inverse-square

fields—stronger near them than farther away. The Moon obviously experiences tidal forces because the

near part to us is in a stronger part of the Earth’s gravitational field than the far part. But even an astronaut

in an orbiting space shuttle strictly speaking experiences tidal forces because parts of her body are closer to

the Earth than other parts. This tidal force, the difference between the forces on near and far parts of her

body, follow an inverse-cube law (in this manual as derived earlier). The micro differences produce

microtides. Farther away in deep space, the differences are less. Put another way, the Earth’s gravitational

field is more uniform farther away. The “deepness” of a deep-space location can in fact be defined in terms

of the amount of microtides experienced by a body there. Or equivalently, by the uniformity of any

gravitational field there. There are no microtides in a body located in a strictly uniform gravitational field.

If there are microtides of an astronaut in orbit, would such microtides are even greater on the Earth’s

surface? The answer is yes. This brings up Exercise 42 that concerns biological tides. Interestingly enough,

microtides in human bodies are popularly attributed to not the Earth, but the Moon. This is because popular

knowledge cites that the Moon raises the ocean an average of 1 meter each 12 hours. Point out that the

reason the tides are “stretched” by 1 meter is because part of that water is an Earth diameter closer to the

Moon than the other part. In terms of fields, the near part of the Earth is in an appreciably stronger part of

the Moon’s gravitational field than the far part. To the extent that part of our bodies are closer to the Moon

than other parts, there would be lunar microtides—but enormously smaller than the microtides produced by

not only the Earth, but massive objects in one’s vicinity.

Is there a way to distinguish between a gravity-free region and orbital free-fall inside the International

Space Station? The answer is yes. Consider a pair of objects placed side by side. If the ISS were floating in

a gravity-free region, the two objects would remain as placed over time. Since the ISS orbits the Earth,

however, each object is in its own orbit about the Earth’s center, in its own orbital plane. All orbital planes

pass through the center of the Earth and intersect, which means that depending on the proximity of the

objects, they may collide by the time the ISS makes a quarter orbit—a little more than 23 minutes! If the

94

pair of objects are placed one in front of the other, with respect to their direction of motion, there will be no

such effect since they follow the same orbital path in the same orbital plane. If the objects are one above the

other, one farther from Earth, they will migrate in seemingly strange ways relative to each other because

they are in distinct orbits with different PEs. Gravity makes itself present to astronauts by secondary effects

,that are not directly related to weight. See the “Bob Biker” Practice Pages 49 and 50 for Chapter 8.

CHECK QUESTION: Consider the tiny tidal forces that DO act on our bodies, as a result of parts

of our bodies experiencing slightly different gravitational forces. What planetary body is most

responsible for microtides in our bodies? [The Earth, by far. When we are standing, there is a

greater difference in Earth gravity on our feet compared to our heads than the corresponding

differences in gravity due to farther away planetary bodies.]

Simulated Gravity in Space Habitats

The tallness of people in outer space compared to the radius of their rotating space habitats is very

important. A gravitational gradient is appreciable in a relatively small structure. If the rim speed is such

that the feet are at Earth-normal one g, and the head is at the hub, then the gravitational gradient is a full 1-

g. If the head is halfway to the hub, then the gradient is 12-g, and so forth. Simulated gravity is directly

proportional to the radius. To achieve a comfortable 1100-g gradient, the radius of the structure must be 100

times that of one’s height. Hence the designs of large structures that rotate to produce Earth-normal gravity.

Tidal forces reach an extreme in the case of a black hole. The unfortunate fate of an astronaut falling into a

black hole is not encountering the singularity, but the tidal forces encountered far before getting that close.

Approaching feet first, for example, his closer feet would be pulled with a greater force than his

midsection, which in turn would be pulled with a greater force than his head. The tidal forces would stretch

him and he would be killed before these forces literally pulled him apart.

Gravitational Fields

Introduce the idea of forcemass for a body, and the gravitational force field. Relate the gravitational field

to the more visible magnetic field as seen via iron filings (Look ahead to Figures 24.2 and 22.4). Since the

field strength of the gravitational field is simply the ratio of force per mass, it behaves as force—it follows

an inverse-square relationship with distance. Pair this with student viewing of my screencast on Earth’s

Gravity, where the field strength inside a planet is treated. Follow this with Tunnel Through Earth.

It’s easy to convince your students that the gravitational force on a body located at the exact center of the

tunnel would be zero—a chalkboard sketch showing a few symmetrical force vectors will do this. Hence

the gravitational field at the Earth’s center is zero. Then consider the magnitude of force the body would

experience between the center of the Earth and the surface. A few more carefully drawn vectors will show

that the forces don’t cancel to zero. The gravitational field is between zero and the value at the surface.

You’d like to easily show that it’s half for an Earth of uniform density, to establish the linear part of the

graph of Figure 9.24. Careful judgment should be exercised at this point. For most classes I would think the

geometrical explanation would constitute “information overload” and it would be best to simply say “It can

be shown by geometry that halfway to the center the field is half that at the surface…” and get on with your

lecture. For highly motivated students it may be best to develop the geometrical explanation (given earlier

in this manual). Then the pedagogical question is raised; how many students profit from your display of the

derivation and how many will not?

Class time might better be spent on speculating further about the hole drilled through the Earth. Show with

the motion of your hand how if somebody fell in such a tunnel they would undergo simple harmonic

motion—and that this motion keeps perfect pace with a satellite in close circular orbit about the Earth. The

time for orbit, nearly 90 minutes, is the time to make a to and fro trip in the tunnel. Consider going further

and explain how ideally the period of oscillation of a body traveling in such a tunnel under the influence of

only gravity would be the same for any straight tunnel—whether from New York to Australia, or from New

York to Hawaii or China. You can support this with the analogy of a pendulum that swings through

different amplitudes with the same period. In non-vertical tunnels, of course, the object must slide rather

than drop without friction. But the period is the same, and timetables for travel in this way would be quite

simple; any one-way trip would take nearly 45 minutes. See the screencast Tunnel Through Earth.

95

Gravitational Field Inside a Hollow Planet

Consider the case of a body at the center of a completely hollow planet. Again, the field at the center is

zero. Then show that the field everywhere inside is zero—by careful

explanation of the following sketch. [Consider sample point P, twice as

far from side A than side B. A solid cone defines area A and area B.

Careful thought shows A has 4 times the area of B, and therefore has 4

times as much mass as B. That would mean 4 times as much

gravitational pull, but being twice as far has only 14 as much pull. 14 of

4 gives the same gravitational pull as the pull toward B. So the forces

cancel out (as they of course do in the center). The forces cancel

everywhere inside the shell provided it is of uniform composition. If you

stress this material (which will likely be on the heavy-duty side for

many students) the following Check Question will measure the worth of

your lecture effort.]

CHECK QUESTION: Sketch a graph similar to that in Figure 9.24 to represent the gravitational

field inside and outside a hollow sphere. (The graphical answers should look like the following: A

thin shelled planet is on the left, a thick shelled one is on the right.)

Speculate about the living conditions of a civilization inside a hollow planet. Expanding on Exercise 54 that

considers a hollow planet, consider what happens to the g field inside when a massive spaceship lands on

the outside surface of the hollow planet. The situation is interesting!

Black Holes

Begin by considering an indestructible person standing on a star, as in Figure 9.27. Write the gravitational

equation next to your sketch of the person on the star, and show how only the radius changes in the

equation as the star shrinks, and how the force therefore increases. Stress that the force on the person who

is able to remain at distance R as the star shrinks experiences no change in force—the field there is constant

as the star shrinks, even to a black hole. It is near the shrinking surface where the huge fields exist.

CHECK QUESTION: Consider a satellite companion to a star that collapses to become a black

hole. How will the orbit of the companion satellite be affected by the star’s transformation to a

black hole? [Answer is not at all. No terms in the gravitation equation change. What does happen,

though, is that matter streams from the visible star to the black hole companion, emitting x-rays as

it accelerates toward the black hole, providing evidence of its existence.]

Cosmological Constant

It was long believed that gravity is only attractive. Newton worried that it would cause the universe to

collapse and assumed God kept that from happening. Einstein sought a natural explanation and added a

constant term to his gravity equation to give a repulsion to stabilize the universe. This was term called the

cosmological constant. A few years later, after discovery that the universe is expanding, Einstein dropped

it, calling it his “biggest blunder.” However, in 1998 it was discovered that the expansion of the universe is

accelerating under the action of some yet unidentified energy field called dark energy, which carries about

68 percent of the energy and mass of the universe. The cosmological constant is thought to be the source of

the dark energy. However, calculations give a result that is fifty orders

,of magnitude greater than what is

observed. This “cosmological constant problem” is one of the biggest unanswered questions in physics.

96

Answers and Solutions for Chapter 9

Reading Check Questions

1. Newton discovered that gravity is universal.

2. The Newtonian synthesis is the union between terrestrial laws and cosmic laws.

3. The Moon falls away from the straight line it would follow if there were no gravitational force acting on it.

4. Every body in the universe attracts every other body with a force that, for two bodies, is directly

proportional to the product of their masses and inversely proportional to the square of the distance

between their centers:

F G

m1m 2

d

2

5. The gravitational force between is 6 10-11 N.

6. The gravitational force is about 10 N, or more accurately, 9.8 N.

7. Actually the mass of Earth could then be calculated, but calling it “weighing of Earth” seemed more

dramatic.

8. The force of gravity is one-fourth as much.

9. Thickness is one-fourth as much.

10. You’re closer to Earth’s center at Death Valley, below sea level, so you weigh more there than on any

mountain peak.

11. Springs would be more compressed when accelerating upward; less compressed when accelerating

downward.

12. No changes in compression when moving at constant velocity.

13. Your weight is measured as mg when you are firmly supported in a gravitational field of g and in

equilibrium.

14. In an upward accelerating elevator your weight is greater than mg, in free fall your weight is zero.

15. The occupants are without a support force.

16. Tides depend on the difference in pulling strengths.

17. One side is closer.

18. Spring tides are higher.

19. Yes, interior tides occur in Earth and are caused by unequal forces on opposite sides of Earth’s interior.

20. At the time of a full or new moon, Sun, Moon, and Earth are aligned.

21. No, for no lever arm would exist between Earth’s gravitational pull and the Moon’s axis.

22. A gravitational field is a force field about any mass, and can be measured by the amount of force on a

unit of mass located in the field.

23. At Earth’s center, its gravitational field is zero.

24. Half way to the center, the gravitational field is half that at the surface.

25. Anywhere inside a hollow planet the gravitational field of the planet is zero.

26. Einstein viewed the curve in a planet’s path as a result of the curvature of space itself.

27. Your weight would increase.

28. Field strength increases as the star surface shrinks.

29. A black hole is invisible because even light cannot escape it.

30. Perturbations of Uranus’ orbit not accounted for by any known planet led to the discovery of Neptune.

Think and Do

31. Open ended.

32. Hold it half way from your eye and it covers the same area of eyesight as the unfolded bill, nicely

illustrating the inverse-square law.

Plug and Chug

33.

F G

m1m 2

d

2 6 .67 10

11

N m 2 /kg

2

(1kg )(6 10 24 kg )

(6 .4 10 6 m)2 9 .8 N .

34.

F G

m1m 2

d

2 6 .67 10

-11

N m2

/kg

2

(1 kg )(6 10

24

kg )

[2 (6 .4 10

6

m )]

2 2 .5 N.

97

35.

F G

m1m 2

d

2

6 .67 10

-11

N m2

/kg

2

(6 .0 10

24

kg )(7 .4 10

22

kg )

(3 .8 10

8

m)

2 2 .1 10

20

N.

36.

F G

m

1

m

2

d

2 6 .67 10

-11

N m2

/kg

2

(6 .0 10

24

kg )(2 .0 10

30

kg )

(1 .5 10

11

m)

2 3 .6 10

22

N.

37.

F G

m

1

m

2

d

2

6 .67 10

-11

N m2

/kg

2

(3 .0 kg )(6 .4 10

23

kg )

(5 .6 10

10

m )

2 4 .1 10

-8

N.

38.

F G

m

1

m

2

d

2 6 .67 10

-11

N m2

/kg

2

(3 .0 kg )(100 kg )

(0 .5 m)

2 8 .0 10

-8

N.

The obstetrician exerts about twice as much gravitational force.

Think and Solve

39. From F = GmM/d2, three times d squared is 9 d2, which means the force is one ninth of surface weight.

40. From F = GmM/d2, (2m)(2M) = 4 mM, which means the force of gravity between them is 4 times greater.

41. From F = G2m2M/(2d2) = 4/4 (GmM/d2), with the same force of gravitation.

42. From F = GmM/d2, if d is made 10 times smaller, 1/d2 is made 100 times larger, which means the force is

100 times greater.

43. g =

GM

d

2 =

(6 .67 10

11

)(6 .0 10

24

)

[(6380 200 )x10

3

]

2 = 9.24 N/kg or 9.24 m/s2; 9.24/9.8 = 0.94 or 94%.

