Instructor’s Solutions Manual
to accompany
COLLEGE PHYSICS
Table of Contents
Chapter 1 Introduction 1
Chapter 2 Force 24
Chapter 3 Acceleration and Newton’s Second Law of Motion 62
Chapter 4 Motion with Constant Acceleration 105
Chapter 5 Circular Motion 144
Review and Synthesis: Chapters 1–5 171
Chapter 6 Conservation of Energy 186
Chapter 7 Linear Momentum 218
Chapter 8 Torque and Angular Momentum 247
Review and Synthesis: Chapters 6–8 281
Chapter 9 Fluids 302
Chapter 10 Elasticity and Oscillations 330
Chapter 11 Waves 361
Chapter 12 Sound 387
Review and Synthesis: Chapters 9–12 408
Chapter 13 Temperature and the Ideal Gas 420
Chapter 14 Heat 447
Chapter 15 Thermodynamics 471
Review and Synthesis: Chapters 13–15 496
Chapter 16 Electric Forces and Fields 507
Chapter 17 Electric Potential 537
Chapter 18 Electric Current and Circuits 566
Review and Synthesis: Chapters 16–18 615
Chapter 19 Magnetic Forces and Fields 627
Chapter 20 Electromagnetic Induction 663
Chapter 21 Alternating Current 692
Review and Synthesis: Chapters 19–21 726
Chapter 22 Electromagnetic Waves 737
Chapter 23 Reflection and Refraction of Light 757
Chapter 24 Optical Instruments 792
Chapter 25 Interference and Diffraction 824
Review and Synthesis: Chapters 22–25 858
Chapter 26 Relativity 867
Chapter 27 Early Quantum Physics and the Photon 893
Chapter 28 Quantum Physics 921
Chapter 29 Nuclear Physics 948
Chapter 30 Particle Physics 974
Review and Synthesis: Chapters 26–30 981
, Chapter 1
INTRODUCTION
Conceptual Questions
1. Knowledge of physics is important for a full understanding of many scientific disciplines, such as chemistry,
biology, and geology. Furthermore, much of our current technology can only be understood with knowledge of
the underlying laws of physics. In the search for more efficient and environmentally safe sources of energy, for
example, physics is essential. Also, many study physics for the sense of fulfillment that comes with learning about
the world we inhabit.
2. Without precise definitions of words for scientific use, unambiguous communication of findings and ideas would
be impossible.
3. Even when simplified models do not exactly match real conditions, they can still provide insight into the features
of a physical system. Often a problem would become too complicated if one attempted to match the real
conditions exactly, and an approximation can yield a result that is close enough to the exact one to still be useful.
4. (a) 3 (b) 9
5. Scientific notation eliminates the need to write many zeros in very large or small numbers. Also, the appropriate
number of significant digits is unambiguous when written this way.
6. In scientific notation the decimal point is placed after the first (leftmost) numeral. The number of digits written
equals the number of significant figures.
7. Not all of the significant digits are precisely known. The least significant digit (rightmost) is an estimate and is
less precisely known than the others.
8. It is important to list the correct number of significant figures so that we can indicate how precisely a quantity is
known and not mislead the reader by writing digits that are not at all known to be correct.
9. The kilogram, meter, and second are three of the base units used in the SI system.
10. The SI system uses a well-defined set of internationally agreed upon standard units and makes measurements in
terms of these units and their powers of ten. The U.S. Customary system contains units that are primarily of
historical origin and are not based upon powers of ten. As a result of this international acceptance and the ease of
manipulation that comes from dealing with powers of ten, scientists around the world prefer to use the SI system.
11. Fathoms, kilometers, miles, and inches are units with dimensions of length. Grams and kilograms are units with
dimensions of mass. Years, months, and seconds are units with dimensions of time.
12. The first step toward successfully solving almost any physics problem is to thoroughly read the question and
obtain a precise understanding of the scenario. The second step is to visualize the problem, often making a quick
sketch to outline the details of the situation and the known parameters.
13. Trends in a set of data are often the most interesting aspect of the outcome of an experiment. Such trends are more
apparent when data is plotted graphically rather than listed in numerical tables.
14. The statement gives a numerical value for the speed of sound in air, but fails to indicate the units used for the
measurement. Without units, the reader cannot relate the speed to one given in familiar units such as km/s.
15. After solving a problem, it is a good idea to check that the solution is reasonable and makes intuitive sense. It may
also be useful to explore other possible methods of solution as a check on the validity of the first.
