Physics Concept and Connections Book 2 Solution Manual

BOOK TWO

Brian Heimbecker

Igor Nowikow

Christopher T. Howes

Jacques Mantha

Brian P. Smith

Henri M. van Bemmel

Solutions Manual

Physics: Concepts and ConnectionsBook Two Solutions Manual

AuthorsBrian HeimbeckerIgor NowikowChristopher T. HowesJacques ManthaBrian P. SmithHenri M. van Bemmel

NELSONDirector of PublishingDavid Steele

PublisherKevin Martindale

Project EditorLina Mockus-O’Brien

EditorKevin Linder

First Folio Resource GroupProject ManagementRobert Templeton

CompositionTom Dart

Proofreading and Copy EditingChristine SzentgyorgiPatricia Trudell

IllustrationsGreg DuhaneyClaire Milne

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Table of Contents iii

Table of Contents

Chapter 1Section 1.3 1

1.4 11.6 11.7 31.8 31.11 41.12 51.13 61.14 61.15 7


2.2 92.3 102.4 112.5 122.6 132.7 142.8 15


3.4 183.5 183.6 203.7 213.8 223.9 22


4.3 244.4 254.5 254.6 26


5.3 285.4 295.5 295.6 305.7 31


6.2 336.3 34


7.3 367.4 367.5 377.6 387.7 387.8 387.9 397.10 397.11 40


8.5 418.6 428.7 438.8 448.9 44



10.3 4710.4 4810.5 48


11.5 4911.6 4911.8 4911.9 5011.10 51


12.3 5212.4 5212.5 5312.6 5312.8 54


13.2 5513.3 5513.4 5613.5 5613.6 5713.7 5713.8 58


14.2 5914.3 5914.4 5914.5 5914.6 6014.7 6014.8 60

I Solutions to Applying the Concepts Questions II Answers to End-of-chapter ConceptualQuestions

Chapter 1 61Chapter 2 63Chapter 3 65Chapter 4 66Chapter 5 67Chapter 6 68Chapter 7 69Chapter 8 71Chapter 9 75Chapter 10 77Chapter 11 79Chapter 12 80Chapter 13 81Chapter 14 83

III Solutions to End-of-chapterProblems

Chapter 1 87Chapter 2 95Chapter 3 107Chapter 4 120Chapter 5 126Chapter 6 134Chapter 7 140Chapter 8 151Chapter 9 160Chapter 10 165Chapter 11 170Chapter 12 178Chapter 13 183Chapter 14 191

Solut ions to Apply ing the Concepts 1

Section 1.3

1. (30 days)� �� 2.6 � 106 s (units cancel to give answer inseconds)

2. (7 furlongs)� �� 1.4 km

(units cancel to give answer in kilometres)

3. (1 quart)� �� 5.5 � 102 mL (units cancel to give answerin millilitres)

Section 1.4

1. Since the question is asking for velocity, theanswer must include a direction. Since thedirection in which the train travels isconstant,

v��avg �

v��avg �

v��avg � 14 m/s [N]2. a) Since the question is asking for average

speed, direction is not required.

vavg �

vavg �

vavg � 1.6 km/hb) Since the question is asking for average

velocity, direction is required.

v��avg �

v��avg �

v��avg �

v��avg �

v��avg � 0.40 km/h [E]

3. a) Since the question asks for the car’svelocity, direction is important. Since thedirection is constant,

v��avg �

v��avg �

v��avg � 1.1 m/s [E]b) The car’s instantaneous velocity at 5 s can

be approximated by the difference betweenthe distance travelled after 6 s and thedistance travelled after 5 s, divided by thetime during that interval:

v��avg �

v��avg �

v��avg � 0 m/s

Section 1.6

1. v22 � v1

2 � 2a�d

�d �

�d �

�d � 9.4 � 103 m2. 10 cm � 1.0 � 10�1 m

�d �

v1 � � v2

v1 � � 0.05 m/s

v1 � 1.7 � 10�2 m/s3. a) Igor: �dI � vI�t

Brian: �dB � �12

�aB�t2

If they meet, �dI � �dB � 8.0 m

vI�t � 8.0 m � �12

�aB�t2

0 � �12

�(2.8 m/s2)�t2 � (7.0 m/s)�t � 8.0 m

0 � (1.4 m/s2)�t2 � (7.0 m/s)�t � 8.0 m

2(1.0 � 10�1 m)��

3.0 s

2�d��t

(v1 � v2)�t��

2

(600 m/s)2 � (350 m/s)2

��2(12.6 m/s2)

v22 � v1

2

�2a

8.0 m [E] � 8.0 m [E]��

6.0 s � 5.0 s

�d��t

9.0 m [E] � 0 m [E]��

8.0 s

�d��t

2.0 km [E]��

5.0 h

�3.0 km [E] � 5.0 km [E]��

5.0 h

3.0 km [W] � 5.0 km [E]��

5.0 h

�d��t

8.0 km�5.0 h

�d��t

2.5 � 104 m [N]��

1.8 � 103 s

�d��t

27.5 mL�

1 oz20 oz�1 quart

1 km��0.63 mile

1 mile��8 furlong

60 s�1 min

60 min�

1 h24 h�1 day

PART 1 Solutions to Applying the Concepts

In this section, solutions have been provided only for problems requiring calculation.

�t �

�t �

�t �

�t � 3.2 s or �t � 1.8 sWe will take the lower value: �t � 1.8 s.

b) �dB � �12

�(2.8 m/s2)(1.8 s)2

�dB � 4.4 m4. 8.0 cm � 8.0 � 10�2 m

v22 � v1

2 � 2a�d

a �

a �

a � �7.7 � 105 m/s2

5. �ttotal � �t1 � �t2 � �t3

�ttotal � 3.0 s � 6.0 s � 10 s�ttotal � 19.0 sIn the first 3.0 s, the truck travels a distanceof:

�d1 � �12

�(v1 � v2)�t1

�d1 � �12

�(0 m/s � 8.0 m/s)(3.0 s)

�d1 � 12 mSince the truck travels at a constant speedover the second interval, �d2 � v2�t2

�d2 � (8.0 m/s)(6.0 s)�d2 � 48 mFor the final interval,

�d3 � v1�t3 � �12

�a�t32

�d3 � (8.0 m/s)(10 s) � �12

�(2.5 m/s2)(10 s)2

�d3 � 2.1 � 102 m�dtotal � �d1 � �d2 � �d3

�dtotal � 12 m � 48 m � 2.1 � 102 m�dtotal � 2.7 � 102 m

vavg �

vavg �

vavg � 14 m/s

6. 100 km/h � 27.8 m/s

�d � v1�t � �12

�a�t2

500 m � (27.8 m/s)�t � �12

�(30 m/s2)�t2

0 � (15 m/s2)�t2 � (27.8 m/s)�t � 500 m

�t �

�t � 4.9 s

7. a) �d � v1�t � �12

�a�t2

80 m � (17 m/s)�t � �12

�(9.8 m/s2)�t2

0 � (4.9 m/s2)�t2 � (17 m/s)�t � 80 m

�t �

�t � 2.7 sb) v2

2 � v12 � 2a�d

v2 � �v12 � 2�a�d�

v2 � �(17 m�/s)2 �� 2(9.8� m/s2)�(80 m�)�v2 � 43 m/s

8. a) a �

�t � (eq.1)

�d � � ��t (eq. 2)

Substituting equation 1 into equation 2,

�d � � �� 2a�d � v2

2 � v12 � v2v1 � v1v2

v22 � v1

2 � 2a�d

b) a �

v1 � v2 � a�t (eq. 1)

�d � � ��t (eq. 2)

Substituting equation 1 into equation 2,

�d � � ��t

�d � v2�t � �12

�a�t2

v2 � v2 � a�t��

2

v2 � v1�2

v2 � v1��t

v2 � v1�a

v2 � v1�2

v2 � v1�2

v2 � v1�a

v2 � v1��t

�17 m/s � �(17 m�/s)2 �� 4(4.9� m/s2)�(�80�m)��

2(4.9 m/s2)

�27.8 m/s � �(27.8�m/s)2�� 4(1�5 m/s2�)(�50�0 m)��

2(15 m/s2)

2.7 � 102 m��

19.0 s

�dtotal��ttotal

(0 m/s)2 � (350 m/s)2

��2(8.0 � 10�2 m)

v22 � v1

2

�2�d

7.0 m/s � 2.05 m/s��

2.8 m/s2

7.0 m/s � �(�7.0� m/s)2� � 4(1�.4 m/s�2)(8.0�m)��

2(1.4 m/s2)

�b � �b2 � 4�ac��

2a

2 Solut ions to Apply ing the Concepts

Section 1.7

1. a) v22 � v1

2 � 2a�dAssuming up is positive,

�d �

�d �

�d � 330 mb) v2 � v1 � a�t

�t �

�t �

�t � 8.16 sc) 2(8.16 s) � 16.3 s

2. a) �d � v1�t � �12

�a�t2

Assuming down is positive,

30.0 m � (4.0 m/s)�t � �12

�(9.8 m/s2)�t2

0 � (4.9 m/s2)�t2 � (4.0 m/s)�t � 30.0 m

�t �

�t �

�t �

�t � 2.1 s

b) �d � v1�t � �12

�a�t2


30.0 m � (�4.0 m/s)�t � �12

�(9.8 m/s2)�t2

0 � (4.9 m/s2)�t2 � (�4.0 m/s)�t � 30.0 m

�t �

�t �

�t �

�t � 2.9 s

3. �d � v1�t � �12

�a�t2


35 m � v1(3.5 s) � �12

�(9.8 m/s2)(3.5 s)2

v1 � �7.2 m/s or 7.2 m/s [up]

Section 1.8

1. a) at�7.0s �

at�7.0s �

at�7.0s � 2.0 m/s2

at�12s �

at�12s � 0 m/s2

at�3.0s �

at�3.0s � 12 m/s2

b) The distance travelled by Puddles fromt � 5.0 s to t � 13 s can be found byfinding the area under the curve betweenthose times. We must consider twoseparate intervals: between 5.0 s and 10 s,and between 10 s and 13 s. The area underthe graph in the first interval can beexpressed as the sum of the areas of atriangle and a rectangle:

d1 � � �t1�v1

d1 �

� (10 s � 5.0 s)(50 m/s)d1 � 275 mThe area under the graph in the secondinterval can be expressed as a rectangle:d2 � �t2�v2

d2 � (13 s � 10 s)(60 m/s � 0 m/s)d2 � 180 mdT � d1 � d2

dT � 275 m � 180 mdT � 455 m

2. a) For Super Dave, Sr.,

vavg �

�t �

�t �

�t � 5.0 s

50 m�10 m/s

�d�vavg

�d��t

(10 s � 5.0 s)(60 m/s � 50 m/s)��

2

�t1�v1�2

32.0 m/s � 8.0 m/s��

4.0 s � 2.0 s

60 m/s � 60 m/s��

13 s � 11 s

55.0 m/s � 51.0 m/s��

8.0 s � 6.0 s

vt2� vt1�

t2 � t1

4.0 m/s � 24.6 m/s��

9.8 m/s2

4.0 m/s � �(�4.0� m/s)2� � 4(4�.9 m/s�2)(�30�.0 m)��

2(4.9 m/s2)

� b � �b2 � 4�ac��

2a

� 4.0 m/s � 24.6 m/s��

9.8 m/s2

�4.0 m/s � �(4.0 m�/s)2 �� 4(4.9� m/s2)�(�30.�0 m)��

2(4.9 m/s2)

� b � �b2 � 4�ac��

2a

0 � 80.0 m/s��

� 9.8 m/s2

v2 � v1�a

0 � (80.0 m/s)2

��2(�9.8 m/s2)

v22 � v1

2

�2a

Solutions to Apply ing the Concepts 3

We can find the acceleration of SuperDave, Jr. from the slope of his v��t graph:

a �

a �

a � 3 m/s2

�d � v1�t �

But v1 � 0 m/s, so

�d �

�t � ��t � ��t � 6 s

b) Super Dave, Sr. wins the race by 1 s.c) Super Dave, Sr.:

vavg �

�t �

�t �

�t � 10 sSuper Dave, Jr.:

�d � v1�t � , where v1 � 0 m/s, so

�d �

�t � ��t � ��t � 8 sSuper Dave, Jr. wins.

3. a) For segment 1,�d1 � 2.0 m � 0.5 m�d1 � 1.5 m�t1 � 0.6 s � 0.0 s�t1 � 0.6 s

vavg1�

vavg1�

vavg1� 2.5 m/s

For segment 2,�d2 � 2.0 m � 2.0 m�d2 � 0 mvavg2

� 0 m/sFor segment 3,�d3 � 1.0 m � 2.0 m�d3 � �1.0 m�t3 � 1.8 s � 1.0 s�t3 � 0.8 s

vavg3�

vavg3�

vavg3� �1.25 m/s

For segment 4,�d4 � 2.2 m � 1.0 m�d4 � 1.2 m�t4 � 2.6 s � 1.8 s�t4 � 0.8 s

vavg4�

vavg4�

vavg4� 1.5 m/s

b) vavg �

vavg �

vavg � 0.65 m/s

Section 1.11

1. a) Forces are unbalanced asthe force provided by thekicker, Fk, will cause theball to accelerate.

Ball

Fn

Fk

Fg

2.2 m � 0.5 m��

2.2 s � 0.0 s

�dtotal��ttotal

1.2 m�0.8 s

�d4��t4

�1.0 m�

0.8 s

�d3��t3

1.5 m�0.6 s

�d1��t1

2(100 m)��

3 m/s2

2�d�

a

a�t2

�2

a�t2

�2

100 m�10 m/s

�d�vavg

�d��t

2(50 m)�3 m/s2

2�d�

a

a�t2

�2

a�t2

�2

6 m/s�

2 s

�v��t


b) The forces arebalanced. Theforce he provideson the gun, Fm,will balance theforce of thebullet.

c) The forces are not balanced, asthe penny still acceleratesdownward, but at a slower rate.

d) These forces are balanced, andthe soldier falls downward at aconstant speed.

Section 1.12

1. a) F1 � m1a1

a1 �

a1 �

a1 � 5.0 m/s2

b) F1 � 2m1a2

a2 �

a1 �

a2 �

a2 �

a2 � 2.5 m/s2

c) � m1a3

F1 � 2m1a3

a3 �

a3 �

a3 � 2.5 m/s2

2. F � maFg � Ff � ma

Ff � m( g � a)Ff � (90 kg)(9.8 m/s2 � 6.8 m/s2)Ff � 270 N

3. v22 � v1

2 � 2a�d

a �

a �

a � �2.5 � 104 m/s2

F � maF � (8.0 � 10�2 kg)(�2.5 � 104 m/s2)F � �2000 N

4. For the first kilometre,

�d � v1�t � �12

�a1�t2

�d � �12

�a1�t2

a1 �

a1 �

a1 � 4.54 m/s2

v22 � v1

2 � 2a�dv2 � �2a�d�v2 � �2(4.54� m/s2)�(1000� m)�v2 � 95.3 m/s

For the last 1.4 km, the car’s acceleration is:v2

2 � v12 � 2a2�d

a2 �

a2 �

a2 � �3.24 m/s2

Ff � ma2

Ff � (600 kg)(�3.24 m/s2)Ff � �1.94 � 103 N

(0 m/s)2 � (9.53 m/s)2

��2(1.40 � 103 m)

v22 � v1

2

�2�d

2(1000 m)��

(21.0 s)2

2�d��t2

(0 m/s)2 � (15 m/s)2

��2(4.5 � 10�3 m)

v22 � v1

2

�2�d

5.0 m/s2

�2

a1�2

F1�2

5.0 m/s2

�2

a1�2

F1�m1

F1�2m1

10 N�2.0 kg

F1�m1

Soldier

Fparachute

Fg

Penny

Fbuoyant

Fg

Gun

Fsupport

Fm FB

Fg


During the first kilometre, the forces acting onthe car are the force due to the engine and thefrictional force:Fengine � Ff � ma1

Fengine � ma1 � Ff

Fengine � (600 kg)(4.54 m/s2) �(�1.94 � 103 N)

Fengine � 4.66 � 103 N5. v2

2 � v12 � 2a�d

a �

a �

a � �1.12 � 103 m/s2

Fmitt � maFmitt � (0.25 kg)(�1.12 � 103 m/s2)Fmitt � �280 N

Section 1.13

1. a) Action: Foot striking the ball eastReaction: Ball pushing west on the foot

b) Action: Paddle pushing backward on thewaterReaction: Water pushing forward on thepaddle

c) Action: Balloon compressing and pushingair outReaction: Air pushing back the other wayon the balloon

d) Action: Earth’s gravity pulling down on theappleReaction: Apple’s gravity pulling up onEarth

e) Action: Gravitational force downward ofthe laptop on the deskReaction: Normal force upward of the deskon the laptop

3. a) FT � mTaFT � (6000 kg � 5000 kg � 4000 kg) �

(1.5 m/s2)FT � 2.25 � 104 N

b) The tension force in the rope betweenbarges 1 and 2 is equal to the forcerequired to accelerate barges 2 and 3 at arate of 1.5 m/s2.

F1-2 � (m2 � m3)aF1-2 � (5000 kg � 4000 kg)(1.5 m/s2)F1-2 � 1.35 � 104 NThe tension force in the rope betweenbarges 2 and 3 can be found two ways:i) The difference between the force

required to accelerate all the barges ata rate of 1.5 m/s2 minus the forcerequired to accelerate the first twobarges at the same rate:F2-3 � FT � F1�2

F2-3 � 2.25 � 104 N �(6000 kg � 5000 kg)(1.5 m/s2)

F2-3 � 6.0 � 103 Nii) The force required to accelerate barge 3

at a rate of 1.5 m/s2:F2-3 � m3aF2-3 � (4000 kg)(1.5 m/s2)F2-3 � 6.0 � 103 N

4. a) FT � mTaFsled � FT � (m1 � m2)a

a �

a �

a � 0.83 m/s2

b) To find the tension force in the ropejoining the two toboggans, we consider theforces acting on the second toboggan:

FT � m2aFrope � Ff � m2a

Frope � m2a � Ff

Frope � (300 kg)(0.83 m/s2) � 100 NFrope � 350 N

Section 1.14

1. a) Friction is the only force acting on thetruck, soFf � ma

a �

Ff �

Ff �

Ff � �6.7 � 103 N

(4000 kg)(0 m/s � 16.7 m/s)��

10 s

m(v2 � v1)��t

v2 � v1��t

700 N � 200 N��

600 kg

Fsled � (Ff1� Ff2

)��

m1 � m2

(0 m/s)2 � (28 m/s)2

��2(0.35 m)

v22 � v1

2

�2�d


b) Ff � k Fn

Fn � mgFf � kmg

k �

k �

k � 0.172. a) Since the toy duck is travelling at a

constant velocity, it is not being acted uponby an unbalanced force. Therefore, theforces must have equal magnitudes andopposite directions.

b) From a), we know that the applied force,Fapp, is equal in magnitude to the force dueto friction, Ff.

Fn � mgFapp � Ff

Fapp � kFn

Fapp � kmg

a �

a � k ga � (0.15)(9.8 m/s2)a � 1.5 m/s2

3. Ff � makFn � makmg � ma

a � k gv2

2 � v12 � 2a�d

�d �

�d �

�d � 42 m

Section 1.15

1. Fg �

Fg �

Fg � 5.5 � 10�67 N

2. Fg �

Fg �

Fg �

Fg � 2.1 � 1020 N

3. a) Fg1�

Fg2� � �

Fg2� (Fg1

)

b) Fg2�

Fg2� � �

Fg2� (Fg1

)

c) Fg2�

Fg2�

Fg2� Fg1

4. (FgEarth) � Fg2

� � �

�

r22 � 2rEarthr2 � rEarth

2 � 0

r2 �

r2 � 2.6 � 106 m

5. Fg �

myou gJupiter �

gJupiter �

gJupiter �

gJupiter � 24 m/s2

(6.67 � 10�11 Nm2�kg2)(1.9 � 1027 kg)��

(7.2 � 107 m)2

GmJupiter�rJupiter

2

GmyoumJupiter��rJupiter

2

GmyoumJupiter��rJupiter

2

�2rEarth � �(2rEarth�)2 � 4�(1)(��rEarth2)�

��2

1��rEarth

2 � 2rEarthr2 � r22

1�2(rEarth

2)

GmyoumEarth��(rEarth � r2)2

GmyoumEarth��rEarth

2

1�2

1�2

Gm1m2�r2

G4m1m2�(2r)2

2�9

Gm1m2�r2

2�9

G(2m1)m2��(3r)2

1�8

Gm1m2�r2

1�8

Gm1m2�r2

(6.67 � 10�11 Nm2�kg2)(0.013)(5.97 � 1024 kg)2

��(3.82 � 108 m)2

G(0.013)mEarth2

��r2

GmEarthmMoon��r2

(6.67 � 10�11 N)(9.11 � 10�31 kg)2

��(0.01 m)2

Gm1m2�r2

(0 m/s)2 � (22.2 m/s)2

��2(0.60)(�9.8 m/s2)

v22 � v1

2

�2k g

Fapp�m

�6.7 � 103 N��(4000 kg)(9.8 m/s2)

Ff�mg


Section 2.1

1. dA � [E35°N] or [N55°E]dB � [S12°E] or [E78°S]dC � [S45°W] or [W45°S]dD � [W80°N] or [N10°W]

2. a) In the N-S direction,�dy � d cos ��d��y � (50 m) cos 14° [S]�d��y � 49 m [S]In the E-W direction,�dx � d sin ��d��x � (50 m) sin 14° [E]�d��x � 12 m [E]

b) In the N-S directionvy � v sin �v��y � (200 m/s) sin 30° [S]v��y � 100 m/s [S]In the E-W direction,vx � v cos �v��x � (200 m/s) cos 30° [W]v��x � 173 m/s [W]

c) In the N-S direction,ay � a sin �a��y � (15 m/s2) sin 56° [N]a��y � 12 m/s2 [N]In the E-W direction,ax � a cos �a��x � (15 m/s2) cos 56° [E]a��x � 8.4 m/s2 [E]

3. Horizontally,vx � v cos �vx � (5.0 m/s) cos 25°vx � 4.5 m/sVertically,vy � v sin �vy � (5.0 m/s) sin 25°vy � �2.1 m/s

4. v��g � v��w � v��b

v��g � 4.0 m/s [forward] � 3.0 m/s [upward]Since vw and vb are perpendicular,

vg � �vw2 ��vb

2�vg � �(4.0 m�/s)2 �� (3.0 m�/s)2�vg � 5.0 m/s

tan � �

� � tan�1 � �� 53°

v��g � 5.0 m/s [up 53° forward]5. a) Component Method:

v��f � v��1 � v��2

For the x components,v��fx � v��1x � v��2x

v��fx � (50 m/s) cos 36° [W] �(70 m/s) cos 20° [E]

vfx � (�50 m/s) cos 36° �

(70 m/s) cos 20°v��fx � 25.3 m/s [E]For the y components,v��fy � v��1y � v��2y

v��fy � (50 m/s) sin 36° [N] �(70 m/s) sin 20° [S]

vfy � (50 m/s) sin 36° �

(70 m/s) sin 20°v��fy � 5.45 m/s [N]vf � �vfx

2 ��vfy

2�vf � �(25.3�m/s)2�� (5.4�5 m/s)�2�vf � 26 m/s

tan � �

� � tan�1 � �� 78°v��f � 26 m/s [N78°E]

Sine/Cosine Method:� � 54° � 90° � 54° � 20° � 16°vf

2 � v12 � v2

2 � 2v1v2 cos vf

2 � (50 m/s)2 � (70 m/s)2 �

2(50 m/s)(70 m/s) cos 16°vf � 26 m/s

To find direction,

�

�

� � 32°

25.9 m/s��sin 16°

50 m/s�

sin �

vf�sin

v1�sin �

25.3 m/s��5.45 m/s

vfx�vfy

4.0 m/s�3.0 m/s

vw�vb


To find �,� �180° � � � � 54°� � 78°v��f � 26 m/s [N78°E]

b) � � 37° (parallel line theorem) � 180° � 53° � 37° (supplementary

angles theorem) � 90°Sine/cosine Method:�df

2 � �d12 � �d2

2 � 2�d1�d2 cos �df

2 � (28 m)2 � (40 m)2 �

2(28 m)(40 m) cos 90°�df � 49 m

To find direction,

�

�

sin � �

� � 35°To find �,� �180° � � � � 37°� � 18°�d��f � 49 m [W18°N]

c) Component Method:F��net � F��1 � F��2 � F��3

For the x components,F��netx � F��1x � F��2x � F��3x

F��netx � 140 N [W] � (200 N) cos 30° [E] �(100 N) sin 35° [W]

Fnetx � �140 N � (200 N) cos 30° �

(100 N) sin 35°Fnetx � �24.15 NF��netx � 24.15 N [W]

For the y components,F��nety � F��1y � F��2y � F��3y

F��nety � (200 N) sin 30° [N] �(100 N) cos 35° [S]

Fnety � (200 N) sin 30° �

(100 N) cos 35°F��nety � 18.08 N [N]Fnet � �Fnetx

2 �� Fnety

2�Fnet � �(24.15� N)2 �� (18.0�8 N)2�Fnet � 30.1 N

tan � �

� � tan�1 � �� 53°

F��net � 30.1 N [N53°W]6. �vx � �v2 sin 40° � v1 sin 15°

�v��x � 25.8 m/s [W]�vy � �v2 cos 40° � (�v1 cos 15°)�v��y � 1.17 m/s [N]v � �(25.8�m/s)2�� (1.1�7 m/s)�2�v � 26 m/s

� � tan�1 � �� 87°�v�� 26 m/s [N87°W]

Section 2.2

1. a) v��og � v��mg � v��om

cos � �

cos � �

� � 76°The ship’s heading is [S76°E].

b) v2og � v2

om � v2mg

vog � �(20 km�/h)2 �� (5.0 k�m/h)2�v��og � 19 km/h [E]

c) t � �dv

�

t �

t � 5.2 h

100 km�19 km/h

5.0 km/h��20 km/h

vmg�vom

25.8 m/s��1.17 m/s

24.15 N�18.08 N

Fnetx�Fnety

d��f

d��2

d��1

γ

β

θϕ

53°37°

28 m�49 m

28 m�sin �

49 m�sin 90°

�d1�sin �

�df�sin

20°

36°β

θϕ

γ

v��2

v��1v��f


2. a) sin � �

� � sin�1� �� 9.6°

The girl’s heading is [N9.6°E].b) The girl:

vg � �(3.0 m�/s)2 �� (0.50� m/s)2�v��g � 2.96 m/s [N]

t �

t �

t � 169 sThe boy:

t �

t �

t � 167 sc) The boy travels an extra distance west of

the girl’s landing point, caused by thehorizontal component of his velocity (equalto the river’s current).d � vtd � (0.50 m/s)(167 s)d � 83 m

d) The time required for the boy to run theextra 83 m at 5.0 m/s is 17 s. The boy’stotal time is 167 s � 17 s � 184 s. Thegirl’s time was 169 s. She wins the race.

3. v��pw � v��sw � v��ps

v2pw � v2

sw � v2ps

v2pw � �(10 km�/h)2 �� (6.0 k�m/h)2�

vpw � 12 km/h

tan � �

� � 59°v��pw � 12 km/h [N59°E]

4. a) v��og � v��om � v��mg

cos � �

� � cos�1� �� 76°

Terry must throw at [S76°E].

b) vom � �(2.0 m�/s)2 �� (0.50� m/s)2�v��om � 1.9 m/s [E]

t �

t �

t � 2.6 s

Section 2.3

1. a) �dy � viy�t � ay�t2

15 m � (0 m/s)�t � (9.8 m/s2)�t2

�t2 �

t � 1.7 s

b) �dx � vix�t � ax�t2

�dx � (25 m/s)(1.7 s) � (0 m/s2)�t2

�dx � 43 m

2. a) ay �

t �

t �

t � 2.3 sb) Since the curve Blasto travels is

symmetrical (a parabola), the time he takesto reach maximum height is the same asthe time he takes to reach the ground.ttotal � 2(2.3 s)ttotal � 4.6 sSolving for horizontal distance,

�dx � vix�t � ax�t2

�dx � (35 m/s) cos 40°(4.6 s)�dx � 120 m

3. a) To find the time required for the bomb toreach the ground,

�dy � viy�t � ay�t2

200 m � (97.2 m/s) cos 25°�t

� (9.8 m/s2)�t2

200 m � (88.1 m/s)�t � (4.9 m/s2)�t2

1�2

1�2

1�2

0 m/s � (35 m/s) sin 40°��

�9.8 m/s2

v2y� v1y�a

v2y� v1y�t

1�2

1�2

30 m�9.8 m/s2

1�2

1�2

5.0 m�1.9 m/s

d�vom

0.50 m/s��2.0 m/s

vmg�vog

10 km/h��6.0 km/h

500 m�3.0 m/s

d�v

500 m��2.96 m/s

d�v

0.50 m/s��3.0 m/s

vmg�vom


0 � (88.1 m/s)�t � (4.9 m/s2)�t2 �

200 m

�t �

�t � 2.0 sTo calculate the horizontal distance,

�dx � vix�t � ax�t2

Since there is no horizontal acceleration,�dx � vix

�t�dx � (97.2 m/s) sin 25°(2.0 s)�dx � 82 m

b) The y component of the final velocity, vfy, isvfy

2 � viy

2 � 2a�dvfy

2 � [(97.2 m/s) cos 25°]2 �

2(9.8 m/s2)(200 m)vfy

� 108 m/svfx

� (97.2 m/s) sin 25°vfx

� 41.1 m/svf � �(108 m�/s)2 �� (41.1�m/s)2�vf � 115.6 m/s

tan � �

� � 21°v��f � 116 m/s inclined at 21° to the vertical

4. Since the time it takes for the ball to hit thegreen is not given, we can find two time-related equations (one for the horizontalcomponent and one for the verticalcomponent), for the golf ball’s velocity, equateboth equations, and solve for horizontalvelocity. For the vertical component,

�dy � viy�t � �

12

�ay�t2

Since the change in height is 0 m,

0 � (vigsin �)�t � �

12

�(�9.8 m/s2)�t2

(4.9 m/s2)�t � vigsin �

�t � (eq. 1)

For the horizontal component,

�dx � vix�t � �

12

�ax�t2

250 m � (vigcos �)�t

�t � (eq. 2)

Equating equations 1 and 2,

�

vig

2 �

vig

2 �

vig� 66 m/s

v��ig� 66 m/s, 17° above the horizontal

Section 2.4

1. F��p � F��1 � F��2

F��p � 200 N [N] � 300 N [W]Fp � �F1

2 ��F22�

Fp � �(200 N�)2 � (3�00 N)�2�Fp � 361 N

tan � �

tan � �

� � 56°

F��p � 361 N [N56°W]For the frictional force,Ff � kmgFf � 0.23(200 kg)(9.8 m/s2)Ff � 451 NThis is the maximum force of friction betweenthe stove and the floor. However, friction onlyacts to oppose motion, so F��f � 361 N [S56°E].F��net � F��p � F��f

F��net � 361 N [N56°W] � 361 N [S56°E]F��net � 361 N[N56°W] � 361 N [N56°W]Fnet � 0 NFnet � ma

a �

a � �20

00Nkg

�

a � 0 m/s2

Since the frictional force is stronger than theforce provided by the people’s pushing, thestove does not move.

Fnet�m

300 N�200 N

F��2

F��f

F��1

F��pθ

F2�F1

1225 m2/s2

��sin 17°cos 17°

(250 m)(4.9 m/s2)��

sin � cos �

250 m�vig

cos �vig

sin ��4.9 m/s2

250 m�vig

cos �

vigsin �

�4.9 m/s2

41.1 m/s�108 m/s

1�2

�88.1 m/s � �(88.1�m/s)2�� 4(4�.9 m/s�2)(�20�0 m)��

2(4.9 m/s2)


2. a) F��net � F��1 � F��2 � F��3

F��net � 25 N [S16°E] � 35 N [N40°E] �45 N [W]

Adding the x components,F��netx � F��1x � F��2x � F��3x

F��netx � (25 N) sin 16° [E] �(35 N) sin 40° [E] � 45 N [W]

Fnetx � (25 N) sin 16° �

(35 N) sin 40° � 45 NFnetx � �15.6 NF��netx � 15.6 N [W]Adding the y components,F��nety � F��1y � F��2y � F��3y

F��nety � (25 N) cos 16° [S] �(35 N) cos 40° [N]

Fnety � (�25 N) cos 16° �

(35 N) cos 40°F��nety � 2.78 N [N]Fnet � �Fnetx

2 �� Fnety

2�Fnet � �(15.6�N)2 �� (2.78� N)2�Fnet � 15.8 N

tan � �

tan � �

� � 80°F��net � 15.8 N [N80°W]

b) F��net � ma��

a��

a��

a�� 0.20 m/s2 [N80°W]3. F��net � ma��

F��net � (0.250 kg)(200 m/s2 [W15°S])F��net � 50.0 N [W15°S]F��net � F��1 � F��2

F��2 � F��net � F��1

F��2 � 50.0 N [W15°S] � 100 N [N25°W]F��2 � 50.0 N [W15°S] � 100 N [S25°E]

Adding the x components,F��2x � (50.0 N) cos 15° [W] �

(100 N) sin 25° [E]F��2x � 6.03 N [W]

Adding the y components,F��2y � (50.0 N) sin 15° [S] � (100 N) cos 25° [S]F��2y � 103.6 N [S]F2 � �F2x

2 �� F2y

2�F2 � �(6.03�N)2 �� (103.6� N)2�F2 � 104 N

tan � �

tan � �

� � 86.7°� � 90° � 86.7°� � 3.3°F��2 � 104 N [S3.3°W]

4. The only two forces in the x direction are Fx

and Ff.F��net � F��x � F��f

Fx � F cos 45°Fx � (250 N) cos 45°Fx � 177 NFf � kFn

F��n � F��g � F��y

Fn � mg � F sin 45°Fn � (20 kg)(�9.8 m/s2) � 177 NFn � �372 NFf � (0.40)(�372 N)Ff � �149 N

Fnet � 177 N � 149 NFnet � 27.9 NFnet � ma

a �

a �

a � 1.38 m/s2

Section 2.5

1. The only two unbalanced forces are F|| and Ff.Fnet � F|| � Ff (eq. 1)F|| � Fg sin 25° (eq. 2)Ff � Fn

Ff � Fg cos 25° (eq. 3)

27.9 N�20 kg

Fnet�m

103.6 N�6.03 N

F2y�F2x

16 N [N80°W]��

80 kg

F��net�m

15.6 N�2.78 N

Fnetx�Fnety


Substituting equations 2 and 3 into equation 1,Fnet � Fg sin 25° � Fg cos 25°Fnet � Fg(sin 25° � cos 25°)Fnet � (2.0 kg)(9.8 m/s2)(sin 25° � cos 25°)Fnet � (19.6 N)(sin 25° � cos 25°)Fnet � 6.51 NFnet � ma6.51 N � (2.0 kg)a

a � 3.26 m/s2

�d � vi�t � a�t2

4.0 m � (3.26 m/s2)�t2

�t � ��t � 1.6 s

2. Since there is no friction, the only force thatprevents the CD case from going upward isthe deceleration due to gravity, F||.Fnet � F||

Fnet � Fg sin 20°Since Fnet � ma,ma � mg sin 20°

a � g sin 20°a � 3.35 m/s2

a �

�t �

�t �

�t � 1.2 s3. To find the distance the skateboarder travels

up the ramp, we need to find the velocity ofthe skateboarder entering the second ramp atv1. Since there is no change in velocity on thehorizontal floor, v1 � v2.For the acceleration on ramp 1,Fnet � F||

ma � mg sin 30°a � g sin 30°a � 4.9 m/s2

v22 � v1

2 � 2adv2

2 � 0 m/s � 2(4.9 m/s2)(10 m)v2 � 9.9 m/s

For the deceleration on ramp 2,Fnet � F|| � Fn

ma � mg sin 25° � (0.1)mg cos 25°a � �5.02 m/s2

For �d,v3

2 � v22 � 2a�d

(0 m/s)2 � (9.9 m/s)2 � 2(�5.02 m/s2)�d�d � 9.8 m

4. Fnet � m(0.60g)Fnet also equals the sum of all forces in theramp surface direction:

F��net � F��|| � F��f � F��engine

m(0.60g) � mg sin 30° � Fn � Fengine

m(0.60g) � mg sin 30° � (0.28)mg cos 30°� Fengine

Fengine � (0.60)mg � mg sin 30° �

(0.28)mg cos 30°Fengine � mg(0.60 � sin 30° �

(0.28) cos 30°)Fengine � 3.36m N

Section 2.6

1. a) For m1,Fnet � m1a

T � m1 g � m1a (eq. 1)For m2,

Fnet � m2am2 g � T � m2a (eq. 2)

Adding equations 1 and 2,m2 g � m1 g � a(m1 � m2)

a �

a �

a�� 5.1 m/s2 [right]Substitute a into equation 2:T � m2 g � m2aT � 71 N

b) For m1,Fnet � m1a

T � m1 g sin 35° � m1 g cos 35° � m1a(eq. 1)For m2,

Fnet � m2am2 g � T � m2a (eq. 2)

(15 kg)(9.8 m/s2) � 0.20(10 kg)(9.8 m/s2)��

25 kg

m2 g � m1 g��

m1 � m2

4.0 m/s��3.35 m/s2

v2 � v1�a

v2 � v1��t

8.0 m��3.26 m/s2

1�2

1�2


Adding equations 1 and 2,m2 g � m1 g sin 35° � m1 g cos 35°� a(m1 � m2)

a �

a �

a�� 3.5 m/s2 [right]Substitute a into equation 2:T � m2 g � m2aT � (5.0 kg)(9.8 m/s2) � (5.0 kg)(3.5 m/s2)T � 32 N

c) For m1,Fnet � m1a

T � m1 g sin 40° � 1m1 g cos 40° � m1a(eq. 1)For m2,

Fnet � m2am2 g sin 60° � T � 2m2 g cos 60° � m2a(eq. 2)Adding equations 1 and 2,m2 g sin 60° � 2m2 g cos 60° �

m1 g sin 40° � 1m1 g cos 40°� a(m1 � m2)

a �

a �

a�� 1.1 m/s2 [right]Substitute a into equation 1:T � m1a � m1 g sin 40° � 1m1 g cos 40°T � (20 kg)(1.1 m/s2) � (20 kg)(9.8 m/s2)

sin 40° � (0.20)(20 kg)(9.8 m/s2)cos 40°

T � 1.8 � 102 Nd) For m1,

Fnet � m1am1 g sin 30° � T1 � m1a (eq. 1)For m2,

Fnet � m2aT1 � T2 � m2a (eq. 2)

For m3,Fnet � m3a

T2 � m3 g � m3a (eq. 3)

Adding equations 1, 2, and 3,m1 g sin 30° � m3 g � a(m1 � m2 � m3)

a �

a �

a�� 0.82 m/s2 [left]Substitute a into equation 3:T2 � m3a � m3 gT2 � (10 kg)(0.82 m/s2) � (10 kg)(9.8 m/s2)T2 � 106 NSubstitute a into equation 2:T1 � m2a � T2

T1 � 106 N � (20 kg)(�0.82 m/s2)T1 � 122 N

Section 2.7

1. ac �

ac �

ac � 21 m/s2

2. v � �dt�

v � 25� �ac �

ac �

ac �

ac � 8.9 m/s2

3. ac �

a) If v is doubled, ac increases by a factor of 4.b) If the radius is doubled, ac is halved.c) If the radius is halved, ac is doubled.