44. (a) Substitute the force of gravity in Newton’s second law:

a

F

m

GmM /d

2

m

G

M

d

2

.

(b) Note that m cancels out. Therefore the only mass affecting your acceleration is the mass M of the

planet, not your mass.

Think and Rank

45. B=C, A, D

46. C, B, A

47 a. B, A=C, D b. D, A=C, B

48. C, B, A

49. B, A, C

Think and Explain

50. Nothing to be concerned about on this consumer label. It simply states the universal law of gravitation,

which applies to all products. It looks like the manufacturer knows some physics and has a sense of

humor.

51. This goes back to Chapter 4: A heavy body doesn’t fall faster than a light body because the greater

gravitational force on the heavier body (its weight), acts on a correspondingly greater mass (inertia).

The ratio of gravitational force to mass is the same for every body—hence all bodies in free fall

accelerate equally.

52. In accord with the law of inertia, the Moon would move in a straight-line path instead of circling both

the Sun and Earth.

53. The force of gravity is the same on each because the masses are the same, as Newton’s equation for

gravitational force verifies.

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54. The force of gravity is the same on each because the masses are the same, as Newton’s equation for

gravitational force verifies. When dropped the crumpled paper falls faster only because it encounters

less air resistance than the sheet.

55. The force decreases as the square of increasing distance, or force increases with the square of

decreasing distance.

56. The forces between the apple and Earth are the same in magnitude. Force is the same either way, but

the corresponding accelerations of each are different.

57. In accord with Newton’s 3rd law, the weight of the Earth in the gravitational field of Larry is 300 N; the

same as the weight of Larry in Earth’s gravitational field.

58. Less, because an object there is farther from Earth’s center.

59. Letting the equation for gravitation guide your thinking, twice the diameter is twice the radius, which

corresponds to 1/4 the astronaut’s weight at the planet’s surface.

60. Letting the equation for gravitation guide your thinking, twice the mass means twice the force, and

twice the distance means one-quarter the force. Combined, the astronaut weighs half as much.

61. Your weight would decrease if the Earth expanded with no change in its mass and would increase if

the Earth contracted with no change in its mass. Your mass and the Earth’s mass don’t change, but

the distance between you and the Earth’s center does change. Force is proportional to the inverse

square of this distance.

62. A person is weightless when the only force acting is gravity, and there is no support force. Hence the

person in free fall is weightless. But more than gravity acts on the person falling at terminal velocity. In

addition to gravity, the falling person is “supported” by air resistance.

63. The high-flying jet plane is not in free fall. It moves at approximately constant velocity so a passenger

experiences no net force. The upward support force of the seat matches the downward pull of gravity,

providing the sensation of weight. The orbiting space vehicle, on the other hand, is in a state of free

fall. No support force is offered by a seat, for it falls at the same rate as the passenger. With no

support force, the force of gravity on the passenger is not sensed as weight.

64. Gravitational force is indeed acting on a person

,who falls off a cliff, and on a person in a space shuttle.

Both are falling under the influence of gravity.

65. In a car that drives off a cliff you “float” because the car no longer offers a support force. Both you and

the car are in the same state of free fall. But gravity is still acting on you, as evidenced by your

acceleration toward the ground. So, by definition, you would be weightless (until air resistance

becomes important).

66. The two forces are the normal force and mg, which are equal when the elevator doesn’t accelerate,

and unequal when the elevator accelerates.

67. The pencil has the same state of motion that you have. The force of gravity on the pencil causes it to

accelerate downward alongside of you. Although the pencil hovers relative to you, it and you are falling

relative to the Earth.

68. The jumper is weightless due to the absence of a support force.

69. You disagree, for the force of gravity on orbiting astronauts is almost as strong as at Earth’s surface.

They feel weightless because of the absence of a support force.

70. In a rotating habitat (as discussed in Chapter 8) rotation provides the required support force. The weight

experienced would be a centrifugal force.

71. Your weight equals mg when you are in equilibrium on a horizontal surface and the only forces acting on

you are mg downward and an equal-and-opposite normal force N upward.

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72. The scale shows the normal force acting on you, which on an incline is less than the normal force that

occurs on a firm horizontal surface. The force of gravity on you is mg whatever the support force. But

for mg to align with the normal force, the scale must be supported on a horizontal surface. If you want

to know how strongly gravity is pulling on you, you need to put your scale on a horizontal surface.

73. The force due to gravity, mg, does not vary with jouncing. Variations in the scale reading are variations

in the support force N, not in mg.

74. Just as differences in tugs on your shirt will distort the shirt, differences in tugs on the oceans distort

the ocean and produce tides.

75. The gravitational pull of the Sun on the Earth is greater than the gravitational pull of the Moon. The

tides, however, are caused by the differences in gravitational forces by the Moon on opposite sides of

the Earth. The difference in gravitational forces by the Moon on opposite sides of the Earth is greater

than the corresponding difference in forces by the stronger pulling but much more distant Sun.

76. No torque occurs when the Moon’s long axis is aligned with Earth because there is no lever arm. A lever

arm exists when the Moon’s CG and CM are not aligned with Earth.

77. No. Tides are caused by differences in gravitational pulls. If there are no differences in pulls, there are

no tides.

78.Ocean tides are not exactly 12 hours apart because while the Earth spins, the Moon moves in its orbit

and appears at its same position overhead about every 25 hours, instead of every 24 hours.

So the two-high-tide cycle occurs at about 25-hour intervals, making high tides about 12.5 hours apart.

79. Lowest tides occur along with highest tides, spring tides. So the tide cycle consists of higher-than-

average high tides followed by lower-than-average low tides (best for digging clams!).

80. Whenever the ocean tide is unusually high, it will be followed by an unusually low tide. This makes

sense, for when one part of the world is having an extra high tide, another part must be donating water

and experiencing an extra low tide. Or as the hint in the exercise suggests, if you are in a bathtub and

slosh the water so it is extra deep in front of you, that’s when it is extra shallow in back of you—

“conservation of water!”

81. Because of its relatively small size, different parts of the Mediterranean Sea and other relatively small

bodies of water are essentially equidistant from the Moon (or from the Sun). So one part is not pulled

with any appreciably different force than any other part. This results in extremely tiny tides. Tides are

caused by appreciable differences in pulls.

82. Tides are produced by differences in forces, which relate to differences in distance from the attracting

body. One’s head is appreciably closer than one’s feet to the overhead melon. The greater

proportional difference for the melon out-tides the more massive but more distant Moon. One’s head is

not appreciably closer to the Moon than one’s feet.

83. In accord with the inverse-square law, twice as far from the Earth’s center diminishes the value of g to

1/4 its value at the surface or 2.5 m/s2.

84. For a uniform-density planet, g inside at half the Earth’s radius would be 5 m/s2. This can be

understood via the spherical shell idea discussed in the chapter. Halfway to the center of the Earth, the

mass of the Earth in the outer shell can be neglected—the gravitational contribution of all parts of the

shell cancels to zero. Only the mass of the Earth “beneath” contributes to acceleration, the mass in the

sphere of radius r/2. This sphere of half radius has only 1/8 the volume and only 1/8 the mass of the

whole Earth (volume varies as r3). This effectively smaller mass alone would find the acceleration due

to gravity 1/8 that of g at the surface. But consider the closer distance to the Earth’s center as well. This

twice-as-close distance alone would make g four times as great (inverse-square law). Combining both

factors, 1/8 of 4 = 1/2, so the acceleration due to gravity at r/2 is g/2.

85. Your weight would be less down in the mine shaft. One way to explain this is to consider the mass of

the Earth above you which pulls upward on you. This effect reduces your weight, just as your weight is

reduced if someone pulls upward on you while you’re weighing yourself. Or more accurately, we see

that you are effectively within a spherical shell in which the gravitational field contribution is zero; and

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that you are being pulled only by the spherical portion below you. You are lighter the deeper you go,

and if the mine shaft were to theoretically continue to the Earth’s center, your weight moves closer to

zero.

86. The increase in weight indicates that the Earth is more compressed—more compact—more dense—

toward the center. The weight that normally would be lost when in the deepest mine shafts from the

upward force of the surrounding “shell” is more than compensated by the added weight gained due to

the closeness to the more dense center of the Earth. (Referring to our analysis of Exercise 49, if the

mine shaft were deep enough, reaching halfway to the center of the Earth, you would, in fact, weigh

less at the bottom of the shaft than on the surface, but more than half your surface weight.)

87. Open-ended.

Think and Discuss

88. Your friend’s misconception is a popular one. But investigation of the gravitational equation shows that

no matter how big the distance, force never gets to zero. If it were zero, any space shuttle would fly off

in a straight-line path!

89. The force of gravity on Moon rocks at the Moon’s surface is considerably stronger than the force of

gravity between Moon distant Earth. Rocks dropped on the Moon fall onto the Moon’s surface. (The

force of the Moon’s gravity is about 1/6 of the weight the rock would have on Earth; but the force of the

Earth’s gravity at that distance is only about 1/3600 of the rock’s Earth-weight.)

90. If gravity between the Moon and its rocks vanished, the rocks, like the Moon, would continue in their

orbital path around the Earth. The assumption ignores the law of inertia.

91. Nearer the Moon, because of its smaller mass and lesser pull at equal distances.

92. The Earth and Moon equally pull on each other in a single interaction. In accord with Newton’s 3rd law,

the pull of the Earth on the Moon is equal and opposite to the pull

,of the Moon on the Earth. An elastic

band pulls equally on the fingers that stretch it.

93.Earth and Moon do rotate around a common point, but it’s not midway between them (which would

require both Earth and Moon to have the same mass). The point around which Earth and Moon rotate

(called the barycenter) is within the Earth about 4600 km from the Earth’s center.

94. For the planet half as far from the Sun, light would be four times as intense. For the planet ten times as

far, light would be 1/100th as intense.

95. By the geometry of Figure 9.4, tripling the distance from the small source spreads the light over 9 times

the area, or 9 m2. Five times the distance spreads the light over 25 times the area or 25 m2, and for 10

times as far, 100 m2.

96. The gravitational force on a body, its weight, depends not only on mass but distance. On Jupiter, this is

the distance between the body being weighed and Jupiter’s center—the radius of Jupiter. If the radius

of Jupiter were the same as that of the Earth, then a body would weigh 300 times as much because

Jupiter is 300 times more massive than Earth. But the radius of Jupiter is about 10 times that of Earth,

weakening gravity by a factor of 100, resulting in 3 times its Earth weight. (The radius of Jupiter is

actually about 11 times that of Earth).

97.If Earth gained mass you’d gain weight. Since Earth is in free fall around the Sun, the Sun contributes

nothing to your weight. Earth gravitation presses you to Earth; solar gravitation doesn’t press you to

Earth.

98. First of all, it would be incorrect to say that the gravitational force of the distant Sun on you is too small

to be measured. It’s small, but not immeasurably small. If, for example, the Earth’s axis were

supported such that the Earth could continue turning but not otherwise move, an 85-kg person would

see a gain of 1/2 newton on a bathroom scale at midnight and a loss of 1/2 newton at noon. The key

idea is support. There is no “Sun support” because the Earth and all objects on the Earth—you, your

bathroom scale, and everything else—are continually falling around the Sun. Just as you wouldn’t be

pulled against the seat of your car if it drives off a cliff, and just as a pencil is not pressed against the

floor of an elevator in free fall, we are not pressed against or pulled from the Earth by our gravitational

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interaction with the Sun. That interaction keeps us and the Earth circling the Sun, but does not press

us to the Earth’s surface. Our interaction with the Earth does that.

99. The gravitational force varies with distance. At noon you are closer to the Sun. At midnight you are an

extra Earth diameter farther away. Therefore the gravitational force of the Sun on you is greater at

noon.

100. As stated in question 98, our “Earth weight” is due to the gravitational interaction between our mass

and that of the Earth. The Earth and its inhabitants are freely falling around the Sun, the rate of which

does not affect our local weights. (If a car drives off a cliff, the Earth’s gravity, however strong, plays no

role in pressing the occupant against the car while both are falling. Similarly, as the Earth and its

inhabitants fall around the Sun, the Sun plays no role in pressing us to the Earth.)

101. The Moon does rotate like a top as it circles Earth. It rotates once per revolution, which is why we

see only the same face. If it didn’t rotate, we’d see the back side every half month.

102. Tides would be greater if the Earth’s diameter were greater because the difference in pulls would be

greater. Tides on Earth would be no different if the Moon’s diameter were larger. The gravitational

influence of the Moon is just as if all the Moon’s mass were at its CG. Tidal bulges on the solid surface

of the Moon, however, would be greater if the Moon’s diameter were larger—but not on the Earth.

103. Earth would produce the largest microtides in your body. Microtides are greatest where the difference

between your head and feet is greatest compared with the distance to the tide-pulling body, Earth.