1
,Chapter 1: Introduction College Physics
Multiple-Choice Questions
1. (b) 2. (b) 3. (a) 4. (c) 5. (d) 6. (d) 7. (b) 8. (d) 9. (b) 10. (c)
Problems
1. Strategy The new fence will be 100% + 37% = 137% of the height of the old fence.
Solution Find the height of the new fence.
1.37 × 1.8 m = 2.5 m
60 s 60 min 24 h
2. Strategy There are × × = 86, 400 seconds in one day and 24 hours in one day.
1 min 1h 1d
Solution Find the ratio of the number of seconds in a day to the number of hours in a day.
86, 400 24 × 3600
= = 3600 1
24 24
3. Strategy Relate the surface area S to the radius r using S = 4π r 2 .
Solution Find the ratio of the new radius to the old.
S1 = 4π r12 and S2 = 4π r22 = 1.160 S1 = 1.160(4π r12 ).
4π r22 = 1.160(4π r12 )
r22 = 1.160r12
2
⎛ r2 ⎞
⎜⎜ ⎟⎟ = 1.160
⎝ r1 ⎠
r2
= 1.160 = 1.077
r1
The radius of the balloon increases by 7.7%.
4. Strategy Relate the surface area S to the radius r using S = 4π r 2 .
Solution Find the ratio of the new radius to the old.
S1 = 4π r12 and S2 = 4π r22 = 2.0S1 = 2.0(4π r12 ).
4π r22 = 2.0(4π r12 )
r22 = 2.0r12
2
⎛ r2 ⎞
⎜⎜ ⎟⎟ = 2.0
⎝ r1 ⎠
r2
= 2.0 = 1.4
r1
The radius of the balloon increases by a factor of 1.4.
5. Strategy To find the factor by which the metabolic rate of a 70 kg human exceeds that of a 5.0 kg cat use a ratio.
Solution Find the factor.
3/4 3/4
⎛ mh ⎞ ⎛ 70 ⎞
⎜⎜ ⎟⎟ =⎜ ⎟ = 7.2
⎝ mc ⎠ ⎝ 5.0 ⎠
2
, College Physics Chapter 1: Introduction
6. Strategy To find the factor Samantha’s height increased, divide her new height by her old height. Subtract 1 from
this value and multiply by 100 to find the percent increase.
Solution Find the factor.
1.65 m
= 1.10
1.50 m
Find the percentage.
1.10 − 1 = 0.10, so the percent increase is 10 % .
7. Strategy Recall that area has dimensions of length squared.
Solution Find the ratio of the area of the park as represented on the map to the area of the actual park.
map length 1 map area
= = 10−4 , so = (10−4 )2 = 10−8 .
actual length 10, 000 actual area
8. Strategy Let X be the original value of the index.
Solution Find the net percentage change in the index for the two days.
(first day change) × (second day change) = [ X × (1 + 0.0500)] × (1 − 0.0500) = 0.9975 X
The net percentage change is (0.9975 − 1) × 100% = −0.25%, or down 0.25% .
9. Strategy Use a proportion.
Solution Find Jupiter’s orbital period.
T 2 R3
T 2 ∝ R3 , so J = J = 5.193. Thus, TJ = 5.193/2 TE = 11.8 yr .
TE2 RE3
10. Strategy The area of the circular garden is given by A = π r 2 . Let the original and final areas be A1 = π r12 and
A2 = π r22 , respectively.
Solution Calculate the percentage increase of the area of the garden plot.
∆A π r 2 − π r12 r2 − r2 1.252 r12 − r12 1.252 − 1
× 100% = 2 × 100% = 2 1 × 100% = × 100% = × 100% = 56%
A π r12 r12 r12 1
11. Strategy The area of the poster is given by A = w. Let the original and final areas be A1 = 1w1 and
A2 = 2 w2 , respectively.
Solution Calculate the percentage reduction of the area.
A2 = 2 w2 = (0.800 1 )(0.800 w1 ) = 0.640 1w1 = 0.640 A1
A1 − A2 A − 0.640 A1
× 100% = 1 × 100% = 36.0%
A1 A1
12. Strategy The volume of the rectangular room is given by V = wh. Let the original and final volumes be
V1 = 1w1h1 and V2 = 2 w2 h2 , respectively.
Solution Find the factor by which the volume of the room increased.