4. a) v � �2

T�r�, where

r � 3.8 � 105 kmr � 3.8 � 108 mT � 27.3 daysT � 2.36 � 106 s

v2

�r

2500�2(1.3 m)��

(60 s)2

2500�2r�

t2

v2

�r

2�r�

t

(25 m/s)2

��30 m

v2

�r

(30 kg)(9.8 m/s2)sin 30° � (10 kg)(9.8 m/s2)��

60 kg

m1 g sin 30° � m3 g��

m1 � m2 � m3

(9.8 m/s2)[(30 kg) sin 60° � 0.30(30 kg) cos 60° � (20 kg) sin 40° � 0.20(20 kg) cos 40°]��

50 kg

g(m2 sin 60° � 2m2 cos 60° � m1 sin 40° � 1m1 cos 40°)��

m1 � m2

(9.8 m/s2)[5.0 kg � (3.0 kg) sin 35° � 0.18(3.0 kg) cos 35°]��

8.0 kg

g(m2 � m1 sin 35° � m1 cos 35°)��

m1 � m2


ac �

ac �

ac �

ac � 2.7 � 10�3 m/s2

b) The Moon is accelerating toward Earth.c) The centripetal acceleration is caused by

the gravitational attraction between Earthand the Moon.

5. r � 60 mmr � 0.06 mac � 1.6 m/s2

ac �

v � �acr�v � 0.31 m/s

6. Since d � 500 m, r � 250 m

v �

f � �T1

�

v � 2�rfac � g

ac �

g � 4�2rf 2

f � ��f � ��f � 0.0315 rotations/sf � (0.0315 rotations/s) �

� �� f � 2724 rotations/day

Section 2.8

1. a) v � �dt�

v �

v � 3.5 m/s

b) Fc � mac

Fc � (10 kg)

Fc � 24 Nc) Friction holds the child to the merry-go-

round and causes the child to undergocircular motion.

2. Tension acts upward and the gravitationalforce (mg) acts downward. Fc � Fnet andcauses Tarzan to accelerate toward the pointof rotation (at this instant, the acceleration isstraight upward).

Fc � mac

T � mg �

T � m� � g�T � (60 kg)� � 9.8 m/s2T � 9.7 � 102 N

3. Both tension and gravity act downward.Fc � mac

T � mg �

When T � 0,

mg �

v � �gr�v � �(9.8 m�/s2)(1.�2 m)�v � 3.4 m/s

4. a)

b) Fc � mg tan 20°

� mg tan 20°

v � �rg tan� 20°�v � �(100 m�)(9.8 m�/s2) ta�n 20°�v � 19 m/s

c) The horizontal component of the normalforce provides the centre-seeking force.

mv2

�r

N�� N�� cos 20°

N�� sin 20°

20°��mg

mv2

�r

mv2

�r

(4 m/s)2

�2.5 m

v2

�r

mv2

�r

v2

�r

20(2�r)�

180 s

24 h�1 day

60 min�

1 h60 s�1 min

9.8 m/s2

��4�2(250 m)

g�4�2r

v2

�r

2�r�

T

v2

�r

4�2(3.8 � 108 m)��

(2.36 � 106 s)2

4�2r�

T 2

v2

�r


d) If the velocity were greater (and the radiusremained the same), the car would slide upthe bank unless there was a frictional forceto provide an extra centre-seeking force.The normal force would not be sufficientto hold the car along its path.

e) Friction also provides a centre-seekingforce.

5. G � 6.67 � 10�11 N·m2/kg2, mE � 5.98 � 1024

kgFc � mMac

�

GmE � v2r

GmE � , where v �

T � ��T � ��T � 1.97 � 106 sT � 22.8 days

6. G � 6.67 � 10�11 N·m2/kg2, mE � 5.98 � 1024 kg, rE � 6.37 � 106 m

Fc � mHac

�

GmE � v2r

v � ��r � height of orbit � rE

r � 6.00 � 105 m � 6.37 � 106 mr � 6.97 � 106 m

v � ��v � ��v � 7.57 � 103 m/s

7. G � 6.67 � 10�11 N·m2/kg2

mM � (0.013)mE

mM � 7.77 � 1022 kgrM � 1.74 � 106 m

Fc � mApolloac

�

GmM � v2r

GmM � , where v �

r � height of orbit � rM

r � 1.9 � 105 m � 1.74 � 106 mr � 1.93 � 106 m

T � ��T � ��T � 7.4 � 104 s

400�2(1.93 � 106 m)3

��(6.67 � 10�11 N m2/kg2)(7.77 � 1022 kg)

400�2r3

�GmM

10(2�r)�

T400�2r3

�T2

mApollov2

�r

GmMmApollo��r 2

(6.67 � 10�11 N m2/kg2)(5.98 � 1024 kg)��

6.97 � 106 m

GmE�r

GmE�r

mHv2

�r

GmEmH�r 2

4�2(3.4 � 108 m)3

��(6.67 � 10�11 N m2/kg2)(5.98 � 1024 kg)

4�2r3

�GmE

2�r�

T4�2r3

�T 2

mMv2

�r

GmEmM�r 2


Section 3.3

1.

Horizontal:Th � T cos 60°Th � (1.0 � 104 N) cos 60°Th � 5.0 � 103 NVertical:Tv � T sin 60°Tv � (1.0 � 104 N) sin 60°Tv � 8.7 � 103 N

2.

Fnet � Tv � TA � TA

Fnet � maFnet � 0Tv � �TA � TA

Tv � �2TA

Tv � �2(�100.0 N) cos 70°Tv � 68.4 N

3. a)

b) dv � (1.5 m) sin 1.5°dv � 0.039 mdv � 3.9 cm

c) Fnet � 2Tv � Fg

Fnet � maFnet � 0Fg � �2Tv

m �

m �

m � 0.45 kg4. a)

tan � �

� � 71.1°Fnet � mg � 2FBv

Fnet � maFnet � 0

0 � mg � 2FB sin �

FB �

FB �

FB � 20.7 Nb) Fh � FB cos �

Fh � (20.7 N) cos 71.1°Fh � 6.71 N

c) Fv � FB sin �Fv � (20.7 N) sin 71.1°F��v �19.6 N [down] (not including theweight of the beams)

6.

F|| � mg sin �Ff � mg cos �Fnet � T � Ff � F||

Fnet � ma

θ

Ff

Fn

F||

F

T

boat

�(4.0 kg)(�9.8 N/kg)��

2 sin 71.1°

�mg�2 sin �

1.90 m�0.650 m

pail

θFB FB

mg

+

Fg

θ

1.90 m

0.65 m

=

�2(85 N) sin 1.5°��

�9.8 N/kg

�2T sin ��

g

bag

Fg mg

+

5°5°

=

TT = 85 N = 85 N

Tv

Ta Ta

+

70°70°= 100.0 N= 100.0 N

60°

Tv

Th

Th = 1.0 × 104 N


Fnet � 0T � �F|| � Ff

T � �mg sin � � mg cos �T � �mg(sin � � cos �)T � �(400.0 kg)(9.8 N/kg)

(sin 30° � (0.25) cos 30°)T � 1.11 � 103 N

Section 3.4

1. a)

b) � � rF sin �� rmg sin �� (1.50 m)(45.0 kg)(9.8 N/kg) sin 40°� � 425 N·m

2. a) � � 2.0 � 103 N·mr � 1.5 m� � 90°F � ?� � rF sin �

F �

F �

F � 1.3 � 103 N3.

a) Vw � 10.0 L

�w � 1000 kg/m3

Vw � (10.0 L)� ��

Vw � 0.0100 m3

mw � �w·Vw

mw � (1000 kg/m3)(0.0100 m3)mw � 10.0 kgFg � mgFg � (10.0 kg)(9.8 N/kg)Fg � 98.0 N

b) Position B provides the greatest torquebecause the weight is directed at 90° to thewheel’s rotation.

c) �A � rF sin ��A � (2.5 m)(10.0 kg)(9.8 N/kg) sin 45°�A � 1.7 � 102 N·m�B � rF sin ��B � (2.5 m)(10.0 kg)(9.8 N/kg) sin 90°�B � 2.4 � 102 N·m�C � �A

�C � 1.7 � 102 N·md) A larger-radius wheel or more and larger

compartments would increase the torque.

Section 3.5

1.

� � 90°r1 � ?m1 � 45.0 kg

m2 � 20.0 kg� �m2 � 5.0 kg

r2 �

r2 � 0.375 mm3 � 20.0 kg � m2

m3 � 15.0 kg

0.75 m�

2

0.75�3.0

20.0 kg P

0.75 m3.0 m

1 m3

��1.00 � 106 cm3

1000 cm3

��1 L

10.0 L

2.5 m

B

C

A

2.0 � 103 N·m��(1.5 m) sin 90°

��r sin �

1.50 m

50°

45.0 kg

Fg mg=


r3 �

r3 � 1.12 m0 � �1 � �2 � �3

0 � r1F1 sin �1 � r2F2 sin �2 � r3F3 sin �3

r1 �

r1 �

r1 � 0.332 m2. a)

�t-t � rF sin �

�t-t �

�t-t � 147 N·mThis torque applies to both sides of theteeter-totter, so the torques balance eachother.

b)

�H � �L � 0�L � �H

rL �

rL �

rL � 2.63 mc)

cos � �

� � 75.5°At the horizontal position:�H � (1.75 m)(45 kg)(9.8 N/kg)�H � 7.7 � 102 N·m

At maximum height:�H � (1.75 m)(45 kg)(9.8 N/kg) sin 75.5°�H � 7.5 � 102 N·m% � � 100

% � 2.6%3.

a) ��1 � ��2 � 0r1F1 sin � � rcmFg sin �

F1 �

F1 �

F��1 � 24.5 Nb) Frv � Fv2 � 0

Frv � �Fv2

Frv � �(5.00 kg)(�9.8 N/kg)F��rv � 49 N [up]

Frh � Fh1 � 0Frh � �Fh1

Frh � �24.5 NF��rh � 24.5 N [left]

The vertical reaction force is 49 N [up]and the horizontal reaction force is 24.5 N[left].

4.

��1 � ��2 � ��3 � 0r1F1 � r2F2 � r3FR3 � 0

r1 �

r1 � 0.375 m

0.75 m�

2

F4 F3

F2

P

0.4 m1.6 m

F1

(0.375 m)(5.00 kg)(9.8 N/kg)��

0.75 m

rcmFg sin ��

r1 sin �

40°

50°

40°

P

Fg

F1

(7.7 � 102 N·m � 7.5 � 102 N·m)��

7.7 � 102 N·m

0.5 m�2.0 m

θ2.0 m

0.50 m

(1.75 m)(45.0 kg)��

30.0 kg

rHmH g�

mL g

rh = 1.75 m rl = ?

(1.0 m)(30.0 kg)(9.8 N/kg)��

2

1.7 m

4.0 m

(1.12 m)(15.0 kg)(9.8 N/kg) � (0.375 m)(5.0 kg)(9.8 N/kg)��

(45.0 kg)(9.8 N/kg)

r3F3 � r2F2��F1

3.0 m � 0.75 m��

2


r2 �

r2 � 1.0 mr3 � 1.60 m� � 90°sin � � 1

F3 �

F3 �

F��3 � 306 N [up]F4 � F1 � F2 � F3

F4 � (120.0 kg)(9.8 N/kg) �(5.0 kg)(9.8 N/kg) � 306 N

F��RP � 919 N [up]Left saw horse: 919 N [up]Right saw horse: 306 N [up]

Section 3.6

1.

� � 45°rw � 48.0 cmrw � 0.480 mmw � 10.0 kg

rL �

rL � 24.0 cmrL � 0.240 mmL � 5.00 kg�� w � ��L � 0

� � ��w � �L

� � �(�rwFw sin 45°) �

(�rLFL sin 45°)� � (0.480 m)(10.0 kg)(9.8 N/kg)

sin 45° � (0.240 m)(5.00 kg)(9.8 N/kg) sin 45°

�� 41.6 N·m [clockwise]

2.

��1 � ��2 � 0�2 � ��1

�2 � �1

r2F2 sin �2 � r1F1 sin �1

F2 �

F2 �

F2 � 529.2 NF2 � 5.3 � 102 N

The angle makes no difference — it cancelsout.

3.

a) ��m � ��b � ��s � 0��m � �b � �s � 0

�m � �b � �s

rmFm sin �m � rbFb sin �b � rsFs sin �s

Fm �

Fm �

Fm �

Fm �

Fm � 5.57 � 103 N (tension)

(75 � 10–2 m)(9.8 N/kg) sin 75°[(0.57)85 kg�19.0 kg]��

(45 � 10–2 m) sin 11°

rg sin �(mb � ms)��rm sin �m

rbmb g sin �b � rsms g sin �s��rm sin �m

rbFb sin �b � rsFs sin �s��rm sin �m

15°11°

30 cm

45 cm

P Fg–s

Fm Fg–b

(8.0 � 10�2 m)(27 kg)(9.8 N/kg) sin ��

(4.0 � 10�2 m) sin �

r1F1 sin �1��r2 sin �2

θ2

θ1

F1 Fn

F2

P

+

8 cm

4 cm =

48.0 cm�

2

Fw

FL

+45°48 m

τ

�(0.375 m)(120.0 kg)(9.8 N/kg) � (1.0 m)(5.0 kg)(9.8 N/kg)��

1.60 m

�r1F1 � r2F2��r3

2.0 m�

2


Reaction forces:0 � F��py � F��my � F��by � F��sy

Fpy � �Fmy � Fby � Fsy

Fpy � �(�5.57 � 103 N)(sin 4°) �

(19.0 kg)(�9.8 N/kg) �

(0.57)(85 kg)(�9.8 N/kg)Fpy � 1049.6 NF��py � 1.05 � 103 N [up]

0 � F��px � F��mx � F��bx � F��sx

Fpx � �Fmx � Fbx � Fsx

Fpx � �(5.57 � 103 N)(cos 4°) � 0 � 0F��px � 5.55 � 103 N [right]

Horizontal force: 1.49 � 103 N [right]; verticalforce: 7.65 � 102 N [up]

Section 3.7

1. a) sin 43° �

htipped �

htipped � 49.8 cm

b) tan 43° �

hstraight �

hstraight � 36.5 cm2. Four-wheeled ATV:

tan �T �

�T � 31.0°

Three-wheeled ATV:

tan � �

� � 25.64°

sin � �

x � (0.55 m)(sin 25.64°)x � 0.237 m

tan �T �

tan �T � 13.3°

0.237 m�1.00 m

x�0.55 m

0.60 m�1.25 m

θ

θ

θ

θ

1.25 m

0.6 m

0.6 m

0.7 m

0.55 m

1.0 m

τ

Back View

x

x

T

T

Top View

0.60 m�1.0 m

θ

θ

1.0

m

0.6 m

34.0 cm�tan 43°

34.0 cm�

hstraight

34.0 cm�sin 43°

34.0 cm�

htipped


Section 3.8

1. k � 16.0 N/m∆x � 30.0 cm∆x � 30.0 � 10�2 ma) F � k∆x

F � (16.0 N/m)(30.0 � 10�2 m)F � 4.80 N

b) F � ma

a � �mF

�

a �

a � 1.78 � 103 m/s2

2. Fg � (67.5 kg)(9.8 N/m)Fg � 661.5 NF � kx

k �

k �

k � 66150 N/mk � 6.61 � 104 N/mFg-truck � mgFg-truck � (2.15 � 103 kg)(9.8 N/kg)Fg-truck � 2.1 � 104 NThis weight is distributed equally over foursprings.

Fs �

Fs � 5267.5 N/spring

x �

x �

x � 0.0796 mx � 8.0 � 10�2 m

3. F � kxF � (120 N/m)(30.0 � 10�2 m)F � 36 N

Section 3.9

1. d � 0.29 mmL � 0.90 m∆L � 0.22 mmEsteel � 200 � 109 N/m2

A �

E �

E �

F �

F �

F �

F � 0.807 NFor nylon,Enylon � 5 � 109 N/m2

d � 2��d � 2��d � 1.83 mmd � 1.83 � 10�3 m

2. Emarble � 50 � 109 N/m2

A � 3.0 m2

m � 3.0 � 104 kg

a) Stress � �AF

�

Stress �

Stress � 9.8 � 104 N/m2

b) E �

Strain �

Strain �

Strain � 2.0 � 10�6

9.8 � 104 N/m2

��50 � 109 N/m2

Stress�

E

Stress�Strain

(3.0 � 104 kg)(9.8 N/kg)��

3.0 m2

4(0.807 N)(0.90 m)��(5 � 109 N/m2)(0.22 � 10�3 m)

4FL��E�L

��0.29 �

210�3 m��

2

(200 � 109 N/m2)(0.22 � 10�3 m)

��4(0.90 m)

��d2

��2E�L

��4L

AE�L�

L

FL�A�L

��AF

��

��

LL��

��d2

��2

�4

5267.5 N/spring��6.6150 � 104 N/m

Fs�k

2.1 � 104 N��

4 springs

661.5 N��1.0 � 10�2 m

F�x

4.80 N��2.7 � 10�3 kg


c) L � 15 m∆L � ?

Strain � ��

LL�

�L � L(Strain)�L � (15 m)(2.0 � 10�6)�L � 3.0 � 10�5 m

3. a) Compressive strength of bone � 17 � 107 N/m2

dbone � 4.0 � 10�2 mBone cross-sectional area is:A � �r2

A � �(2.0 � 10�2 m)2

A � 1.26 � 10�3 m2

Fb �

Fb �

Breakage occurs if � Strength

�

� Strength

m �

m �

m � 4.4 � 104 kg

2(17 � 107 N/m2)(1.26 � 10�3 m2)��

9.8 N/kg

2(Strength)A��

g

��m2

g��

�A

��m2

g��

�A

Fb�A

Fb�A

mg�2

Fg�2

Fg Fg

200 kg

—2

—2


Section 4.2

1. p�� mv��

p�� (8 kg)(16 m/s [W20°N])p�� 128 kg·m/s [W20°N]p�� 1.3 � 102 kg·m/s [W20°N]

2. p�� 9.0 � 104 kg·m/s [E]

v�� (72 km/h [E])� �� v�� 20 m/s [E]

m �

m �

m � 4.5 � 103 kg3. a) p�� mv��

p�� (0.5 kg)(32 m/s [S])p�� 16 kg·m/s [S]Using a scale factor of 1 mm � 1 kg·m/s,

b) p�� mv��

p�� (0.5 kg)(45 m/s [N])p�� 22.5 kg·m/s [N]

c) �p�� p��2 � p��1

�p�� 22.5 kg·m/s [N] � 16 kg·m/s [S]�p�� 22.5 kg·m/s [N] � 16 kg·m/s [N]�p�� 38.5 kg·m/s [N]

Section 4.3

1. a) J� � F��tJ� � (3257 N [forward])(1.3 s)J� � 4234.1 N·s [forward]J� � 4.2 � 103 N·s [forward]

b) J� � F��tJ� � ma��t

J� � m� ��t

J� � m(v��2 � v��1)J� � (0.030 kg)(200 m/s � 0 m/s)J� � 6.0 N·s [out of gun]

c) J� � F��tJ� � ma��tJ� � (0.500 kg)(9.8 N/kg [down])(3.0 s)J� � 14.7 N·s [down]J� � 15 N·s [down]

2. �p�� p��2 � p��1

�p�� mv��2 � mv��1

�p�� m(v��2 � v��1)�p�� (54 kg)(20 m/s [up] � 25 m/s [down])�p�� (54 kg)(20 m/s [up] � 25 m/s [up])�p�� (54 kg)(45 m/s [up])�p�� 2.4 � 103 N·s [up]

3. a) F �

F �

F � 1.3 � 104 Nb) v1 � 0

v2 � 120 km/hv2 � 33.3 m/s

a �

a �

a � 166.7 m/s2

�d � v1�t � �12

�a�t2

�d � (0 m/s)(0.2 s) � �12

�(166.7 m/s2)(0.2 s)2

�d � 3.3 m

33.3 m/s � 0 m/s��

0.2 s

v2 � v1��t

2.5 � 103 N·s��

0.2 s

J��t

v��2 � v��1��t

p = 38.5 kg·m/s [N]

p = 22.5 kg·m/s [N]

p = 16 kg·m/s [S]

9.0 � 104 kg·m/s��

20 m/s

p�v

1 h�3600 s

1000 m�

1 km


4. a) J � �12

�bh

J� � �12

�(5 s)(25 N [S])

J� � 62.5 N·s [S]b) J � Area under triangle � rectangle

J� � �12

�(500 � 250 N [W])(3 s) �

(250 N [W])(6 s)J� � 1875 N·s [W]

c) J � Area above � area below (counting thesquares: approximately)

J � (13 squares above) � (4 squaresbelow)

J � 9 squaresMultiplying 9 by the length and width ofeach square,J� � 9(0.05 s)(100 N [E])J� � 45 N·s [E]

Section 4.4

1. m1 � 1.2 kg, v1o � �6.4 m/s, v1f � �1.2 m/s,m2 � 3.6 kg, v2o � 0, v2f � ?

po � pf

m1v1o � m2v2o � m1v1f � m2v2f

(1.2 kg)(6.4 m/s) � (1.2 kg)(�1.2 m/s) �

(3.6 kg)v2f

v��2f � 2.5 m/s [forward]2. m1 � 30 g � 0.03 kg, v1o � 0, v1f � 750 m/s,

m2 � 1.9 kg, v2o � 0, v2f � ?po � pf


� v2f

v2f �

v��2f � 11.8 m/s [back]3. m1 � 400 g � 0.400 kg,

v��1o � 3.0 m/s [forward], v��1f � 1.0 m/s [forward],m2 � 0.400 kg, v��2o � 0, v��2f � ?

p��o � p��f

m1v��1o � m2v��2o � m1v��1f � m2v��2f

� v��2f

v��2f �

v��2f � 2.0 m/s [forward]4. m1 � m, m2 � 80m, m(1�2) � 81m, v(1�2)o � ?,

v1f � �1.5 � 106 m/s, v2f � 4.5 � 103 m/s,po � �7.9 � 10�17 kg·m/s

po � pf

po � m1v1f � m2v2f

�7.9 � 10�17 kg·m/s � m(�1.5 � 106 m/s) �

80m(4.5 � 103 m/s)�7.9 � 10�17 kg·m/s � m[�1.5 � 106 m/s �

80(4.5 � 103 m/s)]

m �

m � 6.9 � 10�23 kg5. m1 � 5m, v1o � v, v(1�2)f � ?, m2 � 4m, v2o � 0

po � pf

m1v1 � m2v2 � (m1 � m2)v(1�2)f

(5m)(v) � (4m)(0) � (5m � 4m)v(1�2)f

5mv � 9mv(1�2)f

v(1�2)f � �59

�v

Section 4.5

1. m1 � m2 � 2.0 kg, v��1o � 5.0 m/s [W], v��2o � 0,v��1f � 3.0 m/s [N35°W], v��2f � ?p��1o � (2.0 kg)(5.0 m/s [W])p��1o � 10 kg·m/s [W]p��1f � (2.0 kg)(3.0 m/s [N35°W])p��1f � 6.0 kg·m/s [N35°W]

p��o � p��f

p��1o � p��2o � p��1f � p��2f, where p��2o � 0p��1o � p��1f � p��2f

Using the cosine law,p2f

2 � (10 kg·m/s)2 � (6.0 kg·m/s)2 �

2(10 kg·m/s)(6.0 kg·m/s) cos 55°p2f � 8.2 kg·m/s

p � mv

v2f �

v2f � 4.1 m/s

8.2 kg·m/s��

2 kg

35°

θ

θ

p1f = 6.0 kg·m/s

p1o = 10 kg·m/s

p2f

� 7.9 � 10�17 kg·m/s�� 1.5 � 106 m/s � 80(4.5 � 103 m/s)

(0.400 kg)(3.0 m/s [forward]) � (0.400 kg)(0) � (0.400 kg)(1.0 m/s [forward])��

0.400 kg

m1v��1o � m2v��2o � m1v��1f��m2

(0.03 kg)(0) � (1.9 kg)(0) � (0.03 kg)(750 m/s)��

1.9 kg

m1v1o � m2v2o � m1v1f��m2


Using the sine law to find direction,

�

� � 37°v��2f � 4.1 m/s [W37°S]

2. m1 � 85 kg, v��1o � 15 m/s [N], p��1o � 1275 kg·m/s [N], m2 � 70 kg, v��2o � 5 m/s [E], p��2o � 350 kg·m/s [E]

p��o � p��f

p��1o � p��2o � p��f

Using Pythagoras’ theorem to solve for pf,pf

2 � (1275 kg·m/s)2 � (350 kg·m/s)2

pf � 1322 kg·m/s

tan � �

� � 15.4°p��f � mfv��f

v��f �

v��f � 8.5 m/s [N15°E]3. m1 � 0.10 kg, v��1f � 10 m/s [N],

p��1f � 1.0 kg·m/s [N], m2 � 0.20 kg, v��2f � 5.0 m/s [S10°E], p��2f � 1.0 kg·m/s [S10°E], m3 � 0.20 kg, v��3f � ?p��To � 0p��To � p��Tf

0 � p��1f � p��2f � p��3f

Using the cosine law,

p3f2 � (1.0 kg·m/s)2 � (1.0 kg·m/s)2 �

2(1.0 kg·m/s)(1.0 kg·m/s)(cos 10°)p3f � 0.1743 kg·m/s

v3f �

v3f � 0.87 m/sUsing the sine law to find direction,

�

� � 85°v��3f � 0.87 m/s [S85°W] or 0.87 m/s [W5°S]

4. m1 � 0.5 kg, v��1o � 2.0 m/s [R], p��1o � 1.0 kg·m/s [R], m2 � 0.30 kg, v��2o � 0,p��2o � 0, v��1f � 1.5 m/s [R30°U], p��1f � 0.75 kg·m/s [R30°U], v��2f � ?, p��2f � ?

p��To � p��Tf

p��1o � p��2o � p��1f � p��2f, where p��2o � 0p��1o � p��1f � p��2f

Using the cosine law,

p2f2 � (1.0 kg·m/s)2 � (0.75 kg·m/s)2 �

2(1.0 kg·m/s)(0.75 kg·m/s)cos 30°p2f � 0.513 kg·m/sp � mv

v2f �

v2f � 1.7 m/sUsing the sine law to find direction,

�

� � 47°v��2f � 1.7 m/s [R47°D] or 1.7 m/s [D43°R]

Section 4.6

1. a) � 1.5 m from both objects

b) � �(60 cm)

� 17.1 cm from the larger mass

c) � �(20 km)

� 6.67 km from the larger satellite

200�600

2.0 kg��5.0 kg � 2.0 kg

3.0 m�

2

sin 30°��0.513 kg·m/s

sin ��0.75 kg·m/s

0.513 kg·m/s��

0.30 kg

θ

p1f = 0.75 kg·m/s

p1o = 1.0 kg·m/s

30°

p2f

sin 10°��0.1743 kg·m/s


0.17 kg·m/s��

0.2 kg

θ

p3f

p1f = 1.0 kg·m/s

p2f = 1.0 kg·m/s

10°

1322 kg·m/s [N15°E]��

85 kg � 70 kg

350 kg·m/s��1275 kg·m/s

θ

p2o = 350 kg·m/s

p1o = 1275 kg·m/s pf

sin 55°��8.2 kg·m/s



2. a) p1o � (2.0 kg)� �p��1o � 0.22 kg·m/s [S20°E]

p2o � (1.0 kg)� �p��2o � 0.17 kg·m/s [S10°W]

p1f � (2.0 kg)� �p��1f � 0.26 kg·m/s [S5°W]

p2f � (1.0 kg)� �p��2f � 0.15 kg·m/s [S30°E]

pcm � (3.0 kg)� �p��cm � 0.39 kg·m/s [S8°E]

b) i)

ii)

c) The total momentum before and aftercollision is the same as the momentum ofthe centre of mass. The total momentumvectors have the same length and directionas the momentum of the centre of mass.

5°

30°

p1f

p2f

pTf

10°

70°

p1o

p2o

pTo

0.013 m�

0.1 s

0.015 m�

0.1 s

0.013 m�

0.1 s

0.017 m�

0.1 s

0.011 m�

0.1 s


Section 5.2

1. a) W � F�dW � (40 N)(0.15 m) W � 6.0 J

b) W � F�dW � mg�dW � (50 kg)(9.8 N/kg)(1.95 m)W � 9.6 � 102 J

c) W � F�d cos �W � (120 N)(4 m)(cos 25°)W � 4.4 � 102 J

2. 45 km/h � 12.5 m/sTo find �d,v2

2 � v12 � 2a�d

�d �

�d �

�d � 31.25 mW � F�dW � (5000 N)(31.25 m)W � 1.6 � 105 J

3. W � F�d cos �W � (78 N)(10 m)(cos 55°)W � 4.5 � 102 J

4. a �

a �

a � �2.2 m/s2

F � maF � (52 000 kg)(�2.2 m/s2)F � �114 400 N

�d �

�d �

�d � 97.5 mW � F�dW � (�114 400 N)(97.5 m)W � �1.1 � 107 J

5. a) W � F�dW � (175 N)(55 m)W � 9625 J

b) The triangular areas above and below theaxis are identical and cancel out, therefore,W � (0.040 m)(20 N)W � 0.80 J

6. F � maF � (3 kg)(9.8 N/kg)F � 29.4 N

�d �

�d �

�d � 16 m

Section 5.3

1. a) Ek � �12

�mv2

Ek � �12

�(20 000 kg)(7500 m/s)2

Ek � 5.6 � 1011 Jb) 20 km/h � 5.6 m/s

Ek � �12

�mv2

Ek � �12

�(1.0 kg)(5.6 m/s)2

Ek � 15.4 J

c) Ek � �12

�mv2

Ek � �12

�(0.030 kg)(400 m/s)2

Ek � 2.4 � 103 J

2. Ek � �12

�mv2

3900 J � �12

�(245 kg)v2

v � ��v � 5.6 m/s

3. Ek � �12

�mv2

m �

m �

m � 6.5 kg4. p � �2mEk�

p � �2(9.11� � 10��31 kg)(�6000 e�V)(1.6� � 10��19 J/eV�)�p � 4.2 � 10�23 N·s

2(729 J)��(15 m/s)2

2Ek�v2

(3900 J)(2)��

245 kg

480 J�29.4 N

W�F

[(14 m/s)2 � (25 m/s)2]��

2(�2.2 m/s2)

(v22 � v1

2)��

2a

(14 m/s � 25 m/s)��

5.0 s

(v2 � v1)��t

(12.5 m/s)2 � 0��

2(2.5 m/s2)

(v22 � v1

2)��

2a


5. �Ek � Ek2 � Ek1

�Ek � �12

�(60 kg)(5.0 m/s)2 � �12

�(60 kg)(14 m/s)2

�Ek � �5.1 � 103 J

6. a) Ek � �12

�mv2

Ek � �12

�(0.350 kg)(25.0 m/s)2

Ek � 1.1 � 102 J

b) a �

a �

a � �1.3 � 104 m/s2

F � maF � (0.350 kg)(�1.3 � 104 m/s2)F � �4557 NW � F�dW � (�4557 N)(0.024 m)W � �1.1 � 102 J

c) Favg � maFavg � �4557 NFavg � �4.6 � 103 N

Section 5.4

1. a) Eg � mghEg � (3.5 kg)(9.8 N/kg)(1.2 m)Eg � 4.1 � 101 J

b) Eg � mghEg � (2000 kg)(9.8 N/kg)(0)Eg � 0 J

c) Eg � mghEg � (2000 kg)(9.8 N/kg)(1.9 m)Eg � 3.7 � 104 J

2. a) v22 � v1

2 � 2a�d (or use the conservationof energy)

v22 � (0) � 2(9.8 m/s2)(27 m)

v2 � 23 m/sb) Efinal � Einitial

Ekf � Ego � Eko

�12

�(65 kg)vf2 � (65 kg)(9.8 N/kg)(27 m) �

�12

�(65 kg)(3.0 m/s)2

vf � 23 m/s

3. a) Using the law of conservation of energy,Etotal � 5460 J

�12

�mv2 � mgh � 5460 J

�12

�(3.0 kg)v2 � (3.0 kg)(9.8 N/kg)(5.0 m) � 5460 J

v � 60 m/sb) Eg � mgh

5460 J� (3.0 kg)(9.8 N/kg)hh � 185.7 m from the groundh � 180.7 m from the pad

c) v2 � v1 � a�tv2 � (60 m/s) � (9.8 N/kg)(2.0 s)v2 � 40.4 m/s

Ek � �12

�mv2

Ek � �12

�(3.0 kg)(40.4 m/s)2

Ek � 2.4 � 103 JEp � Etotal � Ek

Ep � 5460 J � 2448.24 JEp � 3.0 � 103 J

4. F � kx

k �

k �

k �

k � 1.2 � 106 N/mFor only one spring:

k �

k � 3.0 � 105 N/m

Section 5.5

1. a) k �

k �

k � 200 N/mk � 2.0 � 102 N/m

20 N�0.1 m

rise�run

1 225 000 N/m��

4

(5000 kg)(9.8 N/kg)��

0.04 m

mg�

x

F�x

0 � (25.0 m/s)2

��2(0.024 m)

(v22 � v1

2)��

2�d


b) Maximum elastic potential energy occursat x � 0.1 m.

Ep � �12

�kx2

Ep � �12

�(200 N/m)(0.1 m)2

Ep � 1.0 Jc) �Ee � Ee2 � Ee1

�E � �12

�(200 N/m)(0.04 m)2 �

�12

�(200 N/m)(0.03 m)2

�Ee � 7.0 � 10�2 J2. Fg � Fe

mg � kx(0.500 kg)(9.8 N/kg) � k(0.04 m)

k � 122.5 N/m3. a) W � �E

W � E2 � E1 where E1 � 0W � E2

W � �12

�kx2

W � �12

�(55 N/m)(�0.04 m)2

W � 4.4 � 10�2 Jb) W � �E

W � E2 � E1 where E1 � 0W � E2

W � �12

�kx2

W � �12

�(85 N/m)(0.08 m)2

W � 2.7 � 10�1 J4. Ee � Ek

�12

�kx2 � �12

�mv2

(200 N/m)(0.08 m)2 � (0.02 kg)v2

v � 8.0 m/s5. Ee � Ek

�12

�kx2 � �12

�mv2

(5 � 106 N/m)x2 � (2000 kg)(4.5 m/s)2

x � 9 cm

6. The loss in elastic potential energy is equal tothe gain in kinetic energy.��Ee � �Ek

Let the subscript 1 represent the initialcompressed spring and subscript 2 representthe moment after the spring has been releasedwhen the cart has a velocity of 0.42 m/s.

�(Ee2 � Ee1) � Ek2 � Ek1

kx12 � kx2

2 � mv22 � 0

x2 � ��x2 � ��

x2 � 0.056 mx2 � 5.6 cm

Section 5.6

1. The energy required to heat the water isEw � (4.2 � 103 J/°C/L)(65°C � 10°C)(2.3 L)Ew � 5.31 � 105 JThe energy expended by the stove is

P �

Es � P�tEs � (1000 W)(600 s)Es � 6.0 � 105 JThe energy lost to the environment is�E � Es � Ew

�E � 6.9 � 104 J2. a) Ep � mgh

Ep � (83.0 kg)(9.8 N/m)(13.0 m)Ep � 1.057 � 104 J

P �

P �

P � 590 Wb) Ep � 1.057 � 104 J

Ep � 10 600 J3. Once the radiation of the Sun reaches Earth,

it has spread out into a sphere surroundingthe Sun. This sphere has a surface area of:SA � 4r2

SA � 4(1.49 � 1011 m)2

SA � 2.79 � 1023 m2

1.057 � 104 J��

18.0 s

E��t

E��t

(65 N/m)(0.08 m)2 � (1.2 kg)(0.42 m/s)2

��65 N/m

kx12 � mv2

2

��k

1�2

1�2

1�2


The ratio of this area to the area of Earthexposed to the radiation will be equal to theratio of the power radiated by the Sun to thepower absorbed by Earth.

�

�

�

x �

x � 2 � 1017 WTherefore, Earth intercepts 2 � 1017 J ofenergy from the Sun each second.

4. The total time is 3(20 min)(60 s/min) � 3600 sThe time the player spends on ice is (3600 s)(0.25) � 900 s

P �

E � P�tE � (215 W)(900 s)E � 1.935 � 105 JWhile sitting on the bench, the player expends100 W of power.He spends 3600 s � 900 s � 2700 s on thebench.E � (100 W)(2700 s)E � 2.7 � 105 JET � (1.935 � 105 J) � (2.7 � 105 J)ET � 4.6 � 105 J

Section 5.7

3. a) m1 � 3000 kgv��1o � 20 m/s [W]v��1f � 10 m/s [W]m2 � 1000 kgv2o � 0v2f � ?

pTo � pTf


(3000 kg)(20 m/s) � 0 � (3000 kg)(10 m/s) �

(1000 kg)v2f

v2f � 30 m/s

b) Since Eko � Ekf, the collision is elastic(EkTotal � 6 � 105 J).

c) W � �Ek�truck

W � �12

�(3000 kg)(10 m/s)2 �

�12

�(3000 kg)(20 m/s)2

W � �4.5 � 105 J4. mp � 0.5 kg

mg � 75 kg�dp � 0.03 mvpo � 33.0 m/svgo � 0vgf � 0.30 m/sa) pgo � mv

pgo � (75 kg)(0)pgo � 0Ekgo � 0ppo � mvppo � (0.5 kg)(33.0 m/s)ppo � 16.5 kg·m/s

Ekpo � �12

�(0.5 kg)(33.0 m/s)2

Ekpo � 272.25 Jb) po � pf

ppo � pgo � ppf � pgf

mpvpo � 0 � mpvpf � mgvgf

(0.500 kg)(33.0 m/s) � (0.500 kg)vpf �

(75 kg)(0.30 m/s)vpf � �12 m/s

c) Ek�p � �12

�mpvpf2

Ek�p � �12

�(0.500 kg)(12 m/s)2

Ek�p � 36 J

Ek�g � �12

�mgvgf2

Ek�g � �12

�(75 kg)(0.30 m/s)2

Ek�g � 3.4 Jd) The collision is inelastic due to the loss of

kinetic energy.