104. Tides occur in Earth’s crust and Earth’s atmosphere for the same reason they occur in Earth’s oceans.

Both the crust and atmosphere are large enough so there are appreciable differences in distances to

the Moon and Sun. The corresponding gravitational differences account for tides in the crust and

atmosphere.

105. More fuel is required for a rocket that leaves the Earth to go to the Moon than the other way around.

This is because a rocket must move against the greater gravitational field of the Earth most of the way.

(If launched from the Moon to the Earth, then it would be traveling with the Earth’s field most of the

way.)

106. On a shrinking star, all the mass of the star pulls in a noncanceling direction (beneath your feet)—you

get closer to the overall mass concentration and the force increases. If you tunnel into a star, however,

there is a cancellation of gravitational pulls; the matter above you pulls counter to the matter below

you, resulting in a decrease in the net gravitational force. (Also, the amount of matter “above” you

decreases.)

107. F ~ m1 m2/d

2, where m2 is the mass of the Sun (which doesn’t change when forming a black hole), m1

is the mass of the orbiting Earth, and d is the distance between the center of mass of Earth and the

Sun. None of these terms change, so the force F that holds Earth in orbit does not change.

108. Letting the gravitational force equation be a guide to thinking, we see that gravitational force and

hence one’s weight does not change if the mass and radius of the Earth do not change. (Although

one’s weight would be zero inside a hollow uniform shell, on the outside one’s weight would be no

different than if the same-mass Earth were solid.)

109. Astronauts are weightless because they lack a support force, but they are well in the grips of Earth

gravity, which accounts for them circling the Earth rather than going off in a straight line in outer space.

110. The misunderstanding here is not distinguishing between a theory and a hypothesis or conjecture.

A theory, such as the theory of universal gravitation, is a synthesis of a large body of information

that encompasses well-tested and verified hypotheses about nature. Any doubts about the theory

have to do with its applications to yet untested situations, not with the theory itself. One of the

features of scientific theories is that they undergo refinement with new knowledge. (Einstein’s

general theory of relativity has taught us that in fact there are limits to the validity of Newton’s

theory of universal gravitation.)

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104

10 Projectile and Satellite Motion

Conceptual Physics Instructor’s Manual, 12th Edition

10.1 Projectile Motion

Projectiles Launched Horizontally

Projectiles Launched at an Angle

Hang time Revisited

10.2 Fast-Moving Projectiles—Satellites

10.3 Circular Satellite Orbits

10.4 Elliptical Orbits

World Monitoring by Satellite

10.5 Kepler’s Laws of Planetary Motion

Finding Your Way

10.7 Energy Conservation and Satellite Motion

10.7 Escape Speed

My granddaughter Emily shares good information in opening photo 1. Photos 2 and 3 are of Tenny Lim. As

teachers we are rewarded by the success that some of our students achieve after leaving our tutelage. In my

career, Tenny Lim is the most outstanding of these. It is with great pride that she is featured in the photo

opener to this chapter and the personal profile as well. Photo 3 is of CCSF physics instructor Shruti Kumar

projecting a ball after students have made predictions of the landing point. Photo 5 is Peter Rea, whose

company Arbor Scientific supplies teaching materials to schools. Arbor is the primary supplier of materials

that complement Conceptual Physics. He is

,also the main supplier of my classroom videos, both on DVDs

and more recently via streaming from the Arbor Scientific website.

In editions previous to the ninth edition, projectile motion was treated with linear motion. Kinematics

began the sequence of mechanics chapters. Since then I’ve postponed projectile motion until after

Newton’s laws and energy, and just before satellite motion. When projectiles move fast enough for the

Earth’s curvature to make a difference in range, you’re at the doorstep to satellite motion.

Regarding 45 as the maximum range for projectiles, keep in mind that this is only true when air resistance

can be neglected, and most important and often overlooked, when the launching speed is the same at all

angles concerned. Tilt a water hose up 45 and sure enough, for short distances where air resistance is nil, it

attains maximum range. The same is true for a slowly-bunted baseball. But for a high-speed ball, air

resistance is a factor and maximum range occurs for angles between 39 and 42. For very high speeds

where the lesser air resistance of high altitudes is a consideration, angles greater than 45 produce

maximum range. During World War I, for example, the German cannon “Big Bertha” fired shells 11.5 km

high and attained maximum range at 52. Air resistance is one factor; launching speed is another. When

one throws a heavy object, like a shot put, its launching speed is less for higher angles simply because some

of the launching force must be used to overcome the force due to gravity. (You can throw a heavy boulder a

lot faster horizontally than you can straight up.) Shot puts are usually launched at angles slightly less than

40. The fact that they are launched higher than ground level decreases the angle as well. Screencast Ball

Toss covers much of this.

Interestingly, the maximum height of a projectile following a

parabolic path is nicely given by sketching an isosceles triangle

with the base equal to the range of the projectile. Let the two side

angles be equal to the launch angle , as shown in the figure. The

maximum height h reached by the projectile is equal to one-half

H, the altitude of the triangle. This goodie from Jon Lamoreux

and Luis Phillipe Tosi, of Culver Academies, Culver, IN.

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The interesting fact that projectiles launched at a particular angle have the small range if launched at the

complementary angle is stated without proof in the chapter, and is shown in Figure 10.11. This fact is

shown by the range formula, R = (2v sin cos)g. Because the sine of an angle is the cosine of the

complement of that angle, replacing the angle with its complement results in the same range. So the range

is the same whether aiming at or at (90 - ). As said, maximum range occurs at a projection angle of 45,

where sine and cosine are equal.

The spin of the Earth is helpful in launching satellites, which gives advantage to launching cities closest to

the equator. The launch site closest to the equator is Kaurou, French Guiana, in South America, 5 08', used

by the European Space Agency. The U.S. launches from Cape Canaveral, 28 22', and Vandenberg, 34 38'.

Russia used to launch at Kapustin Yar, 48 31', Plesetsk, 62 42', and Tyuratam (Baikonur) 45 38'. Is

Hawaii, less than 20 in our space launching future?

If you haven’t shown the 15-minute oldie but goodie NASA film, “Zero g,” be sure to show it now. It is of

footage taken aboard Skylab in 1978, narrated by astronaut Owen Garriott. Newton’s laws of motion are

reviewed with excellent and entertaining examples. (It would be a shame for this stimulating movie to fall

through the cracks due to being “dated.”)

Ask your students this question: An Earth satellite remains in orbit because it’s above Earth’s A.

atmosphere. B. gravitational field. C. Both of these. D. Neither of these. Be prepared for most to answer C.

That’s because of the common misconception that no gravitational field exists in satellite territory. Of

course it’s the gravitational field that keeps a satellite from flying off in space, but this isn’t immediately

apparent to many students. Strictly speaking, there IS some atmosphere in satellite territory. That’s why

boosters on the ISS have to periodically fire to overcome the slight drag that exists. So perhaps in your

discussion, qualify your question to “it’s above most of Earth’s”.

Solar Photon Force: To a small extent, sunlight affects satellites, particularly the large disco-ball-like

satellite LAGEOS, which wobbles slightly in its orbit because of unequal heating by sunlight. The side in

the Sun radiates infrared photons, the energy of which provides a small, but persistent, rocket effect as the

photons eject from the surface. So a net force some 100 billion times weaker than gravity pushes on the

satellite in a direction away from its hot end. LAGEOS has 426 prism-shaped mirrors. By reflecting laser

beams off its mirrored surface, geophysicists can make precise measurements of tiny displacements in the

Earth’s surface.

Asteroids: Of particular interest are asteroids that threaten Planet Earth. Asteroid 2004 MN4 is big enough

to flatten Texas and a couple of European countries with an impact equivalent to 10,000 megatons of

dynamite—more than the world’s nuclear weapons. The asteroid is predicted to have a close encounter

with Earth in 2029, which likely won’t be the last of its close encounters. Space missions in the future may

employ “tugboat” spacecraft to near-Earth objects, dock with them and gently alter their speeds to more

favorable orbits.

Space Debris: More than 20,000 pieces of space trash larger than 10 cm are in low-Earth orbit, along with

a half million bits of 1-to-2 cm bits of junk in between. Yuk!

Tunnel through Earth: Neil de Grasse Tyson does a nice job on NOVA describing what would happen if

you fell into a tunnel that goes from one side of Earth, through its center, to the other side. I attempt the

same in the screencast Tunnel Through Earth.

Practicing Physics Book:

• Independence of Horizontal and Vertical • Satellites in Circular Orbit

Components of Motion • Satellites in Elliptical Orbit

• Tossed Ball • Mechanics Overview

Problem Solving Book:

Many problems involve projectile motion and satellite motion

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Laboratory Manual:

• The BB Race Horizontal and Vertical Motion (Demonstration)

• Bull’s Eye A Puzzle You CAN Solve (Experiment)

• Blast Off! Rockets Real and Virtual (Experiment and Tech Lab)

• Worlds of Wonder Orbital Mechanics Simulation (Tech Lab)

Next-Time Questions:

• Ball Toss from Tower • Orbital Speed

• Monkey and Banana • Escape Fuel

• Elliptical Orbit • Moon Face

• Escape Velocity • Dart Gun

• Satellite Speed • Bull’s-Eye

• Satellite Mass • Projectile Speeds

• Earth Satellites

Hewitt-Drew-It! Screencasts: •Sideways Drop •Ball Toss •Tennis-Ball Problem •Satellite Speed

•Circular/Eliptical Orbits

SUGGESTED LECTURE PRESENTATION

Independence of Horizontal and Vertical Motion

Roll a ball off the edge of your lecture table and call attention to the curve it follows. The ball is a

projectile. Discuss the idea of the “downwardness” of gravity, and how there is no “horizontalness” to it,

and therefore no horizontal influence on the projectile. Draw a rendition of Figure 10.2 on the board, with

vectors. You’re going an extra step beyond the textbook treatment.

Pose the situation of the horizontally-held gun and the shooter who drops a bullet at the same time he pulls

the trigger, and ask which bullet hits the ground first. (This is treated in screencast Sideways Drop and also

in a video.)

DEMONSTRATION: Show the independence of horizontal and vertical motion with a spring-gun

apparatus that will shoot a ball horizontally while at the same time dropping another that falls

vertically. Follow this up with the popular “monkey and hunter” demonstration.

CHECK QUESTIONS: Point to some target

,at the far side of your classroom and ask your class to

imagine you are going to project a rock to the target via a slingshot. Ask if you should aim at the

target, above it, or below it. Easy stuff. Then ask your class to suppose it takes 1 second for the

rock to reach the target. If you aim directly at the target, it will fall beneath and miss. How far

beneath the target would the rock hit (if the floor weren’t in the way)? Have your students check

with their neighbors. Then ask how far above should you aim to hit the target. Do a neighbor

check. Now you’re ready to discuss Figure 10.6 (nicely developed in Practicing Physics Book

pages 55 and 56).

My screencast Tennis-Ball Problem features an interesting case involving the independence of horizontal

and vertical motion, highlighting a method for solving problems in general—which is to begin with what is

asked for. This method is as simple and direct as can be, answering the student question, “How do I begin a

problem solution?”

Air Resistance

Acknowledge the large effect of air resistance on fast-moving objects such as bullets and cannonballs. A

batted baseball, for example, travels only about 60 percent as far in air as it would in a vacuum. Its

curved path is no longer a parabola, as Figure 10.13 indicates. What makes its decent steeper than its

ascent is the horizontal slowing due to air resistance. What makes it not as high is air resistance vertically,

which diminishes with height. This is covered in the screencast Ball Toss.

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Hang Time Again

Ask if one could jump higher if on a moving skateboard or in a moving bus. It should be clear that the

answer is no to both. But one can usually jump higher from a running jump. It is a mistake to assume that

the horizontal motion is responsible for the higher jump and longer hang time. The action of running likely

enables a greater force between the foot and floor, which gives a greater vertical lift-off component of

velocity. This greater bound against the floor, and not any holiday by gravity on a horizontally moving

body, is the explanation. Stress that the vertical component of velocity alone determines vertical height and

hang time.

Projectiles

There are several ways to horizontally launch a projectile and have your class predict where it will strike

the floor. Chuck Stone nicely shows one method in Figure 10.5. If this isn’t a lab activity, consider it a

classroom demonstration. If students know the speed at which the projectile is horizontally launched, and

the height of launch, they can predict where the projectile will hit. It’s a fun experience.

This is supported by the relationship of the curved path of Figure 10.6 and the vertical distance fallen, d =

5t2, of Chapter 3. Stress that the projectile is falling beneath the straight line it would otherwise follow. This

idea is important for later understanding of satellite motion. Continue with an explanation of Figure 10.7,

and how the dangling beads of page 187 nicely summarizes projectile motion. A worthwhile class project

can be fashioning such beads from points on a meterstick.