V2 w h (1.50 1 )(2.00w1 )(1.20h1 )
= 2 2 2 = = 3.60
V1 1w1h1 1w1h1
3
to accompany
COLLEGE PHYSICS
Table of Contents
Chapter 1 Introduction 1
Chapter 2 Force 24
Chapter 3 Acceleration and Newton’s Second Law of Motion 62
Chapter 4 Motion with Constant Acceleration 105
Chapter 5 Circular Motion 144
Review and Synthesis: Chapters 1–5 171
Chapter 6 Conservation of Energy 186
Chapter 7 Linear Momentum 218
Chapter 8 Torque and Angular Momentum 247
Review and Synthesis: Chapters 6–8 281
Chapter 9 Fluids 302
Chapter 10 Elasticity and Oscillations 330
Chapter 11 Waves 361
Chapter 12 Sound 387
Review and Synthesis: Chapters 9–12 408
Chapter 13 Temperature and the Ideal Gas 420
Chapter 14 Heat 447
Chapter 15 Thermodynamics 471
Review and Synthesis: Chapters 13–15 496
Chapter 16 Electric Forces and Fields 507
Chapter 17 Electric Potential 537
Chapter 18 Electric Current and Circuits 566
Review and Synthesis: Chapters 16–18 615
Chapter 19 Magnetic Forces and Fields 627
Chapter 20 Electromagnetic Induction 663
Chapter 21 Alternating Current 692
Review and Synthesis: Chapters 19–21 726
Chapter 22 Electromagnetic Waves 737
Chapter 23 Reflection and Refraction of Light 757
Chapter 24 Optical Instruments 792
Chapter 25 Interference and Diffraction 824
Review and Synthesis: Chapters 22–25 858
Chapter 26 Relativity 867
Chapter 27 Early Quantum Physics and the Photon 893
Chapter 28 Quantum Physics 921
Chapter 29 Nuclear Physics 948
Chapter 30 Particle Physics 974
Review and Synthesis: Chapters 26–30 981
, Chapter 1
INTRODUCTION
Conceptual Questions
1. Knowledge of physics is important for a full understanding of many scientific disciplines, such as chemistry,
biology, and geology. Furthermore, much of our current technology can only be understood with knowledge of
the underlying laws of physics. In the search for more efficient and environmentally safe sources of energy, for
example, physics is essential. Also, many study physics for the sense of fulfillment that comes with learning about
the world we inhabit.
2. Without precise definitions of words for scientific use, unambiguous communication of findings and ideas would
be impossible.
3. Even when simplified models do not exactly match real conditions, they can still provide insight into the features
of a physical system. Often a problem would become too complicated if one attempted to match the real
conditions exactly, and an approximation can yield a result that is close enough to the exact one to still be useful.
4. (a) 3 (b) 9
5. Scientific notation eliminates the need to write many zeros in very large or small numbers. Also, the appropriate
number of significant digits is unambiguous when written this way.
6. In scientific notation the decimal point is placed after the first (leftmost) numeral. The number of digits written
equals the number of significant figures.
7. Not all of the significant digits are precisely known. The least significant digit (rightmost) is an estimate and is
less precisely known than the others.
8. It is important to list the correct number of significant figures so that we can indicate how precisely a quantity is
known and not mislead the reader by writing digits that are not at all known to be correct.
9. The kilogram, meter, and second are three of the base units used in the SI system.
10. The SI system uses a well-defined set of internationally agreed upon standard units and makes measurements in
terms of these units and their powers of ten. The U.S. Customary system contains units that are primarily of
historical origin and are not based upon powers of ten. As a result of this international acceptance and the ease of
manipulation that comes from dealing with powers of ten, scientists around the world prefer to use the SI system.
11. Fathoms, kilometers, miles, and inches are units with dimensions of length. Grams and kilograms are units with
dimensions of mass. Years, months, and seconds are units with dimensions of time.
12. The first step toward successfully solving almost any physics problem is to thoroughly read the question and
obtain a precise understanding of the scenario. The second step is to visualize the problem, often making a quick
sketch to outline the details of the situation and the known parameters.
13. Trends in a set of data are often the most interesting aspect of the outcome of an experiment. Such trends are more
apparent when data is plotted graphically rather than listed in numerical tables.
14. The statement gives a numerical value for the speed of sound in air, but fails to indicate the units used for the
measurement. Without units, the reader cannot relate the speed to one given in familiar units such as km/s.
15. After solving a problem, it is a good idea to check that the solution is reasonable and makes intuitive sense. It may
also be useful to explore other possible methods of solution as a check on the validity of the first.