E��t

(3.9 � 1026 W)(1.48 � 1014 m2)��

2.79 � 1023 m2

3.9 � 1026 W��

x2.79 � 1023 m2

��(6.87 � 106 m)2

3.9 � 1026 W��

x2.79 � 1023 m2

��(�

dE

2arth�)2

Sun’s radiation��absorbed radiation

SASun�AEarth


5. m1 � 10 gm2 � 50 gv1o � 5 m/sv2o � 0

v1f � v1o

v1f � (5 m/s)

v1f � �3.3 m/s

v2f � v1o

v2f � (5 m/s)

v2f � 1.7 m/s6. m1 � 0.2 kg

m2 � 0.3 kgv1o � 0.32 m/sv2o � �0.52 m/sChanging the frame of reference,v1o � 0.84 m/sv2o � 0 m/s

v1f � (0.84 m/s)

v1f � �0.168 m/s

v2f � (0.84 m/s)

v2f � 0.672 m/sReturning to the original frame of reference,v1f � �0.168 m/s � 0.52 m/sv1f � �0.69 m/sv2f � 0.672 m/s � 0.52 m/sv2f � 0.15 m/s

8. a) Estored � �12

�bh

Estored � �12

�(0.06 m � 0.02 m)(50 N)

Estored � 1.0 J

b) Elost � 1.0 J � �12

�(0.005 m)(30 N) �

(0.005 m)(20 N) �

�12

�(0.035 m)(20 N)

Elost � 1.0 J � 0.075 J � 0.1 J � 0.35 JElost � 0.475 J

9. a) Counting the squares below the top curve,there are about 16.5 squares, each with anarea of (0.01 m)(166.7 N) � 1.6667 J. Theamount of energy going into the shockabsorber is (16.5)(1.6667 J) � 27.5 J.

b) There are roughly 6 squares below thelower curve. The energy returned to theshock absorber is (6)(1.6667 J) � 10 J

c) % energy lost � � 100

% energy lost � 64%

27.5 J � 10 J��

27.5 J

2(0.2 kg)��0.2 kg � 0.3 kg

0.2 kg � 0.3 kg��0.2 kg � 0.3 kg

2(10 g)��10 g � 50 g

2m1�m1 � m2

10 g � 50 g��10 g � 50 g

m1 � m2�m1 � m2


Section 6.1

1. mE � 5.98 � 1024 kg, mS � 1.99 � 1030 kg, r � 1.50 � 1011 m

a) Ek �

Ek �

Ek � 2.65 � 1033 J

b) Ep �

Ep �

Ep � �5.29 � 1033 Jc) ET � Ek � Ep

ET � 2.65 � 1033 J � (�5.29 � 1033 J)ET � �2.65 � 1033 J

2. ag �

ag �

ag � 7.32 m/s2

3. v1000 km � 6.0 km/s � 6.0 � 103 m/s, h � 1000 km � 1 � 106 m

a) vesc � ��vesc � ��vesc � 10 397 m/sSince the rocket has only achieved6000 m/s, it will not escape Earth.

b) Ek 1000 km � �12

�mv2

Ek 1000 km��12

�m(6000 m/s)2

Ek 1000 km�1.8 � 107m JSince all kinetic energy is converted togravitational potential energy at maximumheight,

Ek � �Ep

Ek � E2 � E1

1.8 � 107m J � � � �

r2 �

r2 ��(6.67 � 10�11 N·m2/kg2)(5.98 � 1024 kg)

1.8 � 107 J�

r2 � 1.1 � 107 mhmax � r2 � rE

hmax � 1.1 � 107 m � 6.38 � 106 mhmax � 4.7 � 106 m

Section 6.2

1. a) MSun � 1.99 � 1030 kg, T � 76.1 a � 2.4 � 109 sT2 � ka3

a � � �13

�

a � � �13

�

a � 2.7 � 1012 mb) 0.97

c) v �

v �

v � 7031 m/s2. raltitude � 10 000 km � 1 � 107 m,

rJupiter � 7.15 � 107 m, mJupiter � 1.9 � 1027 kg

vesc � ��vesc � ��vesc � 56 000 m/s

3. mMoon � 7.36 � 1022 kg, mEarth � 5.98 � 1024 kg, r � 3.82 � 108 m

a) vesc � ��vesc � ��vesc � 1445 m/s

2(6.67 � 10�11 N·m2/kg2)(5.98 � 1024 kg)��

3.82 � 108 m

2GM�

r

2(6.67 � 10�11 N·m2/kg2)(1.9 � 1027 kg)��

7.15 � 107 m � 1 � 107 m

2GM�

r

2(2.69 � 1012 m)��

2.4 � 109 s

d�t

(2.4 � 109 s)2

��42

��(6.67 � 10�11 N·m2/kg2)(1.99 � 1030 kg)

(2.4 � 109 s)2

��

�G4

M

2

�

(6.67 � 10�11 N·m2/kg2)(5.98 � 1024 kg)��

6.38 � 106 m � 1 � 106 m

�GM��1.8 � 107 J � �

GrM1�

�GMm�

r1

�GMm�

r2

2(6.67 � 10�11 N·m2/kg2)(5.98 � 1024 kg)��

(6.38 � 106 m � 1 � 106 m)

2GM�

r

(6.67 � 10�11 N·m2/kg2)(5.98 � 1024 kg)��

(6.38 � 106 m � 1 � 106 m)2

GM�

r2

�(6.67 � 10�11 N·m2/kg2)(1.99 � 1030 kg)(5.98 � 1024 kg)��

(1.50 � 1011 m)

� GMm�

r

(6.67 � 10�11 N·m2/kg2)(1.99 � 1030 kg)(5.98 � 1024 kg)��

2(1.50 � 1011 m)

GMm�

2r


To find the current speed of the Moon,

mv2 �

v � ��v � ��v � 1022 m/s

To find the additional speed required forescape,vadd esc � 1445 m/s � 1022 m/svadd esc � 423 m/s

b) �Ek � mvesc2 � mv2

�Ek � (7.36 � 1022 kg)[(1445 m/s)2 �

(1022 m/s)2]�Ek � 3.84 � 1028 J

c) This value is comparable to a 900-MWnuclear power plant (e.g., Darlington)running for 2.35 � 1011 years!

4. Geostationary Earth satellites orbit constantlyabove the same point on Earth because theirperiod is the same as that of Earth.

5. M � 5.98 � 1024 kg, r � 6.378 � 106 m, v � 25 m/sTo find the semimajor axis,

ET � Ep � Ek

� � mv2

� � v2

� �

�

a �

a �

a � 3.19 � 106 m

To find the period,

T2 � ka3, where k �

T � ��T � 1792 s

Section 6.3

1. a) At the equilibrium point, the bob’s kineticenergy accounts for all the energy in thesystem. This total energy is the same as themaximum elastic potential energy.Ek equil�ET

Ek equil�Epmax

Ek equil��12

�kx2

Ek equil��12

�(33 N/m)(0.23 m)2

Ek equil�0.87 Jb) 0

c) Ek � �12

�mv2

v � ��v � ��v � 1.9 m/s

2. a) To find the period of an object in simpleharmonic motion,

T � 2��T � 2��T � 0.76 s

b) At 0.16 m, the elastic potential energy ofthe bob is

Ep 0.16m� �12

�kx2

Ep 0.16m� �12

�(33 N/m)(0.16 m)2

Ep 0.16m� 0.42 JET � Ek � Ep

Ek � ET � Ep

Ek � 0.87 J � 0.42 JEk � 0.45 J

0.485 kg�33 N/m

m�k

2(0.87 J)��0.485 kg

2Ek�m

42(3.19 � 106 m)3

��(6.67 � 10�11 N·m2/kg2)(5.98 � 1024 kg)

42

�GM

(6.67 � 10�11 N·m2/kg2)(5.98 � 1024 kg)(6.378 � 106 m)��2(6.67 � 10�11 N·m2/kg2)(5.98 � 1024 kg) � (25 m/s)2(6.378 � 106 m)

GMr��2GM � v2r

2GM � v2r��

GMr1�a

v2

�GM

2�r

1�a

�2GM�

r�GM�

a

1�2

�GMm�

r�GMm�

2a

1�2

1�2

1�2

(6.67 � 10�11 N·m2/kg2)(5.98 � 1024 kg)��

3.82 � 108 m

GM�

r

GMm�

2r1�2


Ek � �12

�mv2

v � ��v � ��v � 1.36 m/s

c) Ek � 0.45 J, from part b3.

Dis

plac

emen

t (m

)

Position vs. Time

Time (s)

–0.6–0.4–0.2

0

0.20.4

0.20 0.4 0.6 0.8 1.0 1.2 1.4

2(0.45 J)��0.485 kg

2Ek�m


Section 7.2

1. a) � 0.17 rad

b) � 1.0 rad

c) � 1.6 rad

d) � 3.07 rad

e) � 4.47 rad

2. a) ( rad)(57.3°/rad) � 180°

b) � rad�(57.3°/rad) � 45°

c) (3.75 rad)(57.3°/rad) � 675°d) (11.15 rad)(57.3°/rad) � 639°e) (40 rad)(57.3°/rad) � 2.3 � 103°

3. a) Earth rotates 2 radians every 24 h.

6.0 h � � 1.57 rad

b) Earth moves 2 rad every 365 days.

265 d � � 4.56 rad

c) The second hand moves 2 rad every 60 s.

25 s � � 2.62 rad

d) A runner moves 2 rad for every lap.

25.6 laps � � 161 rad

Section 7.3

2. a) ac �

v � �acr�v � �(9.8 m�/s2)(12�00 m)�v � 108 m/sv � 1.1 � 102 m/s

b) �

�

� 0.090 rad/sThe angular acceleration is zero becausethe angular velocity is constant.

3. a) � � �

� 0.12566 rad/s � 0.13 rad/s

b) r � 1500 mac � r2

ac � (1500 m)(0.12566 rad/s)2

ac � 24 m/s2

c) The angular acceleration is zero becausethe angular velocity is constant.

d) ac-space-station � 24 m/s2

ac-Earth � 9.8 m/s2

� 2.4

Section 7.4

1. a) � (3.35 rev/s)(2 rad/rev) � 21.0 rad/s

�t � �2 min � � � (50 sec)

�t � 170 s

�

�� t�� (21.0 rad/s)(170 s)�� 3.58 � 103 rad

b) � �

� �

� � 44 rad/s2

2. a) �t �

�t �

�t � 8.3 s

b) �� t

�� (8.3 s)

�� 7.3 radc) There are 2 radians in one cycle.

number of cycles �

number of cycles � 1.16number of cycles � 1.2

7.3 rad��2 rad/cycle

(1.75 rad/s � 0)��

2

(1 � 2)��2

(0 � 1.75 rad/s)��

�0.21 rad/s2

��

(22.0 rad/s � 0)��

0.5 s

��t

��t

60 s�1 min

24 m/s2

�9.8 m/s2

1 min�60 s

2 rad�1 rev

1.2 rev�1 min

108 m/s�1200 m

v�r

v2

�r

2 rad�1 lap

2 rad�

60 s

2 rad�365 d

2 rad�

24 h

�4

256°��57.3°/rad

176°��57.3°/rad

90°��57.3°/rad

60°��57.3°/rad

10°��57.3°/rad


d) � 0.58 cycles

�� (0.58 cycles)(2 rad/cycle)�� 3.6 rad

�� 2�t � �12

��t2

3.6 rad � 0 � �12

�(�0.21 rad/s2)�t2

�t � 5.9 s

3. a) �t �

�t � 2

�t � 6.026 s�t � 6.03 s

b) � �

� �

� � �0.266 rad/s2

Section 7.5

2. a) � � I�� (0.045 kg·m2)(�1.90 rad/s2)� � �0.086 N·m

b) For 78 rpm:1 � 0

2 � � �

2 � 8.2 rad/s2

2 � 12 � 2��

��

��

�� 17.69 rad�� 18 rad

number of turns �

number of turns � �2.8For 45 rpm:1 � 0

2 � � �

2 � 4.7 rad/s

22 � 1

2 � 2��

��

��

�� 5.813 rad�� 5.8 rad

number of turns �

number of turns � �0.93

For 33 �13

� rpm:

1 � 0

2 � � � � �

2 � 3.5 rad/s2

2 � 12 � 2��

��

��

�� 3.223 rad�� 3.2 rad

number of turns �

number of turns � �0.51

3. I �

I �

I � 0.693 kg·m2

4. a) I � �12

�mr2 (moment of inertia for a disk)

I � �12

�(5.55 kg)(1.22 m)2

I � 4.13 kg·m2

b) � � rF� � (1.22 m)(15.1 N)� � 18.4 N·m

c) � � ��

I�

� �

� � 4.46 rad/s2

18.4 N·m��4.13 kg·m2

8.45 N·m��12.2 rad/s2

��

�3.223 rad��2 rad/turn

(3.5 rad/s)2 � 0��2(�1.90 rad/s2)

(22 � 1

2)��

2�

1 min�60 s

2 rad�1 rev

�1

300� rev

�1 min

� 5.813 rad��2 rad/turn

(4.7 rad/s)2 � 0��2(�1.90 rad/s2)

(22 � 1

2)��

2�

1 min�60 s

2 rad�1 rev

45 rev�1 min

�17.69 rad��2 rad/turn

(8.2 rad/s)2 � 0��2(�1.90 rad/s2)

(22 � 1

2)��

2�

1 min�60 s

2 rad�1 rev

78 rev�1 min

14.5 rad/s � 16.1 rad/s��

6.026 s

��t

92.2 rad��(16.1 rad/s � 14.5 rad/s)

��

��(1 �

22)��

1.16 cycles��

2


Section 7.6

1. a) � � rF� � (0.20 m)(23.1 N)� � 4.62 N·m� � 4.6 N·mWR � ��

WR � (4.62 N·m)(2 rad)WR � 29 J

b) WR � ��

WR � (4.62 N·m)(1.5 rad)WR � 6.9 J

c) � � 95°� � 1.66 radWR � ��

WR � (4.62 N·m)(1.66 rad)WR � 7.7 J

2. a) � � 45°

� � �

4� rad

WR � ��

WR � rF�

WR � (0.556 m)(12.2 N)��

4� rad�

WR � 5.3 Jb) The work done does not change.

Section 7.7

1. I � �25

�mr2

I � �25

�(0.0350 kg)(0.035 m)2

I � 1.7 � 10�5 kg·m2

Erot � �12

�I2

Erot � �12

�(1.7 � 10�5 kg·m2)(165 rad/s)2

Erot � 0.23 J2. a) � (5.3 rev/s)(2 rad/rev)

� 33.3 rad/s

Erot � 4��12

�I2Erot � 4��

12

�(0.900 kg·m2)(33.3 rad/s)2Erot � 2.0 � 103 J

b) v � r

v � (0.320 m)(33.3 rad/s)v � 10.7 m/s

Ek � �12

�mv2

Ek � �12

�(1000 kg)(10.7 m/s)2

Ek � 5.7 � 104 J

Section 7.8

1. a) � �vr

�

�

� 78 rad/s

Erot � 4��12

�I2Erot � 2(0.900 kg·m2)(78 rad/s)2

Erot � 1.1 � 104 J

b) Ek � �12

�mv2

Ek � �12

�(1300 kg)(25 m/s)2

Ek � 4.1 � 105 Jc) ET � Ek � Erot

ET � (4.1 � 105 J) � (1.1 � 104 J)ET � 4.2 � 105 J

2. v1 � 01 � 0h1 � 12.0 mm � 2.2 kgr � 0.056 mI � mr2 (moment of inertia for a hollowcylinder)a) ET � mgh1

ET � (2.2 kg)(9.8 m/s2)(12.0 m)ET � 2.6 � 102 J

b) To find the gravitational potential energyhalfway down:Eg � mgh2

Eg � mg� �Eg � (2.2 kg)(9.8 m/s2)� �Eg � 1.29 � 102 J

12.0 m�

2

h1�2

25 m/s�0.320 m


To find the velocity halfway down:ET1 � ET2

mgh1 � �12

�mv22 � �

12

�I2 � mgh2

mgh1 � �12

�mv22 � �

12

�mr2� �2

� mgh2

mgh1 � mv22 � mgh2

mv22 � mgh1 � mgh2

v22 � gh1 � g� �

2v22 � 2gh1 � gh1

2v22 � gh1

v2 � ��v2 � ��v2 � 7.67 m/s

c) � �vr

�

�

� 1.9 � 102 rad/s

Section 7.9

1. � � � �

� 1.99 � 10�7 rad/s

I � �25

�mr2 (moment of inertia for a sphere)

L � I

L � �25

�mr2

L � �25

�(5.98 � 1024 kg)(6.38 � 106 m)2

(1.99 � 10�7 rad/s)L � 1.94 � 1031 kg·m2/s

2. � �

� 25.7 rad/s

r �

r � 0.9 m

I � �25

�mr2 (moment of inertia for a sphere)

L � I

L � �25

�mr2

L � �25

�(85 kg)(0.9 m)2(25.7 rad/s)

L � 7.1 � 102 kg·m2/s3. At perihelion,

v � 5472.3 m/sr � 4.4630 � 1012 mm � 1.027 � 1026 kg

� �vr

�

�

� 1.2261 � 10�9 rad/sL � I

L � �25

�mr2

L � �25

�(1.027 � 1026 kg)(4.4630 � 1012 m)2

(1.2261 � 10�9 rad/s)L � 1.003 � 1042 kg·m2/sAt aphelion:v � 5383.3 m/sr � 4.5368 � 1012 mm � 1.027 � 1026 kg

� �vr

�

�

� 1.1866 � 10�9 rad/sL � I

L � �25

�mr2

L � �25

�(1.027 � 1026 kg)(4.5368 � 1012 m)2

(1.1866 � 10�9 rad/s)L � 1.003 � 1042 kg·m2/s

Section 7.10

2. 1 �

1 �

1 � 2.94 � 10�6 rad/s

2 rad��2.14 � 106 s

��t

5383.3 m/s��4.5368 � 1012 m

5472.3 m/s��4.4630 � 1012 m

1.8 m�

2

2 rad�1 cycle

4.5 cycles��

1.1 s

2 rad�1 rev

1 h�3600 s

1 d�24 h

1 rev�365 d

10.8 m/s��0.056 m

(9.8 m/s2)(12.0 m)��

2

gh1�2

h1�2

v2�r


I11 � I22

mr121 � mr2

22

r121 � r2

22

2 �

2 �

2 � 4.69 � 104 rad/s

T2 �

T2 �

T2 � 1.34 � 10�4 s3. ra � 1.52 � 1011 m

rp � 1.47 � 1011 mvp � 30 272 m/s

Iaa � Ipp

mra2� � � mrp

2� �rava � rpvp

va �

va �

va � 2.93 � 104 m/sva � 29.3 km/s

Section 7.11

3. R � 0.040 mr � 0.0070 m

a �

a �g

� � 1�a �

9.8 m/s2

� � 1�a � 0.64 m/s2

�12

�(0.040 m)2

��(0.0070 m)2

�12

�mR2

�mr2

g��

��m

Ir2� � 1�

(1.47 � 1011 m)(30 272 m/s)��

1.52 � 1011 m

rpvp�ra

vp�rp

va�ra

2 rad��4.69 � 104 rad/s

2 rad�

2

(6.95 � 108 m)2(2.94 � 10�6 rad/s)��

(5500 m)2

r121�

r22

2�5

2�5


Section 8.4

1. q1 � 3.7 � 10�6 C, q2 � �3.7 � 10�6 C, d � 5.0 � 10�2 m, k � 9.0 � 109 N·m2/C2

F �

F �

F � �49 NF � 49 N (attraction)

2. F � 2(�49 N)F � �98 N

r � ��r � ��r � 3.5 � 10�2 m

3. a)

b)

c) How close do the dust balls get and what isthe charge on the tethered dust ball?m � 2.0 � 10�10 kg , l � 0.42 m, dwall-1 � 0.35 m, q � 3.0 � 10�6 C, � � 21°dwall-2 � 0.35 m � 0.42 m(sin 21°)dwall-2 � 0.35 m � 0.15 mdwall-2 � 0.20 m

From the force vector diagram we see that,

tan � �

Fe � mg tan �

� mg tan �

q1 �

q1 �

q1 � 1.1 � 10�15 CThe dust balls are 0.20 m apart, and thecharge on the tethered dust ball is1.1 � 10�15 C.

Section 8.5

1. a)

b)

c)

+

(0.20 m)2(2.0 � 10�10 kg)(9.8 N/kg)(tan 21°)��

(9.0 � 109 N·m2/C2)(3.0 � 10�6 C)

r2mg tan ��

kq2

kq1q2�r2

Fe�mg

mgT

Fe

Fe

T

mg

(9.0 � 109 N·m2/C2)(3.7 � 10�6 C)(�3.7 � 10�6 C)��

�98 N

kq1q2�F

(9.0 � 109 N·m2/C2)(3.7 � 10�6 C)(�3.7 � 10�6 C)��

(5.0 � 10�2 m)2

kq1q2�d2


Section 8.6

1. a) q � �1.0 � 10�6 C, �� 1.7 � 106 N/C [right]Let right be the positive direction.F��e � q ��

Fe � (�1.0 � 10�6 C)(1.7 � 106 N/C)Fe � �1.7 NF��e � 1.7 N [left]

b) q � 1.0 � 10�6 C, �� 2(1.7 � 106 N/C) [right]If right is still the positive direction,F��e � q ��

Fe � (1.0 � 10�6 C)[2(1.7 � 106 N/C)]Fe � 3.4 NF��e � 3.4 N [right]

2.

q � �1.0 � 10�6 C, �� 1.7 � 106 N/C [right]Fe � mg tan �F��e � 1.7 N [left]

3. a)

The field lines radiate outward, away fromthe charge.

b) k � 9.0 � 109 N·m2/C2, q � 3.0 � 10�6 CAt 2 cm away from the charge:

�

�

� 6.8 � 107 N/C

At 4 cm away:

�

�

� 1.7 � 107 N/CAt 6 cm away:

�

�

� 7.5 � 106 N/Cc) Doubling the distance,

1 �

1 � �14

�

Tripling the distance,

2 �

2 � �19

�

1 decreases to �14

� and 2 decreases to �19

� of

the original strength.

d) � . The field strength varies as the

inverse square of the distance away fromthe charge.

e) q1 � 1.0 � 10�6 C, q2 � 3.0 � 10�6 C, r � 8.0 � 10�2 m

�

�

� 4.22 � 106 N/CF��e � q ��

Fe � (1.0 � 10�6 C)(4.22 � 106 N/C)F��e � 4.22 N [right]

4. a) q1 � q2 � 1.0 � 10�6 C, r � 0.20 mLet the positive direction be left.At point A:r1 � 0.05 m, r2 � 0.25 m

(9.0 � 109 N·m2/C2)(3.0 � 10�6 C)��

(8.0 � 10�2 m)2

kq1�r2

1�r2

kq�(3r)2

kq�(2r)2

(9.0 � 109 N·m2/C2)(3.0 � 10�6 C)��

(6.0 � 10�2 m)2

kq�r2

(9.0 � 109 N·m2/C2)(3.0 � 10�6 C)��

(4.0 � 10�2 m)2

kq�r2

(9.0 � 109 N·m2/C2)(3.0 � 10�6 C)��

(2.0 � 10�2 m)2

kq�r2

+

mg

Fe

T

Stationary chargecreating a field

––


��TA � ��1 � ��2

TA � �

TA � (9.0 � 109 N·m2/C2)(1.0 � 10�6 C)

� � ��TA � 3.7 � 106 N/C [left]At point B:r1 � �0.10 m, r2 � 0.10 mThe addition of these two distances as wasdone in the previous question will yield azero quantity. TB � 0 N/CAt point C:r1 � �0.15 m, r2 � 0.05 m ��TC � ��1 � ��2

TC � �

TC � (9.0 � 109 N·m2/C2)(1.0 � 10�6 C)

� � ��TC � 3.2 � 106 N/C [left]

b) At the centre point, 1 is equal inmagnitude but opposite in direction to 2,therefore there is no net field strength asthe fields cancel out.

c) For all field strengths to cancel out, the

magnitudes of the ratio of must be

equal and pointing in opposite directions.

Section 8.7

1. a) Ee �

Ee �

Ee � �6.8 � 10�1 J

b) V �

V �

V � �4.5 � 105 V

c) V �

V �

V � �9.0 � 105 V

∆V � V2 � V1

∆V � �9.0 � 105 V � (�4.5 � 105 V)∆V � �4.5 � 105 V

2. a) m1 � m2 � 5.0 � 10�9 g � 5.0 � 10�12 kg,q1 � 4.0 � 10�10 C, q2 � 1.0 � 10�10 COn particle 1:W1 � qVW1 � (4.0 � 10�10 C)(50 V)W1 � 2.0 � 10�8 JOn particle 2:W2 � qVW2 � (1.0 � 10�10 C)(50 V)W2 � 5.0 � 10�9 J

b) W � Ek

W � mv2

v � ��

� The similar masses cancel.

� �� 2

3. a) Extensive: electric force, potential energyIntensive: field strength, electric potential

b) Electric force — Charge and the fieldstrengthPotential energy — Charge and the electricpotential

c) Extensive propertiesProduct cost (per package)MassVolumeLengthForce of gravityEtc.Intensive propertiesUnit product cost (per unit weight or measure)DensityHeat capacity

v1�v2

2.0 � 10�8 J��5.0 � 10�9 J

v1�v2

W1�W2

v1�v2

��2mW

1

1��

��2mW

2

2��v1�v2

2W�

m

1�2

(9.0 � 109 N·m2/C2)(�5.0 � 10�6 C)��

5.0 � 10�2 m

kq�r

�6.8 � 10�1 J��1.5 � 10�6 C

Ee�q

(9.0 � 109 N·m2/C2)(�5.0 � 10�6 C)(1.5 � 10�6 C)��

10 � 10�2 m

kq1q2�r

q�r2

1��(0.15 m)2

1��(0.05 m)2

kq1�r2

1

kq2�r2

2

1��(0.25 m)2

1��(0.05 m)2

kq2�r2

2

kq1�r2

1


Indices of refractionGravitational field strengthEtc.

Section 8.8

1. qA � �2e, qB � �79e, Ek � 7.7 MeV

� (7.7 � 106 eV)(1.602 � 10�19 J)�Ee � �Ek

�Ee �

r �

r �

r � 2.96 � 10�14 mr � 3.0 � 10�14 m

3. q � �1.5 � 10�5 C

�12

�mv2 � q(V2 � V1)

v � ��v � ��v�� 6.0 m/s [left]

4. a) V � 1.5 � 103 V, m � 6.68 � 10�27 kg, q � 2e � 3.204 � 10�19 C

Ek � Ee

�12

�mv2 � Vq

v � ��v � ��v � 3.8 � 105 m/s

b) �12

�mv2 � �12

�Vq

v � ��v � ��v � 2.7 � 105 m/s

5. a) V � 20 kV � 2.0 � 104 V, q � 1.602 � 10�19 C, m � 9.11 � 10�31 kgEk � Ee

Ek � VqEk � (2.0 � 104 V)(1.602 � 10�19 C)Ek � 3.2 � 10�15 J

b) Ek � �12

�mv2

v � ��v � ��v � 8.4 � 107 m/s

Section 8.9

1. W � 2.4 � 10�4 J, q � 6.5 � 10�7 C

�V �

�V �

�V � 3.7 � 102 V2. d � 7.5 � 10�3 m, V � 350 V,

�

�

� 4.7 � 104 N/C3. m � 2.166 � 10�15 kg, V � 530 V,

d � 1.2 � 10�2 mFe � Fg

� mg

q �

q �

q � 4.8 � 10�19 C

(2.166 � 10�15 kg)(9.8 N/kg)(1.2 � 10�2 m)��

530 V

mgd�

V

qV�

d

350 V��7.5 � 10�3 m

V�d

2.4 � 10�4 J��6.5 � 10�7 C

W�q

2(3.2 � 10�15 J)��9.11 � 10�31 kg

2Ek�m

(1.5 � 103 V)(3.204 � 10�19 C)��

6.68 � 10�27 kg

Vq�m

2(1.5 � 103 V)(3.204 � 10�19 C)��

6.68 � 10�27 kg

2Vq�

m

2(�1.5 � 10�5 C)(�12 V)��

(1.0 � 10�5 kg)

2q(V2 � V1)��m

(9.0 � 109 N·m2/C2)(1.602 � 10�19 C)2(2)(79)��

(7.7 � 106 eV)(1.602 � 10�19 J)

kqAqB��Ee

kqAqB�r


Section 9.5

1. L � 0.30 mI � 12 AB � 0.25 T� � 90°F � BIL sin �F � (0.25 T)(12 A)(0.30 m) sin 90°F � 0.90 N

2. L � 0.15 mF � 9.2 � 10�2 NB � 3.5 � 10�2 T� � 90°

I �

I �

I � 18 A3. a) L � 50 m

I � 100 AF � 0.25 N� � 45°

B �

B �

B � 7.1 � 10�5 Tb)

4. B � 3.0 � 10�5 TL � 0.20 mN � 200� � 4 � 10�7 T·m/A

I �

I �

I � 2.4 � 10�2 A

5. a) I � 100 AL � 50 mB � 3.0 � 10�5 T� � 45°

r �

r �

r � 0.67 mb) Referring to the diagram in question 3,

Earth’s field lies in a line that is crossingthe wire at 45° below the horizontal. Themagnetic field would form a circular ringin the clockwise direction (rising on thesouth side of the wire, descending on thenorth with a radius of 0.67 m). Therefore,the field will cancel that of Earth on thesouth side below the wire, as shown in thediagram.

2x2 � (0.67 m)2

x � 0.47 mThe fields will cancel 4.7 � 10�1 m southand 4.7 � 10�1 m below the wire.

6. a) r � 2.4 � 10�3 mI � 13.0 AL � 1 m

F �

F �

F � 1.4 � 10�2 N/m7. q � 20 C

B � 4.5 � 10�5 Tv � 400 m/s� � 90°F � qvB sin �F � (20 C)(400 m/s)(4.5 � 10�5 T) sin 90°F � 0.36 N

(4 � 10�7 T·m/A)(13.0 A)2(1 m)��

2(2.4 � 10�3 m)

�I2L�2r

N

45° x

x

0.67 m

(4 � 10�7 T·m/A)(100 A)��

2(3.0 � 10�5 T)

�I�2B

(3.0 � 10�5 T)(0.20 m)��(4 � 10�7 T·m/A)(200)

BL��N

Tower

Direction of Force

Wire(cross-section) Earth's

MagneticField45°

S N

45°

(0.25 N)��(100 A)(50 m) sin 45°

F�IL sin �

(9.2 � 10�2 N)��(3.5 � 10�2 T)(0.15 m) sin 90°

F�BL sin �


8. q � 1.602 � 10�19 Cv � 4.3 � 104 m/sB � 1.5 T� � 90°F � qvB sin �F � (1.602 � 10�19 C)(4.3 � 104 m/s)(1.5 T) sin 90°

F�� 1.0 � 10�14 N [south]


Section 10.2

1. a) T �

T �

T � 75 min

b) T �

T � 0.67 s

c) T �

T � 1.80 s

d) T �

T � 0.838 s

2. a) f �

f �

f � 60 Hz

b) f �

f � 0.75 Hz

c) f �

f � 0.009 26 Hz

d) f �

f � 1.35 Hz

3. a) i) f �

f �

f � 2.22 � 10�4 Hz

ii) f �

f � 1.49 Hz

iii) f �

f � 0.556 Hz

iv) f �

f � 1.19 Hz

b) i) T �

T �

T � 0.0167 s

ii) T �

T � 1.33 s

iii) T �

T � 108 s

iv) T �

T � 0.74 s5. a) x � (30 cm) cos �

x � (30 cm) cos 30°x � 26 cm

b) x � (30 cm) cos 180°x � �30 cm

c) x � (30 cm) cos 270°x � 0 cm (equilibrium)

d) x � (30 cm) cos 360°x � 30 cm

e) x � (30 cm) cos ��

4��

x � 21 cm

Section 10.3

4. a) v � �f

f �

f �

f � 4.7 � 1014 Hz

b) f �

f � 2.5 � 108 Hz

c) f �

f � 1.5 � 1017 Hz5. a) v � �f

� �

� �

� � 2.0 � 10�5 m

3.0 � 108 m/s��1.5 � 1013 Hz

v�f

3.0 � 108 m/s��

2 � 10�9 m

3.0 � 108 m/s��

1.2 m

3.0 � 108 m/s��640 � 10�9 m

v��

1�1.35 Hz

1��0.009 26 Hz

1�0.75 Hz

1�60 Hz

1�f

1�0.838 s

1�1.80 s

1�0.67 s

1��75 � 60 s

1�T

65�48 s

40��1.2 � 60 � 60 s

45�60 s

120�2.0 s

cycles�

t

57 s�68

60 s�33.3

6.7 s�

10

375 min�

5

t�cycles


b) � �

� � 0.15 m

c) � �

� � 1.0 � 10�14 m

Section 10.4

4. a) n �

v �

v �

v � 2.26 � 108 m/s

b) v �

v � 1.24 � 108 m/s

c) v �

v � 1.99 � 108 m/s

5. a) n �

n �

n � 1.43

b) n �

n � 2.0

c) n �

n � 1.276. a) n1 sin �1 � n2 sin �2

�2 � sin�1 � ��2 � sin�1 � ��2 � 18.5°

b) �2 � sin�1 � ��2 � 10.1°

c) �2 � sin�1 � ��2 � 16.3°

Section 10.5

5. a) n �

vo ray �

vo ray �

vo ray � 1.81 � 108 m/s

ve ray �

ve ray �

ve ray � 2.02 � 108 m/s

b) � � 100%

� 111.6%

Therefore, the speed of the e ray is 11.6%greater than the speed of the o ray.

ve ray�vo ray

2.02 � 108 m/s��1.81 � 108 m/s

ve ray�vo ray

3.0 � 108 m/s��

1.486

c�ne ray

3.0 � 108 m/s��

1.658

c�no ray

c�v

sin 25°�

1.51

sin 25°�

2.42

sin 25°�

1.33

n1 sin �1�n2

3.0 � 108 m/s��0.79(3.0 � 108 m/s)

3.0 � 108 m/s��1.5 � 108 m/s

3.0 � 108 m/s��2.1 � 108 m/s

c�v

3.0 � 108 m/s��

1.51

3.0 � 108 m/s��

2.42

3.0 � 108 m/s��

1.33

c�n

c�v

3.0 � 108 m/s��3.0 � 1022 Hz

3.0 � 108 m/s��2.0 � 109 Hz


Section 11.4

2. d � 5.6 �mx2 � 28 cmL � 1.1 mm � 2

� �

� �

� � 7.13 � 10�7 m� � 713 nm

3. � � 510 nmd � 5.6 �mL � 1.1 m

�x �

�x �

�x � 0.10 m�x � 10 cm

4. m � 3d � 5.6 �mL � 1.1 m� � 713 nm

xm �

x3 �

x3 � 0.42 mx3 � 42 cm

Section 11.5

2. ∆PD � 3ng � 1.52� � 624 nm

t �

t �

t � 1.8 � 10�6 mt � 1.8 �m

Section 11.6

2. m � 22� � 625 nm

t �

t �

t � 6.87 � 10�6 mt � 6.9 �m

3. t � 1.75 � 10�5 m� � 625 nm

2t � �m � ��

m � �

m � �

m � 55.5m � 55

Section 11.8

1. w � 5.5 � 10�6 m� � 550 nmL � 1.10 mm � 2

a) sin m �

sin 2 �

sin 2 � 0.22 � 11.5°

b) xm � L sin m

xm � (1.10 m)(0.2)xm � 0.22 mxm � 22 cm

2. a) x �

x �

x � 0.22 mx � 22 cm

2(5.50 � 10�7 m)(1.10 m)

(5.5 � 10�6 m)

2�L

w

(2)(5.50 � 10�7 m)

5.5 � 10�6 m

m�

w

12

2(1.75 � 10�5 m)(6.25 � 10�7 m)

12

2t�

12

(22)(6.25 � 10�7 m)

2

m�

2

(6.24 � 10�7 m)(3)

2(0.52)

��PD2(ng � 1)

(3)(7.13 � 10�7 m)(1.1 m)

(5.6 � 10�6 m)

m�L

d

(5.10 � 10�7 m)(1.1 m)

5.6 � 10�6 m

�Ld

(5.6 � 10�6 m)(0.28 m)

(2)(1.1 m)

dxmmL


b) sin �12

� �

sin �12

� �

sin �12

� � 0.1

� 11.5°

3. �x �

�x �

�x � 0.11 m�x � 11 cm

6. R � 1 � 10�7 radd � 2.4 m

a) � �

� �

� � 1.97 � 10�7 m� � 197 nm

b) sin = Lx

L �

L � 5000 mL � 5 km

Section 11.9

1. N � 8500w � 2.2 cm� � 530 nm

d �

d �

d � 2.59 � 10�6 m

sin m �

sin 1 �

sin 1 � 0.2051 � 12°

sin m �

sin 2 �

sin 2 � 0.4102 � 24°

sin m �

sin 3 �

sin 3 � 0.6143 � 38°

2. a) m �

m �

m � 4

b) m �

m �

m � 4.7m � 4

c) m �

m �

m � 5.7m � 5

3. m � 22 � 8.41o

� � 614 nm

a) d �

d �

d � 8.396 � 10�6 md � 8.40 �m

b) w � 1.96 cm

N �

N �

N � 2334 slits

1.96 � 10�2 m8.396 � 10�6 m

wd

(2)(6.14 � 10�7 m)

sin 8.41°

m�sin m

2.59 � 10�6 m4.50 � 10�7 m

d�

2.59 � 10�6 m5.50 � 10�7 m

d�

2.59 � 10�6 m6.50 � 10�7 m

d�

3(5.30 � 10�7 m)

2.59 � 10�6 m

m�

d

2(5.30 � 10�7 m)

2.59 � 10�6 m

m�

d

5.30 � 10�7 m2.59 � 10�6 m

m�

d

2.2 � 10�2 m

8500

wN

12

(1.0 � 10�3 m)sin(1 � 10�7 rad)

(1 � 10�7 rad)(2.4 m)

1.22

Rd1.22

(5.50 � 10�7 m)(1.10 m)

(5.5 � 10�6 m)

�Lw

5.50 � 10�7 m5.5 � 10�6 m

�w


Section 11.10

1. � � 300 000 lines/m

� � 100 000 lines/m

Therefore, 3000 lines/cm produces the bestresolution.

3. sin Red �

sin Red �

sin Red � 0.386Red � 22.7°

sin Violet �

sin Violet �

sin Violet � 0.211Violet � 12.2°

sin Green �

sin Green �

sin Green � 0.269Green � 15.6°

This can be similarly proven for the next 3orders using the appropriate m.The sequence is violet, green, red.At the fourth order, green and red maxima areno longer visible.

5. d � 2.5 � 10�10 m � 12o

m � 2

� �

� �

� � 5.198 � 10�11 m� � 52 pm

6. � �

sin �

sin �

sin � 0.208 � 168°, 192°

(5.2 � 10�11 m)(2)2(2.5 � 10�10 m)

�m2d

2d sin

m

2(2.5 � 10�10 m) sin 12°

2

2d sin

m

(1)(5.10 � 10�7 m)

1.89 � 10�6 m

m�

d

(1)(4.00 � 10�7 m)

1.89 � 10�6 m

m�

d

(1)(7.30 � 10�7 m)

1.89 � 10�6 m

m�

d

100 cm

1 m20 000 lines

20 cm

100 cm

1 m3000 lines

1 cm


Section 12.2

1. T � 12 000 Ka) The maximum wavelength can be found

using Wien’s law:

�max �

�max �

�max � 2.4 � 10�7 mThe peak wavelength of Rigel is 2.4 � 10�7 m. It is in the ultravioletspectrum.

b) It would appear violet.c) No: the living cells would be damaged by

the highly energetic UV photons.2. T � 900 K

a) The maximum wavelength can be foundusing Wien’s law:

�max �

�max �

�max � 3.2 � 10�6 mThe peak wavelength of the light is 3.2 � 10�6 m.

b) It would appear in the infrared spectrum.c) Since the peak is in infrared, more energy

is required to produce the light in thevisual spectrum.

Section 12.3

1. V � � � f0 �

eV � hf0 � W0

Choosing two pairs of values from the tableand subtracting,

(1.6 � 10�19 C)(0.95 V) � h(7.7 � 1014 Hz) � W0

(1.6 � 10�19 C)(0.7 V) � h(7.2 � 1014 Hz) � W0

(1.6 � 10�19 C)(0.25 V) � h(0.5 � 1014 Hz)h � 8 � 10�34 J·s

W0 � 4.64 � 10�19 JW0 � 2.9 eV

2. a) Increasing the work function by 1.5 wouldcause a vertical shift of the line. Hence,potential would have to be greater, but thefrequencies would not change.

b) The term he is constant and hence the

slope would not change.3. � � 230 nm � 2.3 � 10�7 m

The energy can be found as follows:

E� � � W0

E� � �

4.64 � 10�19 JE� � 5.79 � 10�19 J

Section 12.4

2. E � 85 eV, � � 214 nm � 2.14 � 10�7 ma) Momentum of the original electron can be

found using:

p �

p �

p � 4.53 � 10�26 N·sb) Momentum of the resultant electron can be

found using:

p �

p �

p � 3.1 � 10�27 N·sc) The energy imparted can be found by:

�E � E �hc�

6.626 � 10�34 J·s

2.14 � 10�7 m

h�

(85 eV)(1.6 � 10�19 C)

3.0 � 108 m/s

Ec

(8 � 10�34 J·s)(3.0 � 108 m/s)

2.3 � 10�7 m

hc�

3

2

1

07 8

Vst

op (

V)

f0 (×1014 Hz)

Vstop vs. f0

9 10 11 12 13

W0e

he

2.898 � 10�3

900 K

2.898 � 10�3

T

2.898 � 10�3

12 000 K

2.898 � 10�3

T


�E � (85 eV)(1.6 � 10�19 C) �

�E � 1.27 � 10�17 JThe energy imparted to the electron was1.27 � 10�17 J.

d) The energy imparted increased the speedof the electron. Hence, it can be foundusing:

�v � ��v � ��v � 5.27 � 106 m/sThe speed increase of the electron is5.27 � 106 m/s.