Discuss Figure 10.15 and ask for the pitching speed if the ball traveled 30 m instead of 20 m. Note the

vertical height is 5 m. If you use any height that does not correspond to an integral number of seconds,

you’re diverting your focus from physics to algebra. This leads to what the screencast Tennis-Ball Problem

that asks for the maximum speed of a tennis ball clearing the net. More interesting is considering greater

horizontal distances—great enough for the curvature of the Earth to make a difference in arriving at the

answer. It’s easy to see that the time the projectile is in the air increases where the Earth curves beneath the

trajectory.

Satellite Motion

Sketch “Newton’s Mountain” and consider the longer time

intervals for greater and greater horizontal speeds. Ask if

there is a “pitching speed” or cannonball velocity large

enough so the time in the air is forever. Not literally “in the

air,” which is why the cannon is atop a mountain that extends

above the atmosphere. The answer of course is yes. Fired fast

enough the cannonball will fall around the world rather than

into it. You’re into satellite motion.

CHECK QUESTION: Why is it confusing to ask why a satellite doesn’t fall? [All satellites are

continuously falling, in the sense that they fall below the straight line they would travel if they

weren’t. Why they don’t crash to Earth is a different question.]

Calculating Satellite Speed

An effective skit (covered in the screencast Satellite Speed) that can have your class calculating the speed

necessary for close Earth orbit is as follows:

Call attention to the curvature of the Earth,

Figure 10.17. Consider a horizontal laser

standing about a meter above the ground with its

beam shining over a level desert. The beam is

straight but the desert floor curves 5 m over an

8000 m or 8 km tangent, which you sketch on

your chalkboard. Stress this is not to scale.

Now erase the laser and sketch in a super cannon positioned so it points along the laser line. Consider a

cannonball fired at say, 2 kms, and ask how far downrange will it be at the end of one second. A neighbor

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check should yield an answer of 2 km, which you indicate with an “X.” But it doesn’t really get to the “X,”

you say, for it falls beneath the “X” because of gravity. How far? 5 m if the sand weren’t in the way. Ask if

2 kms is sufficient for orbiting the Earth. Clearly not, for the cannonball strikes the ground. If the

cannonball is not to hit the ground, we’d have to dig a trench first, as you show on your sketch, which now

looks like this:

Continue by considering a greater muzzle velocity, say 4 kms, so the cannonball travels 4 km in one

second. Ask if this is fast enough to attain an Earth orbit. Student response should indicate that they realize

that the cannonball will hit the ground before 1 second is up. Then repeat the previous line of reasoning,

again having to dig a trench, and your sketch looks like this:

Continue by considering a greater muzzle velocity—great enough so the cannonball travels 6 km in 1

second. This is 6 kms. Ask if this is fast enough not to hit the ground (or equivalently, if it is fast enough

for Earth orbit). Then repeat the previous line of reasoning, again having to dig a trench. Now your sketch

looks like this:

You’re almost there. Continue by considering a muzzle velocity great enough so the cannonball travels 8

km in one second. (Don’t state the velocity is 8 kms here as you’ll diminish your punch line.) Repeat your

previous reasoning and note that this time you don’t have to dig a trench! After a pause, and with a tone of

importance, ask the class with what speed must the cannonball have to orbit the Earth. Done properly, you

have led your class into a “derivation” of orbital speed about the Earth with no equations or algebra.

Acknowledge that the gravitational force is less on satellites in higher orbits so they do not need to go so

fast. This is acknowledged later in the chapter in a footnote. (Since v = √GMd, a satellite at 4 times the

Earth’s radius needs to travel only half as fast, 4 kms.)

You can wind up your brief treatment of satellite motion and catch its essence via the following skit: Ask

your students to pretend they are encountered by a bright youngster, too young to have much knowledge of

physics and mathematics, but who nevertheless asks why satellites seem to defy gravity and stay in orbit.

You ask what answer could correctly satisfy the curiosity of the kid, then pose the following dialogue

between the kid and the students in your class (you’re effectively suggesting how the student might interact

with the bright kid). Ask the kid to observe and then describe what you do, as you hold a rock at arm’s

length and then simply drop it. The kid replies,

,“You dropped the rock and it fell to the ground below,” to

which you respond, “Very good—now what happens this time?”, as you move your hand horizontally and

again drop the rock. The kid observes and then says, “The rock dropped again, but because your hand was

109

moving it followed a curved path and fell farther away.” You continue, “Very good—now again—” as you

throw the rock still farther. The kid replies, “I note that as your hand moves faster, the path follows a wider

curve.” You’re elated at this response, and you ask the kid, “How far away will the rock hit the ground if its

curved path matches the curved surface of the Earth?” The kid at first appears very puzzled, but then

beams, “Oh—I get it! The stone doesn’t hit at all—it’s in Earth orbit.” Then you interrupt your dialogue

and ask the class, “Do YOU get it?” Then back to the kid who asks, “But isn’t it really more complicated

than that?”, to which the answer is NO. The essential idea of satellite motion IS that simple.

Moving Perpendicular vs Moving Nonperpendicular to Gravity

Pose the case of rolling a ball along a bowling alley. Does gravity pull on the ball? [Yes.] Does gravity

speed up or slow down the ball? [No.] Why? [Because all along the horizontal surface, gravity pulls in a

direction downward, perpendicular to the surface. There is no component of gravity pulling horizontally,

not forward and not backward.] This is the topic of Figure 10.22. Then ask if this fact relates to why a

satellite in circular orbit similarly doesn’t speed up or slow down due to gravity’s persistent pull on it.

[Aha! In both the ball on the alley and the satellite above, both “criss-cross” gravity, having no component

of gravitational force in the direction of motion. No change in speed, no work, no change in KE, no change

in PE. Aha! The cannonball and the bowling ball simply coast.]

Discuss the motion of a cannonball fired horizontally from a mountain top. Suppose the cannonball leaves

the cannon at a velocity of say 1 km/s. Ask your class whether the speed when it strikes the ground

will be 1 km/s, more than 1 km/s, or less than 1 km/s (neglecting air resistance). The answer is that it

strikes at more than 1 kms because gravity speeds it up. (Toss your keys horizontally from a one-story

window and catching them would pose no problem. But if you toss them horizontally from the top of a 20-

story building, you wouldn’t want to catch them!) That’s because gravity plays a role on speed. Sketch

“Newton’s Mountain” on the whole world as shown, and sketch a trajectory that meets Earth’s surface.

Suppose the firing speed is now 4 kms. Repeat your question: Will it be traveling faster, slower, or 4 kms

when it hits the ground? Again, faster, because it moves in

the direction of gravity. Caution: Do not draw a trajectory

that meets the Earth’s surface at a point beyond the halfway

mark. (Interestingly, the Zero-g film and other depictions

show a complete orbit when past the half-way point, which is

erroneous. Why? Because the parabolic path is actually a

segment of a Keplerian ellipse, Figure 10.28. Halfway

around puts it all around). Now draw the circular trajectory

that occurs when the firing speed is 8 kms. Ask if the speed

increases, decreases, or remains the same after leaving the

cannon. This time it remains the same. Why?

Neighbor checking time!

Circular Orbits

Erase the mountain from your sketch of the world and draw a huge

elevated bowling alley that completely circles the world (Figure 10.23).

You’re extending Figure 10.22. Show how a bowling ball on such an alley

would gain no speed because of gravity. But now cut part of the alley

away, so the ball rolls off the edge and crashes to the ground below. Does

it gain speed after falling in the gap? [Yes, because its circular path

becomes a parabolic path, no longer moving perpendicular to gravity—

having a component of velocity in the downward direction of the Earth’s

gravity.] Acknowledge that if the ball moves faster it will fall farther

before crashing to the ground. Ask what speed would allow it to clear the

gap (like a motorcyclist who drives off a ramp and clears a gap to meet a

ramp on the other side). [8 kms, of course.] Can the gap be bigger at this speed? Sketch a gap that nearly

circles the world when you ask this question. Then ask, what happens with no alley? And your class sees

that at 8 kms no supporting alley is needed. The ball orbits the Earth.

110

CHECK QUESTION: We say that satellites are falling around the Earth. But communication

satellites remain at one place overhead. Isn’t this contradictory? [Communication satellites fall in

a wider circle than closer satellites. Their periods are 24 hours, which coincides with the period

of the spinning Earth. So from Earth they appear to be motionless.]

Ask your class if they are familiar with an Earth satellite that has an average period of one month.

There certainly is—it’s the Moon, which has been falling around Earth for billions of years! Go further

and ask about stars in the sky that appear motionless. Are they motionless, just hovering in space? The

answer is NO. Stars in our galaxy, the Milky Way, are falling around the center of the galaxy. How

intriguing that everything is falling! (This information should elicit interest even in the dullest of your

students :-)

CHECK QUESTION: Why is it advantageous to launch rockets close to the equator? [The

tangential speed at the equator is 1000 miles per hour, which can be subtracted from the speed

needed to put a satellite in orbit. The closer the launch site to the equator, the closer it is to the

1000 mph free ride.]

Elliptical Orbits

Back to Newton’s Mountain. Fire the cannonball at 9 kms. It overshoots a circular path. Your sketch looks

like this. Ask, at the position shown, is the cannonball moving at 9

kms, more than 9 kms, or less than 9 kms. And why? After a

neighbor check, toss a piece of chalk upward and say you toss it upward

with an initial speed of 9 ms. When it’s halfway to the top of its path, is

it moving 9 ms, more than 9 ms, or less than 9 ms? Equate the two

situations. [In both cases the projectile slows because it is going against

gravity.]

Continue your sketch and show a closed path—an ellipse. As you draw

the elliptical path, show with a sweeping motion of your arm how the satellite slows in receding from the

Earth, moving slowest at its farthermost point, then how it speeds in falling towards the Earth, whipping

around the Earth and repeating the cycle over and over again. Move to a fresh part of the chalkboard and

redraw with the mountain at the bottom, so your sketch is more like Figure 10.27. (It is more comfortable

seeing your chalk moving slowest when farthest coincides with the direction “up” in the classroom. I quip

that Australians have no trouble seeing it the first way.)

Sketch in larger ellipses for still greater cannon speeds, with the limit being 11.2 kms, beyond which the

path does not close—escape speed.

State that Newton’s equation was deducted from Kepler’s laws.

Kepler’s Laws

Briefly discuss Kepler’s laws. Sketch an elliptical path of a planet about the Sun as in Figure 10.29. Show

how the equal areas law means that the planet travels slowest when farthest from the Sun, and fastest when

closest. State that Kepler had no idea why this was so. Walk to the side of your room and toss a piece of

chalk upward at a slight angle so the class can see the parabolic path it traces. Ask where the chalk is

moving slowest? Fastest? Why is it moving slowest at the top? [Because it has been traveling against

gravity all the way up!] Why is it moving fastest when it is thrown and when it is caught? [It’s moving

fastest when it is caught because it has been traveling in the direction of gravity all the way down!]

Speculate how amazed Kepler would have been if the same questions were asked of him, and relate this to

,the speeds of the planets around the Sun—slowest where they have been traveling against the gravity of the

Sun, and fastest where they have been falling back toward the Sun. Kepler would have been amazed to see

the physics of a body tossed upward is essentially the physics of satellite motion! Kepler lacked this simple

model to guide his thinking. What simple models of tomorrow do we lack today, that finds us presently

blind to the common sense of tomorrow?

Work-Energy Relationship for Satellites

111

You already have sketches on the board of circular and elliptical orbits. Draw sample satellites and then

sketch in force vectors. Ask the class to do likewise, and then draw component vectors parallel and

perpendicular to instantaneous directions of motion. Then show how the changes in speed are consistent

with the work-energy relationship.

Draw a large ellipse on the board with a planet in various positions and ask your class for a comparison of

the relative magnitudes of KE and PE along the orbit. You can do this with different size symbols for KE

and PE. Stress that the two add up to be the same. (This is treated in the screencast Circular/Elliptical

Orbit.)

Escape Speed

Distinguish between ballistic speed and sustained speed, and that the value 11.2 kms refers to ballistic

speed. (One could go to the Moon at 1 kms, given a means of sustaining that speed and enough time to

make the trip!) Compare the escape speeds from different bodies via Table 10-1.

Maximum Falling Speed

The idea of maximum falling speed, footnoted on page 199, is sufficiently interesting for elaboration.

Pretend you throw your car keys from ground level to your friend at the top of a building. Throw them too

fast and they pass beyond your friend; throw them too slow and they never reach your friend. But if you

throw them just right, say 11 ms, they just barely reach her so she has only to grab them at their point of

zero speed. Question: It took a speed of 11 ms to get the keys up to her—if she simply drops them, how

fast will they fall into your hands? Aha! If it takes a speed of 11.2 kms to throw them to her if she is

somewhat beyond Pluto, and she similarly drops them, how fast will they fall into your hands? Now your

students understand maximum falling speed.

