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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
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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.
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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.
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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.
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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!
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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.
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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).
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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.
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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