Section 12.5

1. v � 1 km/s � 1000 m/sThe wavelength can be found usingde Broglie’s equation:

� �

� �

� � 7.27 � 10�7 mHence, the wavelength of the electron is7.27 � 10�7 m.

Section 12.6

2. We shall first compute the change in energiesand the wavelength of spectral lines emitted ineach case. From that, the wavelengthseparation can be computed.The energy change when the electrontransfers from 8 to 1 is:�E8�1 � E8 � E1

�E8�1 � �

�E8�1 � 2.15 � 10�18 J

The wavelength of the spectral lines is:

�8�1 �

�8�1 �

�8�1 � 9.25 � 10�8 mSimilarly, the energy change when theelectron transfers from 7 to 2 is:�E7�2 � E7 � E2

�E7�2 � �

�E7�2 � 5 � 10�19 J

�7�2 �

�7�2 �

�7�2 � 3.98 � 10�7 mHence the wavelength separation is∆� � �7�2 � �8�1

∆� � 3.98 � 10�7 m � 9.25 � 10�8 m∆� � 3.05 � 10�7 m

3. The change in energy can be computed using:�E � Ef � Ei

�E � �

For the Lyman series, the lower boundary iswhen the electron jumps from the second tothe first orbital:

�Emin � �

�Emin � 10.2 eVThe higher boundary for the Lyman series iswhen the electron jumps from infinity to thefirst orbital:

�Emax � �

�Emax � 13.6 eVFor the Balmer series, the lower boundary iswhen the electron jumps from the third to thesecond orbital:

�Emin � �

�Emin � 1.89 eV

13.6 eV

22

�13.6 eV

32

13.6 eV

12

�13.6 eV

�2

13.6 eV

12

�13.6 eV

22

13.6 eV

ni2

�13.6 eV

nf2

(6.26 � 10�34 J·s)(3.0 � 108 m/s)

5 � 10�19 J

hc�E7�2

2.18 � 10�18 J

22

� 2.18 � 10�18 J

72

(6.626 � 10�34 J·s)(3.0 � 108 m/s)

2.15 � 10�18 J

hc�E8�1

2.18 � 10�18 J

12

�2.18 � 10�18 J

82

6.626 � 10�34 J·s(9.11 � 10�31 kg)(1000 m/s)

hmv

2(1.27 � 10�17 J)9.11 � 10�31 kg

2Em

(6.626 � 10�34 J·s)(3.0 � 108 m/s)

2.14 � 10�7 m


The higher boundary for the Balmer series iswhen the electron jumps from infinity to thesecond orbital:

�Emax � �

�Emax � 3.4 eVFor the Paschen series, the lower boundary iswhen the electron jumps from the fourth tothe third orbital:

�Emin � �

�Emin � 0.66 eVThe higher boundary for the Paschen series iswhen the electron jumps from infinity to thethird orbital:

�Emax � �

�Emax � 1.51 eVFor the Brackett series, the lower boundary iswhen the electron jumps from the fifth to thefourth orbital:

�Emin � �

�Emin � 0.31 eVThe higher boundary for the Brackett series iswhen the electron jumps from infinity to thefourth orbital:

�Emax � �

�Emax � 0.85 eVThus, the boundaries for the four series are:Lyman: 10.2 eV to 13.6 eVBalmer: 1.89 eV to 3.4 eVPaschen: 0.66 eV to 1.51 eVBrackett: 0.31 eV to 0.85 eV

Section 12.8

1. ∆v � 1 �m/s � 1 � 10�6 m/s, mp � 1.673 � 10�27 kgThe uncertainty in position can be foundusing:

�y

�y

�y 6.3 � 10�2 m

Hence, the uncertainty in position is6.3 � 10�2 m.

2. In the equation ∆E∆t ≥ h– , the units are J·s.

This coincides with the units of h in h– � ,

where 2� is a constant.6. Ek � 1.2 keV � 1.92 � 10�16 J,

mp � 1.673 � 10�27 kgFirst we shall find the velocity using:

v � ��v � ��v � 4.8 � 105 m/sThe uncertainty in position can be foundusing:

�y

�y

�y 1.32 � 10�13 mThe uncertainty in the position is 1.32 � 10�13 m.

7. The uncertainty does not affect the objectat a macroscopic level.

1.0546 � 10�34 J·s(1.673 � 10�27 kg)(4.8 � 105 m/s)

h–m�v

2(1.92 � 10�16 J)1.673 � 10�27 kg

2Ekmp

h2�

1.0546 � 10�34 J·s(1.673 � 10�27 kg)(1 � 10�6 m/s)

h–m�v

13.6 eV

42

�13.6 eV

�2

13.6 eV

42

�13.6 eV

52

13.6 eV

32

�13.6 eV

�2

13.6 eV

32

�13.6 eV

42

13.6 eV

22

�13.6 eV

�2


Section 13.1

1. For Nadia:mR�Lv0 � mRvR2 � mLvR2

(6 kg)(0 m/s) � (3 kg)(2 m/s) �

(3 kg)(�2 m/s)0 kg·m/s � 0 kg·m/s

For Jerry:mR�Lv0 � mRvR2 � mLvR2

(6 kg)(�2 m/s) � (3 kg)(�2 � 2 m/s) �

(3 kg)(2 � 2 m/s)�12 kg·m/s � �12 kg·m/s

Section 13.2

1. v � 0.5c or 1.5 � 108 m/s2. The classical addition of velocities gives:

kvp � kvu � uvp

kvp � 0.5c [R] � 0.6c [R]

kvp � 1.1cThis answer violates the second postulate ofspecial relativity.

3. Ek-gained � Ee-lost

12

mv2 � Vq

v � ��v � ��v � 5.93 � 108 m/s

This value is almost double the speed of light.

Section 13.3

1. The muon travels farther due to the timedilation from 2.2 �s to 3.1 �s that occurs atits speed of v � 0.7c. The extra path length is:∆d � d2 � d1

∆d � vt2 � vt1

∆d � v(t2 � t1)∆d � (0.7c)(3.1 �s � 2.2 �s)∆d � 189 m

2. For Phillip, at rest relative to the experiment:

t �

t0 �

t0 �

t0 � 2.0 � 10�8 sFor Barb, the stationary observer watching theexperiment travel by at v � 0.6c:

t �

t �

t � 2.5 � 10�8 s3. For Marc, the time for one beat is:

� 1.1538 s

The dilated time for the earthly observers is:

t �

t �

t � 1.2019 sThe new rate is:

� 49.9 bpm

4. The contracted distance L, measured byKatrina, is given by:L � 0.5L0

L � L0�1 ��0.5L0 � L0�1 ��0.25 � 1 �

� �0.75�

v � 0.866cv � 2.60 � 108 m/s

vc

v2

c2

v2

c2

v2

c2

60 s/min1.20 s/beat

1.1538 s

�1 � (0�.2

c28c)2

�

t0

�1 � vc2

2

�

60 s/min52 beats/min

2.0 � 10�8 s

�1 � (0�.

c62c)2

�

t0

�1 � vc2

2

�

2(3.0 m)3.0 � 108 m/s

2h

c

dv

2(1.00 � 106 V)(1.6 � 10�19 C)

(9.11 � 10�31 kg)

2Vq

m


6. L0 � 5.75 � 1012 mThe time you take:

t0 �

t0 �

(vt0)2 � L02�1 � �

�cLt0

0�2

v2 � c2 � v2

v � ��v � ��v � 2.95 � 108 m/s

Section 13.4

1. L0 � 7 ca

t � Lv

0

7 a � 3 a � 7

vca

v � 170caa

v � 0.7c2. The age or time difference for the twins is:

5 a � tS � tT

5 a � dv

S � dvT

5 a � �

5 a � �

v � 2c � 2c�1 ��v � 2c � �2c�1 ��

v2 � 4vc � 4c2 � 4c2 � 4v2

v(5v � 4c) � 0v � 0.8c

since v ≠ 0

3. L0 � 200 cav � 0.9986c

t0 �

t0 �

t0 � 10.59 a6. For Rashad:

(∆s)2 � c2(∆t)2 � (∆x)2

(∆s)2 � (3 � 108 m/s)2(1.5 s)2 � 02

(∆s)2 � 2.05 � 1017 m2

For Kareem:(�s)2 � c2(�t)2 � (�x)2

�x � �c2(�t)2� � (�s�)2��x � �(3 � 1�08 m/s�)2(2 s)�2 � (2�.025 �� 1017 m�2)��x � 3.97 � 108 m

Section 13.5

1. m0 � 5.98 � 1024 kgv � 2.96 � 104 m/s

m �

m �

m � 5.980 000 03 � 1024 kg

2. m �

At 0.9c:

m �

m � 2.294m0

At 0.99c:

m �

m � 7.089m0

At 0.999c:

m �

m � 22.366m0

Therefore, there is a much greater increase inmass when accelerating from 0.99c to 0.999c.

m0�1 � (0�.999)2�

m0�1 � (0�.99)2�

m0�1 � (0�.9)2�

m0

�1 � vc2

2

�

5.98 � 1024 kg

�1 � ((2�3

.9.0

6�

��11008

4

mm

/�/ss))2

2

�

m0

�1 � vc2

2

�

200 ca�1 � (0�.9986)�2�

0.9986c

L0�1 � vc2

2

�

v

v2

c2

v2

c2

10 ca�1 � vc2

2

�

v10 ca

v

2L0�1 � vc2

2

�

v2L0

v

(3.0 � 108 m/s)2

1� (3.0 � 108 m/s)(3600 s)

(5.75 � 1012 m)

c2

1 �

cLt0

0

v2

c2

L0�1 � vc2

2

�

v

distance measured

velocity


3. Using the low-speed mass dilation approximation,

�m �

�m �

�m � 3.3 � 10�14 kg4. Since cost, C, is proportional to energy3, E3,

� � �3

C2 � ($100 million)� �3

C2 � $100 billion5. The radius for charges moving at right angles

to a magnetic field is r � mBq

v. The ratio of the

radius of a fast to slow electron is rr

s

f � mm

s

fvv

f

s.

Assuming ms � m0 (its rest mass), and

mf � , the ratio becomes .

6. As in question 5, the ratio of radii is:

� and since vp � ve:

rp �

rp �

rp � 1833re

7. m0 � 1.67 � 10�27 kgv � 0.996cB � 5.0 � 10�5 T

r �

r �

r � 6.98 � 105 m

Section 13.62. Using the relativistic formula for velocity

addition:

�vL �

�vL �

�vL � c

3. The speed of the bullet relative to Earth is:

bvE �

bvE �

3c �

2c

�1� �bvE �

bvE �

bvE � 0.714cTherefore, the bullet will never reach thebandits because its speed is less than 0.75c.

4. Putting the limiting velocity v � c intoHubble’s law:v � Hrgives the limiting case of:

r �

r �

r � 1.76 � 1010 ca

Section 13.7

1. For momentum dilation,

p �

At v � 0.2c:

p �

p � 0.204m0cAt v � 0.5c:

p �

p � 0.577m0cAt v � 0.8c:

p �

p � 1.33m0c

m0(0.8c)

�1 � (0�.

c82c)2

�

m0(0.5c)

�1 � (0�.

c52c)2

�

m0(0.2c)

�1 � (0�.

c22c)2

�

m0v

�1 � vc2

2

�

3.0 � 108 m/s1.7 � 10�2 m/s/ca

cH

5c7

56c

�1 � 16

�

3c �

2c

c2

bvc � cvE

�1 � bvc

c�2cvE�

c � 0.999c1 �

(c)(0c.9299c)

�vN � NvL1 � �

vN

c�2NvL

(1.67 � 10�27 kg)(0.996c)(1.6 � 10�19 C)(5.0 � 10�5 T)�1 � (0�.996)2�

m0v

qB�1 � vc2

2

�

(1.67 � 10�27 kg)re9.11 � 10�31 kg

mpreme

mpvpmeve

rpre

vf

vs�1 � vc2

2

�m0

�1 � vc2

2

�

5000 MeV500 MeV

E2E1

C2C1

(60 kg)(10 m/s)2

(3.0 � 108 m/s)2

12

m0v2

c2

12


2. E � m0c2 � Ek

Case A:125 J � m0c2 � 87 J

m0 � 38c�2 JCase B:54 J � m0c2 � 15 J

m0 � 39c�2 JTherefore, B has the greater rest mass.

3. The energy used by the bulb is:E � mc2

E � Pt

m �

m �

m � 2.80 � 10�8 kgm � 2.80 � 10�5 g

4. E � mc2

E � (65 kg)(3 � 108 m/s)2

E � 5.85 � 1018 J

Section 13.8

1. E � mc2

E � (106 MeV/c2)c2

E � 106 MeV

E � 106 MeV � �

E � 1.696 � 10�11 JThe equivalent mass is:

m �

m �

m � 1.88 � 10�28 kg2. A mass, m, is equivalent to an energy:

E � mc2

E � (1.67 � 10�27 kg)(3 � 108 m/s)2

E � 1.503 � 10�10 J1 eV � 1.6 � 10�19 J

m �

m � 939.37 � 106 eV/c2

m � 939.4 MeV/c2

3. E2 � (pc)2 � (m0c)2

(mvc)2 � (mc2)2 � (m0c)2

mc2 � (m0c2 � Ek)mc2 � m0c2 � 5m0c2

mc2 � 6m0c2

(mvc)2 � (6m0c2)2 � (m0c2)2

(mvc)2 � 35m02c4

m2v2 � 35m02c2

Since m � ,

v2 � 35�1 � �c2

v2 � 35c2 � 35v2

36v2 � 35c2

v � ��c2�v � 0.986cv � 2.96 � 108 m/s

4. Given the dilated mass of the proton,m � 4 � 106m0

m �

4 � 106 �

Since v2 ≈ c2, we can use the high-speedapproximation:

�1 �� 2��1 � vc

�4 � 106 �

�1 � vc

� �

1 � � 3.13 � 10�14

c � v � 3.13 � 10�14cc � v � 9.38 � 10�6 m/s

The protons are travelling 9.38 � 10�6 m/sslower than c.

vc

1(4 � 106)�2�

1

�2��1 � vc

�

v2

c2

1

�1 � vc2

2

�

m0

�1 � vc2

2

�

3536

v2

c2

m0

�1 � vc2

2

�

1.503 � 10�10 J1.6 � 10�19 J/eV

1.696 � 10�11 J(3.0 � 108 m/s)2

Ec2

1.6 � 10�19 J

1 eV1 � 106 eV

1 MeV

(80 W)(365 � 24 � 60 � 60 s)

(3.0 � 108 m/s)2

Ptc2


Section 14.1

3. a) Binding energy is:B � [Zm(1H) � Nmn � m(2H)]c2

B � 938.78 MeV � 939.57 MeV �1876.12 MeV

B � 2.23 MeV

b) � � 1.12 MeV/nucleon

4. Average atomic mass of Cl is 0.758(35 u) � 0.242(37 u) � 35.48 u,compared to 35.453 u in the periodic table.

Section 14.2

2. Since AZ X → A�4

Z�2 Y��:a) 234

90Th b) 24494Pu c) 219

84Po d) 24092U e) 60

27Co3. Since A

Z X → AZ�1Y�e�:

a) 3216S b) 11

23Na c) 3517Cl d) 21

45Sc e) 6430Zn

4. Since AZ X → A

Z�1Y�e�:a) 19

9F b) 1022Ne c) 46

23V d) 23992U e) 64

28Ni

Section 14.3

1. The amount eaten is:

� � � � � �

� . The amount left is 1 � �

or � �8

.

2. T12

� 1.28 � 109 a, N0 � 5 mg, N � 1 mg

N � N0� �log � � � log � �

t � T12

t � (1.28 � 109 a)

t � 2.97 � 109 a

3. T235 � 7.04 � 108 a, T238 � 4.45 � 109 a,

� 0.0044, � �0

� 0.030

�

0.0044 � (0.030)�12

�t(7.04 �

1108 a �

4.45 �

1109 a)

log � � � t(1.196 � 10�9 a�1) log � �t �

t � 2.3 � 109 a

Section 14.4

1. Bismuth or 20983Bi

2. � �

1800 doses4. annual dose � dose equivalent � activity �

timeD � (1.3 � 106 eV)(1.602 � 10�19 J/eV)(1) �

(29 000 Bq/kg) � (365 � 24 � 60 � 60 s)D � 0.1905 J/kgD � 191 mSv

Section 14.5

2. In a head-on elastic collision with the target,3H at rest, the recoil velocity is:

v� � v� �v� � v� �v� � �0.5vTritium is 50% effective in slowing down thefast neutrons.

4. power � amount of energy/mole �number of moles used/12 h �(12 � 3600)�1 h/s

P � (1699 GJ/mol)� or ��

P � 7.87 GW

143200 s

600 g3 g/mol

400 g2 g/mol

1 u � 3 u1 u � 3 u

mn � mxmn � mx

360 mSv0.20 mSv

unobservable annual dosage

dosage per dental x-ray

�0.8337(�0.3010)(1.196 � 10�9 a�1)

12

0.00440.030

(235N)0�12

�T

t

235

(238N)0�

12

�T

t

238

235N238N

235N238N

235N238N

log �15mm

gg

�

log �12

�

log �NN

0�

log �

12

�

12

tT

12

NN0

tT

1

2

12

12

1256

255256

255256

1256

1128

132

116

18

14

12

2.23 MeV2 nucleons

BA


5. Since 1 neutrino is created along with1 deuterium atom, and 2 deuterium atoms areneeded to create an 4He ion, 2 neutrinos arecreated.

Section 14.6

1. Using Einstein’s energy triangle:(mvc)2 � (m0c2�Ek)2 � (m0c2)2

mvc � �(0.511� MeV�� 310�0 MeV�)2 � (0�.511 M�eV)2�

mvc � (3100.5 MeV)(1.602 � 10�13 J/MeV)

mv �

mv � 1.6557 � 10�18 N·sThe de Broglie wavelength is:

� �

� �

� � 4.0 � 10�16 m

2. f �

f �

f � 11 kHz3. a) In Einstein’s energy triangle,

(mc2)2 � (m0c2)2 � (mvc)2 [see Chapter 13]m0c2 � 938.27 MeVmc2 � m0c2 � Ek

mc2 � 938.27 MeV � 10 MeVmc2 � 948.27 MeVIn the triangle,

cos �

cos �

� 8.328°

� sin

� sin 8.328°

v � 0.1448cv � 4.35 � 107 m/s

b) r �

r �

r � 0.216 m

Section 14.7

2. a) uud � � � � �1

b) u� u� d� � � � � � �1

c) ud� � � � �1

d) udd � � � � 0

e) su� � � � � �1

3. a) proton (baryon)b) antiproton (baryon)c) pion (meson)d) neutron (baryon)e) kaon (meson)

4. udd � � � � 0

5. The mass “defect” of a �0 meson is:md � mb � m� � (8 � 4700 � 5279) MeV/c2

� �571 MeV/c2

Section 14.8

1. i) An electron and a positron annihilate eachother, releasing two gamma rays.

ii) A neutron undergoes �� decay to anantineutrino, a positron, and an electron.

iii)A planet orbits the Sun via the exchange ofa graviton.

13

13

23

23

13

13

13

23

13

23

13

23

23

13

23

23

4.35 � 107 m/s2�(32 � 106 Hz)

v2�f

vc

mvcmc2

938.27 MeV948.27 MeV

m0c2

mc2

3.0 � 108 m/s2�(4300 m)

v2�r

6.63 � 10�34 J·s1.6557 � 10�18 N·s

hmv

4.9670 � 10�10 J

3.0 � 108 m/s


PART 2 Answers to End-of-chapter Conceptual Questions

Answers to End-of-chapter Conceptual Quest ions 61

Chapter 11. It is possible for an object to be accelerating

and at rest at the same time. For example, con-sider an object that is thrown straight up inthe air. During its entire trajectory it is accel-erating downward. At its maximum height ithas a speed of zero. Therefore, at that point itis both accelerating and at rest.

2. A speedometer measures a car’s speed, not itsvelocity, since the speedometer gives no indi-cation as to the direction of the car’s motion.

3.

4. Displacement, velocity, and acceleration areall vector quantities. Therefore, a negative dis-placement, velocity, or acceleration is a nega-tive vector quantity, which indicates that thevector’s direction is opposite to the directiondesignated as positive.

5. The seconds are squared in the standard SIunit for acceleration, m/s2, because accelera-tion is the change in velocity per unit of time.Therefore, the standard SI unit for accelera-tion is (m/s)/s, which is more convenientlywritten as m/s2.

6. Assume for all cases that north is positive andsouth is negative.a) Position�time graph: The object sits

motionless south of the designated zeropoint. The object then moves northwardwith a constant velocity, crossing the zeropoint and ending up in a position north ofthe zero point.Velocity�time graph: The object movessouthward with a constant velocity. Theobject then slows down while still movingsouthward, stops, changes direction, andspeeds up northward with a constantacceleration.

b) Position�time graph: The object starts atthe zero point and speeds up while movingnorthward, then continues to move north-ward with a constant velocity.Velocity�time graph: The object starts atrest and speeds up with an increasingacceleration while moving northward. Theobject then continues to speed up with aconstant acceleration northward.

c) Position�time graph: The object startsnorth of the zero point and moves south-ward past the zero point with a constantvelocity. The object then abruptly slowsdown and continues to move southwardwith a new constant velocity.Velocity�time graph: The object slowsdown while moving northward, stops,changes direction, and speeds up south-ward with a constant acceleration. Theobject then abruptly reduces the magni-tude of its acceleration and continues tospeed up southward with a new constantacceleration.

d) Position�time graph: The object starts atthe zero point and moves northward andslows down to a stop, where it sits motion-less for a period of time. The object thenquickly speeds up southward and moves

m

Vel

ocity

in M

etre

s pe

r S

econ

d

Time in Seconds1

–1

–2–3

–4

–5–6

–7

–8

–9

–10

2 3 4 5 6 7 t

1

1 2 3 4 5 6 7 t

m

23

4

5

Pos

ition

in M

etre

s

Position-Time

Time in Seconds

southward with a constant velocity, goingpast the zero point.Velocity�time graph: The object starts atrest and speeds up while moving north-ward. The acceleration in this time periodis decreasing. The object then continues tomove northward with a constant velocity.The object then slows down while movingnorthward, stops, changes direction, andspeeds up southward with a constant accel-eration.

7. a) vavg �

�

� 3.3 m�s

b) vavg �

�

� 4.2 m�s

c) vavg �

�

� 3.7 m�sd) The answer for c) is the average speed of

the bus over the whole trip, whereas halfthe sum of its speed up the hill and itsspeed down the hill is an average of theaverage speeds up and down the hill.

8. In flying from planet A to planet B, you wouldneed to burn your spacecraft’s engines whileleaving planet A in order to escape its gravita-tional pull and then to make any necessarycourse corrections, and while arriving atplanet B in order to slow down and stop.Assuming there were no forces acting on thecraft in between, it would travel with constantvelocity once the engines were turned off.

9. A free-body diagram shows the forces actingon an object, as these are the only forces thatcan cause the body to accelerate. Since, byNewton’s third law, for every action forcethere is a reaction force, equal in magnitudeand opposite in direction, then each of the

forces acting on an object is half of anaction�reaction pair. If both the action forcesand the reaction forces were included in afree-body diagram, then all the forces wouldcancel. For example, a free-body diagram for aball being kicked must not include the reac-tion force provided by the ball on the foot, orelse the forces would cancel and the ballwould not accelerate.

10.

11. Dear Cousin,You asked me to explain Newton’s first law ofmotion to you. Newton’s first law of motionstates that an object will keep moving at aconstant speed in the same direction unless aforce makes it slow down, speed up, or changedirection. Here’s an example. Suppose you’repushing a hockey puck across the carpet.When you let go, the puck quickly stops mov-ing. This is because the carpet is not very slip-pery; we say that it has a lot of friction. Theforce of friction is making the puck slowdown. What if you slide the puck across a sur-face with less friction, like ice? The puck willtake longer to stop moving, because the forceof friction is much less than on the carpet.Now suppose you slide the puck across an airhockey table. The force of friction is so smallthat the puck will slide for a much, muchlonger time. So, you can imagine sliding apuck on a surface with no friction at all. Thepuck never stops, because there is no force toslow it down! Perhaps you’re wonderingabout a motionless object that isn’t experienc-ing a force — why isn’t it moving at a con-stant speed in the same direction? But it is!Zero is a constant speed.

Motor-cycle

Fn

Fm Ff

Fg

2000 m��(9)(60 s)

dtot�ttot

1000 m��(4)(60 s)

dtot�ttot

1000 m��(5)(60 s)

dtot�ttot

62 Answers to End-of-chapter Conceptual Quest ions

12. The gravitational force applied by the Moonon Earth does not cancel with the gravita-tional force applied by Earth on the Moonbecause these forces act on different bodies.Only forces applied on the same body can pos-sibly cancel one another.

13. When you fire a rifle, the forces applied to thebullet and the rifle make up an action�reac-tion pair. By Newton’s third law, the forceapplied to the bullet is equal and opposite toits reaction force, the force applied to the rifle.This reaction force causes the rifle and you torecoil in the opposite direction.

14. While in the air, the ball’s vertical accelerationis constant and equal to g � �9.8 m/s2. Theball travels the same distance upward asdownward, and therefore the ball’s speed isthe same when it reaches the ground as whenit leaves the ground, since its acceleration isconstant. Suppose the lengths of time it takesthe ball to travel upward and downward are t1

and t2, respectively. We can use the equations

d1 � and d2 �

for the distances travelled upward and down-ward, respectively, where v1i and v1f are theinitial and final velocities during the upward

flight, respectively, and v2i and v2f are the ini-tial and final velocities downward, respec-tively. Since d1 � �d2, we can write the following equation:

� �

On the left side, the final velocity upward, v1f,is equal to zero. On the right side, the initialvelocity downward, v2i, is equal to zero. Theequation simplifies:

� �

But v1i is equal to �v2f and is not zero, andtherefore t1 � t2.

15. The ball is undergoing uniform circularmotion, as it is travelling in a circle at a con-stant speed. Because its trajectory is curved, itcannot be undergoing uniform motion, whichrequires an object to be travelling at a con-stant speed in a straight line.

Chapter 21. Frictional forces are forces that oppose motion.

A frictional force will only try to prevent anobject from moving, it will not actually causean object to move.

2. It is not possible to swing a mass in a horizon-tal circle above your head. Since gravity isalways pulling down on the mass, an upwardcomponent of the tension force is required tobalance gravity. As the speed of rotationincreases, the angle relative to the horizontalmay approach 0° but will never reach 0°.

3. If the gravitational force downward and thenormal force upward are the only two verticalforces acting on an object, we can be certainthat they are balanced if the object is not accel-erating. If one of these forces were greaterthan the other, the object would accelerate inthe direction of the greater force.

4. The most common way to describe directionsin three dimensions is by the use of three unitvectors (and their opposites). Traditionally,the three unit vectors used are labelled as i�, j�,and k��. One of these unit vectors will representright, one will represent up, and one will

t2(v2f)�2

t1(v1i)�2

t2(v2i � v2f)��2

t1(v1i � v1f)��2

t2(v2i � v2f)��2

t1(v1i � v1f)��2

Fn

Fg

Puckon

frictionlesssurface

Fn

Ff

Fg

Puckon

air table

Fn

Ff

Fg

Puckonice

Fn

Ff

Fg

Puckon

carpet


represent coming out of the plane of the pagetoward you.

5. The bullets reach the ground in the sameamount of time. Recall that the horizontal andvertical motions of each bullet are independentof each other. Since both identical objects areaccelerating downward at the acceleration dueto gravity and they are both dropped from thesame height, it takes the same time for them toreach the ground.

6. Dear Wolfgang,You asked whether the time it takes to paddlea canoe across a river depends on the strengthof the current. When you are paddling a canoeacross a river, the variables that determinehow long it takes are the width of the riverand the forward velocity of the canoe due toyour paddling. The canoe’s forward velocityand the current velocity are perpendicular toeach other, so they don’t affect each other. Asa result, the current does not affect the lengthof time required to cross the river. The onlyeffect of the current on the motion of thecanoe is to cause it to move downstream fromwhere it would otherwise have landed.

7. The student who wants to apply the forceabove the horizontal has the better idea. Thehorizontal component of the applied force inthe direction of motion will be the sameregardless of whether the force is appliedabove or below the horizontal. It is in the stu-dents’ best interest to minimize the amount offriction. Recall that the frictional force isdirectly proportional to the normal force. Ifthey apply the force above the horizontal, thiswill reduce the magnitude of the normal forceneeded to be supplied by the floor on the sofa,which will therefore reduce the frictional forceand make it easier to move the sofa. On theother hand, if they apply the force below thehorizontal, this will increase the normal forcerequired and thereby increase the frictionalforce, making it harder to move the sofa.

8. a) The baseball’s velocity will be upward with amagnitude less than its initial velocity. Theacceleration will be downward at 9.8 m/s2.

b) The baseball’s velocity will be zero. Theacceleration will be downward at 9.8 m/s2.

c) The baseball’s velocity will be downwardwith the same magnitude as in a). Theacceleration will be downward at 9.8 m/s2.

9. You would still need a pitcher’s mound on theMoon because the ball would still acceleratedownward due to gravity. Since the Moon hasa smaller mass than Earth, the accelerationdue to gravity on the Moon is less than that onEarth. As a result, the height of the moundwould not have to be as great as that on Earth.

10. She could jump twice as far on a planet thathas one-half the gravity of Earth. If we assumethat her initial speed and the direction forlaunch are the same, and that her initial vertical displacement is zero, we can write the following.

�dy � v1y�t � �

12

�ay�t2

0 � v1y� �

12

�ay�t

�t � �

If the acceleration, ay, is halved, then the timein flight, �t, will be doubled. Therefore, thehorizontal distance travelled will also be dou-bled, assuming that her horizontal speed isconstant.

11. As your bicycle’s rear tire spins, it takes waterwith it due to adhesion. Inertia causes thewater to try to move in a straight line. As aresult, the water leaves the wheel with a veloc-ity tangential to the tire and may spray yourback if your bicycle does not have a protectiverear fender.

12. Inertia causes the water in your clothing to tryto move in a straight line. If the drum in thewashing machine were solid, it would apply acentripetal force on the water, which wouldkeep it moving in a circle. Since the drum hasholes in it, however, the water is able to leavethe drum as it spins.

2v1y�

ay


13. The aircraft can be flown in one of two ways,or a combination of these, to provide “weight-lessness.” If the aircraft accelerates downwardat the acceleration due to gravity, the astro-nauts inside the aircraft will experience“weightlessness.” The other possibility is totravel in a vertical arc. If the aircraft flies in avertical arc at such a speed that at the top ofthe arc the gravitational force provides all thecentripetal force required to keep the aircraftand its occupants travelling in a circle, theywill experience “weightlessness.”

Chapter 31. Hydro lines and telephone cables cannot be

run completely horizontally because the forceof gravity acts downward on the entire wireand there is very little means of counterbal-ancing this force using supports.

2. a) The ladder is pushing directly into the wallon which it is resting, normal to the sur-face of the wall. With no friction, there isno force to prevent the ladder from slidingdown the wall.

b) The force exerted by the ladder on theground is exactly equal to the force of grav-ity (weight) of the ladder because there isno vertical force due to friction. The onlyforce that acts vertically, upward or down-ward, is the force of gravity.

3. Standing with your feet together or wide apartmakes no difference to the condition of staticequilibrium, since in both cases all forces arebalanced. In terms of stability, the widerstance is more stable. A wider stance means alower centre of mass and a wider “footprint.”This means there is a greater tipping angle forthis wider stance.

4. High-heeled shoes force the centre of mass ofthe person wearing to move forward from itsnormal position. To maintain balance, the person must move the centre of mass backagain, usually by leaning the shoulders back-ward. This effort can cause fatigue in the backmuscles.

5. Line installers allow a droop in their lineswhen installing them because the droopallows a moderate upward vertical applicationof force as the wire curves upward to the sup-port standards. This allows an upward forceto support the wire when loaded with freezingrain and ice buildup. This droop means thatthe tension to support the load can be muchless because of the greater angle.

6. A wrench can be made to more easily open arusty bolt by adapting the wrench so as toapply more torque. More torque can be appliedby the same force by adding length to thewrench handle.

7. The higher up on a ladder a person is, the far-ther he is from the pivot point, which is thepoint where the ladder touches the ground.Therefore, the ladder will be more likely toslide down the wall if the person stands on ahigher rung.

8. The torque varies as sin �, where � is theangle between the pedal arm and the appliedforce. The torque is at a minimum (zero)when the pedals are vertical (one on top ofthe other), because the force (weight) isapplied at 0° to the pedal arm, and sin 0° � 0.The maximum torque is applied when thepedals are horizontal, because the anglebetween the pedal arm and the applied force is90°, and sin 90° � 1.

9. There is no extra benefit for curls to be doneto their highest position. As the forearm israised, the angle of the force of gravity vectordecreases at the same rate as the anglebetween the muscle of effort and the arm. Asthe forearm is raised, the effort required to liftthe arm decreases, but so does the muscle’sability to provide the effort.

10. Your textbook is sitting in stable equilibriumwhen flat on your desk. When the book is bal-anced on its corner, it is in unstable equilib-rium. Motion in any direction will cause alowering of the centre of mass and a release ofgravitational potential energy, making the tip-ping motion continue and thus making thebook fall.


11. In terms of stability, a walking cane provides awider base (footprint) over which the personis balancing. It is harder to force the person’scentre of mass outside this wider support base.

12. When standing up from a sitting position, wefirst must lean forward to move our centre ofmass over our feet to maintain stability.Unless we first lean forward, our centre ofmass is already outside our support base andit is impossible to stand up.

13. A five-legged chair base is more stable becauseof the wider support base (footprint). Theextra leg effectively increases the tippingangle, making the chair more stable.

14. Tall fluted champagne glasses must have awide base to improve the stability of the glass.Recall that the tipping angle is given by the

expression � � tan�1 .

Therefore, the taller the glass, the greater theheight of the centre of mass, and the smallerthe tipping angle. A wider base increases thetipping angle by compensating for the tallerglass.

15. The extra mass helps to mimic the mass ofthe cargo and lowers the centre of mass of theship. Without this extra mass, the ship wouldbe top-heavy and more prone to capsizing,especially in rough weather.

16. This figure is so stable because the design ofthe toy places the effective centre of massbelow the balance point. A gentle push actu-ally raises the centre of mass like a pendulum,which increases the gravitational potentialenergy, which tends to return the toy to itsstable equilibrium position.

17. The bone that has the smaller length will frac-ture first if the same twisting stress is appliedto two bones of equal radius but differentlengths. This is due to the fact that the strain

� � on the longer bone will be much

smaller than that on the shorter bone, becausethe length term appears in the denominator ofthe expression for strain.

18. Lumber is used this way to support greaterspans because of the greater dimensions ofwood in the vertical direction. More woodprovides a means of supporting a greaterweight through a tension force throughout thewood.

19. Concrete would not be an ideal material for acantilevered structure because of the differ-ence in the way that this material deals withtension and compression forces. A cantileverwould require a great tensile strength in theupper layer and a great compressive strengthin the lower layer. Concrete has great com-pressive strength but poor tensile strength.

Chapter 41. Momentum is the product of mass and velocity;

p�� mv��. Since velocity is a vector quantity, sois momentum.

2. A system represents all the objects involved ina collision. In a closed system, the boundary isclosed (that is, there are no interactions withthe external environment) and therefore thenet external force acting on the system’sobjects as a group is zero. In an open isolatedsystem, the boundary is not closed but the netexternal force acting on the system is zero.

3. The net force is used in the calculation ofimpulse; J � F∆t.

4. Impulse is the change in momentum; J � ∆p.5. In an isolated system, the net external force, F,

acting on the system is zero. Therefore, theimpulse, J, is zero (J � F∆t), and the change inmomentum, ∆p, is zero (J � ∆p).

6. The law of conservation of (linear) momen-tum states that the total momentum of an isolated system before a collision is equal tothe total momentum of the system after thecollision. This can be expressed algebraicallyas ptotalinitial � ptotalfinal. Equivalently, in an isolatedsystem the change in momentum is zero; ∆p � 0.

7. Yes, a ball thrown upward loses momentum asit rises because there is a net external forcedownward (gravity) acting on the ball, slow-ing it down.

�L�

L

(0.5)(width of base)��height of centre of mass


8. Assuming that the net external force acting onthe grenade during the explosion is zero(ignoring gravity), the sum of the 45 momen-tum vectors after the explosion is equal to themomentum vector of the grenade before theexplosion, since ptotalinitial � ptotalfinal.

9. Assume that the astronaut’s initial momen-tum is zero as he floats in space. By throwingthe monkey wrench in the opposite directionof the space station, he would be propelledtoward the space station. This is an exampleof Newton’s third law: The total momentumof the astronaut–wrench system would still bezero after he threw the wrench.

10. A rocket can change its course in space byejecting any object or matter such as a gas.Assuming that the total momentum of therocket–gas system is conserved, the momen-tum of the rocket will change as the gas isejected. This change in momentum will corre-spond to an impulse, which will change thecourse of the rocket.

11. Assume that the total momentum of the sys-tem is conserved:

pTo � pTf

p1o � p2o � p1f � p2f

mv1o � mv2o � mv1f � mv2f

mv1o � m(–v1o)� mv1f � mv2f

(substituting v2o � �v1o)0� m(v1f � v2f)

Therefore, the general equation for the totalmomentum before and after the collision ispTo � 0 � m(v1f � v2f) � pTf.

12. As rain falls into the open-top freight car, thecar will slow down. Assuming that momen-tum is conserved as the rain falls into the car,the combined mass of the car and the waterwill move along the track at a slower speed.

13. Object A is moving faster before the collision.Assuming that the momentum of the A-B system is conserved, the final velocity of theobjects, vf, is equal to the average of their initial velocities, vAo and vBo:

pTo � pTf

mvAo � mvBo � mvAf � mvBf

vAo � vBo � vf � vf

� vf

Since the angle between vBo and vf is greaterthan the angle between vAo and vf, the magni-tude of vAo is greater than the magnitude of vBo.

14. The component method would be preferredfor solving momentum problems in whichtrigonometry could not be used readily — forinstance, problems involving more than twoobjects colliding, or non-linear problems.

15. a) Grocery clerks lean back when carryingheavy boxes so that their centres of massstay in line with their feet.

b) The centre of mass of a system of masses isthe point where the masses could be con-sidered to be concentrated or “balanced”for analyzing their motion. This conceptcan simplify momentum problems sincethe momentum of the centre of mass isequal to the total momentum before, andafter, a collision, and is conserved duringthe collision.

Chapter 51. When you are holding your physics book

steady in your outstretched arm, there is nowork done because there is no displacement(W � F∆d).

2. The momentum, p, of an object with mass mis related to its kinetic energy, Ek, according tothe equation p � �2mEk�. If a golf ball and afootball have the same kinetic energy then thefootball has the greater momentum, since themass of the football is greater than the mass ofthe golf ball.