CHECK QUESTIONS: This reviews several chapters of mechanics; draw an elliptical orbit about

a planet as shown on the board. Pose the following questions (from the Practicing Physics Book):

At which position does the satellite experience the maximum

(a) gravitational force on it?

(b) speed?

(c) momentum?

(d) kinetic energy?

(e) gravitational potential energy?

(f) total energy (KE + PE)?

(g) acceleration?

(h) angular momentum?

Don’t be surprised to find many of your students miss (g), acceleration, even though they answer the first

about force correctly. If they use either equation for acceleration as their “guide,” the answer is at hand;

that is, from a - Fm, the acceleration is seen to be maximum where the force is maximum—at A. Or from

a = (change in v)t, acceleration is seen to be greatest where most of the change occurs—where the satellite

whips around A. This Check Question summarizes important ideas in four chapters. Go over the answers

carefully.

112

Answers and Solutions for Chapter 10

Reading Check Questions

1. A projectile is any object that is projected by some means and continues in motion by its own inertia.

2. The vertical component moves with or against gravity, while the horizontal component moves with no

horizontal force acting.

3. With no air resistance the horizontal component of velocity remains constant, both in rising and falling.

4. Neglecting air resistance, the vertical component of velocity decreases as the stone rises, and increases

as it descends, the same as with any freely-falling object.

5. In 1 second it falls 5 m beneath the line; For 2 seconds, 20 m beneath.

6. No, the falling distance beneath the line makes no difference whether or not the line is at an angle.

7. An angle of 15° would produce the same range, in accord with Figure 4.19.

8. The projectile would return at the same speed of 100 m/s, as indicated in Figure 4.22.

9. A projectile can fall around the Earth if it has sufficient tangential speed so that its curve downward is no

sharper than that of Earth’s curvature.

10. The speed must be enough so that the path of the projectile matches Earth’s curvature.

11. A satellite must remain above the atmosphere because air resistance would not only slow it down, but

incinerate it at its high speed. A satellite must not have to contend with either of these.

12. Speed doesn’t change because there is no component of gravitational force along the ball’s direction of

motion when the bowling ball is moving horizontally.

13. As with the previous question, speed doesn’t change when there’s no component of gravitational force in

the direction of its motion.

14. The time for a complete close orbit is about 90 minutes.

15. The period for satellites at higher altitudes is more than 90 minutes.

16. In an elliptical orbit there is a component of force in the direction of motion.

17. A satellite has the greatest speed when nearest Earth, and least when farthest away.

18. Tycho Brahe gathered the data, Kepler discovered elliptical orbits, and Newton explained them.

19. Kepler discovered the period squared was proportional to the radial distance cubed.

20. Kepler thought the planets were being pulled along their orbits. Newton realized they were being pulled

toward the Sun.

21. KE is constant because no work is done by gravity on the satellite.

22. The sum of KE and PE is constant for all orbits.

23. Yes, escape speed can be at speeds less than 11.2 km/s if that speed is sustained.

Think and Do

24. A worthwhile activity, and holding the stick at different angles nicely illustrates that distance of “fall”

doesn’t depend on angle of launch. If using a meterstick, at the 25 cm mark a 5-cm string can be

attached; at the 50-cm mark, a 20 cm string; at the 75-cm mark, a 45 cm string; and at the end of the

stick, the 100-cm mark, an 80 cm string. (Consider this as a classroom activity.)

25. Physics is about connections in nature. Discovering the connection between

falling water in a swung bucket and falling satellites was an “aha” moment for

PGH while whirling a water-filled bucket during a rotational-motion classroom

demonstration—on a day when a much-publicized satellite launch was being

discussed. How exhilarating to discover connections in nature!)

Think and Solve

26. One second after being thrown, its horizontal component of velocity is 10 m/s, and

its vertical component is also 10 m/s. By the Pythagorean theorem, V = √(102 + 102) = 14.1 m/s. (It

is moving at a 45° angle.)

27. (a) From y = 5t2 = 5(30)2 = 4,500 m, or 4.5 km high (4.4 km if we use g = 9.8 m/s

2

).

(b) In 30 seconds; d = vt = 280 m/s 30 s = 8400 m.

113

(c) The engine is directly below the airplane. (In a more practical case, air resistance is overcome for

the plane by its engines, but not for the falling engine. The engine’s speed is reduced by air resistance,

covering less than 8400 horizontal m, landing behind the plane.)

28. Time during which the bullet travels is 200 m/400 m/s = 0.5 s. (a) So distance fallen is = ½ gt2 = ½ (10

m/s2)(0.5 s)2 = 1.25 m. (b) The barrel must be aimed 1.25 m above the bullseye to match the falling

distance.

29. At the top of its trajectory, the vertical component of velocity is zero, leaving only the horizontal

component. The horizontal component at the top or anywhere along the path is the same as the initial

horizontal component, 100 m/s (the side of a square where the diagonal is 141).

30. The distance wanted is horizontal velocity time. We find the time from the vertical distance the ball

falls to the top of the can. This distance y is 1.0

,Various Course Designs

You’ll teach more physics in your course if you spend less time on topics that are more math than

physics. Topics I suggest you remove from your front seat include units conversions, graphical

analysis, measurement techniques, error analysis, overtime on significant figures, and the wonderful

and seductive time-consuming toys for kinematics instruction.

Very few one-semester and virtually no one-quarter courses will include all the material presented in

the text. The wide variety of chapters provides a selection of course topics to suit the tastes of

individual instructors. Most begin their course with mechanics, and treat other topics in the order

presented in the text. Some will go immediately from mechanics to relativity. Many will begin with a

study of light and treat mechanics later. Others will begin with the atom and properties of matter

before treating mechanics, while others will begin with sound, then go to light, and then to electricity

and magnetism. Others who wish to emphasize modern physics will skim through Chapters 11, 19, 30

and 31, to then get into Parts 7 and 8. Some will cover many chapters thereby giving students the

widest possible exposure of physics, while others will set the plow deeper and treat fewer chapters.

The following breakdown of parts and chapters is intended to assist you in selecting a chapter

sequence and course design most suited to your objectives and teaching style. You should find that

the chapters of Conceptual Physics are well suited to stand on their own.

PART 1: MECHANICS After the first chapter, About Science, Mechanics begins with forces, rather

than kinematics as in earlier editions. Newton’s first law kicks off by featuring the concept of

mechanical equilibrium. Force vectors are introduced. After this chapter, kinematics is treated, which

I urge you to go through quickly. The important concepts of velocity and acceleration are developed

in further chapters, which makes prolonged time in Chapter 3 a poor policy. Certainly avoid

kinematics problems that are more math than physics, and that many students encounter in their math

courses anyway. Chapter 4 goes to Newton’s second law, followed by a separate chapter for the third

law. There is more treatment of vectors in this 12-th Edition. They use no trig beyond the

Pythagorean Theorem. There are no sines, cosines, or tangents, for the parallelogram method is used.

(Trig is introduced in the Problem Solving Book, however.) Chapters, 2-5, are central to any treatment

of mechanics. Only Chapters 2, 4, and 9 have a historical flavor. Note in the text order that

momentum conservation follows Newton’s 3rd law, and that projectile motion and satellite motion

are combined in Chapter 10. My recommendation is that all the chapters of Part I be treated in the

order presented. To amplify the treatment of vectors, consider the Practice Book and Appendix D. For

an extended treatment of mechanics consider concluding your treatment with Appendix E,

Exponential Growth and Doubling Time.

PART 2: PROPERTIES OF MATTER The very briefest treatment of matter should

be of Chapter 11, atoms, which is background for nearly all the chapters to follow in the text.

Much of the historical development of our understanding of atoms is extended in Chapter 32, which

could well be coupled to Chapter 11. Chapters 12, 13 and 14 are not prerequisites to chapters that

follow. Part 2, with the exception of the brief treatment of kinetic and potential energies in the

Bernoulli’s principle section of Chapter 14 may be taught before, or without, Part 1. With the

exception noted, Part 1 is not prerequisite to Part 2.

xviii

PART 3: HEAT Except for the idea of kinetic energy, potential energy, and energy conservation

from Part 1, the material in these chapters is not prerequisite to the chapters that follow, nor are Parts

1 and 2 prerequisites to Part 3.

PART 4: SOUND Material from these chapters (forced vibrations, resonance, transverse and

standing waves, interference) serves as a useful background for Chapters 26, 29 and 31. Parts 1-3 are

not prerequisites to Part 4.

PART 5: ELECTRICITY AND MAGNETISM Part 1 is prerequisite to Part 5. Also helpful are

Chapters 11, 14, and 19. The chapters of Part 5 build from electrostatics and magnetism to

electromagnetic induction—which serve as a background for the nature of light.

PART 6: LIGHT Parts 4 and 5 provide useful background to Part 6. If you begin your course with

light, then be sure to discuss simple waves and demonstrate resonance (which are treated in Part 4). If

you haven’t covered Part 5, then be sure to discuss and demonstrate electromagnetic induction if you

plan to treat the nature of light. The very briefest treatment of light can cover

Chapters 26-28. A very brief treatment of lenses is in Chapter 28. A modern treatment of light

should include Chapters 30 and 31.

PART 7: ATOMIC AND NUCLEAR PHYSICS Chapter 11 provides a good background for Part

7. Chapter 33 is prerequisite to Chapter 34. Otherwise, Part 7 can stand on its own.

PART 8: RELATIVITY This part can stand on its own and will nicely follow immediately from

Part 1, if the ideas of the Doppler effect and wave frequency are treated in lecture. A thorough

treatment of only Parts 1 and 8 should make a good quarter-length course.

1

1 About Science

Conceptual Physic Instructor’s Manual, 12th Edition

1.1 Scientific Measurements

How Eratosthenes Measured the Size of Earth

Size of the Moon

Distance to the Moon

Distance to the Sun

Size of the Sun

Mathematics—The Language of Science

1.2 Scientific Methods

The Scientific Attitude

1.3 Science, Art, and Religion

Pseudoscience

1.4 Science and Technology

Risk Assessment

1.5 Physics—The Basic Science

1.6 In Perspective

Much of this introductory chapter, like most introductions, can be regarded as a personal essay by the

author. While many physics instructors may discuss somewhat different topics in a somewhat different

way, the comments made here may prove to be useful as a background for further comments of your own.

The chapter opens with a pair of photos of my wife beneath a tree in front of our residence in San

Francisco. They replace the rendered photos of a partial eclipse of the previous edition. This pair of photos

is real, and others taken by Dean Baird and Paul Doherty are in Chapter 26. The pair of photos lead to the

profile of Eratosthenes, and his early measurements of the Earth. Such merits further explanation and a

good way to kick off your course. Follow this up with the early measurement of the Moon and Sun by

Aristarchus. More on these early measurements is found in the excellent book Physics for the Inquiring

Mind, by Eric Rogers, originally published in 1960 by Princeton University Press.

You may consider elaborating on the idea about the possible wrongness versus rightness of ideas; an idea

that characterizes science. This is generally misunderstood, for it is not generally a criterion in other

disciplines. State that it is the prerogative of science, in contrast to the speculative procedures of philosophy

and meta-physics, to embrace only ideas that can be tested and to disregard the rest. Ideas that can’t be

tested are not necessarily wrong—they are simply useless insofar as advancement in scientific knowledge

is concerned. Ideas must be verifiable by other scientists. In this way science tends to be self-correcting.

Expand on the idea that honesty in science is not only a matter of public interest, but is a matter of self-

interest. Any scientist who misrepresents or fudges data, or is caught lying about scientific information, is

ostracized by the scientific community. There are no second chances. The high standards for acceptable

performance in science, unfortunately, do not extend to other fields that

,m – 0.2 m = 0.8 m. The time is found using g = 10 m/s2

and y = 0.8 m = ½ gt2. Solving for t we get t = √2y/g = √[2(0.8m)/10 m/s2] =0.4 s. Horizontal travel is

then d = vt = (8.0 m/s)(0.4 s) = 3.2 m. (If the height of the can is not subtracted from the 1.0-m vertical

distance between floor and tabletop, the calculated d will equal 3.6 m, the can will be too far away, and

the ball will miss!)

31. Total energy = 5000 MJ + 4500 MJ = 9500 MJ. Subtract 6000 MJ and KE = 3500 MJ.

32. In accord with the work-energy theorem (Chapter 7) W = ∆KE the work done equals energy gained.

The KE gain is 8 - 5 billion joules = 3 billion joules. The potential energy decreases by the same

amount that the kinetic energy increases, 3 billion joules.

33. Hang time depends only on the vertical component of initial velocity and the corresponding vertical

distance attained. From d = 5t2 a vertical 1.25 m drop corresponds to 0.5 s (t = √2d/g = √2(1.25)/10 =

0.5 s). Double this (time up and time down) for a hang time of 1 s. Hang time is the same whatever the

horizontal distance traveled.