1
,Chapter 1: Introduction College Physics
Multiple-Choice Questions
1. (b) 2. (b) 3. (a) 4. (c) 5. (d) 6. (d) 7. (b) 8. (d) 9. (b) 10. (c)
Problems
1. Strategy The new fence will be 100% + 37% = 137% of the height of the old fence.
Solution Find the height of the new fence.
1.37 × 1.8 m = 2.5 m
60 s 60 min 24 h
2. Strategy There are × × = 86, 400 seconds in one day and 24 hours in one day.
1 min 1h 1d
Solution Find the ratio of the number of seconds in a day to the number of hours in a day.
86, 400 24 × 3600
= = 3600 1
24 24
3. Strategy Relate the surface area S to the radius r using S = 4π r 2 .
Solution Find the ratio of the new radius to the old.
S1 = 4π r12 and S2 = 4π r22 = 1.160 S1 = 1.160(4π r12 ).
4π r22 = 1.160(4π r12 )
r22 = 1.160r12
2
⎛ r2 ⎞
⎜⎜ ⎟⎟ = 1.160
⎝ r1 ⎠
r2
= 1.160 = 1.077
r1
The radius of the balloon increases by 7.7%.
4. Strategy Relate the surface area S to the radius r using S = 4π r 2 .
Solution Find the ratio of the new radius to the old.
S1 = 4π r12 and S2 = 4π r22 = 2.0S1 = 2.0(4π r12 ).
4π r22 = 2.0(4π r12 )
r22 = 2.0r12
2
⎛ r2 ⎞
⎜⎜ ⎟⎟ = 2.0
⎝ r1 ⎠
r2
= 2.0 = 1.4
r1
The radius of the balloon increases by a factor of 1.4.
5. Strategy To find the factor by which the metabolic rate of a 70 kg human exceeds that of a 5.0 kg cat use a ratio.
Solution Find the factor.
3/4 3/4
⎛ mh ⎞ ⎛ 70 ⎞
⎜⎜ ⎟⎟ =⎜ ⎟ = 7.2
⎝ mc ⎠ ⎝ 5.0 ⎠
2
, College Physics Chapter 1: Introduction
6. Strategy To find the factor Samantha’s height increased, divide her new height by her old height. Subtract 1 from
this value and multiply by 100 to find the percent increase.
Solution Find the factor.
1.65 m
= 1.10
1.50 m
Find the percentage.
1.10 − 1 = 0.10, so the percent increase is 10 % .
7. Strategy Recall that area has dimensions of length squared.
Solution Find the ratio of the area of the park as represented on the map to the area of the actual park.
map length 1 map area
= = 10−4 , so = (10−4 )2 = 10−8 .
actual length 10, 000 actual area
8. Strategy Let X be the original value of the index.
Solution Find the net percentage change in the index for the two days.
(first day change) × (second day change) = [ X × (1 + 0.0500)] × (1 − 0.0500) = 0.9975 X
The net percentage change is (0.9975 − 1) × 100% = −0.25%, or down 0.25% .
9. Strategy Use a proportion.
Solution Find Jupiter’s orbital period.
T 2 R3
T 2 ∝ R3 , so J = J = 5.193. Thus, TJ = 5.193/2 TE = 11.8 yr .
TE2 RE3
10. Strategy The area of the circular garden is given by A = π r 2 . Let the original and final areas be A1 = π r12 and
A2 = π r22 , respectively.
Solution Calculate the percentage increase of the area of the garden plot.
∆A π r 2 − π r12 r2 − r2 1.252 r12 − r12 1.252 − 1
× 100% = 2 × 100% = 2 1 × 100% = × 100% = × 100% = 56%
A π r12 r12 r12 1
11. Strategy The area of the poster is given by A = w. Let the original and final areas be A1 = 1w1 and
A2 = 2 w2 , respectively.
Solution Calculate the percentage reduction of the area.
A2 = 2 w2 = (0.800 1 )(0.800 w1 ) = 0.640 1w1 = 0.640 A1
A1 − A2 A − 0.640 A1
× 100% = 1 × 100% = 36.0%
A1 A1
12. Strategy The volume of the rectangular room is given by V = wh. Let the original and final volumes be
V1 = 1w1h1 and V2 = 2 w2 h2 , respectively.
Solution Find the factor by which the volume of the room increased.
V2 w h (1.50 1 )(2.00w1 )(1.20h1 )
= 2 2 2 = = 3.60
V1 1w1h1 1w1h1
3