3. A negative area under a force–displacementgraph represents negative work, which meansthat the displacement is in the opposite direc-tion of the force applied. For example, whenfriction is slowing down a car, there is a posi-tive displacement but a negative force.

4. After work is done on an object, it has gainedenergy.

vAo � vBo�2


5. When a spring diving board is compressed bya diver jumping on it, the diving board pos-sesses elastic potential energy. As the divingboard straightens out, it transfers its elasticpotential energy to the diver, who gainskinetic and gravitational potential energy. Asthe diver rises in the air, her kinetic energy istransformed into potential energy until sheonly has gravitational potential energy as shereaches her highest point. As she descendstoward the pool, her potential energy is trans-formed into kinetic energy and she increasesher speed as she falls. As she enters the pooland slows down in the water, her kineticenergy is transferred to the water as kineticenergy, potential energy, and heat energy.

6. Ek � mv2

( J) � (kg)[(m�s)2]( J) � (kg·m2�s2)( J) � (kg·m�s2·m)( J) � (N·m)( J) � ( J)

7. The equation �∆Ee � ∆Ek means that a lossof elastic potential energy becomes a gain inkinetic energy.

8. Yes, since gravitational potential energy ismeasured relative to a point which couldchange. That point could be the ground level,the basement level, or any other arbitrarypoint.

9. In an elastic collision the total kinetic energyis conserved, whereas in an inelastic collisionthe total kinetic energy is not conserved. Anexample of an (almost) elastic collision is acollision between two billiard balls. An exam-ple of an inelastic collision is a collisionbetween two vehicles in which their kineticenergy is transferred to heat energy, soundenergy, and energy used to permanentlydeform the vehicles.

10. No, the equation Ek � shows that if an

object has momentum then it must havekinetic energy. The converse is also true, asthe equation also shows.

Chapter 61. We do not require the more general form of

Newton’s law of universal gravitation becausefor situations on or near the surface of Earth,the values of G, M, and r can be assumed tobe specified constants. After these simplifica-tions are made, the general form becomesequivalent to the simpler form.

2. Due to the direction in which Earth rotates,more energy would be required to reach thesame orbit if a spacecraft was launched westward, since an eastward launch aids thespacecraft.

3. The near side of the Moon is more massivethan the far side, possibly due to impactedmeteors. Over time this side was moreattracted to Earth, so that eventually the moremassive side came to face Earth all the time.This is also true for the moons of Jupiter andSaturn relative to their planets.

4. The force of gravity is the derivative of gravi-tational potential energy, Ep. Equivalently, theforce of gravity is the slope of the graph of Ep

versus x.5. Assuming that the spacecraft is initially in

orbit and that jettisoning a large piece of itselfdoes not significantly alter its momentum, itwill continue in the same orbit.

6. The velocity of a spacecraft in orbit is con-stantly changing due to the centripetal forceacting on it. Therefore, if one spacecraftpoints toward another and rockets in thatdirection, the two spacecraft will not meetbecause the “added velocity” vector of the firstspacecraft does not change as is required forconvergence.

7. a) The escape speed required to leave Earthis approximately 11 km/s. The necessaryupward acceleration, a, of a spacecraft during firing from an 80-m cannon is given

p2

�2m

1�2


by a � � 756 250 m/s2.

This is more than 77 000 times the magni-tude of the acceleration due to gravity, and would be experienced for about

∆t � � 0.015 s. The

mission would not be survivable.b) The downward force of the gun’s recoil

would be roughly equal to the upwardforce on the spacecraft. If the spacecrafthad a mass of 5000 kg, the force of therecoil would be approximately(5000 kg)(756 250 m/s2) � 3.781 109 N.

8. Given:hmax � 2 mk � 500 N/mxmax � 0.45 mm � 80 kg

First, calculate the maximum energy that ourknees can absorb without damage.

Vg � mghVg � (80 kg)(9.8 m/s2)(2 m)Vg � 1568 J

Next, calculate the maximum energy that thesprings can absorb.

Ve � �12

�kx2

Ve � �12

�(500 N/m)(0.45 m2)

Ve � 51 JFinally, calculate the maximum height fromwhich we could survive a fall without damage.

Vg � mgh

h �

h �

h � 2.07 mWith the springs attached, we could survive afall of at most 2.07 m without damage.

9. The force of gravity would be (9.8 m/s2)(80 kg) � 1837 N downward,whereas the force of the springs would beonly (500 N/m)(0.45 m) � 225 N upward.The net force acting on us would act down-ward, so we would not bounce off the ground.

10. Three everyday examples of SHM are: anidling engine, as periodic power from combus-tion keeps piston movement in a state ofSHM; someone rocking in a rocking chair,where periodic “foot pushes” or shifts in thecentre of mass counteract dampening; themotion of a toy bird that “drinks” water, pro-vided that there is a constant supply of water.

11. Three examples of damping in oscillatory sys-tems are: engine braking (desired) — as thefuel supply to the cylinders is lessened, so isthe power, which dampens piston movement;swinging on a swing (undesired) — theheight of successive swings becomes smallerand smaller due to friction and air resistance;air bags (desired) — when deployed, theygradually dampen the effects of a collision ona person’s body, as opposed to a steeringwheel or dashboard, which do so almostinstantaneously.

Chapter 71. All objects on Earth that are stationary rela-

tive to Earth’s surface have the same angularvelocity, since they all complete one rotationabout Earth’s axis in the same amount oftime. However, they do not all have the sametangential velocity, since they are not all thesame distance from Earth’s axis of rotation. If is the angular velocity of an object onEarth’s surface and r is the object’s distancefrom Earth’s axis of rotation then the object’stangential velocity, v, is given by v � r.

2. A differential mechanism is necessary toallow a car to turn smoothly. The wheels onthe inside of a turn move through a smallerradius than the wheels on the outside, thustravelling a smaller arc distance in the sameamount of time. Therefore, the inside wheelsrotate at a smaller angular speed. In theabsence of a differential, however, the insideand outside drive wheels (connected to themotor) must rotate at the same angular speed.To turn, you would have to lock up the insidedrive wheel, causing an uncontrolled turn.

1568 J � 51 J��(80 kg)(9.8 m/s2)

Vg�mg

11 000 m/s��756 250 m/s2

(11 000 m/s)2

��2(80 m)


The differential allows the drive wheels toturn at different angular speeds.

3. The top of the CN Tower, with the towerlocated at the highest altitude on the equator,would have the greatest tangential speed,because in that case the top of the towerwould be the greatest distance from the axisof rotation of Earth. However, since the heightof the CN Tower is negligible compared to theradius of Earth, the variation in tangentialspeed among different parts of the tower isnegligible.

4. A larger car tire has a greater moment ofinertia (greater radius and mass), thus inprinciple more energy would be needed tostart turning the tire. Once the tire was mov-ing, the law of inertia would apply and agreater force would be needed to slow andstop the tire, thus less energy would beneeded to keep the tire moving.

5. a) Yes, changing the tire size affects theodometer reading. For example, a tire witha larger radius than the calibrating tire cov-ers a greater distance in the same numberof turns. In that case, the car will travel agreater distance than what the odometerindicates.

b) Yes, the speedometer reading is affected,for the same reason. For example, a carwith larger tires will travel at a greaterspeed than what the speedometer indicates.

6. The angular equivalents to force and displace-ment are torque and angular displacement. Nolinear work is done on an object if an appliedforce does not change the displacement of theobject in the direction that the force isapplied. No rotational work is done if anapplied torque does not result in a change inangular displacement.

7. No, angular momentum is conserved becausethe diver is in fact still rotating as she entersthe water. There is no external torque appliedto the diver after she leaves the diving board.Because the diver increases her moment ofinertia by extending out straight from atuck, her angular spin decreases. This is

not visually apparent as the diver then entersthe water out of sight of the spectators andjudges.

8. No, the centripetal force acting on a ridervaries depending on the radius of turn: thelarger the radius, the larger the centripetalforce. The riders on the outer part of the rideswing out farther than the inner ridersbecause of the larger centripetal force.

9. According to the law of conservation of angu-lar momentum, the total angular momentumbefore the tape recorder was turned on highspeed was equal to the total angular momen-tum after. When the tape recorder was turnedon high speed, the angular momentum of thesystem had an added component in the angu-lar direction of the turning tape. Voyager 2rotated in the opposite direction to compen-sate, although not as fast, since its moment ofinertia was much larger than the taperecorder’s.

10. a) The hollow cylinder has a greater momentof inertia than the solid one because thehollow cylinder’s mass is concentrated far-ther from the axis of rotation. However,since there is no friction, there is no forceavailable to create the torque necessary toturn the cylinders. Translational motiondoes not depend on the distribution ofmass, so both objects accelerate at the samerate and reach the base of the incline at thesame time.

b) As in part a), in the absence of friction thecylinder does not roll. Therefore, bothobjects slide down the ramp, accelerating atthe same rate (ignoring the effect of windresistance on the different shapes).Translational motion does not depend onthe distribution of mass, so both objectsreach the ground at the same time.

11. A spinning projectile behaves like a gyroscope.The spin means that the object possessesangular momentum about its axis of rotation.This allows the object to resist forces actingon it as it travels, which in turn allows theobject to maintain its projectile motion.


Without the spin, uneven airflow over the sur-face of the object would make it tumble, expe-rience greater air resistance, and travel ashorter distance.

12. If the wheel does not slip as it rolls then thetranslational distance, d, that the axle moves isequal to the arc length, s, along the outside ofthe wheel. This is not true in the case of“squealing your tires.”

13. Rotation axes can be anywhere, but for sim-plicity’s sake consider only some symmetricones. Ranked from least to greatest moment ofinertia, the rotation axis can pass through thecentre of the top and bottom (shown),through the centre of the spine, through thecentre of the front and back cover, or rundiagonally from one corner to another.

14. The angular momentum of a Sun–planet sys-tem is conserved. The force acting on theplanet is that of gravity due to the Sun. At anyinstant in time, this force acts through the axisabout which the planet instantaneouslyrotates. This means that the moment arm iszero and no torque acts on the planet.Therefore, the angular momentum of theplanet remains constant and the total momen-tum of the system does not change.

15. It is easier to balance on a moving bike thanon a stationary one because of a combinationof the aspect called “trail” and gyroscopicaction.

16. The law of conservation of angular momen-tum applies when a motorcycle is in mid-air.In the absence of an external torque, theincreased angular momentum of the faster-spinning rear wheel causes the entire motor-cycle to rotate in the other direction in orderto keep the total angular momentum the sameas it was when the motorcycle left the ground.

Chapter 81. A neutral object is attracted to a charged

object because the charged object induces acharge separation in the neutral object. Theelectrons in the neutral object are forced awayfrom or toward the charged object, depending

on whether the charged object has a negativeor positive charge, inducing an oppositecharge which acts to attract the two objects byway of the law of electric forces.

2. The function of an electroscope is to detect anelectric field. An electric field will cause themovement of electrons within an electroscope,inducing similar charges to cluster at each ofthe two pieces of dangling foil. The two piecesof foil will repel each other, indicating thepresence of the electric field.

3. Rubbing the balloon against your dry haircharges the balloon electrostatically. When theballoon approaches the wall, the negativecharge forces the electrons in the ceiling away,leaving the positive charges close to the sur-face. The result is that the negatively chargedballoon attracts the now positively chargedceiling surface.

4. The electrostatic series identifies silk as hav-ing a greater affinity for electrons than acetatedoes. When acetate and silk are rubbedtogether, electrons move from the acetate tothe silk because of the different affinity thematerials have for electrons.

5. Choose two materials listed at either end ofthe electrostatic series, such as acetate andsilk, and rub them together to place the pre-dictable negative charge on the silk.Neutralize the acetate and then rub it with themystery substance. Place the mystery sub-stance next to the silk and judge whether themystery substance has a negative charge(repulsion) or a positive charge (attraction).A negative charge would place the mysterysubstance below acetate in the electrostaticseries. Similarly, rubbing the mystery sub-stance with silk would help to place the

+

–

–

–

– –

–– – –+ + +

+

–

+

–

+

–

+

–

+

–

+

–

+

–

+

–

Force of Attraction

Balloon

Ceiling


mystery substance in the series compared tosilk. By selectively choosing different sub-stances, you could narrow down the appropri-ate spot for the mystery substance in theelectrostatic series.

6. Computer technicians touch the metallic partof a computer before repair, assuming it is stillplugged into the wall outlet, so that theyground themselves from any excess charge.Otherwise, a static electric discharge coulddamage the computer’s micro-circuitry.

7. Criterion Newton’s law Coulomb’s law of universal gravitation of electrostatic forces

Equation F � �Gm

r21m2� F � �

kqr12

q2�

Constant of G � 6.67 10�11 N·m2/kg2 k � 9.0 109 N·m2/C2

proportionalityType of Attraction only Attraction and repulsionforce(s)Conditions Acts between any Acts between any two for use two masses electrostatic charges

8. Field lines show the direction of the net forceon a test charge in an electric field. Twocrossed field lines would mean that therewould be two net forces acting on a testcharge in two different directions at the sametime. This is impossible, since there is onlyone net force at any point, which is only oneforce in one direction by definition.

9. In an electric field, charges always move alongthe direction described by the field lines. Thedirection in which a charge moves along afield line depends on the sign of the charge.A positive charge will move in the directiondescribed by the arrows in a field diagram,whereas a negative charge will move in theopposite direction.

10. a) When a polar charged rod is placed perpen-dicular to electric field lines, the rod willtend to rotate such that it will become par-allel to the field lines. The positive end ofthe rod will point in the same direction inwhich the field lines are oriented.

b) When a polar charged rod is placed parallelto electric field lines, the rod will tend tostay in the same orientation if its positiveend is pointing in the same direction as the

field lines. Otherwise, the rod will tend torotate 180° and point in the opposite direc-tion (still parallel to the field lines).

11. Each point charge experiences an identicalforce of repulsion from all of the other pointcharges, so that they are all repelled symmetri-cally outward from the centre of the orienta-tion. A test charge placed outside of the circlewould experience a net force directed alongradial lines inward to the centre of the circle,as shown in the diagram. A test charge placedinside of the circle would experience no netforce, and therefore there would be no electricfield inside the circle at all.

This charge distribution models the electricfield inside a coaxial cable because the outerbraided conductor in a coaxial cable acts asthe site modelled by the ring of chargedescribed above. This ring acts to eliminatethe field within the entire cable.

12. By definition, the electric potential is the sameat any point along an equipotential line.Therefore, no force is required, and no workis done, to move a test charge along this line.In a situation like this, a constant force causesthe constant acceleration of the test charge.

13. We use the term “point charge” to imply thatthe charge has no larger physical dimensions.Larger dimensions would mean that thecharge would exist within a region of spaceinstead of at a specific location. This implica-tion reduces the number of variables and simplifies questions that deal with the distri-bution of charges within a three-dimensionalspace. Any other approach would requiresome way of accounting for the variability ofdistances between charges.

–

–––

––

––


14. Statement: In each case the field getsstronger as you proceed from left to right.FalseReasoning: The field lines remain the samedistance apart as you move from left to rightin the field in (b), so the field does not changein strength.Statement: The field strength in (a) increasesfrom left to right but in (b) it remains thesame everywhere. TrueReasoning: The field lines become closertogether as you move from left to right in thefield in (a), so the field does increase instrength, whereas the field lines in (b) are par-allel, so the field strength does not change.Statement: Both fields could be created by aseries of positive charges on the left and nega-tive ones on the right. FalseReasoning: Although true for (b), (a) mustbe created by a single positive point charge atthe base of the four arrows.Statement: Both fields could be created by asingle positive point charge placed on theright. FalseReasoning: As described above, a pointcharge could be responsible for (a), but (b)would require rows of parallel opposites suchas those in oppositely charged parallel plates.

15. Electric fields are more complicated to workwith because the forces that charges exert oneach other are all significant. In contrast, thegravitational force between small masses isnegligible compared with the gravitationalforce exerted on them by large masses likeEarth.

16. The field shape around a single negative pointcharge is exactly like that around a single posi-tive point charge with the exception that for anegative point charge, the arrows are all point-ing inward instead of outward, as shown inthe following diagram.

17. Doubling the value of the test charge will donothing to the measurement of the strength ofthe electric field. The force on the test chargewill double because of the change to the testcharge, but the field strength is measured as

the force experienced per unit charge, �� qF��

t�.

Therefore, the doubling of the test charge andthe doubling of the force will cancel, leavingthe measurement of the field strengthunchanged.

18. The stronger an electric field is, the closertogether the field lines are. Therefore, a weakelectric field has field lines that are fartherapart than the field lines of a strong electricfield.

19. Both gravitational fields and electric fields aremade up of lines of force that are directed in away that a test “item” would be forced.Gravitational fields are created by and influ-ence masses, whereas electric fields involvecharges. Gravitational fields are always attrac-tive. Electric fields can be attractive or repul-sive, since they can exert forces in oppositedirections depending on the charge of theobject that is experiencing the field.

20. The direction of an electric field between apositive charge and a negative charge is fromthe positive charge toward the negative charge,since electric fields are always directed theway that a positive test charge would beforced.

21. The electric potential energy is greaterbetween two like charges than between twounlike charges the same distance apartbecause of the differing sign of the electricpotential energy. The calculation of the

–


electric potential energy involves multiplyingthe two charges. The product of two likecharges is positive and therefore greater thanthe product of two unlike charges, which isnegative.

22. A high-voltage wire falling onto a car pro-duces a situation in which there is a high-potential source (the wire) very close to alow-potential region (the ground). The peoplein the car will be safe from electrocution aslong as they do not complete a circuit betweenthis high and low electric potential. Theyshould not open the car door, for example, andstep to the ground while maintaining contactwith the car.

23. Although opposite electric charges occur atthe two plates of a parallel-plate apparatuswhen it is connected to a power supply, theoverall charge on the apparatus remains zero.For every charge at one plate, there is an oppo-site charge at the other plate, which balancesthe overall charge to zero.

24. a) If the distance between the plates is dou-bled then the field strength between theplates will be halved.

b) If the charge on each plate is doubled thenthe field strength will double.

c) If the plates are totally discharged and neu-tral then the field strength will drop tozero.

25. Two point charges of like charge and equalmagnitude should be placed side by side sothat both the electric field strength and theelectric potential will be zero at the midpointbetween the charges. If one of the two likecharges were doubled, the field strength andthe potential would both be zero at a pointtwo-thirds the separation distance away fromthe doubled charge.

26. In the presence of electric fields, a fieldstrength and a potential of zero would exist ata point where the sum of all electric forceswas zero. In question 25, the sum of the repul-sive forces from each of the two like charges iszero at some point between the two charges.

27. If a proton and an electron were released at adistance and accelerated toward one another,the electron would reach the greater speed justbefore impact. The reason is that both parti-cles would be acted upon by the same force ofattraction, but the electron has less mass. Theacceleration of each particle is

described by the formula a�� mF��

�, which

shows that for the same force, the smallermass would have the greater acceleration overthe same time period and therefore the greaterfinal speed.

28.

This type of motion is like upside-down pro-jectile motion, since the charge moves in aparabolic path. This is the type of motion thatan object would take if it were thrown hori-zontally in Earth’s gravitational field. Theonly difference here is that this charge appearsto be “falling upward” instead of downward.

29. No, a parallel-plate capacitor does not haveuniform electric potential. It does have uni-form field strength between the two plates,but the potential varies in a linear fashionfrom one plate to the other. By definition, theelectric potential is uniform along any equipo-tential line, which in this case is any line par-allel to the two plates.

30. Charge Distribution Equipotential Lines(a) (iii)(b) (i)(c) (ii)

31. a) The electrostatic interaction responsiblefor the large potential energy increase atvery close distances is the repulsionbetween the two positive nuclei.

b) This repulsion of the nuclei, and the asso-ciated increase in electric potential energy,is one of the main stumbling blocks forgenerating energy through nuclear fusion.This repulsion between nuclei means that

q


a very large amount of energy is requiredto begin the reaction process.

c) The smaller increase in electric potentialenergy upon separation of the two atoms iscaused by the attraction between the posi-tively charged nucleus in each atom andthe negatively charged electron in the otheratom.

d) A stable bond is formed when two hydro-gen atoms are about 75 pm apart becausethis is the distance at which the electricpotential energy is minimized — any closerand the repulsion between nuclei pushesthe atoms apart, any farther away and thenucleus-electron attraction draws theatoms closer together.

32. A positive test charge moving along a linebetween two identical negative point chargeswould experience a topography similar to avehicle moving up a hill (away from onecharge), increasing the vehicle’s gravitationalpotential energy, and then rolling down theother side of the hill (toward the othercharge).a) If the two identical point charges were

both positive, the hill would change to avalley with the lowest part in the middle.

b) If a negative test charge was placedbetween the two identical positive charges,the topography would still resemble a val-ley but now there would be a very deepcrater at the lowest part of the valley.

Chapter 91. The law of magnetic forces states that like

(similar) magnetic poles repel one anotherand different (dissimilar) poles attract oneanother, even at a distance.

2. A magnet can attract non-magnetic materialsas long as they are ferromagnetic in nature.The magnet causes the internal domains(small magnets) of a ferromagnetic substanceto line up in such a way that a new magnet isinduced in the substance such that there areopposite magnetic poles which attract oneanother.

3. A material that is attracted to a magnet or thatcan be magnetized is called ferromagnetic.Examples of ferromagnetic materials includematerials made from iron, nickel, or cobalt.These materials are ferromagnetic becausethey have internal domains that can be readilyaligned, due to the fact that these materialshave unpaired electrons in their outermostelectron energy level.

4. Magnets can lose their strength over timebecause their domains, which initially arealigned (pointing in the same direction), canbecome randomized and point in other direc-tions. This randomizing of the domainsreduces the overall strength of the entire mag-net.

5. When a magnet is dropped or heated up, thedomains of the magnet, which initially arealigned (pointing in the same direction), canbe disrupted and forced to point in other, ran-dom directions. This randomizing of thedomains reduces the overall strength of theentire magnet.

6. a)

b)

7. The electrons in the beam that is illuminatingyour computer monitor’s screen are directedfrom the back of the monitor forward to thefront of the screen, toward your face.Therefore, conventional (positive) currentpoints in the opposite direction, away fromyour face and back into the computer monitor.This is the direction of the thumb of the right

F F

Currents in opposite directions—wires forced apart

Tx

F F

Currents in the same direction—wires forced together

T


hand when applying right-hand rule #1 forcurrent flow. From your perspective, the mag-netic field forms circular formations in theclockwise direction in and around the com-puter monitor. Relative to the direction of theelectron beam, the magnetic field is directed inthe counterclockwise direction around thebeam.

8. A wire possessing an eastbound conventional(positive) current has an associated circularmagnetic field that points upward on thenorth side of the wire and downward on thesouth side.

9. The magnetic field strength of a coil (an insu-lated spring) varies inversely with the lengthof the coil. Therefore, a reduction in the coillength to half its original length will cause adoubling of the magnetic field strength. Thisall depends on the assumption that the lengthof the coil is considerably larger than its diam-eter.

10. a) For the force applied to a current-carryingconductor to be at a maximum, the mag-netic field must cross the conductor at anangle of 90°.

b) For the force applied to a current-carryingconductor to be at a minimum, the mag-netic field must cross the conductor at anangle of 0°.

11. According to right-hand rule #3 for the motorprinciple, the direction of the force on the con-ductor will be to the north.

12. An electron moving vertically downward thatenters a northbound magnetic field will beforced toward the west.

13. A current-carrying solenoid produces a mag-netic field coming directly out of one end ofthe coil and into the other end. An electronpassing by either end of this coil experiences aforce that is at right angles to its motion. Asthis force changes the direction of motion (acentripetal force), the electron takes on acurved path (circular motion). Application ofthe appropriate right-hand rules predicts thatthe electron’s motion will curve in the same

direction as the direction of conventional cur-rent flow through the coil.

14. The cathode rays will be deflected away fromthe current-carrying wire, moving in a planethat contains the wire.

15. Current passing through a helical spring willproduce a situation very similar to having twoparallel conductors with a current flowing inthe same direction. Application of the appro-priate right-hand rules predicts that the mag-netic field interaction between each pair ofthe helical loops will force the spring to com-press, reducing its length.

16. Current passing through a highly flexible wireloop will tend to result in magnetic field inter-actions that will force apart nearby sections ofthe wire, so that the wire loop will most likely(if the proper conditions exist) straighten out.

17. Faraday’s principle states that a magnetic fieldthat is moving or changing in intensity in theregion around a conductor causes or induceselectrons to flow in the conductor. To improvethe electromotive force induced in a conduc-tor, we can increase the magnetic fieldstrength, the length of the conductor, and thestrength of the current flowing through theconductor.

18. Current can be induced to flow in a conductorif the conductor is moving with respect to amagnetic field. The maximum induced currentoccurs if the conductor and the field crosseach other at right angles.

19. Lenz’s law states that the direction of theinduced current creates an induced magneticfield that opposes the motion of the inducingmagnetic field. Lenz derived this law by rea-soning that a decrease in kinetic energy in theinducing magnetic field must compensate forthe increase in the electric potential energy ofthe charges in the induced current, accordingto the law of conservation of energy. Thisdecrease in kinetic energy is felt as an opposi-tion to the inducing magnetic field by aninduced magnetic field.


20. The induced current can only create a mag-netic field that opposes the action of move-ment (conductor or field) in order to followthe law of conservation of energy and Lenz’slaw. If the motion were not opposed and theinduced magnetic field instead “boosted” themotion, this would increase the kinetic energyof the moving conductor or magnet, whichwould violate the law of conservation ofenergy.

21. a) Electromagnetic brakes might work byusing the undesirable motion of the vehicleto provide the energy to induce currentflow in a conductor. The resulting creationof electrical energy would be at the expenseof the kinetic energy of the vehicle, whichwould slow down. This would be a case ofenergy being transformed from one form toanother, following the law of conservationof energy.

b) Electromagnetic induction brakes would becapable of recovering some of the kineticenergy of a vehicle that is normally lost asheat in conventional brakes, thereby savingmoney. The electrical energy generatedcould be used to recharge the battery for anelectric vehicle/hybrid.

Chapter 101. The motion of a vibrating spring can be mod-

elled mathematically by a sine wave, whichresembles (visually) an electromagnetic wave.As well, both waves are periodic.

2. The magnetic field is induced by the electricfield and thus they would both decrease. Ifone component vanishes then the electromag-netic radiation ceases to exist.

3. “Visible light” is relative to the human beingperceiving it. Also, some other animals seein other regions such as the infrared andultraviolet.

4.

5. When metallic objects are placed in amicrowave oven, they can absorb electromag-netic microwaves, which dislocate loose elec-trons in the metal and allow charges to buildup on the surfaces, until the cumulativecharge is large enough to “jump” across an airgap to another conductive material in theoven, causing a spark.

6. Simple harmonic motion refers to a physical“state” where the restoring force, acting on anobject when it is pulled away from some equi-librium position, is proportional to the dis-placement of the object from the equilibriumposition. Since there is a net force acting onthe object, it experiences an acceleration, andthus the speed cannot be constant.

7. If a circle is viewed edge-on, with a dotpainted on the edge, and the circle is spun, thedot will seem to exhibit simple harmonicmotion as it moves around the circle. Fromthe edge it will seem as if the dot is movingback and forth, constantly passing the equilib-rium position.

8. Electron oscillators absorb energy from theincoming wave, causing it to be retarded.When this secondary wave interferes with theincident wave, a phase lag is created retardingthe wavefront, slowing it down.

9. Newton’s theory of refraction predicts thatlight speeds up as it changes direction. This isincorrect since light decreases its speed whenbending toward the normal. You can show histheory by rolling a marble across a boundarybetween a flat area and an incline. As themarble crosses the boundary, it bends towarda line drawn perpendicular to the edge but itspeeds up.

θrθi

Normal


10. One example of an invisible medium is a vac-uum. The refractive index of a vacuum is1.00. When the refractive index is 1.00, thereis no component of an incoming light ray thatis reflected. Since no light is reflected, themedium is invisible. Another possibility isthat the medium is of the same refractiveindex as the environment.

11. Using a laser, which is a powerful coherentsource of visible light, you can measure therefraction of the ray as it enters a medium, orthe extent of polarization upon reflectionand/or transmission, all of which can be com-bined to calculate the optical density of themedium.

12. Because the refractive index is wavelengthdependent, when white light refracts througha material, each component of light bendsslightly differently. This separates the light.If the separation is great enough, dispersionoccurs.

13. As light passes through a prism, both refrac-tions cause the light to refract in the same spa-tial direction. This accentuates the spreadingof the colours.

14. No, sound waves cannot be polarized. Soundwaves are mechanical waves and refer to com-pressions and rarefactions within a medium.Sound waves have only one component, nottwo like electromagnetic waves, and thuspolarization is impossible.

15. A polarizer and an analyzer are both thinpieces of film. They are given different namesbased on the order in which a wave entersthem. If two pieces of thin film are positionedside by side, the first one struck by the wave isknown as the polarizer and the second onethe analyzer. If the two are flipped, the ana-lyzer will become the polarizer and the polar-izer will become the analyzer.

16. The lenses in polarized sunglasses are nor-mally oriented in such a way as to restrict thepassage of plane-polarized light reflecting offthe surface of the ground and water (glare). Ifthe lenses are rotated, they will no longerblock the glare.

17. Yes, the effectiveness of Polaroid sunglassesvaries as the relative positions of the Sun andthe horizon vary, since the distribution ofscattered angles varies as well. The amount ofpolarization is angle dependent, hence theeffectiveness of the glasses varies.

18. No, Polaroid sunglasses are not effective oncircularly polarized light, which is composedof the two polarization directions combined ina specific phase relationship causing the direc-tion of the electric field vector to rotatearound. The linear polarizer cannot block outboth components, hence light is transmitted.

19. With a powerful light source, you can easilynotice that light reflects off dust particles inthe air. Sometimes, depending on the size ofthe particles, certain frequencies of the elec-tromagnetic spectrum are deflected/reflectedmore than others. This effect causes a certaincolour to appear in the medium (for example,the blue colour of the sky). By noting thecolour, you can determine the frequency andthus the wavelength of light associated withthat colour. Once you know the wavelength,you can calculate the approximate size of theparticles that would deflect waves of those fre-quencies. Also, you can use the intensity ofthe colour to estimate the density of the parti-cles in the air.

20. Polarization: Electric fields of electromagneticradiation behave sinusoidally. The direction ofthese fields is randomly oriented in any direc-tion for unpolarized light. Two componentsare obtained by using plane polarizers. Thetwo components can be combined using thewave equation (� sin t) to form circularly orelliptically polarized light.Scattering: The wavelength of light, �, compa-rable to the size of particles in the air createsthe maximum scattering. The extent of scat-tering of light by air molecules is proportionalto ��4.Refraction: Using wavefronts, Snell’s law ofrefraction is derived. Based on phase relation-ships between the incident wave and the


transmitted wave, light is bent and sloweddown in different mediums.

Chapter 111. Refraction, polarization, interference, and

diffraction2. Refraction, diffraction, and interference can

be demonstrated using water waves in rippletanks. Polarization cannot.

3. The film on a soap bubble is thicker at thebottom than at the top, forming a wedgeshape, since gravity pulls the soap down. Asthe film’s thickness changes, the interferencechanges (destroys some wavelengths) and thecolours change.

4. As the gasoline evaporates, it becomes thin-ner, changing the interference pattern and thecolours.

5. A camera lens has a thickness and materialdesigned to block out certain colours, whereasa car windshield does not. These properties ofa lens produce interference patterns and acolour change. Camera lenses are designed tocorrect chromatic aberration caused by differ-ent wavelengths bending at different angleswhile being refracted.

6. a) Newton believed that light was a particle.b) Changing people’s environments through

innovation can leave people feeling not incontrol, especially in cases where a newtechnology has the possibility of replacingpeople in jobs.

c) Accepting theories prematurely hindersprogress, since it discourages research.

8. No, there are no interference patterns becausethe two car headlights are not coherent lightsources and do not form a double slit.

9. Any imperfections are in the order of magni-tude of the wavelengths of light used for theexperiment. This washes out the effect withits own random interference patterns.

10. Sound waves are comparable in wavelengthsize to the openings, increasing the diffractiveeffect. Light waves have much smaller wave-lengths and hence do not show these effects.

11. The resolving power of your eyes restrictsyour ability to distinguish between objects atgreat distances. This is because your pupilsare circular, allowing diffraction to occur.

12. No, diffraction patterns place a limit onresolving power as well as the magnitude ofthe wavelength of light used.

13. Both spectroscopes separate white light intoits colour components, but the prism spectro-scope uses refraction and dispersion while thegrating spectroscope uses diffraction.

14. Continuous spectra involve an extensive rangeof frequencies (example: sunlight spectrum).With line spectra, on the other hand, discretefrequencies are observed (example: moleculargas spectrum).

15. Each piece of a hologram contains the com-plete interference pattern of the object fromwhich the hologram was created, whereas apiece of a normal photograph contains onlylocal information and nothing about the com-plete photograph.

16. Diffraction gratings and interference gratingsare really the same thing. Diffraction gratingsactually use the interference superpositionformula. Gratings show both effects—thosedue to the width of a single opening and thecombination of all the openings.

17. Close spacing in a grating provides strongmutual coupling, increasing the effect of inter-ference. The separation of the maxima increases.

18. Gratings with many slits have high resolvingpower. This means that the individual max-ima become sharper.

19. Yes, because increasing the number of slitsdecreases the slit separation. If the slit separa-tion is reduced beyond what is comparable tothe wavelength of light, no light will getthrough.

20. A single slit has a double central maximum,with the intensities of the maxima droppingoff dramatically with order number. A diffrac-tion grating has a single central maximum andthe intensities do not drop off as dramatically.


21. Diffraction occurs as light enters the pupil.This places a limit on the eye’s resolvingpower. As you move away from the picture,sooner or later you cannot distinguishbetween the dots and they blend together toform a continuous picture.

22. Electrons have a smaller wavelength tothat of visible light, and therefore have ahigher resolution. This also minimizes diffraction. In fact, the beams of electronshave an effective wavelength that is 10�5 timesthat of visible light. This is a 100 000-foldincrease in resolution.

Chapter 121. A photon is a unit particle (as opposed to

wave) of electromagnetic radiation that movesat the speed of light. Its energy is proportionalto the frequency of the radiation.

2. Ultraviolet radiation from the Sun is veryenergetic due to its high frequency. The pho-tons that possess this energy are the cause ofsunburn. These photons are energetic enoughto remove electrons from our body cells, caus-ing a change in our skin biology and in severecases causing cancer.

3. Visual light is mostly in the infrared-visualspectrum. The energy of these photons is notsufficient to damage skin cells.

4. If h � 0, quantization would not exist. Therewould be no energy levels in atoms. Electronsin atoms would therefore not attain any realvalue for energy, resulting in the absence oforbitals in atoms.

5. The electron volt (eV) corresponds to theenergy of an electron at a potential of onevolt. Hence, one electron volt is the energyequalling the charge of an electron multipliedby the potential of one volt: 1 eV � qe 1 V.

6. Wien’s law relates the wavelength of photonsto the temperature of the black body.

7. W0, the work function, is the amount of energyrequired to produce the photoelectric effect in agiven metal. It is the minimum energy requiredto liberate electrons from a metal.

8. Since the photons have detectable linearmomentum, their mass equivalence can becomputed. Momentum is an intrinsic propertyof matter, therefore we can assume that massequivalency is correct.

9. An empirical relationship is a relationshipthat is determined experimentally. It is notbacked up by theory.

10. Determinacy is a condition of a measurementbeing characterized definitely. An example ofan everyday event could be a repetitive meas-urement of the length of a table. Each time themeasurement is made, errors are encountered.If determinacy existed at the macroscopiclevel, we would get the same length everytime.

11. The computation of uncertainties usingHeisenberg’s uncertainty principle yieldsminute values for speed and position. The lim-itations of human perception prevent us fromexperiencing such minute variances at themacroscopic level.

12. Another device besides the STM that operatesusing the principle of quantum tunnelling isthe electron tunnelling transistor, which is anon-off switch that uses the ability of an elec-tron to pass through impenetrable energyobstacles.

13. The energy of an orbital varies as the inversesquare of the radius. Hence, the spectral linesare closer together farther away from thenucleus.

14. a) The peak wavelength emitted by a mercurylamp lies in the visual spectrum. However,this implies that there is a tail in the ultra-violet spectrum. The ultraviolet photonsare energetic enough to damage skin cells.

b) An appropriate shielding that blocks ultra-violet light but allows photons in the visualspectrum to pass through could be used.

15. Consider two particles that have the same deBroglie wavelength � and masses m1 and m2

such that m1 m2. According to de Broglie’s

equation, � � and � � , where v1

and v2 are the velocities of the two particles.

h�m2v2

h�m1v1


Since � is the same for both particles, the fol-lowing equation can be written:

�

This equation can be simplified:m1v1 � m2v2

Since m1 m2, it follows that v2 v1. If themass of the first particle is much greater thanthat of the second particle, the velocity of thesecond particle must be much greater thanthat of the first particle.

16. According to Planck, the energy is quantized.The angular momentum is certainly related tothe energy. Hence, the angular momentumneeds to be quantized as well. To quantize L,Bohr had to quantize both the velocity, v, andthe radius, r.

17. Although the initial and the final speed andthe scatter angles are known, the manner inwhich the actual collision occurs cannot beprecisely predicted, and the exact position ofthe particles during the collision is notknown. Hence, the uncertainty principle isnot violated.

Chapter 131. Your car is in an inertial frame when it is

stopped, or when it is moving at a steadyspeed in a straight line. Your car is in anon-inertial frame when it is accelerating,such as when you are braking, or when youare making a turn.

2. Donovan’s reference frame is inertial becausethe 100-m dash is in a straight line. Leah’sframe is non-inertial because the 400-m ovalrequires her to constantly change direction.

3. No, without reference to the outside world,it would be difficult to determine whetherthe cruise ship was at rest or moving with aconstant velocity.

4. Suppose v � swim speed and w � waterspeed. To swim upstream and back down,it would take a total time of:

� � .

To swim straight across the stream (perpendi-cular to the current) and back, it would take atotal time of:

2� � � .

But v �v2 � w�2�, so

.

Therefore it would take longer to swimupstream and back down than to swim acrossthe stream and back.

5. The Michelson-Morley null result led to thedevelopment of special relativity, a tool neededin the understanding of high-energy physics.

6. Analogous to the Doppler shift of sound, theconstant speed of light in a vacuum, c, requiresthe wavelength of the approaching amber lightto shorten or become more yellowish.

7. In terms of Einstein’s first postulate involvingrelative motion, the two situations are equiva-lent. The same physics occurs whether a mag-net is moved into a stationary coil or a coil ismoved around a stationary magnet.

8. Proper time is the time measured by onewatch between the beginning and the end ofthe experiment. This is the time measured bya watch moving with the muon. The scientistsof Earth would require at least two watches,one at the birth of the muon and the other atits disintegration.

9. The relativity equation for length is

L � L0�1 ��. If v > c, then �1 � � is

negative and L becomes imaginary, which isnot physically reasonable.

10. Since the electrons would have a greater rela-tive velocity than the protons, the spacebetween the electrons would be more con-tracted. As a result, the concentration of elec-trons would exceed that of the protons, andthe wire would seem negatively charged. Forthis reason magnetism is a result of special rel-ativity.

v2

�c2

v2

�c2

2d�v2 � w�2��

v2 � w2

2dv�v2 � w2

2d�v2 � w�2��

v2 � w2

d��v2 � w�2�

2dv�v2 � w2

d�v � w

d�v � w

h�m2v2

h�m1v1


11. No, you have not travelled faster than thespeed of light. Instead, you have measured theEarth�star distance to be contracted and thusit took less time to travel at your speed ofv � 0.98c.

12. No, although time is dilated (events seemlonger at relativistic speeds), it cannot slowto a complete standstill unless v � c.