34. (a) We’re asked for horizontal speed, so we write,

v x

d

t

, where d is horizontal distance traveled in

time t. The time t of the ball in flight is as if we drop it from rest a vertical distance y from the top of the

net. At highest point in its path, its vertical component of velocity is zero.

From y 1/2gt2 t2

2 y

g

t

2y

g

. So v

d

2 y

g

.

(b ) v

d

2y

g

12 .0m

2(1 .00 m )

10m / s

2

26 .8 m / s 27 m / s.

(c) Note mass of the ball doesn’t show in the equation, so mass is irrelevant.

Think and Rank

35. a. B, C, A, D b. B, D, A, C c. A=B=C=D (10 m/s2)

36. a. A=B=C b. A=B=C c. A=B=C d. B, A, C

37. a. A, B, C b. C, B, A

38. a. A, B, C, D b. A, B, C, D c. A, B, C, D d. A, B, C, D e. D, C, B, A f. A=B=C=D g. A, B, C, D

Think and Explain

39. Divers can orient their bodies to change the force of air resistance so that the ratio of net force to mass

is nearly the same for each.

40. In accord with the principle of horizontal and vertical projectile motion, the time to hit the floor is

independent of the ball’s speed.

114

41. Yes, it will hit with a higher speed in the same time because the horizontal (not the vertical) component

of motion is greater.

42. No, because while the ball is in the air its horizontal speed doesn’t change, but the train’s speed does.

43. The crate will not hit the Porsche, but will crash a distance beyond it determined by the height and

speed of the plane.

44. The path of the falling object will be a parabola as seen by an observer off to the side on the ground.

You, however, will see the object fall straight down along a vertical path beneath you. You’ll be directly

above the point of impact. In the case of air resistance, where the airplane maintains constant velocity

via its engines while air resistance decreases the horizontal component of velocity for the falling object,

impact will be somewhere behind the airplane.

45. (a) The paths are parabolas. (b) The paths would be straight lines.

46. There are no forces horizontally (neglecting air resistance) so there is no horizontal acceleration,

hence the horizontal component of velocity doesn’t change. Gravitation acts vertically, which is why

the vertical component of velocity changes.

47. Minimum speed occurs at the top, which is the same as the horizontal component of velocity anywhere

along the path.

48. The bullet falls beneath the projected line of the barrel. To compensate for the bullet’s fall, the barrel is

elevated. How much elevation depends on the velocity and distance to the target. Correspondingly, the

gunsight is raised so the line of sight from the gunsight to the end of the barrel extends to the target. If

a scope is used, it is tilted downward to accomplish the same line of sight.

49.Both balls have the same range (see Figure 10.9). The ball with the initial projection angle of 30°,

however, is in the air for a shorter time and hits the ground first.

50.The monkey is hit as the dart and monkey meet in midair. For a fast-moving dart, their meeting place is

closer to the monkey’s starting point than for a slower-moving dart. The dart and monkey fall equal

vertical distances—the monkey below the tree, and the dart below the line of sight—because they both

fall with equal accelerations for equal times.

51. Any vertically projected object has zero speed at the top of its trajectory. But if it is fired at an angle,

only its vertical component of velocity is zero and the velocity of the projectile at the top is equal to its

horizontal component of velocity. This would be 100 m/s when the 141-m/s projectile is fired at 45°.

52. Hang time depends only on the vertical component of your lift-off velocity. If you can increase this

vertical component from a running position rather than from a dead stop, perhaps by bounding harder

against the ground, then hang time is also increased. In any case, hang time depends only on the

vertical component of your lift-off velocity.

53. The hang time will be the same, in accord with the answer to the preceding exercise. Hang time is

related to the vertical height attained in a jump, not on horizontal distance moved across a level floor.

54. The Moon’s tangential velocity is what keeps the Moon coasting around the Earth rather than crashing

into it. If its tangential velocity were reduced to zero, then it would fall straight into the Earth!

55. From Kepler’s third law, T

2

~ R

3

, the period is greater when the distance is greater. So the periods of

planets farther from the Sun are longer than our year.

56. Yes, the satellite is accelerating, as evidenced by its continual change of direction. It accelerates due

to the gravitational force between it and the Earth. The acceleration is toward the Earth’s center.

57. Speed does not depend on the mass of the satellite (just as free-fall speed doesn’t).

58. Neither the speed of a falling object (without air resistance) nor the speed of a satellite in orbit depends

on its mass. In both cases, a greater mass (greater inertia) is balanced by a correspondingly greater

gravitational force, so the acceleration remains the same (a = F/m, Newton’s 2nd law).

115

59. Gravitation supplies the centripetal force on satellites.

60. The initial vertical climb lets the rocket get through the denser, retarding part of the atmosphere most

quickly, and is also the best direction at low initial speed, when a large part of the rocket’s thrust is

needed just to support the rocket’s weight. But eventually the rocket must acquire enough tangential

speed to remain in orbit without thrust, so it must tilt until finally its path is horizontal.

61. Gravity changes the speed of a cannonball when the cannonball moves in the direction of Earth

gravity. At low speeds, the cannonball curves downward and gains speed because there is a

component of the force of gravity along its direction of motion. Fired fast enough, however, the

curvature matches the curvature of the Earth so the cannonball moves at right angles to the force of

gravity. With no component of force along its direction of motion, its speed remains constant.

62. Upon slowing it spirals in toward the Earth and in so doing has a component of gravitational force in its

direction of motion which causes it to gain speed. Or put another way, in circular orbit the

perpendicular component of force does no work on the satellite and it maintains constant speed. But

when it slows and spirals toward Earth there is a component of gravitational force that does work to

increase the KE of the satellite.

63. A satellite travels faster when closest to the body it orbits. Therefore Earth travels faster about the Sun

in December than in June.

64. Yes, a satellite

,are as important to the human

condition. For example, consider the standards of performance required of politicians.

Distinguish between hypothesis, theory, fact, and concept. Point out that theory and hypothesis are not the

same. A theory applies to a synthesis of a large body of information. The criterion of a theory is not

whether it is true or untrue, but rather whether it is useful or nonuseful. A theory is useful even though the

ultimate causes of the phenomena it encompasses are unknown. For example, we accept the theory of

gravitation as a useful synthesis of available knowledge that relates to the mutual attraction of bodies. The

theory can be refined, or with new information it can take on a new direction. It is important to

acknowledge the common misunderstanding of what a scientific theory is, as revealed by those who say,

“But it is not a fact; it is only a theory.” Many people have the mistaken notion that a theory is tentative or

speculative, while a fact is absolute.

2

Impress upon your class that a fact is not immutable and absolute, but is generally a close agreement by

competent observers of a series of observations of the same phenomena. The observations must be testable.

Since the activity of science is the determination of the most probable, there are no absolutes. Facts that

were held to be absolute in the past are seen altogether differently in the light of present-day knowledge.

By concept, we mean the intellectual framework that is part of a theory. We speak of the concept of time,

the concept of energy, or the concept of a force field. Time is related to motion in space and is the

substance of the Theory of Special Relativity. We find that energy exists in tiny grains, or quanta, which is

a central concept in the Quantum Theory. An important concept in Newton’s Theory of Universal

Gravitation is the idea of a force field that surrounds a material body. A concept envelops the

overriding idea that underlies various phenomena. Thus, when we think “conceptually” we envelop a

generalized way of looking at things.

Prediction in science is different than prediction in other areas. In the everyday sense, one speaks of

predicting what has not yet occurred, like whether or not it will rain next weekend. In science, however,

prediction is not so much about what will happen, but about what is happening and is not yet noticed, like

what the properties of a hypothetical particle are and are not. A scientist predicts what can and cannot

happen, rather than what will or will not happen.

In biology, for example, you explain events once you see them. In a sense you’re looking at the historical

behavior and then you explain patterns. In physics you’re more likely to predict patterns before they’re

seen.

Max Born, Nobel-prize recipient and one of the most outstanding physicists of the twentieth century, is

quoted in the insight box of page 12. It was to a letter to Max by his close friend Albert Einstein in 1926

that Einstein made his famous remark regarding quantum mechanics, often paraphrased as “God does not

play dice with the universe.” Max Born died in 1970, and was the maternal grandfather of the popular

singer Olivia Newton-John.

Science and Technology

In discussions of science and technology and their side effects, a useful statement is: You can never do just

one thing. This is similar to “there is never just one force” in discussions of Newton’s third law.

With regard to risk, you can prove something to be unsafe, but you can never prove something to be

completely safe.

Engineering is the practical application of science to commerce or industry. The tripartite arrangement of

science, technology, and engineering has always been the combination for successful advancement.

“Any sufficiently advanced society is indistinguishable from magic.” Arthur C. Clark

One can quip that a first stage of scientific discovery is to deny that it’s true, the second is to deny that it’s

important, and the third is to credit the wrong person.

The medieval philosopher, William of Occam, wisely stated that when deciding between two competing

theories, choose the simpler explanation—don’t make more assumptions than are necessary when

describing phenomena.

Physicists have a deep-seated need to know “Why?” and “What if?”. Mathematics is foremost in the

toolkits they develop to tackle these questions. Galileo stated that the book of nature is written in

mathematics. (Tidbit: Galileo and Shakespeare were born in the same year, 1564.)

Science is never a closed book, for its conclusions are based on evidence, and new evidence can contradict

old conclusions and lead to a better understanding of nature. Anytime anybody tells you that they are

“absolutely certain” of some general idea, you can be assured that their conclusion is not scientific, because

science never produces absolute certainty. - Art Hobson

3

Science and Religion

Do the two contradict each other—must one choose between them? These questions are foremost among

many students, yet physics texts usually sidestep such questions, for religion is very personal for so many

people. I hope the very brief treatment in the text presents a satisfactory answer to these questions. Your

feedback on this matter will be appreciated.

With regard to science courses and liberal arts courses, there is a central factor that makes it difficult for

liberal arts students to delve into science courses the way that science students can delve into liberal arts

courses—and that’s the vertical nature of science courses. They build upon each other, as noted by their

prerequisites. A science student can take an intermediate course in literature, poetry, or history at any time,

and in any order. But in no way can a humanities student take an intermediate physics or chemistry course

without first having a foundation in elementary physics and mathematics. Hence the importance of this

conceptual course.

Except of the measurements by early Greek scientists, I do not lecture about Chapter 1 material and instead

assign it as reading. It can be omitted without interfering with the following chapters.

Measuring Solar Diameter: One of my very favorite class assignments is the task of measuring the

diameter of the Sun with a ruler or tape measure. This makes sense by first explaining the physics of a

pinhole camera. The pinhole image technique is described on Practice Page 2. Hold a meterstick up and

state to the class that with such a measuring device, a strip of measuring tape or a simple ruler, they can

measure the diameter of the Sun. Call attention to Figure 1.6, then sketch a simple pinhole camera on the

board thusly.

Tell of how a small hole poked in a piece of cardboard will show the

image of the Sun when the card is placed in sunlight. You can explain this

without referring to Figure 1.6 in the text because the figure gives the ratio

you wish them to determine. I find this first assignment very successful, in

that simple measurements yield a most impressive value—a confidence

builder. For those who don’t succeed, or succeed partially, I urge them to

try again for full credit.

Practicing Physics Book:

• Making Hypotheses

• Pinhole Formation

Next-Time Questions (in the Instructor Resource DVD):

• Scientific Claims

• Pinhole Image of the Sun

• Solar Image

• Cone, Ball, and a Cup

Laboratory Manual:

There are no labs for Chapter 1

4

Answers for Chapter 1

Reading Check Questions

1. Science is the product of human curiosity about how the world works—an organized body of knowledge

that describes order and causes within nature and an ongoing human activity dedicated to gathering and

organizing knowledge about the world.

2. The general reaction has been to forbid new ideas.

3. Alexandria was farther north, at a higher latitude.

4. The shadow tapers because of the large size of the Sun, certainly not a point source of

,light.

5. Like the Sun, the Moon’s diameter is 1/110 the distance between Earth and the Moon.

6. The Sun’s diameter is 1/110 the distance between Earth and the Sun.

7. At the time of a half moon he knew the angle between a line joining the Moon and Earth was at 90° to the

line joining the Moon and the Sun.

8. The circular spots are pinhole images of the Sun.

9. The equations are guides to thinking that show the connections between concepts in nature.

10. First, observe; 2. Question; 3. Predict; 4. Test predictions; 5. Draw a conclusion.

11. The answer is as stated in the Summary of Terms.

12. Competent scientists must be experts at changing their minds.

13. A scientific hypothesis must be testable.

14. Whereas mistakes or misrepresentations are given second chances in daily life, second chances are not

given to scientists by the scientific community.

15. See if you can state the position of an antagonist to the antagonist’s satisfaction, and compare it to how

well the antagonist can state your position. If you can, and your antagonist can’t, the likelihood is that you

are correct in your position.

16. To know more than what’s in your bag of beliefs and attitudes is to expand your education.

17. No. Science and religion can work well together, and even complement each other. (Religious

extremists, however, may assert that the two are incompatible).

18. One benefit is an open and exploring mind.

19. Science is gathering knowledge and organizing it; technology puts scientific knowledge to practical use

and provides the instruments scientists need to conduct their investigations.

20. The other sciences build upon physics, and not the other way around.

Think and Do

21. The triangle coin image-coin distance is similar to the larger triangle Sun diameter-Sun distance, so the

numbers of coins and Suns are the same. The number of Suns that would fit between Earth’s surface

and the Sun is 110.

22. Open ended, as lists will vary.

Think and Explain

23.The penalty for fraud is professional excommunication.

24. (a) This is a scientific hypothesis, for there is a test for wrongness. For example, you can extract

chlorophyll from grass and note its color.

(b) This statement is without a means of proving it wrong and is not a scientific hypothesis. It is

speculation.

(c) This is a scientific hypothesis. It could be proved wrong, for example, by showing tides that do not

correspond to the position of the Moon.

25. Aristotle’s hypotheses was partially correct. Plant material comes partly from the soil, but mainly from the

air and water. An experiment would be to weigh a pot of soil with a small seedling, then weigh the potted

plant later after it has grown. The fact that the grown plant will weigh more is evidence that the plant is

composed of more material than the soil offers. Keep a record of the weight of water used to water the

plant, and cover the soil with plastic wrap to minimize evaporation losses. Then the weight of the grown

plant can be compared with the weight of water it absorbs. How can the weight of air taken in by the

plant be estimated?

26. The Sun’s radius is approximately 7 108 m. The distance between the Earth and Moon is about 4 108

m. So the Sun’s radius is much larger, nearly twice the distance between the Earth and Moon. The

Earth and Moon at their present distance from each other would easily fit inside the Sun. The Sun is

really big—surprisingly big!

5

27.What is likely being misunderstood is the distinction between theory and hypothesis. In common usage,

“theory” may mean a guess or hypothesis, something that is tentative or speculative. But in science a

theory is a synthesis of a large body of validated information (e.g., cell theory or quantum theory). The

value of a theory is its usefulness (not its “truth”).

28.Yes, there is a geometric connection between the two ratios.

As the sketch shows, they are approximately equal.

Pole shadow

Pole height

=

Alexandria - Syene distance

Earth radius

From this pair of ratios, given the distance between

Alexandria and Syene, the radius of the Earth can be calculated!

29. The shadow would be longer because on the smaller planet the angle of the pole would be greater

relative to the sunlight. The ratio of the shadow to pole height would be greater than 1 to 8 as in the

previous answer.

Think and Discuss

30. To publicly change your mind about your ideas is a sign of strength rather than a sign of weakness. It

takes more courage to change your ideas when confronted with counter evidence than to hold fast to

your ideas. If a person’s ideas and view of the world are no different after a lifetime of varied

experience, then that person was either miraculously blessed with unusual wisdom at an early age, or

learned nothing. The latter is more likely. Education is learning that which you don’t yet know about. It

would be arrogant to think you know it all in the later stages of your education, and stupid to think so at

the beginning of your education.

31. The examples are endless. Knowledge of electricity, for example, has proven to be extremely useful. The

number of people who have been harmed by electricity who understood it is far fewer than the number

of people who are harmed by it who don’t understand it. A fear of electricity is much more harmful than

useful to one’s general health and attitude.

32. Your advice will depend on your own views about questioning authority. Would you suggest that your

young congregate with a smaller number of friends who have reasonable doubts than ones who are

absolutely certain about everything? If the concern is for the largest numbers of potential friends, groups

in the United States that feature non-questioning of authority have enormously large memberships.

6

2 Newton’s First Law of Motion—Inertia

Conceptual Physics Instructor’s Manual, 12th Edition

2.1 Aristotle on Motion

Aristotle

Copernicus and the Moving Earth

2.2 Galileo’s Experiments

Leaning Tower

Galileo Galilei

Inclined Planes

2.3 Newton’s First Law of Motion

Personal Essay

2.4 Net Force and Vectors

2.5 The Equilibrium Rule

2.6 Support Force

2.7 Equilibrium of Moving Things

2.8 The Moving Earth

The little girl with the Newton’s Cradle apparatus in the Part One opener is Charlotte Ackerman, of San

Francisco. She appears later as a chapter opener in Chapter 20. Photo openers begin with a recent inertia

demonstration of me on my back with a blacksmith’s anvil resting on my body, and friend Will Maynez

who swings the sledge hammer. The photo of the balanced rock is by my friend Howie Brand, who retired

to Thailand. Sweden friends Cedric and Anne Linder, profiled on the next page, pose with the vector

demonstration. Karl Westerberg of CCSF shows one of my favorite demos with the suspended ball and

strings.

Galileo was introduced in Chapter 1 and is also featured in this chapter. On August 25 in 1609 (405 years

before 2014) he demonstrated his newly constructed telescope to the merchants of Venice, and shortly

thereafter, aimed it on the skies. And as we know, his findings had much to do with the advent of science in

those intellectual scary times.

Whereas the study of mechanics in earlier editions began with kinematics, we begin our study with a much

easier concept for your students—forces. We postpone what I call the black hole of physics instruction—

overemphasis on kinematics. You should find that starting a course off with forces first will lessen the

initial roadblock that kinematics poses. Of particular interest to me is the Personal Essay in the chapter,

which relates to events that inspired me to pursue a life in physics—my meeting with Burl Grey on the

sign-painting stages of Miami, Florida. Relative tensions in supporting cables are what first caught my

interest in physics,

,and I hope to instill the same interest with your students with this opening chapter. This

story is featured on the first of the Hewitt-Drew-It screencasts, Equilibrium Rule, which nicely introduces

vectors.

Force vectors are easier to grasp than velocity vectors treated in the following chapter. (More on vectors in

Appendix A.)

Note that in introducing force I first use pounds—most familiar to your students. A quick transition,

without fanfare, introduces the newton. I don’t make units a big deal and don’t get into the laborious task of

unit conversions, which is more appropriate for physics majors.

The distinction between mass and weight will await the following chapter, when it’s needed in Newton’s

second law. I see the key to good instruction as treating somewhat difficult topics only when they are used.

For example, I see as pedagogical folly spending the first week on unit conversions, vector notation,

graphical analysis, and scientific notation. How much better if the first week is a hook to promote class

interest, with these things introduced later when needed.

Practicing Physics Book:

7

• Static Equilibrium

• The Equilibrium Rule: F = 0

• Vectors and Equilibrium

Laboratory Manual:

• Walking the Plank Equilibrium Rule (Experiment)

Next-Time Questions (in the Instructor Resource DVD):

• Ball Swing

• Pellet in the Spiral

• Falling Elephant and Feather

Hewitt-Drew-It! Screencasts:

• Equilibrium Rule • Equilibrium Problems

•Net Force and Vectors •Nellie’s Rope Tensions

•Nellie’s Ropes •Force Vector Diagrams

•Force Vectors on an Incline

SUGGESTED LECTURE PRESENTATION

Newton’s 1st Law

Begin by pointing to an object in the room and stating that if it started moving, one would reasonably look

for a cause for its motion. We would say that a force of some kind was responsible, and that would seem

reasonable. Tie this idea to the notion of force maintaining motion as Aristotle saw it. State that a

cannonball remains at rest in the cannon until a force is applied, and that the force of expanding gases

drives the ball out of the barrel when it is fired. (I have a 10-cm diameter solid steel sphere, actually a

huge ball bearing, that I use in this lecture. Use one, or a bowling ball, if available.) But what keeps the

cannonball moving when the gases no longer act on it? This leads you into a discussion of inertia. In the

everyday sense, inertia refers to a habit or a rut. In physics it’s another word for laziness, or the resistance

to change as far as the state of motion of an object is concerned. I roll the ball along the lecture table to

show its tendency to keep rolling. Inertia was first introduced not by Newton, but by Galileo as a result of

his inclined-plane experiments.

DEMONSTRATION: Show that inertia refers also to objects at rest with the classic tablecloth-

and-dishes demonstration. [Be sure to pull the tablecloth slightly downward so there is no upward

component of force on the dishes!] I precede this demo with a simpler version, a simple block of

wood on a piece of cloth—but with a twist. I ask what the block will do when I suddenly whip the

cloth toward me. After a neighbor check, I surprise the class when they see that the block has been

stapled to the cloth! This illustrates Newton’s zeroth law—be skeptical. Then I follow up with the

classic tablecloth demo. Don’t think the classic demo is too corny, for your students will really

love it.

Of course when we show a demonstration to illustrate a particular concept, there is almost always more

than one concept involved. The tablecloth demo is no exception, which also illustrates impulse and

momentum (Chapter 6 material). The plates experience two impulses; one that first involves the friction

between the cloth and dishes, which moves them slightly toward you. It is brief and very little momentum

builds up. Once the dishes are no longer on the cloth, a second impulse occurs due to friction between the

sliding dishes and table, which acts in a direction away from you and prevents continued sliding toward

you, bringing the dishes to rest. Done quickly, the brief displacement of the dishes is hardly noticed. Is

inertia really at work here? Yes, for if there were no friction, the dishes would strictly remain at rest.

DEMONSTRATION: Continuing with inertia, do as Jim Szeszol does and

fashion a wire coat hanger into an m shape as shown. Two globs of clay

are stuck to each end. Balance it on your head, with one glob in front of

your face. State you wish to view the other glob and ask how you can do

8

so without touching the apparatus. Then simply turn around and look at it. It’s like the bowl of

soup you turn only to find the soup stays put. Inertia in action! (Of course, like the tablecloth

demo, there is more physics here than inertia; this demo can also be used to illustrate rotational

inertia and the conservation of angular momentum.)

A useful way to impart the idea of mass and inertia is to place two objects, say a pencil and a piece of

chalk, in the hands of a student and ask for a judgment of which is heavier. The student will likely respond

by shaking them, one in each hand. Point out that in so doing the student is really comparing their inertias,

and is making use of the intuitive knowledge that weight and inertia are directly proportional to each other.

In Chapter 4 you’ll focus more on the distinction between mass and weight, and between mass and volume.

CHECK QUESTION: How does the law of inertia account for removing snow from your shoes by

stamping on the floor, or removing dust from a coat or rug by shaking it?

DEMONSTRATION: Do as Marshall Ellenstein does and place a metal hoop atop a

narrow jar. On top of the hoop balance a piece of chalk. Then whisk the hoop away

and the chalk falls neatly into the narrow opening. The key here is grabbing the hoop

on the inside, on the side farthest from your sweep. This elongates the hoop

horizontally and the part that supports the chalk drops from beneath the chalk. (If you

grab the hoop on the near side, the elongation will be vertical and pop the chalk up

into the air!)

Units of Force—Newtons:

I suggest not making a big deal about the unfamiliar unit of force—the newton. I simply state it is the unit

of force used by physicists, and if students find themselves uncomfortable with it, simply think of “pounds”

in its place. Relative magnitudes, rather than actual magnitudes, are the emphasis of conceptual physics

anyway. Do as my influential pal Burl Grey does in Figure 2.13 and suspend a familiar mass from a spring

scale. If the mass is a kilogram and the scale is calibrated in newtons, it will read 10 N (more precisely, 9.8

N). If the scale is calibrated in pounds it will read 2.2 pounds. State that you’re not going to waste good

time in conversions between units (students can do enough of that in one of those dull physics courses

they’ve heard about).

CHECK QUESTION: Which has more mass, a 1-kg stone or a 1-lb stone? [A 1-kg stone has more

mass, for it weighs 2.2 lb. But we’re not going to make a big deal about such conversions. If the

units newtons bugs you, think of it as a unit of force or weight in a foreign language for now!]

Net Force

Discuss the idea of more than one force acting on something, and the resulting net force. Figure 2.10 or

Figure 2.12 captures the essence.

Support Force (Normal Force)

Ask what forces act on a book at rest on your lecture table. Then discuss Figure 2.15, explaining that the

atoms in the table behave like tiny springs. This upward support force is equal and opposite to the weight of

the book, as evidenced by the book’s state of rest. The support force is a very real force. Without it, the

book would be in a state of free fall.

Statics and the Equilibrium Rule: Cite other static examples, where the net force is zero as evidenced by

no changes in motion. Hold the

,1-kg mass at rest in your hand and ask how much net force acts on it. Be

sure they distinguish between the 10-N gravitational force on the object and the zero net force on it—as

evidenced by its state of rest. (The concept of acceleration is introduced in the next chapter.) When

suspended by the spring scale, point out that the scale is pulling up on the object, with just as much force as

the Earth pulls down on it. Pretend to step on a bathroom scale. Ask how much gravity is pulling on you.