13. In both cases of the Doppler shift for sound,there is a shift to higher frequencies. Howeverthe physics of sound waves generated by amoving vibrating source colliding with airmolecules and perceived by a stationaryreceiver is different from that of sound createdby a stationary source and perceived by amoving source. For light, the frequency shiftdepends only on the relative speed of thesource and receiver because the speed of lightis always c, according to the second postulate.

14. Only Barb is correct in saying that Phillip’sclock ran slow, because his time was theproper time that was at the beginning and fin-ish of his experiment. If Phillip observed sta-tionary Barb doing a similar experimentbeside the train, he would be correct in sayingthat her clock ran slow.

15. If the charge of an electron depended on itsspeed then the neutrality of atoms would beupset by the motion of electrons within theatoms. Experiments have shown that thecharge on an electron is the same at all speeds.

16. The radius of the orbit becomes smaller as themagnetic field is increased because the radius

is equal to �mqB

v�, where B is the magnetic field

strength.

17. No, because mass dilates as �1 ��, but

density � �vo

mluas

mse

�, and volume contracts as

�1 ��. Therefore, density dilates as

�1 � �.

18. The starlight will pass you at a speed of caccording to the second postulate of specialrelativity.

19. The occupants of the spacecraft would saythat they observed the same things about us,due to relative motion.

20. No, according to the second postulate of spe-cial relativity, the light leaving the recedingmirror travels with speed c relative to you.

21. Tachyons or particles that travel with a speedgreater than c would seem to require infiniteenergy. Experiments do not support their exis-tence.

22. Particle A would have the greater speedbecause its total energy due to mass dilation(mc2) is three times its rest energy, whereasparticle B has a total energy dilated by a factorof only two.

23. Since the ice and the water have the samemass, they have the same total energy(m0c2 � Ek). However, the kinetic energy, Ek,of the water is higher than that of the ice andfor that reason the rest energy of the water isless than that of the ice.

24. If you consider that energy is equivalent tomass (E � mc2), then electromagnetic energyin the form of light could be considered tohave an equivalent mass.

25. A 100-eV electron has a dilated mass accord-ing to:mc2 � m0c2 � qVmc2 � (0.511 � 0.000 100) MeVmc2 � 0.5111 MeVThis means that its mass is less than 0.02%greater than its rest mass. A 100-MeV electronhas a mass equivalent to (0.511 � 100) MeVof energy, which means that its mass is about197 times its rest mass.

26. When we say that the rest mass of a muon is106 MeV/c2, we mean that its rest energy isequivalent to the kinetic energy of an electronaccelerated from rest through 106 million volts.

27. When a particle is travelling at an extremelyhigh speed, say 90% of the speed of light, a lotof energy is needed to increase the particle’svelocity by a few percent. As a result, the

v2

�c2

v2

�c2

v2

�c2


mass of the particle increases by a largeamount. Therefore, it may be more accurate tosay that a particle accelerator increases themass of electrons rather than their speed.

Chapter 141. Every atom of the same element has the same

number of protons, and the number of protonsin the nucleus, Z, determines the chemicalproperties of the atom. However, atoms of dif-ferent isotopes of the same element have dif-ferent numbers of neutrons (and thus differentA values), which results in different physicalproperties such as nuclear stability or decay.

2. Many elements are composed of several natu-rally occurring isotopes, each with a differentatomic mass number, A. The weighted averageof the isotopes’ mass numbers often results ina non-integral value for the atomic mass ofthat element.

3. Each nuclear isotope has a unique total bind-ing energy determined by its nuclear struc-ture. This binding energy is equivalent to themass difference between the nucleus and itsconstituent nucleons (protons and neutrons)according to E � mc2.

4. The missing mass was converted to energyof various forms such as gamma radiationemitted during the formation of the deuteriumatom.

5. Your body, composed of many elements, likelyhas more neutrons than protons, since stableatoms with A > 20 have more neutrons thanprotons.

6. During a nuclear reaction, nucleons may beconverted from one type to another, such asneutrons to protons in beta decay. However,the total nucleon number is conserved orremains constant. On the other hand, variousforms of energy may be absorbed or emitted,resulting in an equivalent change in mass.

7. The average binding energy per nucleon isgreater in the more stable isotopes because itis the “glue” holding the nucleons together, orthe average amount of energy needed to breakthem apart.

8. During alpha decay of a uranium-238 nucleus,

for example, the �NZ

� ratio of the parent nucleus

is �19426

� or about 1.59, and the ratio of the

daughter nucleus, , is or about

1.60. This leads to greater nuclear stability byreducing the electrical repulsion of the pro-tons relative to the nuclear attraction of

nucleons. During beta decay, the �NZ

� ratio of

the parent nucleus, �19426

� or about 1.59, is

greater than the ratio of the daughter nucleus,

, which is or about 1.56.

Although the greater ratio of protons to neu-trons in the daughter tends to increase theelectrical repulsive forces, the beta-decayprocess can lead to greater nuclear stabilitythrough the pairing of previously unpairedneutrons or protons in the nuclear shells.

9. During alpha decay, the daughter nucleus hasa mass, M, that is much larger than the massof the alpha particle, m. Since momentum isconserved, the velocity of the daughternucleus, v, is much smaller than the velocityof the alpha particle, V (Mv � mV).Therefore, the kinetic energy of the alpha par-ticle, 0.5mV2, is much greater than that of thedaughter nucleus, 0.5Mv2.

10. If an alpha particle had enough initial kineticenergy to contact a gold nucleus then anuclear process such as fusion or fission couldoccur, because at that closeness the short-range nuclear force would overpower the electrical force of proton repulsion that isresponsible for scattering.

11. a) p b) � c) � d) � e) �

12. The strong nuclear force differs from the elec-trical force in that: (i) the strong nuclear forceis very short-range, acting over distances ofonly a few femtometres (10�15 m); (ii) thestrong nuclear force is much stronger than theelectrical force over nuclear distances of 1 or2 fm; (iii) the strong nuclear force does

145�93

N � 1�Z � 1

144�90

N� 2�Z � 2


not vary with distance r as �r1

2� as does the

electrical force; and (iv) the strong nuclearforce is attractive only, acting between allnucleons (proton�proton, proton�neutron,and neutron�neutron).

13. The rate of decay of radioactive isotopes wasnot affected by combining them in differentmolecules or by changing the temperature.These changes usually affect the rate of chem-ical reactions, thus radioactivity must befound deeper within the atom (in thenucleus).

14. The nuclear force only binds nucleons thatare neighbours. This short-range energy isproportional to the number of nucleons, A, inthe nucleus. On the other hand, the electricalrepulsion of protons is long-range and actsbetween all proton pairs in the nucleus. Theelectrical energy is therefore proportional toZ2. Repulsion would overcome attraction in alarger nucleus if there were not more neutronsthan protons to keep the forces balanced andthe nucleus stable.

15. Alpha particles are ions, since they are heliumatoms stripped of their electrons.

16. If human life expectancy were a randomprocess like radioactive decay then youwould expect 25% of the population to liveto 152 years. However, this is not the case.As humans age, their expected number ofyears left to live decreases.

17. Carbon-14 undergoing beta decay results inthe daughter isotope nitrogen-14.

18. Industrialization and automobile emissionshave effected changes in our atmosphere suchas global warming and ozone-layer depletion.Such changes in the past 100 years may bealtering the 14C:12C ratio in the air.

19. Potassium salts are rapidly absorbed by braintumours, making them detectable. The shorthalf-life of potassium-42 means that thedosage decays to a safe, insignificant levelquickly. The transmutation to a stable calciumsalt by beta decay is not harmful to the body.

20. Aquatic creatures do not respire or breatheatmospheric gases directly. The 14C:12C ratio inthe ocean is different than in the air.

21. Relics that are more than 60 000 years oldhave lasted more than 10.5 half-lives of car-bon-14. The 14C:12C ratio in these relics isabout 1500 times smaller now and is difficultto determine.

22. The more massive lead atoms scatter the radi-ation particles more effectively than do theless massive water molecules, and may alsopresent a larger “target” for a high-speed elec-tron or alpha particle.

23. Transmutation involves a change in the pro-ton number, Z. This occurs during alpha andbeta decay but does not occur during gammadecay, in which a nucleus merely becomes lessenergized.

24. Alpha particles are more massive than beta orgamma particles and transfer more energy toa molecule of the body during a collision. Thishas a much more disastrous effect upon thecells of the body.

25. Yes, 4.2 MeV of kinetic energy is sufficient foran alpha particle to overcome the electricalrepulsion of the positively charged nuclei(see problem 72) and contact the nitrogen-14nucleus, thus a nuclear interaction or processis possible.

26. The matches are as follows: gases�wind; liq-uids�water; plasmas�fire; and solids�earth.

27. For fission to occur in naturally occurringdeposits of uranium, a minimum concentrationof uranium would be needed in order to sustaina source of slow neutrons necessary to maintainthe fission process. This concentration is notpresent in uranium deposits.

28. The huge inward pull of the Sun’s gravita-tional field confines the solar plasma. Lackingthis huge confining force on our less massiveEarth, scientists instead use strong electro-magnetic fields to confine plasmas.

29. In a fusion reactor, the major problem is tocreate the exact and difficult conditions ofhigh temperatures and plasma concentrations


needed to initiate a fusion process. Themoment these conditions are not met, theprocess stops.

30. The high temperature in fusion means thatthe ions have a very high speed, which allowsthem to approach one another very closelyduring collisions. If the ions’ kinetic energy issufficient to overcome the electrical repulsionof the nuclei, and the nuclei touch, thenfusion is possible.

31. Critical mass in fission involves the existenceof enough fuel so that the fast neutrons emit-ted during fission are slowed and absorbedwithin the fuel itself before they escape. Inthis way the reaction is sustained by a contin-ual source of slow neutrons.

32. Natural uranium is not concentrated enough(“it’s too wet”) to provide the critical massneeded to slow down any fast neutrons (“thespark needed”) and capture them to create asustainable reaction.

33. A bubble chamber is superheated almost to thepoint of instability. When a charged particlepasses through, it triggers the formation of afine stream of bubbles in its wake. Neutral par-ticles such as neutrons carry no electric fieldand leave no visible tracks in the chamber.

34. High-energy accelerators provide ions withenough kinetic energy that each ion’s totalenergy, E � mc2, becomes many times greaterthan its rest mass. In a collision there is aprobability that this energy could be convertedto a massive elementary particle.

35. In the high-energy accelerator at UBC, thestrong nuclear force, acting over a very briefperiod of time (10�23 s) during collisions, pro-duces pi-mesons or pions.

36. The weak nuclear force is usually maskedby the stronger (by a factor of 103) electro-magnetic force or (by a factor of 105) strongnuclear force, unless these forces are forbid-den. Any process involving the neutrino, suchas beta decay, involves the weak force. Theneutrino reacts rarely, or weakly, with otherelementary particles over a longer time span(10�8 s) compared with the shorter interaction

times (10�23 s) of the strong nuclear force.37. Gravitational interactions are the weakest of

the four forces. At elementary particle dis-tances the gravitational force is 10�40 times asgreat as the strong nuclear force. For this rea-son the graviton or messenger of the gravita-tional force is extremely difficult to detect.

38. The weak force is 10�3 times as great as theelectromagnetic force at elementary particledistances. The weak force is mainly involvedin neutrino interactions or processes wherethe electric and strong forces are forbidden.The exchange bosons of the weak force areW and Z bosons of mass 80 GeV/c2 and91 GeV/c2 respectively, as compared to thephotons of the electric force. The range of theweak force is about 10�18 m, compared toinfinity for electromagnetism. The weak forceacts on both leptons (particles not affectedby the strong force, such as electrons) andhadrons (particles affected by the strongforce), whereas electromagnetism acts onlyon charged particles.

39. Strong nuclear processes are the fastest (orshortest), with a lifetime of about 10�23 s.

40. A high-energy particle travels close to thespeed of light, c � 3 108 m/s. Thus, in astrong nuclear interaction of 10�23 s, thecloud-chamber track would be 3 10�15 m,too small to measure.

41. No, a heavier, unstable version of theelectron, the tauon or tau, �, has a massof 1777 MeV/c2, greater than the mass ofthe proton or neutron (931.5 MeV/c2).

42. Since gluons, the quanta of the quark forcefield, carry one colour and one anti-colour,there should be 32 � 9 possible combinations(rR, rB, rG, bR, bB, bG, gR, gB, and gG).However, the three colour-neutral gluons (rR,bB, and gG) must be handled differentlybecause of what are known as symmetry laws.For this reason only two possible neutral cou-plings exist, not three, making a total of eightcolour gluons to act as the source ofquark�quark interactions.



Chapter 116. a) Distance is a scalar, so your total distance

travelled would be 10 � 20 m � 200 m.b) Displacement is a vector, and since you

end up 0 m from where you started, yourdisplacement is 0 m.

17. a) Since distance is a scalar, his total distancetravelled would be:�d � |15 m [E]| � |6.0 m [W]| � |2.0 m [E]|�d � 15 m � 6.0 m � 2.0 m�d � 23 m

b) Since displacement is a vector, his total displacement would be:�d�� 15 m [E] � 6.0 m [E] � 2.0 m [E]�d�� 11 m [E]

18. g � � � �

g � 32 ft/s2

19. a) 10 knots � � � �

�

10 knots � 18.5 km/h

b) from a),10 knots � � �

10 knots � 5.14 m/s20. To find the number of centimetres in one light

year, simply express the speed of light in cen-timetres per year:

� � �

� �

� 9.5 � 1017 cm/yTherefore, there are 9.5 � 1017 cm in one lightyear.

21. Catwoman:

vavg �

vavg �

vavg � 6.5 m/sRobin:

vavg �

vavg �

vavg � 7.1 m/s22. a) The speed of the sweep second hand at the

6 o’clock position is the same as anywhereelse on the clock:

vavg �

vavg �

vavg �

vavg � 2.1 � 10�3 m/sb) The velocity of the second hand at the 6

o’clock position is 2.1 � 10�3 m/s [left]because the velocity is always tangent tothe face and perpendicular to the hand.

23. a) The time it would take the shopper to walkup the moving escalator is:

t �

where vt is the sum of the velocity of theescalator and the woman:

vt � �

vt �

Therefore, t �

t � 5.2 s

d�

��12230ds

��

23d�120 s

d�8.0 s

d�15 s

�d�vt

2�(0.02 m)��

60 s

2�r��t

�d��t

200 m�28.0 s

�d��t

100 m�15.4 s

�d��t

365 d�

1 y24 h�1 d

60 min�

1 h

60 s�1 min

100 cm�

1 m3.0 � 108 m��

1 s

2.78 � 10�4 h��

1 s

1000 m�

1 km18.5 km�

1 h

1.0 � 10�5 km��

1 cm2.54 cm�

1 in

12 in�1 ft

6080 ft�

1 nm10 nm�

1 h

0.083 ft�

1 in0.394 in�

1 cm100 cm�

1 m9.8 m�

1 s2

Solut ions to End-of-chapter Problems 87

PART 3 Solutions to End-of-chapter Problems

b) In this case, vt is equal to her walkingspeed minus the speed of the escalator:

vt � �

vt �

Since this velocity is positive, she couldwalk down the escalator. Also, intuitively,if the escalator takes 15 s to go the samedistance that the woman can in 8 s, thenshe is faster and will therefore make itdown the escalator. To find how long itwill take her, solve for time:

t �

t � 17 s24. Because the rabbit accelerates at a constant rate,

v2 � v1 � a�tv2 � (0.5 m/s) � (1.5 m/s2)(3.0 s)v2 � 5.0 m/s

25. Mach 1 � 332 m/sMach 2 � 2(332 m/s)Mach 2 � 664 m/sBecause the jet accelerates at a constant rate:v2 � v1 � a�t

�t �

�t �

�t � 6.6 s

26. a��

a��

a��

a�� 400 m/s2 [E]27. Let t be the time when the two friends meet.

Let x be the distance travelled by the secondfriend to reach the first friend.For the first friend:

v �

�d � v�t

and�d � 50 � xTherefore:

x � 50 � v�tFor the second friend:

x � v1�t �

x �

Now we set these two expressions for x equalto each other and solve for time:

50 � v�t �

�t2 � �t � 100 � 0

�t �

�t � 9.5 s28. a) v2 � v1 � a�t

and v1 equals zero, sov2 � a�t

�t �

�t �

�t � 6.0 sb) To find Batman’s distance travelled, we

must first convert his acceleration intostandard SI units:

� �

� 2.78 m/s2

Now:

�d �

�d �

�d � 50 mc) Robin’s speed in SI units is:

� �

� 16.7 m/sWhen Batman catches up with Robin,Robin will have travelled:�d � v1�t�d � (16.7 m/s)�trelative to Batman’s initial position.

1000 m�

1 km1 h

�3600 s

60 km�

1 h

(2.78 m/s)(6.0 s)2

��2

a�t2

�2

1000 m�

1 km1 h

�3600 s

10 km�1 hs

60 km/h��10 km/h/s

v2�a

� 1 �1 � 4(� � 100�)��

2

a�t2

�2

a�t2

�2

a�t2

�2

�d��t

(25 m/s [E] � 15 m/s [E])��

0.10 s

(25 m/s [E] � 15 m/s [W])��

0.10 s

(v��2 � v��1)��t

(664 m/s � 332 m/s)��

50 m/s2

(v2 � v1)�a

d�

��1270d

s��

7d�120 s

d�15 s

d�8.0 s

88 Solut ions to End-of-chapter Problems

Similarly, Batman will have travelled:

�d �

�d � (1.39 m/s2)�t2

Setting these two expressions equal andsolving for �t gives:( 16.7 m/s)�t � (1.39 m/s2)�t2 � 0

�t � 12.0 s29. If the child catches the truck, she will have

travelled 20 m � �d, and the truck will havetravelled �d in the same amount of time, �t.For the truck:

�d �

�d � (0.5 m/s2)�t2

For the child:

vavg �

�d � (4.0 m/s)�t � 20 mSetting these two equations equal and solvingfor �t gives:(4.0 m/s)�t � 20 m � (0.5 m/s2)�t2

�t2 � 8�t � 40 � 0This expression has no real roots, thereforethe child will not catch the truck.

30. a) After ten minutes, the runner has gone(4000 m � 800 m) � 3200 m, at a speed of:

vavg �

vavg �

vavg � 5.33 m/sIf she then accelerates at 0.40 m/s2 for thefinal 800 m, it will take:

�d � v1�t �

800 m � (5.33 m/s)�t � (0.20 m/s2)�t2

�t �

�t � 51 sb) Since she had two minutes to go, she will

finish under her desired time.31. We can use the information given to find the

speed of the flower pot at the top of the win-dow, and then use the speed to find the height

above the window from which the pot musthave been dropped.Since the pot accelerates at a constant rate of9.8 m/s2, we can write:

�d � v1�t �

v1 � �

v1 � �

v1 � 8.5 m/sNow we can find the distance above the window:v1

2 � vo2 � 2a�d

�d �

�d �

�d � 3.7 m32. a) The only force acting on the ball while it is

falling is that of gravity, so its accelerationis 9.8 m/s2 downward.

b) Since the ball is being constantly accelerateddownward, it cannot slow down.

c) �d � v1�t �

�d � (8.0 m/s)(0.25 s) �

�d � 2.3 m

33. �d � v1�t �

(�4.0 m) � (4.0 m/s)�t �

(4.9 m/s2)�t2 � (4.0 m/s)�t � 4.0 m � 0

�t �

�t � 1.4 s34. For the first stone, the distance it falls before

reaching the second stone is:

h � �d �

�d � (4.9 m/s2)�t2 � h

a�t2

�2

4 �16 � 4�(4.9)(��4)��

9.8

(�9.8 m/s2)�t2

��2

a�t2

�2

(9.8 m/s2)(0.25 s)2

��2

a�t2

�2

((8.5 m/s)2 � (0 m/s)2)��

2(9.8 m/s2)

(v12 � vo

2)��

2a

(9.8 m/s2)(0.20 s)��

219 m�0.20 s

a�t�

2�d��t

a�t2

�2

�5.33 �(5.33)�2 � 4(�0.20)(��800)��

2(0.20)

a�t2

�2

3200 m�

600 s

�d��t

(�d � 20 m)��

�t

a�t2

�2

a�t2

�2


For the second stone, its distance travelled isgiven by:

�d � vi�t �

�d � vi�t � 4.9�t2

Setting these expressions equal to each otherand solving for �t gives:4.9�t2 � h � vi�t � 4.9�t2

�t �

35. a) Because the jackrabbit’s distance vs. time ischanging at a constant rate during seg-ments B, C, and D, he is undergoing uni-form motion at these times.

b) Because the jackrabbit’s distance vs. time isnot changing at a constant rate during seg-ment A but is increasing exponentially, hisvelocity vs. time must be increasing at aconstant rate, and he is undergoing uni-form acceleration during this segment.

c) The average velocity during segment B is:

vavg �

vavg �

vavg � 5 m/sDuring segment C:

vavg �

vavg �

vavg � 0 m/sDuring segment D:

vavg �

vavg �

vavg � �10 m/s

d) vavg �

vavg �

vavg � 9.1 m/s

e) Since the jackrabbit’s displacement isnot changing between 20 s and 30 s, hisvelocity over this interval, and at 25 s,is 0 m/s.

f) The jackrabbit is running in the oppositedirection.

g) The jackrabbit’s displacement is:�d � 100 m � 50 m � 0 m � 120 m�d � 30 m

36. a) The car’s acceleration for each segmentcan be found by taking the slope of thegraph during that segment:During segment A:

a1 �

a1 �

a1 � 1 m/s2

During segment B:

a2 �

a2 �

a2 � 2 m/s2

During segment C:

a3 �

a3 �

a3 � �2 m/s2

b) The car is slowing down, or decelerating.c) To find the distance travelled by the car,

we must find the area under the graph,which can be approximated by the sum ofrectangles and triangles:

A: d1 �

d1 �

d1 � 12.5 m

B: d2 � � v1�t

d2 � �

(5 m/s)(9 s � 5 s)d2 � 35 m

(12.5 m/s � 5 m/s)(9 s � 5 s)��

2

�v�t�

2

(5 m/s � 0 m/s)(5 s � 0 s)��

2

�v�t�

2

(1 m/s � 13 m/s)��

(15 s � 9 s)

�v��t

(13 m/s � 5 m/s)��

(9 s � 5 s)

�v��t

(5 m/s � 0 m/s)��

(5 s � 0 s)

�v��t

150 m � 0 m��17.5 s � 1.0 s

�d��t

� 50 m � 150 m��

50 s � 30 s

�d��t

150 m � 150 m��

30 s � 20 s

�d��t

150 m � 100 m��

20 s � 10 s

�d��t

h�vi

a�t2

�2


C: d3 � � v2�t

d3 � �

(1 m/s)(15 s � 9 s)d3 � 40.5 m

dtotal � d1� d2� d3

dtotal � 12.5 m � 35 m � 40.5 mdtotal � 88 m

NOTE: The solutions to problem 37 are based onthe velocity axis of the graph reading 60, 40, 20, 0,�20, �40, �60.37. a) Because the skateboarder has a positive

velocity between 0 and 5 seconds, this por-tion of the graph must describe his upwardmotion.

b) Since the skateboarder has a negativevelocity from 5 to 10 seconds on the graph,he must be descending during this portionof the graph.

c) The skateboarder is undergoing uniformacceleration.

d) The skateboarder is at rest when his veloc-ity equals zero, at t � 5 s. When his veloc-ity equals zero, he is at the top of the sideof the swimming pool, or ground level.

e) The skateboarder’s acceleration can befound from the slope of the graph. Itshould be equal to g:

a �

a �

a � �10 m/s2

38. a) At t � 4.0 s, each Stooge’s acceleration is:Curly:

a �

a � 0 m/s2

Larry:

a �

a �

a � 2.5 m/s2

Moe:

a �

a �

a � 5.0 m/s2

b) To find their distance travelled, we takethe area under the graph for each Stooge:Curly:d � �v�td � (25 m/s)(4.0 s)d � 100 mLarry:

d �

d �

d � 20 mMoe:

d �

d �

d � 40 mc) Since Curly is travelling at a constant

velocity:

v �

�t �

�t �

�t � 24 sLarry accelerates for the first 18 s of therace. His distance travelled at this point is:

�d �

�d �

�d � 405 m

(45 m/s)(18 s)��

2

v�t�

2

(600 m)�(25 m/s)

�d�

v

�d��t

(20 m/s)(4.0 s)��

2

�v�t�

2

(10 m/s)(4.0 s)��

2

�v�t�

2

(20 m/s � 0 m/s)��

(4.0 s � 0 s)

�v��t

(10 m/s � 0 m/s)��

(4.0 s � 0 s)

�v��t

�v��t

(�50 m/s � 50 m/s)��

(10 s � 0 s)

�v��t

(1 m/s � 12.5 m/s)(15 s � 9 s)��

2

�v�t�

2


Then, travelling at a constant velocity of45 m/s, he will traverse the last 195 m in:

�t �

�t �

�t � 4.3 sHis total time is: 18 s � 4.3 s � 22.3 sMoe accelerates for the first 8 s of the race.His distance travelled in this time is:

�d �

�d �

�d � 160 mThen, travelling at a constant velocity of40 m/s, he will traverse the last 440 m in:

�t �

�t �

�t � 11 sHis total time is: 8 s � 11 s � 19.0 sTherefore, Moe wins with the fastest timeof 19.0 s.

39.

40.

41.

42.

43. a) The gravitational force down-ward is equal in magnitude tothe tension in the elevatorcable.

b) The gravitational force down-ward is equal in magnitude tothe tension in the elevator cable.

Elevator

F��g

F��tension

Elevator

F��g

F��tension

Textbook F��f

F��g

F��n

Baby

F��support

F��g

Box #2F��f F��2,1

F��g

F��n

Box #1F��1,2

F��applied

F��n

F��f

F��g

Ball

F��g

F��bat

(440 m)�(40 m/s)

�d�

v

(40 m/s)(8.0 s)��

2

v�t�

2

(195 m)�(45 m/s)

�d�

v


c) The acceleration downward is9.8 m/s2.

d) The gravitational force down-ward is equal in magnitude tothe normal force upward.

e) The gravitationalforce is equal inmagnitude to thenormal force, andthe force due to thecatapult must belarge in order toaccelerate the jet.

44. The driver’s initial velocity is the same as thatof the car:

v1 � � � �

v1 � 13.9 m/sHis final velocity is zero, and the distance hetravels is 0.6 m:v2

2 � v12 � 2a�d

a �

a �

a � �161 m/s2

a � 16.4 g

45. Fnet � ma, and

a � , so

Fnet �

Fnet �

Fnet � 6.2 � 104 N46. Fnet � ma

a �

a �

a � �2.0 m/s2

We also know that:

�t �

�t �

�t � 0.25 s47. Since

F � ma(6.0 m/s2) andF � mb(8.0 m/s2), it follows that:ma(6.0 m/s2) � mb(8.0 m/s2)

mb � 0.75ma

If the same force were used to accelerate bothmasses together, we would have:F � (ma � mb)aF � (ma � 0.75ma)aF � 1.75ama

But we already know thatF � ma(6.0 m/s2),so we now have:ma(6.0 m/s2) � 1.75ama

a � 3.4 m/s2

48. The force applied by the hammer is given by:Fnet � maBut we also know that:

�d �

a �2�d��t2

a�t2

�2

(0 m/s � 0.5 m/s)��

�2.0 m/s2

(v2 � v1)�a

� 400 N��

200 kg

Fnet�m

(10 000 kg)[(150 m/s)2 � (100 m�s)2]��

2(1000 m)

m(v22 � v1

2)��

2�d

(v22 � v1

2)��

2�d

((0 m/s)2 � (13.9 m/s)2)��

2(0.6)

(v22 � v1

2)��

2�d

1000 m�

km1 min�60 s

1 h�60 min

50 km�

1 h

F-14

F��n

F��catapult

F��g

Car

F��n

F��g

Elevator

F��g


Therefore:

Fnet �

Fnet �

Fnet � 4.7 NTherefore, the force applied to the nail by thehammer is 4.7 N and the force applied to thehammer by the nail is �4.7 N.

49. The force due to the cows on the plate is:Fnet � (m1 � m2 � m3 � m4 � m5)gFnet � 5(200 kg)(�9.8 m/s2)Fnet � �9800 NFrom Newton’s third law, the steel plate exertsa force of 9800 N upward.

50. a) The acceleration of the water skiers can befound using:Fnet � ma

a �

a �

a � 39.2 m/s2

b) The force applied by the first skier on thesecond two skiers is equal to the sum oftheir masses times their acceleration:Fnet � mtaFnet � (75 kg � 80 kg)(39.2 m/s2)Fnet � 6.1 � 103 NThe force applied by the third skier on thefirst two skiers is equal to his mass timeshis acceleration:Fnet � mtaFnet � (75 kg)(39.2 m/s2)Fnet � 2.9 � 103 NFrom Newton’s third law, the forcesapplied by the second skier on the firstand third skiers are 6.1 � 103 N and 2.9 � 103 N, respectively.

51. The forces in the vertical direction are balanced:Fg � mgFg � Fn, therefore:Ff � �Fn

Ff � �mgFf � (0.16)(2.0 kg)(�9.8 m/s2)Ff � �3.1 N

52. Fnet � Ff

Fnet � maFnet � �Fn

We also know that a � , therefore:

� �Fn

� �mg

�d �

�d �

�d � 6.8 � 10�2 m53. F��net � F��engine � F��friction

Fengine � Fnet � Ffriction

Fnet � maFfriction � �Fn

Fengine � ma � �Fn

Fengine � ma � �mg

a � , therefore:

Fengine � � �mg

Fengine � �

(0.3)(800 kg)(�9.8 m/s2)Fengine � 4.2 � 103 N

54. Fg �

Fg �

Fg � 6.0 � 10�6 NTheir acceleration would be:

a �

a �

a � 2.0 � 10�11 m/s2

(6.0 � 10�6 N)��(300 000 kg)

Fg�m

(6.67 � 10�11 N m2/kg2)(300 000 kg)2

��(1000 m)2

Gm1m2�r 2

(800 kg)(27.8 m/s � 13.9 m/s)��

6.0 s

m(v2 � v1)��t

(v2 � v1)��t

[(0 m/s)2 � (2.0 m/s)2]��

2(3.0)(�9.8 m/s2)

(v22 � v1

2)��

2�g

m(v22 � v1

2)��

2�d

m(v22 � v1

2)��

2�d

(v22 � v1

2)��

2�d

10 000 N��(75 kg � 80 kg � 100 kg)

Fnet��(m1 � m2 � m3)

2(1.8 kg)(0.013 m)��

(0.10 s)2

2m�d�

�t2


55. If the mass of Earth where doubled, the accel-eration due to gravity would be:

Fg �

m1 g �

g �

g �

g � �19.6 m/s2

56. The net gravitational force on planet Z wouldbe equal to the sum of the gravitational forcescaused by each planet:F��net � F��x � F��y

Fnet � �

Fnet � (6.67 � 10�11 N m2/kg2)(5.0 � 1024 kg)

� �

�Fnet � 6.16 � 1017 N

57. The astronaut’s weight, or his mass times theacceleration due to gravity, is:

W �

W �

W � 894 N

Chapter 214. a) Horizontal: dx � (25 km) cos 20°

d��x � 23 km [E]Vertical: dy � (25 km) sin 20°

d��y � 8.6 km [N]b) Horizontal: Fx � (10 N) sin 30°

F��x � 5.0 N [E]Vertical: Fy � (10 N) cos 30°

F��y � 8.7 N [S]c) Horizontal: ax � (30 m/s) cos 45°

a��x � 21 m/s2 [W]Vertical: ay � (30 m/s) sin 45°

a��y � 21 m/s2 [S]d) Horizontal: px � (42 kg·m/s) sin 3°

p��x � 2.2 kg·m/s [W]Vertical: py � (42 kg·m/s) cos 3°

p��y � 42 kg·m/s [N]15. a) l � (10 m) cos 40°

l � 7.7 mb) h � (10 m) sin 40°

h � 6.4 m16. Horizontal: ax � (4.0 m/s2) cos 35°

ax � 3.3 m/s2

Vertical: ay � (�4.0 m/s2) sin 35°ay � �2.3 m/s2

17. Adding by components:

�dx � (�2.0 km) � (3.0 km) cos 20°

�d��x � 4.8 km [W]

�dy � (3.0 km) sin 20°

�d��y � 1.0 km [N]

�d � �(4.8 k�m)2�(�1.0 km�)2�

�d � 4.9 km

� � tan�1 � �� 12°

Therefore, d�� 4.9 km [W12°N].

18. |vi| � �(10 m�/s)2 �� (20 m�/s)2�|vi| � 22 m/s

� � tan�1 � �� 63°Therefore, v��i � 22 m/s inclined 63° to the hor-izontal.

20 m/s�10 m/s

1.0 km�4.8 km

(6.67 � 10�11 N m2/kg2)(100 kg)(5.98 � 1024 kg)��

(6.38 � 106 m � 3.0 � 105 m)2

GmAmE�rAE

2

4.0 � 1024 kg��(5.0 � 1010 m)2

3.0 � 1024 kg��(6.0 � 1010 m � 5.0 � 1010 m)2

Gmzmy�rzy

2

Gmzmx�rzx

2

2(6.67 � 10�11 N m2/kg2)(5.97 � 1024 kg)��

(6.38 � 106 m)2

G2mE�r 2

Gm12mE�r 2

Gm12mE�r 2


19. Adding by components:

�dx � (50 cm) sin 35° � (100 cm) cos 15°

�dx � �67.9 cm

�dy � (20 cm) � (50 cm) cos 35° �

(100 cm) sin 15°

�dy � �46.8 cm

�d � �(67.9�cm)2��(46.8�cm)2�

�d � 82 cm

� � tan�1 � �� 55°

Therefore, �d�� 82 cm [S55°W].

20. �v�� v��f � v��i

�v�� 28 m/s [N30°W] � 30 m/s [S]�v�� 28 m/s [N30°W] � 30 m/s [N]Adding the vectors by components, �v�� 56 m/s [N15°W]

21. �v�� v��f � v��i

�v�� 1.8 m/s [N30°E] � 2.0 m/s [S30°E]�v�� 1.8 m/s [N30°E] � 2.0 m/s [N30°W]Adding the vectors by components, �v�� 3.3 m/s [N2°W]

a��

a��

a�� 33 m/s2 [N2°W]22. a) The current velocity has no effect on the

vertical component of the swimmer’s veloc-ity, which is needed for crossing the river.Therefore:

t �

t �

t � 0.44 hb) The current velocity determines how far

the swimmer travels downstream, there-fore:dd � (vc)(t)dd � (0.50 km/h)(0.44 h)dd � 0.22 km

c) vg � �vs2 � v�c

2�vg � �(1.8 k�m/h)2� � (0.�5 km/�h)2�vg � 1.9 km/h

tan � �

� � tan�1 � �� 16°

The ground velocity is v��g � 1.9 km/h[N16°E].

23. a) In order to go north, his ground velocitymust be north.

Since vs and vc are known,

sin � �

� � sin�1 � �� 16°

The swimmer must swim [N16°W] inorder to go straight north.

b) vg � �vs2 � v�c

2�vg � �(1.8 k�m/h)2� � (0.�5 km/�h)2�vg � 1.7 km/hHis ground velocity is v��g �1.7 km/h [N].

c) t �

t �

t � 0.46 h

0.8 km��1.7 km/h

d�vg

0.5 km/h��1.8 km/h

vc�vs

θ

v��g

v��c

v��s

0.5 km/h��1.8 km/h

vc�vs

0.80 km��1.8 km/h

d�vs

3.3 m/s [N2°W]��

0.10 s

�v��

t

67.9 cm�46.8 cm


24. The time it takes the sandwich to reach theroad is:

t �

t �

t � 5.0 sThe distance of the pick-up truck when thesandwich is released is:dt � (vt)(t)dt � (60 km/h)(5.0 s)dt � (17 m/s)(5.0 s)dt � 83 m

25.

cos � �

� � cos�1 � �� 82°

The pilot must fly [N82°E] or [E7.7°N].26.

Use the sine law to find �:

�

� � sin�1 � �� 6.8°

Use the sum of the interior angles of a triangleto find �:� � � � � � 180°

� � 180° � � � �

� � 180° � 135° � 6.8°� � 38°

The ship’s required heading is [N38°E].

27. a) Since the velocities are given relative to thedeck, they are the velocities relative to theship.Walking towards stern: v�� 0.5 m/s [S];walking towards port, v�� 0.5 m/s [W].

b) � �

v��pw � v��ps � v��sw

Walking towards stern:v��pw � v��ps � v��sw

v��pw � 0.5 m/s [S] � 10 km/h [N]v��pw � � 0.5 m/s [N] � 2.78 m/s [N]v��pw � 2.3 m/s [N]Walking towards port:v��pw � v��ps � v��sw

v��pw � 0.5 m/s [W] � 2.78 m/s [N]vpw � �vps

2 ��vsw2�

vpw � �(0.5 m�/s)2 �� (2.78� m/s)2�vpw � 2.8 m/s

tan � �

� � tan�1 � �� 10°

Walking towards stern, v�� 2.3 m/s [N];walking towards port, v�� 2.8 m/s [N10°W].

28. a) vf �

t �

t �

t � 1.2 sTo reach the pail, the quarterback must be1.2 s away from reaching the garbage pail,therefore:dqg � (vq)(t)dqg � (4.0 m/s)(1.2 s)dqg � 4.8 mThe quarterback must release the ball4.8 m in advance.

b) 1.2 s as calculated in part a.c) vg � �vf

2 � v�q2�

vg � �(5.0 m�/s)2 �� (4.0 m�/s)2�vg � 6.4 m/s

6.0 m�5.0 m/s

dg�vf

dg�t

0.5 m/s��2.78 m/s

vps�vsw

velocity ofship relativeto water

velocity ofpass. relativeto ship

velocity ofpass. relativeto water

vc sin ��

vs

sin ��

vs

sin ��

vc

θ

β

γ

v��s

v��c

v��g

45°

20 km/h��150 km/h

vw�vh

θv��h

v��g

v��w

10 m�2.0 m/s

dp�vs


tan � �

� � tan�1 � �� 51°

The ground velocity is v��g � 6.4 m/s[E51°N].

29. a) In order for the football to reach the garbagepail, the football’s ground velocity must bepointing north at the time of release.

cos � � �vv

q

f�

� � cos�1 � �� 37°

The ball must be thrown [W37°N].b) Calculate the magnitude of v��g:

vf2 � vg

2 � vq2

vg2 � vf

2 � vq2

vg � �vf2 � v�q

2�vg � �(5.0 m�/s)2 �� (4.0 m�/s)2�vg � 3.0 m/s

vg �

t �

t �

t � 3.3 sc) The ball is thrown such that its direction is

north.The ground velocity is v��g � 3.0 m/s [N].

30. The time it takes the ball to reach the ground is:

h � viyt � �

12

�ayt2

�10 m � (0 m/s)t � �12

�(�9.8 m/s2)t2

t � 1.4 sThe horizontal distance travelled in 1.4 s is:

dx � vixt � �

12

�axt2

dx � (3.0 m/s)(1.4 s) � �12

�(0 m/s2)(1.4 s)2

dx � 4.2 mThe friend must be 4.2 m away to catch theball at ground level.

31. a) Find the time it takes the rock to reach theground:

dx � vixt � �

12

�axt2

20.0 m � (10.0 m/s)t � �12

�(0 m/s2)t2

t � 2.00 sFind the height of the water tower:

h � viyt � �

12

�ayt2

h � (0 m/s)(2.00 s)� �12

�(�9.8 m/s2)(2.00 s)2

h � �19.6 mb) In the horizontal direction,

vfx� vix

� axtvfx

� 10.0 m/s � (0 m/s2)(2.00 s)vfx

� 10.0 m/sIn the vertical direction,vfy

� viy� ayt

vfy� 0 m/s � (�9.8 m/s2)(2.00 s)

vfy� �19.6 m/s

vf � �vfx

2 ��vfy

2�vf � �(10.0�m/s)2�� (19�.6 m/s�)2�vf � 22.0 m/s

tan � �

� � tan�1 � �� 63.0°

The rock’s final velocity is 22.0 m/s, 63°below the horizontal.

32. Find the time it takes the mail to reach the second building:

dx � vixt � �

12

�axt2

100 m � [(20 m/s) cos 15°]t � �12

�(0 m/s2)t2

t � 5.2 sFind the drop in height during the 5.2 s:

�h � viyt � �

12

�ayt2

�h � [(20.0 m/s) sin 15°](5.2 s) �

�12

�(�9.8 m/s2)(5.2 s)2

�h � 26.9 m � 132.5 m�h � �105 m

19.6 m/s��10.0 m/s

vfy�vfx

10 m�3.0 m/s

d�vg

d�t

4.0 m/s�5.0 m/s

5.0 m/s�4.0 m/s

vf�vg


Find the height of the second building:h2nd building � h1st building � �hh2nd building � 200 m � 105 mh2nd building � 95 mThe second building is 95 m high.

33. a) h � viyt � �

12

�ayt2

�1.3 m � (0 m/s)t � �12

�(�9.8 m/s2)t2

t � 0.52 sb) The cup lands at the tourist’s feet, since

both the cup of coffee and tourist are notmoving horizontally relative to the train.

c) dx � (vtrain)(t)dx � (180 km/h)(0.52 s)dx � (50 m/s)(0.52 s)dx � 26 mThe train is 26 m closer to Montreal.

34. Find the time it takes the Humvee to dropdown to the other ramp:

h � viyt � �

12

�ayt2

Since both ramps are the same height, h � 0 m.

0 m � [(30 m/s) sin 20°]t �

�12

�(�9.8 m/s2)t2

[(�30 m/s) sin 20°]t � �12

�(�9.8 m/s2)t2

t � 2.1 sFind the maximum horizontal distance theHumvee can travel in 2.1 s:

dx � vixt � �

12

�axt2

dx � [(30 m/s) cos 20°](2.1 s) �

�12

�(0 m/s2)(2.1 s)2

dx � 59 mThe maximum width of the pool is 59 m.

35. Find the time required to reach maximumheight:

h � viyt � �

12

�ayt2

h � (vi sin �)t � �12

�ayt2 (eq. 1)

Since the ball’s motion is symmetrical, it willtake twice the time for the soccer ball to reachthe ground:

0 m � (vi sin �)2t � �12

�ay(2t)2

�ay2t2 � (vi sin �)2t�ayt � vi sin �

ay � (eq. 2)

For the range, R:

R � vix2t � �

12

�ax(2t)2

R � vix2t � �

12

�(0 m/s2)(2t)2

R � vix2t

R � (vi cos �)2t

t � (eq. 3)

Substitute equation 1 into equation 2:

h � (vi sin �)t � �12

�� t2

h � (vi sin �)t � �12

�(vi sin �)t

h � �12

�(vi sin �)t (eq. 4)

Substitute equation 3 into equation 4,

h � �12

�(vi sin �)� �h � �

14

�� R

h � �14

�(tan �)R

h � 0.25R tan �36. If the ball clears the 3.0-m wall 130 m

from home plate, then the ball rises(3.0 m � 1.3 m) � 1.7 m during this time.Thus, for the vertical height:

h � viyt � �

12

�ayt2

1.7 m � vi (sin 45°)t � �12

�(�9.8 m/s2)t2 (eq. 1)

Find the time it takes the ball to clear the wall:

dx � vixt � �

12

�axt2

130 m � vi (cos 45°)t

t � (eq. 2)130 m

��vi cos 45°

sin ��cos �

R�2vi cos �

�vi sin ��

t

R�2vi cos �

�vi sin ��

t


Substitute equation 2 into equation 1:

1.7 m � (vi sin 45°)� � �

�12

�(�9.8 m/s2)� �2

1.7 m � (tan 45°)(130 m) �

(4.9 m/s2)� �vi � 36 m/s

The player strikes the ball at 36 m/s, 45°above the horizontal.

37. a) Fnet � �(30 N�)2 � (1�0 N)2�Fnet � 32 N

� � tan�1 � �� 72°So F��net � 32 N [N72°E]

b) Horizontal components:

�Fx � �(60 N) sin 40°

�F��x � 38.6 N [W]

Vertical components:

�Fy � (60 N) cos 40° � 80 N

�F��y � 34.0 N [S]

F��net � 51 N [S49°W]

c) Horizontal components:

�Fx � (50 N) cos 60° � 10 N

�F��x � 15 N [E]

Vertical components:

�Fy � (50 N) sin 60° � 60 N

�F��y � 16.7 N [S]

F��net � 22 N [S42°E]38. a) F��net � F��a � F��f

The sum of the x components is:F��ax

� F��1x� F��2x

� F��3x

F��ax� (100 N) cos 20° [W] �

(200 N) cos 40° [E]F��ax

� 59 N [E]The sum of the y components is:F��ay

� F��1y� F��2y

� F��3y

F��ay� (100 N) sin 20° [N] �

(200 N) sin 40° [S] � 300 N [S]F��ay

� 394 N [S]

Fa � �Fax

2 �� Fay

2�Fa � �(59 N�)2 � (3�94 N)�2�Fa � 399 N

tan � �

� � tan�1 � �� 8.5°

F��a � 399 N [S8.5°E]Ff � �kFn

Ff � �kmgFf � (0.10)(300 kg)(9.8 m/s2)Ff � 294 NF��f � 294 N [N8.5°W]F��net � F��a � F��f

F��net � 399 N [S8.5°E] � 294 N [N8.5°W]F��net � 399 N [S8.5°E] � 294 N [S8.5°E]F��net � 105 N [S8.5°E]The net force is F��net � 105 N [S8.5°E]

b) F��net � ma��

a��

a��

a�� 0.35 m/s2 [S8.5°E]39. F��net � F��applied force in the x direction�F��kinetic friction

F��net � F��ax� F��k

Find the kinetic frictional force, Fk:Fk � �kFn

Fk � (0.30)(Fg � Fa sin 50°)Fk � (0.30)[(20 kg)(9.8 m/s2) �

(100 N) sin 50°]Fk � 36 N

Fnet � (100 N) cos 50° � 36 NFnet � 28 NFnet � ma

28 N � ma

a �

a � 1.4 m/s2

28 N�20 kg

105 N [S8.5°E]��

300 kg

F��net�m

394 N�59 N

Fax�Fay

30 N�10 N

33 800 m��

vi2

130 m��vi cos 45°

130 m��vi cos 45°


40. The hockey stick provides the only force onthe puck, therefore it is the net force acting onthe puck:Fs � Fnet

F��net � ma��

a��

a��

a�� 1200 m/s2 [N25°E]Find v��f:

a��

a��t � v��f � v��i

v��f � a��t � v��i

v��f � (1200 m/s2 [N25°E])(0.20 s) � 12 m/s [S]v��f � 240 m/s [N25°E] � 12 m/s [S]The vector sum of the x components is:v��fx � (240 m/s) sin 25° [E]v��fx � 101 m/s [E]The vector sum of the y components is:v��fy � (240 m/s) cos 25° [N] � 12 m/s [N]v��fy � 206 m/s [N]vf � �Ffx

2 ��Ffy

2�vf � �(101 m�/s)2 �� (206�m/s)2�vf � 229 m/s

tan � �

� � tan�1 � �� 26°

The final velocity is v��f � 229 m/s [N26°E].41. a) Fk � �kFn

Fk � (0.50)(100 kg)(9.8 m/s2)Fk � 4.9 � 102 N

b) The frictional force is the only force actingon the baseball player, therefore it is alsothe net force.Fnet � Fk

ma � Fk

a �

a � 4.9 m/s2

Find vi:

a �

at � �vi

vi � �(�4.9 m/s2)(1.3 s)vi � 6.4 m/s

42. Fnet � Fk

ma � �k Fn

ma � �kmga ��k ga � (0.3)(�9.8 m/s2)a � �2.94 m/s2

Find distance, d:vf

2 � vi2 � 2ad

�vi2 � 2ad

d �

d �

d � 0.68 mThe key will slide 0.68 m across the dresser.

43. Fnet � Fa � Fk

The horizontal acceleration of 1.0 m/s2 is thenet acceleration of the mop, therefore:

Fnet � max

max � Fax� �kFn

max � (30 N) cos 45° �

[(0.1)(Fg � Fa sin 45°)]max � 21.2 N �

[(0.1)(mg � 21.2 N)]max � 19.09 N � 0.1mg

(1.0 m/s2)m � 19.09 N � 0.1mgm(1.0 m/s2 � 0.1g) � 19.09 N

m(1.98 m/s2) � 19.09 Nm � 9.6 kg

44. Let � be the angle of the inclined plane whenthe box starts to slide.At this angle,Fs � �sFn

Fs � (0.35)(mg cos �) (eq. 1)Fx � mg sin � (eq. 2)

�(2.0 m/s)2

��2(�2.94 m/s2)

�vi2

�2a

vf � vi�t

490 N�100 kg

206 m/s�101 m/s

vfx�vfy

v��f � v��i�t

300 N [N25°E]��

0.25 kg

F��net�m


Set equation 1 equal to equation 2:(0.35)(mg cos �) � mg sin �

0.35 �

tan � � 0.35� � tan�1 (0.35)� � 19°

The minimum angle required is 19°.45. a) The acceleration for child 1:

Fnet � Fx

m1a1 � m1 g sin �a1 � g sin �a1 � (9.8 m/s2) sin 30°a1 � 4.9 m/s2

The acceleration for child 2:Fnet � Fx

m2a2 � m2 g sin �a2 � g sin �a2 � 4.9 m/s2

Both children accelerate downhill at4.9 m/s2

.

b) They reach the bottom at the same time.46. a) Fnet � Fx � Fk

ma � mg sin � � �kFn

ma � mg sin � � �k(mg cos �)a � g sin � � �k g cos �a � (9.8 m/s2) sin 25° �

(0.45)(9.8 m/s2)cos 25°a � 0.14 m/s2

The acceleration of the box is 0.14 m/s2.b) vf

2 � vi2 � 2ad

vf2 � 2(0.14 m/s2)(200 m)

vf � 7.6 m/sThe box reaches the bottom of the hill at7.6 m/s2.

c) a �

t � �va

f�

t �

t � 53 sIt takes the box 53 s to reach the bottom ofthe hill.

47. Find his final speed, vf, at the bottom of theramp by first finding his acceleration:Fnet � Fx

ma � mg sin �a � g sin �a � (9.8 m/s2) sin 35°a � 5.6 m/s2

His final speed at the bottom of the ramp is:vf

2 � vi2 � 2ad

vf2 � 2(5.6 m/s2)(50 m)

vf � 23.6 m/svf will be the initial speed, vi2, for the horizon-tal distance to the wall of snow.Find the deceleration caused by the snow:Fnet � Fk

ma � �kFn

ma � (0.50)(�9.8 m/s2)ma � �4.9 m/s2

Find the distance Boom-Boom will go into thewall of snow:

vf2 � vi

2 � 2ad0 � vi

2 � 2ad�vi

2 � 2ad

d �

d �

d � 57 mBoom-Boom will go 57 m into the wall ofsnow.

48. Find the net force on Spot, then solve for thenet acceleration:Fnet � Fr � Fx

Fnet � 2000 N � mg sin �ma � 2000 N � (250 kg)(9.8 m/s2)(sin 20°)ma � 2000 N � 838 Nma � 1162 N

a �

a � 4.6 m/s2

1162 N�250 kg

�(23.6 m/s)2

��2(�4.9 m/s2)

�vi2

�2a

7.6 m/s��0.14 m/s2

vf � vi�t

sin ��cos �


Find time, t:

d � vit � �12

�at2

d � �12

�at2

250 m � �12

�(4.6 m/s2)t2

t2 � 108 s2

t � 10 s49. a) (a) For m1:

Fnet1 � TT � m1a (eq. 1)

For m2:Fnet2 � Fg � Tm2a � m2 g � T (eq. 2)

Substitute equation 1 into equation 2:m2a � m2 g � m1a

(m1 � m2)a � m2 g(40 kg)a � (20 kg)(9.8 m/s2)

a�� 4.9 m/s2 [left]For tension T, substitute accelerationinto equation 1:T � m1aT � (20 kg)(4.9 m/s2)T � 98 N

(b) Assume the system moves towards m3:For m1:

Fnet1 � T1 � F1g

m1a � T1 � m1 g (eq. 1)For m2:

Fnet2 � T2 � T1

m2a � T2 � T1 (eq. 2)For m3:

Fnet3 � F3g � T2

m3a � m3 g � T2 (eq. 3)Add equations 1, 2, and 3:

m1a � T1 � m1 g (eq. 1)m2a � T2 � T1 (eq. 2)m3a � m3 g � T2 (eq. 3)

(m1 � m2 � m3)a � m3 g � m1 g(10 kg � 10 kg � 30 kg)a

� (30 kg)(9.8 m/s2) �(10 kg)(9.8 m/s2)

(50 kg)a � 196 Na�� 3.9 m/s2 [right]

Find T1:m1a � T1 � m1 g (eq. 1)

T1 � (10 kg)(3.9 m/s2) �(10 kg)(9.8 m/s2)

T1 � 137 NFind T2:

m3a � m3 g � T2 (eq. 3)T2 � (30 kg)(9.8 m/s2) �

(30 kg)(3.9 m/s2)T2 � 176 N

(c) For m1:Fnet1 � T � Fx

m1a � T � mg sin � (eq. 1)For m2:

Fnet2 � F2g � Tm2a � m2 g � T (eq. 2)

Add equations 1 and 2:m1a � T � mg sin � (eq. 1)m2a � m2 g � T (eq. 2)

(m1 � m2)a � m2 g � m1 g sin �(25 kg)a � (15 kg)(9.8 m/s2) �

(10 kg)(9.8 m/s2) sin 25°a�� 4.2 m/s2 [right]

For tension T, substitute accelerationinto equation 2:

m2a � m2 g � TT � (15 kg)(9.8 m/s2) �

(15 kg)(4.2 m/s2)T � 84 N

b) (a) For m1:Fnet1 � T � Fk

m1a � T � �km1 g (eq. 1)For m2:

Fnet2 � Fg � Tm2a � m2 g � T (eq. 2)

Add equations 1 and 2:m1a � T � �km1 g (eq. 1)m2a � m2 g � T (eq. 2)

m1a � m2a � m2 g � �km1 ga(m1 � m2) � g(m2 � �km1)

(20 kg � 20 kg)a � 9.8 m/s2[20 kg �0.2(20 kg)]

a�� 3.9 m/s2 [left]


For tension T, substitute accelerationinto equation 2:m2a � m2 g � T

T � (20 kg)(9.8 m/s2) �(20 kg)(3.9 m/s2)

T � 118 N(b) Assume the system moves towards m3:

For m1:Fnet1 � T1 � F1g

m1a � T1 � m1 g (eq. 1)For m2:

Fnet2 � T2 � T1 � Fk

m2a � T2 � T1 � �km2 g (eq. 2)For m3:


m3a � m3 g � T2 (eq. 3)Add equations 1, 2, and 3:

m1a � T1 � m1 g (eq. 1)m2a � T2 � T1 �

�km2 g (eq. 2)m3a � m3 g � T2 (eq. 3)

(m1 � m2 � m3)a � m3 g � �km2 g �

m1 g(10 kg � 10 kg � 30 kg)a

� 9.8 m/s2[30 kg �0.2(10 kg) �10 kg]

a�� 3.5 m/s2 [right]Find T1:

m1a� T1 � m1 g (eq. 1)T1 � (10 kg)(3.5 m/s2) � (10 kg)

(9.8 m/s2)T1 � 133 N

Find T2:m3a� m3 g � T2 (eq. 3)

T2 � (30 kg)(9.8 m/s2) �(30 kg)(3.5 m/s2)

T2 � 188 N(c) For m1:

Fnet1 � T � Fx � Fk

m1a � T � m1 g sin � �

�km1 g cos � (eq. 1)For m2:

Fnet2 � F2g � Tm2a � m2 g � T (eq. 2)

Add equations 1 and 2:m1a � T � m1 g sin � �

�km1 g cos � (eq. 1)m2a � m2 g � T (eq. 2)

(m1 � m2)a � m2 g � m1 g sin � �

�km1 g cos �(25 kg)a � (9.8 m/s2)[15 kg �

(10 kg) sin 25° �

0.2(10 kg) cos 25°]a�� 3.5 m/s2 [right]

For tension T, substitute accelerationinto equation 2:m2a � m2 g � T

T � (15 kg)(9.8 m/s2) �(15 kg)(3.5 m/s2)

T � 94 N50. For m1:

Fnet1 � T � Ff1

m1a � T � �kFn

m1a � T � �km1 g (eq. 1)For m2:


m2a � m2 g � T1 (eq. 2)Add equations 1 and 2:

m1a � T � �km1 g (eq. 1)m2a � m2 g � T (eq. 2)

(m1 � m2)a � m2 g � �km1 g (eq. 3)(9.0 kg)a � (4.0 kg)(9.8 m/s2) �

(0.10)(5.0 kg)(9.8 m/s2)a�� 3.8 m/s2 [right]

51. For the system to be NOT moving, the acceler-ation of the whole system must be 0.Using equation 3:

(m1 � m2)a � m2 g � �km1 g (eq. 3)0 � m2 g � �km1 g

�km1 g � m2 g�k(5.0 kg) � 4.0 kg

�k � 0.8052. First find the system’s acceleration:

For Tarzana:FnetTA � TmTAa � T (eq. 1)

For Tarzan:FnetTZ � FTZg � TmTZa � mTZ g � T (eq. 2)


Add equations 1 and 2:mTAa � T (eq. 1)mTZa � mTZ g � T (eq. 2)

(mTA � mTZ)a � mTZ g(65 kg � 80 kg)a � (80 kg)(9.8 m/s2)

a �

a � 5.4 m/s2

To find time t:

d � vit � �12

�at2

15 m � �12

�at2

� t2

t � t � 2.4 s

53. ac � Assuming ac is a constant,

T � a) If the radius is doubled, the period

increases by a factor of �2�.b) If the radius is halved, the period decreases

by a factor of �2�.

54. a) ac �

ac �

ac � 78 m/s2

b) The clothes do not fly towards the centrebecause the wall of the drum applies thenormal force that provides the centripetalforce. When the clothes are not in contactwith the wall, there is no force acting onthem. The clothes have inertia and wouldcontinue moving at a constant velocity tan-gential to the drum. The centripetal forceacts to constantly change the direction ofthis velocity.

55. ac �

T � 365 days � 3.15 � 104 s

ac �

ac � 6.0 � 10�3 m/s2

56. Fc � Ff

mac � �Fn

� �mg

v � ��gr�v � 21 m/s

It is not necessary to know the mass.57. Vertically: Fn cos � � mac

Fn �

Horizontally: Fc � Fn sin �mac � Fn sin �

� � sin � � mac

g tan � �

v � �rg tan� 25°�v � 19 m/s

58. Fc � Fg

mac � mg

g �

v � �gr�v � 9.9 m/s

59. a) T � mgT � (0.5 kg)gT � 4.9 N

b) T � mg �

T � � mg

T � �

(0.5 kg)(9.8 m/s2)T � 9.7 N

60. Maximum tension occurs when the mass is atits lowest position. Tension acts upward, andgravity acts downward. The differencebetween these forces is the centripetal force:

Tmax � mg �

Tmax � � mg

Tmax � � (2.0 kg)(9.8 m/s2)

Tmax � 49 N

(2.0 kg)(6.6 m/s)2

��3.0 m

mv2

�r

mv2

�r

(0.5 kg)(2.4 m/s)2

��(0.6 m)

mv2

�r

mv2

�r

v2

�r

v2

�r

mg�cos �

mg�cos �

mv2

�r

4�2(1.5 � 1011 m)��

(3.15 � 107 s)

4�2r�

T 2

4�2(0.35 m)��

(0.42 s)2

4�2r�

T 2

4�2r�

ac

4�2r�

T2

30 m�5.4 m/s2

30 m�

a

(80 kg)(9.8 m/s2)��(65 kg � 80 kg)


The tension is minimized when the mass is atthe top of its arc. Tension and gravity both actdownward, and their sum is the centripetalforce:

Tmin � mg �

Tmin � � mg

Tmin � �

(2.0 kg)(9.8 m/s2)Tmin � 9.4 N

61. a) Fnet � maFn � mg � m(9g)

Fn � 9mg � mgFn � 10mgFn � 5.9 � 103 N

b) ac �

9g �

r �

r �

r � 95 m62. a) G � 6.67 � 10�11 N·m2/kg2,

T � 365 days � 3.15 � 107 s

� mE� �mS �

mS �

mS � 2.0 � 1030 kg

b) Density of the Sun � �mV

�

�

� 1.4 � 103 kg/m3

mEarth � 5.98 � 1024 kg

Density of Earth �

� 5.5 � 103 kg/m3

The Sun is about �14

� as dense as Earth.

63. On mass 2:

Fc � m2� �T2 � m2� �T2 � m2� �On mass 1:

Fc � m1� �T1 � T2 � m1� �T1 � m1� � � m2� �T1 � � �(m1L1 � m2(L1 � L2))

4�2

�T 2

4�2(L1 � L2)��T 2

4�2L1�T 2

4�2r�

T 2

4�2r�

T2

4�2(L1 � L2)��T2

4�2r�

T2

4�2r�

T2

5.98 � 1024 kg��

�43

��r3

2.0 � 1030 kg��

�43

��r3

4�2(1.5 � 1011 m)3

��(6.67 � 10�11 N·m2/kg2)(3.15 � 107 s)2

4�2r3

�GT 2

4�2r�

T 2

GmEmS�r 2

(91.67 m/s)2

��9(9.8 m/s2)

v2

�9g

v2

�r

v2

�r

(2.0 kg)(6.6 m/s)2

��3.0 m

mv2

�r

mv2

�r


Chapter 3

21.

sin 30° �

T �

T �

T �

T � 196 N

22. tan � �

Fs �

Fs �

Fs � 169.7 NFs � 170 N

23.

T��1 � T��2 � T��

cos 30° �

T �

� �T � ��

(cos 30°)T � 566 N

24.

sin 30° �

m �

m �

m � 128 kg25.

cos 12° �

Tcable �

Tcable �

Tcable � 5009.5 NTcable � 5.01 � 103 N

tan 12° �

Frope � mg tan 12°Frope � (500 kg)(9.8 N/kg) tan 12°Frope � 1.04 � 103 N

Frope�mg

(500 kg)(9.8 N/kg)��

cos 12°

mg�cos 12°

mg�Tcable

mg12°

500 kg

Trope

Trope

Tcable

Fg

12°

Tcable

(2500 N) sin 30°��

9.8 N/kg

Fs sin 30°��

g

mg�Fs

mg

mg30°

60°

60°

m

2500 N Fs = 2500 N

Fs = 2500 N

60°

T

T

(100 kg)(9.8 N/kg)��

2

��F2

g��(cos 30°)

��F2

g��

T

T2T1

Fg—2Fg

30°30°

30°

60°

30°

T2

T1

98 N�tan 30°

Fg�tan �

Fg�Fs

(10 kg)(9.8 N/kg)��

sin 30°

mg�sin 30°

Fg�sin 30°

Fg�T

T

Tcable

Fg

Fg

Fstrut

Fs

30°

flower pot

30°


26. a)

� � 0.63Fapp � Ff � 0Fapp � Ff

Fapp � �Fn

Fapp � �mgFapp � 0.63(100 kg)(9.8 N/kg)Fapp � 617.4 N

b)

Using similar triangles, find T first:T2 � (mg)2 � Fapp

2

T � �[(250�kg)(9.�8 N/kg�)]2 � (�617.4�N)2�T � 2526.6 NT � 2.53 � 103 N

�

�d �

�d �

�d � 2.4 m

27.

tan � �

tan � � 0.12� � 6.8°

sin � �

T �

T �

T � 3.59 � 103 NThe rope pulls with a force of 3.59 � 103 N.

28.

tan � �

tan � � 0.058� � 3.3°

0.52 m�9.0 m

18.0 m

0.52 m

9.0 m

θ

Bird

mBg

T T

425 N�sin 6.8°

Fapp�sin �

Fapp�T

1.5 m��

��25.20 m��

Fapp

Fn425 N

1.5 m

car

25.0 mT

θ

1.5 m

25.0 m———2

θθ

(617.4 N)(10 m)��

2526.6 N

FappL�

T

Fapp�T

�d�L

mg

mg

250 kg10 mFapp

Fapp = 617 N

T

Tθ

d

θ

250kg

L = 10 m

d

100 kgFf

Fn

Fapp

mg


sin � �

mB �

mB �

mB � 1.1 kg29.

T � (5.0 kg)(9.8 N/kg)T � 49 N

F��app � T��1 � T��2

cos 40° �

Fapp � 2(T cos 40°)Fapp � 2mg cos 40°Fapp � 2(5.0 kg)(9.8 N/kg) cos 40°Fapp � 75 N [left]

30.

T��h � F��f � 0With left taken to be the positive direction,Th � Ff � 0

Th � Ff

Th � �Fn

Th �

From Pythagoras’ theorem:

T2 � Th2 � � �

2

T2 � �2� �2

� � �2

T2 � � �2

(�2 � 1)

T � � ��(�2 ��1)�

From similar triangles:

�

�

x �ThL�

T

Th�T

x�L

Th�T

��2x

��

��L2

��

mg�2

mg�2

mg�2

mg�2

mg�2

�mg�

2

Pulley

mg

FfTh

Th

Tv

T

T

T

L–2

x–2

L–2

��F2app��

�T

40°

40°

Fapp

T1T2

T1

T2

Leg

40°80° 40°

T1

mLg

T2

2(90 N) sin 3.3°��

9.8 N/kg

2T sin ��

g

��m

2Bg��

�T

mBg

T

T

θθ = 3.3°


Substituting for Th and T,

x �

x �

x �

31. a)

centre of mass � ?��net � 0With clockwise as the positive rotation,�1 � �2 � 0

�1 � �2

r1m1 g sin � � r2m2 g sin �

r1 � r2� �r1 � r2� �r1 �

But r2 � r1 � rT

3r1 � rT � r1

4r1 � rT

r1 �

r1 �

r1 � 0.5 mThe centre of mass is 0.5 m from m1 and1.5 m from m2.

b)

T�� ?If up is positive,T � mTgT � (4.0 kg)(9.8 N/kg)T�� 39.2 N [up]

32. �T � 0The pivot is the left support.�1 � 0��2 � ��Board � ��Duck � 0

�2 � ��B � �D

�2 � �rBFgB � rDFgD

�2 � �rBmBg � rDmDg�2 � �(2.0 m)(50 kg)

(9.8 N/kg) � (4.0 m)(8.5 kg)(9.8 N/kg)

�2 � �1313.2 N/m

F2 �

F2 � �1641.5 NF��2 � 1.6 � 103 N [up]For F1:FT � 0With down as positive,0 � �F1 � F2 � FB � FD

F1 � FB � FD � F2

F1 � (mBg) � (mDg) � F2

F1 � (50 kg)(9.8 N/kg) �(8.5 kg)(9.8 N/kg) � 1.6 � 103 N

F1 � �1068.2 NF��1 � 1.1 � 103 N [down]andF��2 � 1.6 � 103 N [up]

� 1313.2 Nm��

0.8 m

mT = m1 + m2

mTg

T

2.0 m�

4

rT�4

r2�3

m2�m1

m2g�m1g

3 kg 1 kg

+

1 2

2.0 m

P

�L��2 � 1�

�L��2 � 1�

��m2

gL��

��m2

g��2 � 1�


33.

xcm �

xcm �

xcm � 1.5 m [right]

ycm �

ycm �

ycm � 0.75 m [up]Centre of mass � 1.5 m [right], 0.75 m [up]

34.

Let F1 be the pivot.�T � 0

��2 � ��3 � ��L � 0With clockwise as positive,

2�23 � �L � 0

r23� m�g � rLmg

r23 �

r23 �

r23 �

r23 � 3.75 m

x � 5.0 m � 3.75 mx � 1.25 m

35. �T � 0��man � ��L(left) � ��L(right) � ��rock � 0With clockwise as the positive direction ofrotation,

0 � ��man � �L(left) � �L(right) � �rock

�rock � �man � �L(left) � �L(right)

rrockmrock g sin � � rmanmman g sin � �

rL(left)mL(left) g sin � �

rL(right)mL(right) g sin �rrockmrock � [(1.90 m)(86 kg)] �

�� (2.0 kg) ��

�� (2.0 kg) ��rrockmrock � 163.4 kg·m � 1.504 kg·m �

0.104 kg·mrrockmrock � 164.8 kg·m

mrock �

mrock � 329.6 kgmrock � 3.3 � 102 kg

36. a)

�T � 0��1 � ��2 � ��3 � ��TL � ��TR � 0With clockwise as the positive rotation,��1 � �2 � �3 � �TL � �TR � 0

�3 � �2 � �1

r3m3g � r2m2 g � r1m1 g

r3m3 � �� (27 kg)� �

�� (17 kg)�r3m3 � 51.3 kg·m � 32.3 kg·mr3m3 � 19 kg·m

r3 �

r3 � 0.95 m

19 kg·m�

20 kg

3.8 m�

2

3.8 m�

2

17 kg 20 kg

3.8 kg

27 kg1 3 2

+

164.8 kg·m��

0.50 m

0.50 m�2.40 m

0.5 m�

2

1.90 m�2.40 m

1.90 m�

2

15.0 m�

4

3��5.02

m��

��2

3rL�2

2�3

5.0 m

x

x

2.5 mF23 F1

Fg

P

0.5 m � 1.0 m��

2

y1 � y2�2

0.5 m � 2.5 m��

2

x1 � x2�2

X2 = 2.5 m

y 1 =

1.0

m

X1 = 0.5 m

y 2 =

0.5

m


The third child of mass 20 kg must sit0.95 m from the centre of the teeter-totterand on the same side as the 17.0-kg child.

b) No, the mass of the teeter-totter does notmatter.

37.

Let F2 be pivot.�net � 0

��1 � ��B � ��C � 0With clockwise as the positive rotation,�1 � �p � �c � 0

�1 � �p � �c

r1F1 � rpFgp � rcFgc

F1 �

F1 �

F1 � 29.4 NBut Fnet � 0F��1 � F��gB � F��gC � F��2 � 0With up as the positive direction,0 � F1 � FgB � FgC � F2

F2 � FgB � FgC � F1

F2 � mB g � mC g � F1

F2 � (2.0 kg)(9.8 N/kg) � (5.0 kg)(9.8 N/kg)� 29.4 N

F2 � 39.2 NThe man farthest from the cement bag (F1)lifts with 29.4 N and the second man liftswith 39.2 N of force.

38. Take front two and back two legs as singlesupports.��net � 0 with front legs as pivot��D � ��Back � 0

Let clockwise be positive.�D � �Back � 0

�Back � �D

rBFB � rDFgD

rBFB � rDmD grBFB � (0.30 m)(30 kg)(9.8 N/kg)rBFB � 88.2 N·m

FB �

FB � 88.2 NFB � 8.8 � 101 N

But F��net � 0F��F � F��D � F��B � 0Let up be positive.0 � FF � FD � FB

FF � FD � FB

FF � mDg � FB

FF � (30 kg)(9.8 N/kg) � 88.2 NFF � 205.8 NFF � 2.1 � 102 NFront legs: 1.05 � 102 N each; back legs:4.4 � 101 N each (each divided by 2).

39. a)

F��net � 0F��T � F��D � 0Taking up to be positive,0 � FT � FD

FT � mD gFT � (20 kg)(9.8 N/kg)F��T � 196 N [up]

20 kg

2.4 m

0.8 m

C of m

P

88.2 N·m��

1.0 m

��2.52

m��(2.0 kg)(9.8 N�kg)� � [(2.5 m � 1.5 m)(5.0 kg)(9.8 N/kg)]

��2.5 m

rpFgp � rcFgc��r1

mp = 2.0 kg

5.0kg +

1.5 m

P

2.5 m

F2F1


b)

� � tan�1 � �� 18.4°Assume the upper hinge is the pivot.

��B � ��door � 0��B � �door � 0

�B � �door

rBFB sin �B � rDmD g sin �D

FB �

FB �

F��B � 34.2 N [out horizontally]40.

�p � 90° � 65°�p � 25°Choose bottom as pivot.

��net � 0��wall � ��p � 0Taking right (horizontally) as positive,

�wall � �p � 0�wall � �p

rwFw sin �w � rpmp g sin �p

Fw �

Fw �

F��w � 272.6 N [horizontal]Fw � 2.7 � 102 N

But Fh(bottom) � �Fh(top) so 2.7 � 102 N isrequired to keep the ladder from sliding.

41. a)

F��net � 0F��app-h � F��f � 0

Taking the direction of force application tobe positive,Fapp-h � Ff

Fapp-h � �Fn

Fapp-h � �mgFapp-h � 0.42(75 kg)(9.8 N/kg)F��app-h � 308.7 N [horizontally]F��app-h � 3.1 � 102 N [horizontally]

b)

Just to the tip the box,��net � 0

��a � ��box � 0Taking the direction of force application tobe positive,�a � �box � 0

�a � �box

Take bottom corner as pivot.

tan �a �

�a � 58°raFa sin �a � rboxmbox g sin �box

ra �

ra �

ra � 1.40 mBut:h � ra sin 58°h � 1.2 m

�(0.8 m�)2 � (0�.5 m)2�(75 kg)(9.8 N/kg) sin (90° � 58°)��

(308.7 N) sin 58°

rboxmbox g sin �box��Fa sin �a

��1.62

m��

�

��1.02

m��

θboxθa

P

P

C of m

75 kg

1.0 m

1.6 m

[(7.0 m � 1.2 m)(72 kg)(9.8 N/kg)(sin 25°)]��

(7.0 m) sin 65°

rpmp g sin �p��rw sin �w

7.0 m

+

P

θp

65°

1.2 m

72 kg

(1.26 m)(20 kg)(9.8 N/kg) sin 18.4°��

(2.4 m) sin (90° � 18.4°)

rDmD g sin �D��rB sin �B

0.4 m�1.2 m

θD1.2 m

0.4 mC of m

P

rD


42.

��net � 0��muscle � ��arm � ��water � 0With clockwise as the direction of positiverotation,��m � �a � �w � 0

�m � �a � �w

rmFm sin � � ramag sin � � rwmwg sin �

Fm �

Fm �

Fm � 780.1 NF��m � 7.8 � 102 N [up]

43.

The total of all three torques must be equiva-lent to the total torque through the centre ofmass.

��cm � ��ua � ��fa � ��hand

rcmmTg � ruamua g � rfamfa g � rhandmhand g

rcm �

rcm �

rcm � 0.29 m from shoulder44.

Let the contact point of F2 be the pivot P.��1 � ��w � 0With clockwise being the positive torquedirection,�1 � �w � 0

�1 � �w

r1F1 � rwmg

F1 �

F1 �

F1 � 1911 NF��1 � 1.9 � 103 N [up]

F��net � 0F��1 � F��2 � F��g � 0

With up taken to be the positive direction,F2 � �F1 � Fg

F2 � �1911 N � (65.0 kg)(9.8 N/kg)F2 � �2548 NF��2 � 2.5 � 103 N [down]

45.

Use pivot P as the point of contact of F1.��net � 0

��F2 � ��F � 0With clockwise taken to be the positive torquedirection,�F2 � �F � 0

�F2 � �F

rF2F2 � rFF

F2 �

F2 �

F2 � 0.25 N

(0.01 m)(0.5 N)��

0.02 m

rFF�rF2

0.01 m

0.02 m

P

F = 0.5 N

F2

+

(0.12 m)(65 kg)(9.8 N/kg)��

0.04 m

rwmg�

r1

12 cm4.0 cm

WP

F1 +

(0.15 m)(1.9 kg) � (0.40 m)(1.2 kg) � (0.60 m)(0.4 kg)��

3.5 kg

ruamua � rfamfa � rhandmhand��mT

1.9 kg 1.2 kg

0.15 m

0.40 m

0.60 m

0.4 kg

(0.16 m)(3.0 kg)(9.8 N/kg) � (0.35 m)(10 kg)(9.8 N/kg)��

0.050 m

ramag � rwmwg��

rm

P

+

5.0

16 cm

10 kg

35 cm

cm


F��net � 0F�� F��

1� F��2 � 0

With right taken to be the positive direction,F1 � F � F2

F1 � 0.5 N � 0.25 NF1 � 0.75 NF��1 � 0.75 N [left], and F��2 � 0.25 N [right]

46.

Set P at elbow joint.��net � 0

��T � ��arm � ��sp � 0With clockwise taken to be the positive torquedirection,��T � �arm � �sp � 0

�T � �arm � �sp

rTFT � rarmFg(arm) � rspFg(sp)

FT �

FT �

FT � 9.5 � 102 N47.

tan � �

� � 26°The tipping angle is 26° from the horizontal.

48.

tan � �

hcm �

hcm �

hcm � 0.8660 mBut:h � 2hcm

h � 2(0.8660 m)h � 1.73 mNOTE: The solution to problem 49 is basedon the pivot point of the glass being at the cor-ner of the base.

49.

tan � �

� � 21.8°

sin � �

x � (0.14 m � 0.050 m) sin 21.8°x � 0.033 md � x � rd � 0.033 m � 0.020 md � 0.053 m

x��h � 0.050 m

0.020 m�0.050 m

θ

θ

θ

0.14 m

0.050 m

0.020 m

x

d

��1.020 m��

��tan 30°

��L2

��tan �

��L2

��hcm

θ

θ

1.00 m1.00 m

h

hcm L—2

0.3 m�0.6 m

θ

θ

0.6 m

0.3 m

(0.11 m)(2.7 kg)(9.8 N/kg) � (0.280 m)(7.25 kg)(9.8 N/kg)��

0.024 m

rarmmarmg � rspmspg��

rT

28 cm

2.4 cm7.25 kg

11 cm

P

FT

C of m

+


50.

tan � �

tan � �

� � 26.6°51.

F��s � F��g � 0With up taken to be the positive direction,Fs � mgBut:Fs � kxSo:kx � mg

k �

k �

k � 1.6 � 103 N/m52.

Use hinge as pivot.��net � 0

��s � ��m � 0With clockwise taken to be the positive torquedirection,

��s � �m � 0�s � �m

rsFs sin �s � rmFm sin 90°

Fs �

Fs �

Fs � 752.5 NBut:Fs � kx

k �

k �

k � 1.88 � 104 N/m53.

x1 � �(0.50�m)2 �� (0.50� m)2�x1 � 0.7071 mx2 � �(0.50�m)2 �� (1.50� m)2�x2 � 1.58 m

tan � �

� � 71.56°∆x � 1.58 m � 0.7071 m∆x � 0.874 m

bar

T

Fg

Tθθ

1.50 m�0.50 m

x10.50 m

0.50 m

45°

x2 1.5 m

0.50 m

θ

752.5 N��4.0 � 10�2 m

Fs�x

(1.0 m)(10.0 kg)(9.8 N/kg)��

(0.75 m) sin 10°

rmmm g�rs sin �s

0.75 m

Fs

Fgm

P

+

10 kg

1.0 m

10°

(3.0 kg)(9.8 N/kg)��

1.8 � 10�2 m

mg�

x

nails

Fs

Fg

��2.52

m��

�2.5 m

��ba

2se��

�hcm

θ

θ2.5 m

2.5 m


T � Fs

T � k∆xT � (1.5 � 102 N/m)(0.874 m)T � 1.311 � 102 N

sin � �

m �

m �

m � 25.4 kg54. L � 20 m

r � 2.0 � 10�3 mLimit FL � 6.0 � 107 N/m2

a) Stress � �AF

�

F � A(Stress)F � �r2(Stress)F � �(2.0 � 10�3 m)2(6.0 � 107 N/m2)F � 753.6 NF � 7.5 � 102 N

b) E for Al is:EAl � 70 � 109 N/m2

E �

E �

�L �

�L �

�L � 0.017 m�L � 1.7 � 10�2 m

55. a) A � 0.1 m2

Stress � �AF

�

Stress �

Stress � 9.8 � 103 N/m2

Strain Eiron � 100 � 109 N/m2

E �

Strain �

Strain �

Strain � 9.8 � 10�8

b) ∆L � ?L � 2.0 m∆L � L(Strain)∆L � (2.0 m)(9.8 � 10�8)∆L � 1.96 � 10�7 m∆L � 2.0 � 10�7 m

c) Maximum stress is 17 � 107 N/m2.Fmax � Stress(A)Fmax � (17 � 107 N/m2)(0.1 m2)Fmax � 1.7 � 107 Nmg � Fmax

m �

m � 1.7 � 106 kg56. Maximum stress for femur is 13 � 107 N/m2.