This is evident by the scale reading. Then ask what the net force is that acts on you. This is evident by your

absence of motion change. Consider two scales, one foot on each, and ask how each scale would read. Then

ask how the scales would read if you shifted your weight more on one than the other. Ask if there is a rule

9

to guide the answers to these questions. There is; F = 0. For any object in equilibrium, the net force on it

must be zero. Before answering, consider the skit in my Personal Essay.

Sign painter Skit: Draw on the board the sketch below, which shows two painters on a painting rig

suspended by two ropes.

Step 1: If both painters have the same weight and each stands next to a rope, the supporting force

of the ropes will be equal. If spring scales were used, one on each rope, the forces in the ropes

would be evident. Ask what the scale reading for each rope would be in this case. [The answer is

each rope will support the weight of one man + half the weight of the rig—both scales will show

equal readings.]

Step 2: Suppose one painter walks toward the other as shown in the second sketch above, which

you draw on the chalkboard (or show via overhead projector). Will the tension in the left rope

increase? Will the tension in the right rope decrease? Grand question: Will the tension in the left

rope increase exactly as much as the decrease in tension of the right rope? You might quip that is

so, how does either rope “know” about the change in the other rope? After neighbor discussion, be

sure to emphasize that the answers to these questions lie in the framework of the equilibrium rule:

F = 0. Since there is no change in motion, the net force must be zero, which means the upward

support forces supplied by the ropes must add up to the downward force of gravity on the two men

and the rig. So a decrease in one rope must necessarily be met with a corresponding increase in the

other. (This example is dear to my heart. Burl and I didn’t know the answer way back then

because neither he nor I had a model for analyzing the problem. We didn’t know about Newton’s

first law and the Equilibrium Rule. How different one’s thinking is when one has or does not have

a model to guide it. If Burl and I had been mystical in our thinking, we might have been more

concerned with how each rope “knows” about the condition of the other—an approach that

intrigues many people with a nonscientific view of the world.)

Inertia and the Moving Earth:

Stand facing a wall and jump up. Then ask why the wall does not smash into you as the Earth rotates under

you while you’re airborne. Relate this to the idea of a helicopter ascending over San Francisco, waiting

motionless for 3 hours and waiting until Washington, D.C. appears below, then descending. Hooray, this

would be a neat way to fly cross-country! Except, of course, for the fact that the “stationary” helicopter

remains in motion with the ground below. “Stationary” relative to the stars means it would have to fly as

fast as the Earth turns (what jets attempt to do).

Forces at an Angle:

This chapter introduces vectors as they relate to tensions in ropes at

an angle. Other cases are developed in the Practicing Physics Book.

As a demonstration, support a heavy weight with a pair of scales as

shown. Show that as the angles are wider, the tensions increase.

This explains why one can safely hang from a couple of strands of

vertical clothesline, but can’t when the clothesline is horizontally

strung. Interesting stuff.

10

Answers and Solutions for Chapter 2

Reading Check Questions

1. Aristotle classified the motion of the Moon as natural.

2. Aristotle classified the motion of the Earth as natural.

3. Copernicus stated that Earth circles the Sun, and not the other way around.

4. Galileo discovered that objects in fall pick up equal speeds whatever their weights.

5. Galileo discovered that a moving object will continue in motion without the need of a force.

6. Inertia is the name given to the property of matter that resists a change in motion.

7. Newton’s law is a restatement of Galileo’s concept of inertia.

8. In the absence of force, a moving body follows a straight-line path.

9. The net force is 70 pounds to the right.

10. A description of force involves magnitude and direction, and is therefore a vector quantity.

11. The diagonal of a parallelogram represents the resultant of the vector pair.

12. The resultant is √2 pounds.

13. The tension in each rope would be half Nellie’s weight.

14. Yes, although science texts favor the newton.

15. The net force is zero.

16. The net force is zero.

17. All the forces on something in mechanical equilibrium add vectorally to zero.

18. F = 0.

19. The support force is 15 N. The net force on the book is zero.

20. Weight and support force have equal magnitudes.

21. Yes. The ball moving at constant speed in a straight-line path is in dynamic equilibrium.

22. An object in either static or dynamic equilibrium has a zero net force on it.

23. The force of friction is 100 N.

24. They had no understanding of the concept of inertia.

25. The bird still moves at 30 km/s relative to the Sun.

26. Yes, like the bird of Figure 2.18, you maintain a speed of 30 km/s relative to the Sun, in accord with the

concept of inertia.

Think and Solve

27. Since each scale reads 350 N, Lucy’s total weight is 700 N.

28. 800 N on one scale, 400 N on the other. (2x + x = 1200 N; 3x = 1200 N; x = 400 N)

29.From the equilibrium rule, F = 0, the upward forces are 800 N, and the downward forces are 500 N +

the weight of the scaffold. So the scaffold must weigh 300 N.

30.From the equilibrium rule, F = 0, the upward forces are 800 N + tension in the right scale. This sum

must equal the downward forces 500 N + 400 N + 400 N. Arithmetic shows the reading on the right

scale is 500 N.

Think and Rank

31. C, B, A

32. C, A, B, D

33. a. B, A, C, D

b. B, A, C, D

34. a. A=B=C (no force)

b. C, B, A

35. B, A, C

36. (a)

Think and Explain

37. Aristotle favored philosophical logic while Galileo favored experimentation.

38. The tendency of a rolling ball is to continue rolling—in the absence of a force. The fact that it slows

down is likely due to the force of friction.

11

39.Copernicus and others of his day thought an enormous force would have to continuously push the Earth

to keep it in motion. He was unfamiliar with the concept of inertia, and didn’t realize that once a body is

in motion, no force is needed to keep it moving (assuming no friction).

40.Galileo discredited Aristotle’s idea that the rate at which bodies fall is proportional to their weight.

41.Galileo demolished the notion that a moving body requires a force to keep it moving. He showed that a

force is needed to change motion, not to keep a body moving, so long as friction was negligible.

42.Galileo proposed the concept of inertia before Newton was born.

43.Nothing keeps asteroids moving. The Sun’s force deflects their paths but is not needed to keep them

moving.

44.Nothing keeps the probe moving. In the absence of a propelling or deflecting force it would continue

moving in a straight line.

45.If you pull the cloth upward, even slightly, it will tend to lift the dishes, which will disrupt the demonstration

to show the dishes remaining at rest. The cloth is best pulled horizontally for the dishes to remain at

rest.

,m – 0.2 m = 0.8 m. The time is found using g = 10 m/s2 and y = 0.8 m = ½ gt2. Solving for t we get t = √2y/g = √[2(0.8m)/10 m/s2] =0.4 s. Horizontal travel is then d = vt = (8.0 m/s)(0.4 s) = 3.2 m. (If the height of the can is not subtracted from the 1.0-m vertical distance between floor and tabletop, the calculated d will equal 3.6 m, the can will be too far away, and the ball will miss!) 31. Total energy = 5000 MJ + 4500 MJ = 9500 MJ. Subtract 6000 MJ and KE = 3500 MJ. 32. In accord with the work-energy theorem (Chapter 7) W = ∆KE the work done equals energy gained. The KE gain is 8 - 5 billion joules = 3 billion joules. The potential energy decreases by the same amount that the kinetic energy increases, 3 billion joules. 33. Hang time depends only on the vertical component of initial velocity and the corresponding vertical distance attained. From d = 5t2 a vertical 1.25 m drop corresponds to 0.5 s (t = √2d/g = √2(1.25)/10 = 0.5 s). Double this (time up and time down) for a hang time of 1 s. Hang time is the same whatever the horizontal distance traveled. 34. (a) We’re asked for horizontal speed, so we write, v x dt, where d is horizontal distance traveled in time t. The time t of the ball in flight is as if we drop it from rest a vertical distance y from the top of the net. At highest point in its path, its vertical component of velocity is zero. From y 1/2gt2 t2 2 yg t 2yg. So v d2 yg. (b ) v d2yg12 .0m2(1 .00 m )10m / s2 26 .8 m / s 27 m / s. (c) Note mass of the ball doesn’t show in the equation, so mass is irrelevant. Think and Rank 35. a. B, C, A, D b. B, D, A, C c. A=B=C=D (10 m/s2) 36. a. A=B=C b. A=B=C c. A=B=C d. B, A, C 37. a. A, B, C b. C, B, A 38. a. A, B, C, D b. A, B, C, D c. A, B, C, D d. A, B, C, D e. D, C, B, A f. A=B=C=D g. A, B, C, D Think and Explain 39. Divers can orient their bodies to change the force of air resistance so that the ratio of net force to mass is nearly the same for each. 40. In accord with the principle of horizontal and vertical projectile motion, the time to hit the floor is independent of the ball’s speed. 114 41. Yes, it will hit with a higher speed in the same time because the horizontal (not the vertical) component of motion is greater. 42. No, because while the ball is in the air its horizontal speed doesn’t change, but the train’s speed does. 43. The crate will not hit the Porsche, but will crash a distance beyond it determined by the height and speed of the plane. 44. The path of the falling object will be a parabola as seen by an observer off to the side on the ground. You, however, will see the object fall straight down along a vertical path beneath you. You’ll be directly above the point of impact. In the case of air resistance, where the airplane maintains constant velocity via its engines while air resistance decreases the horizontal component of velocity for the falling object, impact will be somewhere behind the airplane. 45. (a) The paths are parabolas. (b) The paths would be straight lines. 46. There are no forces horizontally (neglecting air resistance) so there is no horizontal acceleration, hence the horizontal component of velocity doesn’t change. Gravitation acts vertically, which is why the vertical component of velocity changes. 47. Minimum speed occurs at the top, which is the same as the horizontal component of velocity anywhere along the path. 48. The bullet falls beneath the projected line of the barrel. To compensate for the bullet’s fall, the barrel is elevated. How much elevation depends on the velocity and distance to the target. Correspondingly, the gunsight is raised so the line of sight from the gunsight to the end of the barrel extends to the target. If a scope is used, it is tilted downward to accomplish the same line of sight. 49.Both balls have the same range (see Figure 10.9). The ball with the initial projection angle of 30°, however, is in the air for a shorter time and hits the ground first. 50.The monkey is hit as the dart and monkey meet in midair. For a fast-moving dart, their meeting place is closer to the monkey’s starting point than for a slower-moving dart. The dart and monkey fall equal vertical distances—the monkey below the tree, and the dart below the line of sight—because they both fall with equal accelerations for equal times. 51. Any vertically projected object has zero speed at the top of its trajectory. But if it is fired at an angle, only its vertical component of velocity is zero and the velocity of the projectile at the top is equal to its horizontal component of velocity. This would be 100 m/s when the 141-m/s projectile is fired at 45°. 52. Hang time depends only on the vertical component of your lift-off velocity. If you can increase this vertical component from a running position rather than from a dead stop, perhaps by bounding harder against the ground, then hang time is also increased. In any case, hang time depends only on the vertical component of your lift-off velocity. 53. The hang time will be the same, in accord with the answer to the preceding exercise. Hang time is related to the vertical height attained in a jump, not on horizontal distance moved across a level floor. 54. The Moon’s tangential velocity is what keeps the Moon coasting around the Earth rather than crashing into it. If its tangential velocity were reduced to zero, then it would fall straight into the Earth! 55. From Kepler’s third law, T2 ~ R3, the period is greater when the distance is greater. So the periods of planets farther from the Sun are longer than our year. 56. Yes, the satellite is accelerating, as evidenced by its continual change of direction. It accelerates due to the gravitational force between it and the Earth. The acceleration is toward the Earth’s center. 57. Speed does not depend on the mass of the satellite (just as free-fall speed doesn’t). 58. Neither the speed of a falling object (without air resistance) nor the speed of a satellite in orbit depends on its mass. In both cases, a greater mass (greater inertia) is balanced by a correspondingly greater gravitational force, so the acceleration remains the same (a = F/m, Newton’s 2nd law). 115 59. Gravitation supplies the centripetal force on satellites. 60. The initial vertical climb lets the rocket get through the denser, retarding part of the atmosphere most quickly, and is also the best direction at low initial speed, when a large part of the rocket’s thrust is needed just to support the rocket’s weight. But eventually the rocket must acquire enough tangential speed to remain in orbit without thrust, so it must tilt until finally its path is horizontal. 61. Gravity changes the speed of a cannonball when the cannonball moves in the direction of Earth gravity. At low speeds, the cannonball curves downward and gains speed because there is a component of the force of gravity along its direction of motion. Fired fast enough, however, the curvature matches the curvature of the Earth so the cannonball moves at right angles to the force of gravity. With no component of force along its direction of motion, its speed remains constant. 62. Upon slowing it spirals in toward the Earth and in so doing has a component of gravitational force in its direction of motion which causes it to gain speed. Or put another way, in circular orbit the perpendicular component of force does no work on the satellite and it maintains constant speed. But when it slows and spirals toward Earth there is a component of gravitational force that does work to increase the KE of the satellite. 63. A satellite travels faster when closest to the body it orbits. Therefore Earth travels faster about the Sun in December than in June. 64. Yes, a satellite