A � 6.40 � 10�4 m2

Stress � �AF

�

Fmax � A(Stress)Fmax � (6.40 � 10�4 m2)(13 � 107 N/m2)Fmax � 8.32 � 104 N

57. Fc � 200 NA � 1 � 10�5 m2

L � 0.38 mE � 15 � 109 N/m2

E �

�L �

�L �

�L � 5.067 � 10�6 m

(200 N)(0.38 m)��(1 � 10�3 m2)(15 � 109 N/m2)

FL�AE

��AF

��

��

LL��

1.7 � 107 N��

9.8 N/kg

9.8 � 103 N/m2

��100 � 109 N/m2

Stress�

E

Stress�Strain

(100 kg)(9.8 N/kg)��

0.1 m2

(20 m)(6.0 � 107 N/m2)��

70 � 109 N/m2

L��AF

��

E

��AF

��

��

LL��

Stress�Strain

2(1.311 � 102 N) sin 71.56°��

9.8 N/kg

2T sin ��

g

��m2

g��

�T

T

mg—2 T

θ


k �

k �

k � 3.95 � 107 N/m58.

Gsteel � 80 � 109 N/m2

� � rFC � 2�rC � 2�(20 m)C � 125.6 m

�

� �

� � 5.73°

cos � �

R � �L �

R � �L �

R � �L � 20.1005 m�L � 20.1005 m � R�L � 20.1005 m � 20 m�L � 0.1005 m

G �

Arod � �(0.01 m)2

F �

F �

F � 1 262 920 NF � 1.26 � 106 N�rod � rF sin ��rod � 2.0 m(1 262 920 N) sin 90°�rod � 2.52 � 106 N·mThe torque on rod is 2.5 � 106 N·m.

59. Stress is 10% of Tmax.Stress � 0.10(50 � 107 N/m2)Stress � 5.0 � 107 N/m2

a) A � ?

Stress �

A �

A �

A �

A � 1.96 � 10�3 m2

r � r � r � 0.025 mr � 2.5 � 10�2 m

b) a � 2.0 m/s2

Fnet � Fapp � mgFnet � ma � mgFnet � m(a � g)Esteel � 200 � 109 N/m2

E �

Strain �

Strain �

Strain �

Strain �

Strain � 3.01 � 10�4

60. �L � ?Epine � 10 � 109 N/m2

L � 3.0 mA � (10 � 10�2 m)(15 � 10�2 m)A � 1.5 � 10�3 m2

Fg � 1000 N

[(1.00 � 104 kg)(2.0 m/s2 � 9.8 m/s2)]��

��210.906�

�

11009

�

N

3

/mm

2

2��

[m(a � g)]��

��AE

��

��AF

��

E

Stress�

E

Stress�Strain

1.96 � 10�3 m2

��

A��

(1.00 � 104 kg)(9.8 N/kg)��

5.0 � 107 N/m2

mg�Stress

F�Stress

F�A

(80 � 109 N/m2)(0.1005 m)[�(0.01 m)2]��

2.0 m

G�LA�

L

��AF

��

��

LL��

20 m��cos 5.73°

R�cos �

R�R � �L

(360°)(2 m)��

125.6 m

2.0 m�125.6 m

��360°

2.0 m

RR

θτ

L

200 N��5.067 � 10�6 m

F��x


a) E �

E �

Stress �

Stress �

Stress � 6.67 � 105 N/m2

Strain �

Strain �

Strain � 6.67 � 10�5

b) ∆L � L(Strain)∆L � (3.0 m)(6.67 � 10�5)∆L � 2.0 � 10�4 m

61. m � 2.5 � 104 kgFapp � (2.5 � 104 kg)(9.8 N/kg)Fapp � 2.45 � 105 NA1 � �r2

A1 � �� 2

A1 � 0.785 m2

A2 � �r22

A2 � �� 2

A2 � 0.5024 m2

Emarble � 50 � 109 N/m2

∆L1 � ?

�

�

�L1 �

�L1 �

�L1 � 1.37 � 10�4 mColumn 1 final loaded:Loaded � 22.0 m � �L1

Loaded � 22.0 m � 1.37 � 10�4 mLoaded � 21.999863 m

For column 2:

Stress �

Stress �

Stress �

Stress � 9.75 � 10�6

� 9.75 � 10�6

�L2 � L2 (9.75 � 10�6)But: L2 � �L2 � 21.999 863 mL2 � L2(9.75 � 10�6 m) � 21.999 863 m

L2 �

L2 � 22.000 0775 mThe narrower column needs to be only7.8 � 10�5 m longer than the wider column.

21.999 863 m��1 � 9.75 � 10�6 m

�L2�L2

2.45 � 105 N��(0.5024 m2)(50 � 109 N/m2)

F�A2E

Strain�

E

(2.45 � 105 N)(22.0 m)��(0.785 m2)(50 � 109 N/m2)

FL1�A1E

��AF

1��

�E

�L1�L1

Stress�

E�L1�L1

0.80 m�

2

1.00 m�

2

6.67 � 105 N/m2

��10 � 109 N/m2

Stress�

E

1000 N��1.5 � 10�3 m2

F�A

Stress�Strain

��AF

��

��

LL��


Chapter 416. p � mv

p � (7500 kg)(120 m/s)p � 9.0 � 105 kg·m/s

17. p � mvp � (0.025 kg)(3 m/s)p � 0.075 kg·m/s

18. 90 km/h � 25 m/s, m � 25 g � 0.025 kgp � mvp � (0.025 kg)(25 m/s)p � 0.63 kg·m/s

19. v � 500 km/h � 138.89 m/s, p � 23 000 kg·m/s

m � �pv

�

m �

m � 165.6 kg

20. v � �mp�

v �

v � 6.00 � 1026 m/s, which is much greaterthan the speed of light.

21. p�� mv��

p�� (0.050 g)(10 m/s [down])p�� 0.5 kg·m/s [down]

22. v � (300 km/h)� �� 83.3 m/s

p�� mv��

p�� (6000 kg)(83.3 m/s [NW])p�� 5 � 105 kg·m/s [NW]

23. m � 50 kg, F�� 250 N [forward], ∆t � 3.0 s, v��1 � 0

F��t � m�v��

(250 N [forward])(3.0 s) � (50 kg)(v��2 � v��1)

� v��2

v��2 � 15 m/s [forward]24. m � 150 kg, v1 � 0, a � 2.0 m/s2,

∆t � 4.0 sa) v2 � v1 � a∆t

v2 � 0 � (2.0 m/s2)(4.0 s)v2 � 8.0 m/sp � mvp � (150 kg)(8.0 m/s)p � 1200 kg·m/s

b) J � ∆pJ � m2v2 � m1v1

J � (150 kg)(8.0 m/s) � (150 kg)(0)J � 1200 kg·m/s

25. m � 1.5 kg, ∆h � �17 m, v1 � 0, a � �9.8 N/kg

a) �h � v1�t � �12

�a�t2

�17 m � 0 � �12

�(�9.8 m/s2)�t2

�t � 1.86 sb) F � ma

F � (1.5 kg)(�9.8 N/kg)F � �14.7 N

c) J � F∆tJ � (�14.7 N)(1.86 s)J � �27.3 kg·m/s

26. F � 700 N, ∆t � 0.095 sa) J � F∆t

J � (700 N)(0.095 s)J � 66.5 kg·m/s

b) J � ∆p∆p � 66.5 kg·m/s

27. m � 0.20 kg, v1 � �25 m/s, v2 � 20 m/s∆p � m2v2 � m1v1

∆p � (0.2 kg)(20 m/s) � (0.2 kg)(�25 m/s)∆p � 9.0 kg·m/s

28. F∆t � m∆v(N)(s) � (kg)(m/s)

(kg·m/s2)(s) � (kg)(m/s)kg·m/s � kg·m/s

740 N [forward]��

50 kg

45°

p = 5 x 105 kg·m/s [NW]

1 h�3600 s

1000 m�

1 km

p = 0.5 kg·m/s [down]

1.00 kg·m/s��1.6726 � 10�27 kg

23 000 kg·m/s��

138.89 m/s


29. �p�� p��2 � p��1

30. v1 � 0, v2 � 250 m/s, m � 3.0 kg, F � 2.0 � 104 Na) J � ∆p

J � m2v2 � m1v1

J � (3.0 kg)(250 m/s) � 0J � 750 kg·m/s

b) J � F�t

�t �

�t �

�t � 0.038 s31. m � 7000 kg,

v1 � 110 km/h � 30.56 m/s, ∆t � 0.40 s, v2 � 0

a) F �

F �

F �

F � �5.3 � 105 N

b) F �

F �

F �

F � �2.7 � 104 N32. m � 30 g � 0.03 kg, v1 � 360 m/s,

∆d � 5 cm � 0.05 ma) p � mv

p � (0.03 kg)(360 m/s)p �11 kg·m/s

b) v22 � v1

2 � 2a∆d02 � (360 m/s)2 � 2a(0.05 m)a � �1.3 � 106 m/s2

c) F � maF � (0.03 kg)(�1.3 � 106 m/s2)F � �3.9 � 104 N

d) a �

�t �

�t �

�t � 2.8 � 10�4 se) J � �p

J � m2v2 � m1v1

J � (0.03 kg)(0) � (0.03 kg)(360 m/s)J � �11 kg·m/s

f)

33. a)

b) Area � �12

�h(a � b)

J � �12

�(15 s)(5 � 106 N � 8 � 106 N)

J � 9.8 � 107 N·s34. J � area under the curve

J � �12

�(�90 N)(0.3 s) � (120 N)(0.2 s) �

�12

�(75 N)(0.4 s)

J � (�13.5 N·s) � (24 N·s) � (15 N·s)J � 25.5 N·s

35. J � area under the graphCounting roughly 56 squares,J � 56(0.5 � 103 N)(0.05 s)J � 1.4 � 103 N·s

F

123

4

5

6

7

8

(×1

06 N

)

0 105 15t (s)

F

–4–3–2

–1

0

(×1

04 N

)

1 2 3t (×10–4 s)

0 � 360 m/s��1.3 � 106 m/s2

v2 � v1�a

v2 � v1��t

0 � (7000 kg)(30.56 m/s)��

8.0 s

m2v2 � m1v1��t

�p��t

0 � (7000 kg)(30.56 m/s)��

0.40 s

m2v2 � m1v1��t

�p��t

750 kg·m/s��

2 � 104 N

J�F

33°p2

p1

p


36. J � ∆pJ � mv2 � mv1, where v1 � 0,

1.4 � 103 N·s � (0.250 kg)(v2)v2 � 5.6 � 103 m/s

37. p��To � p��Tf

m1v��1o � m2v��2o � (m1 � m2)v��f, where v��2o � 0

(5000 kg)(5 m/s [S]) � (10 000 kg)(v��f)v��f � 2.5 m/s [S]

38. p��To � p��Tf

m1v��1o � m2v��2o � (m1 � m2)v��f, where v��2o � 0(45 kg)(5 m/s) � (47 kg)(v��f)

v��f � 4.8 m/s [in the same direction as v��1o]

39. p��To � p��Tf

m1v��1o � m2v��2o � m1v��1f � m2v��2f

(65 kg)(15 m/s) � (100 kg)(�5 m/s)

� (65 kg)��13

��(15 m/s) � (100 kg)(v2f)

(975 � 500 � 325) kg·m/s � (100 kg)(v2f)v2f � 1.5 m/s

40. p��To � p��Tf

m1v��1o � m2v��2o � (m1 � m2)v��f, where v2o � 0

(0.5 kg)(20 m/s) � 0 � (30.5 kg)(vf)vf � 0.33 m/s

41. p��To � p��Tf

m1v��1o � m2v��2o � m1v��1f � m2v��2f

(0.2 kg)(3 m/s) � (0.2 kg)(�1 m/s)� (0.2 kg)(2 m/s) � (0.2 kg)(v2f)

0.4 kg·m/s � 0.4 kg·m/s � (0.2 kg)(v2f)v2f � 0

42. v1o � 90 km/h � 25 m/s, vf � 80 km/h � 22.2 m/s

pTo � pTf

m1v1o � m2v2o � (m1 � m2)vf, where v2o � 0

m1(25 m/s) � 0 � (m1 � 6000 kg)(22.2 m/s)

m1(25 m/s) � m1(22.2 m/s) �

133 333.3 kg·m/sm1(25 m/s � 22.2 m/s) � 133 333.3 kg·m/s

m1 �

m1 � 4.8 � 104 kg

43. F1 � �F2

ma1 � �ma2

m� � � �m� �m(v1f � v1o) � �m(v2f � v2o)mv1f � mv1o � �mv2f � mv2o

mv1f � mv2f � mv1o � mv2o

pTf � pTo

pTf � pTo � 0�p � 0

44. m1 � 1.67 � 10�27 kg, m2 � 4m1, v1 � 2.2 � 107 m/s

pTo � pTf

m1v1o � m2v2o � (m1 � m2)vf, where v2o � 0

m1(2.2 � 107 m/s) � (5m1)vf

vf �

vf � 4.4 � 106 m/s45. m1 � 3m, m2 � 4m, v1o � v

pTo � pTf

m1v1o � m2v2o � (m1 � m2)vf, where v2o � 0(3m)v � (7m)vf

vf � �37

�v

46. m1 � 99.5 kg, m2 � 0.5 kg, v1f � ?, v2f � �20 m/spTo � pTf

0 � (99.5 kg)(v1f) � (0.5 kg)(�20 m/s)

v1f �

v1f � 0.1 m/s

�t �

�t �

�t � 2 � 103 s47. p��1o � 375 kg·m/s [E],

p��2o � 450 kg·m/s [N45°E]a) p��To � p��1o � p��2o

θ

pTo p2o = 450 kg·m/s

45°

p1o = 375 kg·m/s

200 m�0.1 m/s

�d�

v

10 kg·m/s��

99.5 kg

2.2 � 107 m/s��

5

v2f � v2o��t

v1f � v1o��t

133 333.3 kg·m/s��

2.8 m/s


b) p��Tf � p��To

Using the cosine and sine laws,�p��To�2 � (375 kg·m/s)2 � (450 kg·m/s)2 �

2(375 kg·m/s)(450 kg·m/s)cos 135°

�p��To� � 762.7 kg·m/s�p��Tf� � 762.7 kg·m/s

�

� � 24.7°Therefore, p��Tf � 763 kg·m/s [E24.7°N]

48. m1 � 3.2 kg, v��1o � 20 m/s [N], p��1o � 64 kg·m/s [N], m2 � 0.5 kg, v��2o � 5 m/s [W], p��2o � 2.5 kg·m/s [W]


p��1o � p��2o � p��Tf

m1v��1o � m2v��2o � (m1 � m2)v��f

Using the diagram and Pythagoras’ theorem,

�p��Tf� � �(2.5 kg�·m/s�)2 � (6�4 kg·�m/s)2��p��Tf� � 64.05 kg·m/s

tan � �

� � 2.2°p��T � (m1 � m2)v��f

64.05 kg·m/s [N2.2°W] � (3.7 kg)v��f

v��f � 17 m/s [N2.2°W]49. m1 � 3000 kg, v��1o � 20 m/s [N],

p��1o � 60 000 kg·m/s [N], m2 � 5000 kg, v��2o � ? [E], p��2o � ? [E], v��f � ? [E30°N], p��f � ? [E30°N]


p��1o � p��2o � p��Tf

Using the following momentum diagram,

tan 60° �

p2o � (60 000 kg·m/s)(tan 60°)p2o � 103 923 kg·m/s

m2v2o � 103 923 kg·m/s

v2o �

v��2o � 20.8 m/s [E]50. mo � 1.2 � 10�24 kg, v��o � 0, p��o � 0,

m1 � 3.0 � 10�25 kg, v��1 � 2.0 � 107 m/s [E],p��1 � 6 � 10�18 kg·m/s [E],m2 � 2.3 � 10�25 kg, v��2 � 4.2 � 107 m/s [N],p��2 � 9.66 � 10�18 kg·m/s [N]m3 � 1.2 � 10�24 kg � 3.0 � 10�25 kg �

2.3 � 10�25 kgm3 � 6.7 � 10�25 kgp��To � 0p��To � p��Tf

0 � p��1 � p��2 � p��3

Drawing a momentum vector diagram andusing Pythagoras’ theorem,

�p��3�2 � (9.66 � 10�18 kg·m/s)2 �

(6 � 10�18 kg·m/s)2

�p��3� � 1.1372 � 10�17 kg·m/s

�v��3� �

�v��3� �

�v��3� � 1.7 � 107 m/s

tan � �

� � 31.8°Therefore, v��3 � 1.7 � 107 m/s [S32°W]

51. m1 � m2 � m, v��1o � � 12.5 m/s

[R], v��2o � 0, v��2f � 1.5 m/s [R25°U]p��To � p��Tf

m1v��1o � m2v��2o � m1v��1f � m2v��2f

Since m1 � m2 and v��2o � 0,v��1o � v��1f � v��2f

60 m�4.8 s

6 � 10�18 kg·m/s��9.66 � 10�18 kg·m/s

1.1372 � 10�17 kg·m/s��

6.7 � 10�25 kg

�p��3��m3

θp3 = m3v3 p2o = 9.66 × 10–18 kg·m/s

p1o = 6.0 × 10–18 kg·m/s

103 923 kg·m/s��

5000 kg

p2o��60 000 kg·m/s

pTfp1o = 60 000 kg·m/s

p2o = m2 v2o

30°

2.5 kg·m/s��64 kg·m/s

θ

pTf

p1o = 64 kg·m/s

p2o = 2.5 kg·m/s

sin 135°��762.7 kg·m/s

sin ��450 kg·m/s


Drawing a vector diagram and usingtrigonometry,

�v��1f�2 � (1.5 m/s)2 � (12.5 m/s)2 �

2(1.5 m/s)(12.5 m/s) cos 25°�v��1f� � 11.16 m/s

�

� � 3.3°Therefore, the first stone is deflected 3.3° or[R3.3°D].

52. m1 � 10 000 kg,v��1 � 3000 km/h [E] � 833.3 m/s [E],p��1 � 8.333 � 106 kg·m/s [E], m2 � ?,v��2 � 5000 km/h [S] � 1388.9 m/s [S],p��2 � m2(1388.9 m/s) [S],m3 � 10 000 kg � m2, v��3 � ? [E10°N],p��3 � (10 000 kg � m2)(v3) [E10°N]p��1 � p��2 � p��3

Drawing a momentum diagram and usingtrigonometry,

tan 10° �

�p��2� � (8.33 � 106 kg·m/s)(tan 10°)�p��2� � 1.47 � 106 kg·m/sp��2 � m2(1388.9 m/s) [S]

m2 �

m2 � 1057.6 kgThe mass of the ejected object is 1.058 � 103 kg.

53. m1 � m2 � m, v��1o � 6.0 m/s [U], v��2o � 0, v��2f � 4 m/s [L25°U], v��1f � ?


m1v��1o � m2v��2o � m1v��1f � m2v��2f

Since m1 � m2 and v��2o � 0,v��1o � v��1f � v��2f

Using the vector diagram and trigonometry,

�v��1f�2 � (6.0 m/s)2 � (4.0 m/s)2 �

2(6.0 m/s)(4.0 m/s) cos 65°�v��1f� � 5.63 m/s

�

� � 40°Therefore v��1f � 5.63 m/s [U40°R]

54. 2m1 � m2, v��1o � 6.0 m/s [U], v��2o � 0, v��2f � 4 m/s [L25°U], v��1f � ?


m1v��1o � m2v��2o � m1v1f � m2v2f, since 2m1 � m2 and v��2o � 0

v��1o � v��1f � 2v��2f

Using the vector diagram and trigonometry,

�v��1f�2 � (6.0 m/s)2 � (8.0 m/s)2 �

2(6.0 m/s)(8.0 m/s) cos 65°�v��1f� � 7.7 m/s

�

� � 70°Therefore, v��1f � 7.7 m/s [R20°U]

55. Counting ten dots for a one-second intervaland measuring the distance with a ruler andthe angle with a protractor gives:

a) �v��1o� �

�v��1o� � 0.033 m/s�v��2o� � 0

�v��1f� �

�v��1f� � 0.033 m/s

�v��2f� �

�v��2f� � 0.033 m/s

33 mm�

1 s

33 mm�

1 s

33 mm�

1 s

sin 65°�7.7 m/s

sin ��8.0 m/s

θ

v2f = 8.0 m /s

V1o = 6.0 m /s

v1f

25°

sin 65°��5.63 m/s

sin ��4.0 m/s

v2f = 4.0 m / s

v1o = 6.0 m / s

v1f

25°

θ

1.47 � 106 kg·m/s��

1388.8 m/s

p��2�p��1

p1 = 8.33 × 106 kg·m/s

p2 p3

10°

sin 25°�11.6 m/s

sin ��1.5 m/s

v2f = 1.5 m/s

v1o = 12.5 m/s

v1f θ

25°


b) v��1o � 0.033 m/s [E]v��2o � 0v��1f � 0.033 m/s [E45°S]v��2f � 0.033 m/s [E45°N]

c) p��1o � (0.3 kg)(0.033 m/s [E])p��1o � 9.9 � 10�3 kg·m/s [E]p��2o � (0.3 kg)(0) � 0p��To � 9.9 � 10�3 kg·m/s [E]p��1f � (0.3 kg)(0.033 m/s [E45°S])p��1f � 9.9 � 10�3 kg·m/s [E45°S]p��2f � (0.3 kg)(0.033 m/s [E45°N])p��2f � 9.9 � 10�3 kg·m/s [E45°N]The vector diagram for the final situationis shown below.

Using Pythagoras’ theorem,�p��Tf�2 � (9.9 � 10�3 kg·m/s)2 �

(9.9 � 10�3 kg·m/s)2

p��Tf � 1.4 � 10�2 kg·m/s [E]d) p1oh � � 9.9 � 10�3 kg·m/s

p1ov � 0p2oh � 0p2ov � 0p1fh � �(9.9 � 10�3 kg·m/s)(cos 45°)p1fh � 7.0 � 10�3 kg·m/sp1fv � �(9.9 � 10�3 kg·m/s)(sin 45°)p1fv � �7.0 � 10�3 kg·m/sp2fh � �(9.9 � 10�3 kg·m/s)(cos 45°)p2fh � 7.0 � 10�3 kg·m/sp2fv � �(9.9 � 10�3 kg·m/s)(sin 45°)p2fv � 7.0 � 10�3 kg·m/s

e) Momentum is not conserved in this colli-sion. The total final momentum is about1.4 times the initial momentum.

56. m1 � 0.2 kg, v��1 � 24 m/s [E], p��1 � 4.8 kg·m/s [E], m2 � 0.3 kg, v��2 � 18 m/s [N], p��2 � 5.4 kg·m/s [N],m3 � 0.25 kg, v��3 � 30 m/s [W],p��3 � 7.5 kg·m/s [W], m4 � 0.25 kg,v��4 � ?, p��4 � ?

p��To � 0p��1 � p��2 � p��3 � p��4 � 0

Drawing a vector diagram and usingtrigonometry,

Using triangle ABC,

tan � �

� � 26.6°�p��4�2 � (2.7 kg·m/s)2 � (5.4 kg·m/s)2

�p��4� � 6.037 kg·m/s

�v��4� �

�v��4� � 24.1 m/sTherefore, v��4 � 24.1 m/s [S26.6°E]

57. a) Masstotal � 5000 kg � 10 000 kgMasstotal � 15 000 kg

b) � �13

� the distance to the larger

truck, or �13

�(400 m) � 133.3 m from the

larger truck.58. a) m1 � 2000 kg, v��1o � 200 m/s [E],

p��1o � 4 � 105 kg·m/s [E]

b) m2 � 1000 kg, v��2o � 200 m/s [S30°E],p��2o � 2 � 105 kg·m/s [S30°E]

c) p��cmo � p��1o � p��2o

d) p��cmf � p��cmo

59. a) p��cmo � p��To

p��cmo � 9.9 � 10�3 kg·m/s [E] (see 55c)b) p��cmf � p��Tf

p��cmf � 1.4 � 10�2 kg·m/s [E] (see vectordiagram for 55c)

pcmf

p2o

p1o

pcmo

30°p2o = 2 × 105 kg·m/s [S30°E]

p1o = 4 × 105 kg·m/s [E]

5000 kg��15 000 kg

6.037 kg·m/s��

0.25 kg

2.7 kg·m/s��5.4 kg·m/s

A

C

B = 7.5 kg·m/s

p2 = 5.4 kg·m/s

p1 = 4.80 kg·m/s

p3

p4θ

p2f = 9.9 × 10–3 kg·m/sp1f = 9.9 × 10–3 kg·m/s

pTf

45°

45°


Chapter 511. a) W � F�d

W � (4000 N)(5.0 m)W � 2.0 � 104 J

b) W � (570 N)(0.08 m)W � 46 J

c) W � �Ek

W � Ek2 � Ek1

W � �12

�mv2 � 0

W � �12

�(9.1 � 10�31 kg)(1.6 � 108 m/s)2

W � 1.2 � 10�14 J12. a) W � F�d

W � (500 N)(5.3 m)W � 2.7 � 103 J

b) W � F�d cos �W � (500 N)(5.3 m) cos 20°W � 2.5 � 103 J

c) W � (500 N)(5.3 m) cos 70°W � 9.1 � 102 J

13.

W � �Eg

F�d � mg�h(350 N)(25.0 m) � (50.0 kg)(9.8 N/kg)h

h � 18 m

sin � �

sin � �

� � 46°14. Using the plow’s speed, in 1 s, the plow will

push a “block” of snow that is 0.35 m deep,4.0 m wide and 10.0 m long.This snow has a mass of:(0.35 m)(4.0 m)(10.0 m)(254 kg/m3) � 3556 kg

The force that the plow applies in 1 s is:Fapp � Fg

Fapp � mgFapp � (3556 kg)(9.8 N/kg)Fapp � 34 848.8 NThis force is applied over a distance of 5 m:W � F�dW � (34 848.8 N)(5.0 m)W � 174 244 JTo find the number of seconds it takes toplow the road:

t �

t �

t � 800 sWT � (800 s)(174 244 J/s)WT � 1.4 � 108 J

15. The two campers must overcome 84 N of fric-tion, or 42 N each in the horizontal directionsince both are at the same 45° angle.

The horizontal component of Fc, called Fh,must be equal to 42 N.W � Fh�dW � (42 N)(50 m)W � 2.1 � 103 N·mEach camper must do 2.1 � 103 J of work toovercome friction.

16. W � F�d ,where �d for each revolution iszero. Therefore, W � 0 J.

17. �350 J indicates that the force and the dis-placement are in the opposite direction. Anexample would be a car slowing down becauseof friction. Negative work represents a flow ortransfer of energy out of an object or system.

45°

45°

Fc

Fh

8000 m�10 m/s

d�v

18 m�25.0 m

h�25.0 m

θ

25.0 mh


18. �dramp � 5 mm � 35 kg�dheight � 1.7 ma) F � ma

F � (35 kg)(9.8 N/kg)F � 343 NF � 3.4 � 102 N

b) W � F�dW � (3.4 � 102 N)(1.7 m)W � 583.1 JW � 5.8 � 102 N

c) W � F�dramp

583.1 J � F(5 m)F � 116.62 NF � 1.2 � 102 N

19. W � Area under the graph

W � � (10 m)(200 N) �

� (20 m)(200 N) �

(10 m)(800 N) � �

(20 m)(800 N) � (10 m)(1200 N)W � 5.4 � 104 J

20. a) W � area under the graph

W � � �

(1 m)(100 N) � (2 m)(300 N)W � 8.5 � 102 J

b) The wagon now has kinetic energy (andmay also have gained gravitational poten-tial energy).

c) W � Ek

Ek � �12

�mv2

850 J� �12

�(120 kg)v2

v � 3.8 m/s

21. a) Ek � �12

�mv2

Ek � �12

�(45 kg)(10 m/s)2

Ek � 2.3 � 103 J

b) v �

v �

v � 0.628 m/s

Ek � �12

�mv2

Ek � �12

�(0.002 kg)(0.628 m/s)2

Ek � 3.9 � 10�4 J

c) v � � �

v � 27.7778 m/s

Ek � �12

�mv2

Ek � �12

�(15 000 kg)(27.7778 m/s)2

Ek � 5.8 � 106 J

22. v �

v �

v � 2.5 m/s

Ek � �12

�mv2

450 J � �12

�m(2.5 m/s)2

m � 1.4 � 102 kg

23. Ek � �12

�mv2

5.5 � 108 J � �12

�(1.2 kg)v2

v � ��v � 3.0 � 104 m/s

24. Ek-gained � �Eg

Ek-gained � mgh2 � mgh1

Ek-gained � mg(h2 � h1)Ek-gained � (15 kg)(9.8 N/kg)(200 m � 1 m)Ek-gained � 2.9 � 104 J

25. p � �2mEk�kg·m/s � �(kg)(J�)�kg·m/s � �(kg)(N�·m)�kg·m/s � �kg(kg� m/s2�)m�kg·m/s � �kg2m�2/s2�kg·m/s � kg·m/s

2(5.5 � 108 J)��

1.2 kg

5.0 m�2.0 s

�d��t

1000 m�

1 km1 h

�3600 s

100 km�

1 h

2(0.1 m)��

1 s

2r��t

(1 m)(200 N)��

2(1 m)(100 N)��

2

(20 m)(400 N)��

2

(20 m)(600 N)��

2

(20 m)(200 N)��

2


26. 5 keV � �

� 8 � 10�16 J

Ek � �12

�mv2

8 � 10�16 J � �12

�(9.1 � 10�31 kg)v2

v � 4.2 � 107 m/sAs a percentage of the speed of light:

� 100 � 14%

27. a) a �

a �

a � �1.86 � 107 m/s2

F � maF � (0.015 kg)(�1.86 � 107 m/s2)F � �2.8 � 105 N

b) F � �force of bulletF � 2.8 � 105 N

28. For 1 m:W � (50 N)(1 m)W � 50 JW � Ek

Ek � �12

�mv2

50 J � �12

�(1.5 kg)v2

v � 8 m/sFor 2 m:

W � 50 J � (50 N)(1 m) � �12

�(250 N)(1 m)

W � 225 J

Ek � �12

�mv2

225 J � �12

�(1.5 kg)v2

v � 17.3 m/sFor 3 m:

W � 225 J � (50 N)� m� �

(300 N)� m� � (350 N)� m�W � 425 J

Ek � �12

�mv2

425 J � �12

�(1.5 kg)v2

v � 23.8 m/s29. p � �2mEk�

p � �2(5 kg�)(3.0 �� 102 J)�p � 55 N·s

30. m1 � 0.2 kgm2 � 1 kgv1o � 125 m/sv1f � 100 m/sv2o � 0v2f � ?�d2 � 3 ma) pTo � pTf


(0.2 kg)(125 m/s) � 0 � (0.2 kg)(100 m/s) �

(1 kg)v2f

v2f � 5 m/s

b) Ek � �12

�mv2

Ek � �12

�(1 kg)(5 m/s)2

Ek � 12.5 Jc) This collision is not elastic since some

kinetic energy is not conserved. Someenergy may be lost due to the deformationof the apple.

d) v22 � v1

2 � 2a�d0 � (5 m/s)2 � 2a(3.0 m)a � �4.1667 m/s2

F � maF � (1.0 kg)(�4.1667 m/s2)F � �4.2 N

31. a) Eg � mg�hEg � (2.0 kg)(9.8 N/kg)(1.3 m)Eg � 25 J

b) Eg � mg�hEg � (0.05 kg)(9.8 N/kg)(3.0 m)Eg � 1.5 J

c) Eg � mg�hEg � (200 kg)(9.8 N/kg)(469 m)Eg � 9.2 � 105 J

d) Eg � mg�hEg � (5000 kg)(9.8 N/kg)(0)Eg � 0 J

5�6

1�2

1�6

1�6

1�2

0 � (350 m/s)2

��2(0.0033 m)

(v22 � v1

2)��

2d

4.2 � 107 m/s��3 � 108 m/s

1.6 � 10�19 J��

1 eV1000 eV�

1 keV


32. a) m �

m �

m � 4.5 � 102 kgb) W � F�d

W � (4410 N)(3.5 m)W � 1.5 � 104 J

33. Using conservation of energy:ETo � ETf

mgh � �12

�mvo2 � �

12

�mvf2

(9.8 m/s2)(1.8 m) � �12

�(4.7 m/s)2 � �12

�v2

17.64 m2�s2 � 11.045 m2�s2 � �12

�v2

v � 7.6 m/s34. Ee � Eg

�12

�kx2 � mg�h

�12

�(1200 N/m)x2 � (3.0 kg)(9.8 N/kg)(0.80 m)

x � 0.2 mx � 20 cm

35. m � 0.005 kgh � 2.0 mInitial:E � mg�hE � (0.005 kg)(9.8 N/kg)(2.0 m)E � 0.098 JAt half the height:E � (0.005 kg)(9.8 N/kg)(1.0 m)E � 0.049 JAfter the first bounce:E � (0.80)(0.098 J)E � 0.0784 JAfter the second bounce:E � (0.80)(0.0784 J)E � 0.062 72 JAfter the third bounce:E � (0.80)(0.062 72 J)E � 0.050 176 JAfter the fourth bounce:E � (0.80)(0.050 176 J)E � 0.040 140 9 JTherefore, after the fourth bounce, the ballloses over half of its original height.

36. a) The greatest potential energy is at point A(highest point) and point F represents thelowest amount of potential energy (lowestpoint).

b) Maximum speed occurs at F when most ofthe potential energy has been converted tokinetic energy.

Eg-lost � Ek-gained

mg�h � mv2

(1000 kg)(9.8 N/kg)(75 m) � �12

�(1000 kg)v2

v � 38 m/sc) At point E, 18 m of Ep is converted to Ek.

mg�h � mv2

(1000 kg)(9.8 N/kg)(18 m) � �12

�(1000 kg)v2

v � 19 m/sd) Find the acceleration, then use F � ma.

v22 � v1

2 � 2a�d0 � (38 m/s)2 � 2a(5 m)a � �144.4 m/s2

F � maF � (1000 kg)(144.4 m/s2)F � 1.4 � 105 N

37. Ee � Ek

�12

�kx2 � �12

�m�v2

(890 N/m)x2 � (10 005 kg)(5 m/s)2

x � 16.8 mx � 17 m

38. �dh �

15 m �

v1 � 12.1 m/sEk � Ee

�12

�mv2 � �12

�kx2

(0.008 kg)(12.1 m/s) � (350 N/m)x2

x � 0.058 mx � 5.8 cm

v12 sin 90°

��9.8 m/s2

v12 sin 2�

��g

1�2

1�2

4410 N�9.8 N/kg

F�a


39. 85% of the original energy is left after thefirst bounce, therefore,(0.85)mg�htree � mg�hbounce

(0.85)(2 m) � �hbounce

�h � 1.7 m40. Ee � Ek

�12

�kx2 � �12

�mv2

(35 000 N/m)(4.5 m)2 � (65 kg)v2

v � 104.4 m/s

�dh �

�dh �

�dh � 1.1 � 103 m41. k � slope

k �

k � �Fx

�

k �

k � 5.3 � 102 N/m42. W � area under the graph

a) W � �12

�(0.05 m)(2 � 103 N)

W � 50 Jb) W � 50 J � (0.02 m)(2 � 103 N) �

(0.02 m)(4.5 � 103 N)

W � 135 JW � 1.4 � 102 J

43. Ek � Ee

�12

�mv2 � �12

�kx2

(0.05 kg)v2 � (400 N/m)(0.03 m)2

v � 2.7 m/s44. Ek � Ee

�12

�mv2 � �12

�kx2

(2.5 � 103 kg)(95 m/s)2 � k(35 m)2

k � 1.8 � 104 N/m45. Ee � Ek

kx2 � mv2

(5 � 107 N/m)(0.15 m)2 � (1000 kg)v2

v 34 m/s

46. a) Ek � Ee

�12

�mv2 � �12

�kx2

(3 kg)v2 � (125 N/m)(0.12 m)2

v � 0.77 m/sb) Ff � �Fn

Ff � (0.1)(3 kg)(�9.8 N/kg)Ff � �2.94 NF � ma

a � �mF

�

a �

a � �0.98 m/s2

v22 � v1

2 � 2a�d0 � (0.77 m/s)2 � 2(�0.98 m/s2)�d

�d � 0.3 m�d � 30 cm

47. Ek � Ee

mv2 � kx2

(3.0 kg)v2 � (350 N/m)(0.1 m)2

v � 1.1 m/s48. F � kx

F � (4000 N/m)(0.15 m)F � 600 N

49. F � maF � (100 kg)(9.8 N/kg)F � 980 NDivided into 20 springs:

F �

F � 49 N per springF � kx

49 N � k(0.035 m)k � 1.4 � 103 N/m

50. F � kxmg � kx

(10 kg)(9.8 N/kg) � k(1.3 m)k � 75.3846 N/m

Ee � kx2

2 � 106 J � (75.3846 N/m)x2

x � 2.3 � 102 m

1�2

1�2

980 N�

20

1�2

1�2

�2.94 N��

3 kg

1�2

1�2

1�2

120 N�0.225 m

rise�run

(104.4 m/s)2 sin 90°��

9.8 m/s2

v12 sin 2�

��g


51. 3 d � � � � 259 200 s

8 h � � � 28 800 s

15 min � � 900 s

�t � 259 200 s � 28 800 s � 900 s�t � 288 900 s

P �

E � P�tE � (60 W)(288 900 s)E � 1.7 � 107 J

1.7 � 104 kJ � � 4.8 kWh

52. a) E � �Eg

E � mghE � (3500 kg)(9.8 m/s2)(13.4 m)E � 459 620 J

P �

P �

P � 19 983 W

PE �

PE � 4.3 � 104 W

b) 4.3 � 104 W � � 58 hp

54. a) P � FvP � FgvP � mgvP � (4400 kg � 2200 kg)(9.8 m/s2)(2.4 m/s)P � 1.6 � 105 W

55. Since the cyclist’s speed is 2.78 m/s, thecyclist travels 2.78 m up the hill per second.The cyclist’s change in height per second is:h � d sin �h � (2.78 m) sin 7.2°h � 0.348 mThe increase in potenial energy is:Ep � mghEp � (75 kg)(9.8 m/s2)(0.348 m)Ep � 255.78 J

In 1 s:

P �

P � 256 W56. Using the conservation of momentum:


m1v1o � m1v1f � m2v2f

m1(v1o � v1f) � m2v2f (eq. 1)Using the conservation of kinetic energy:

m1(v1o2 � v1f

2) � m2v2f2 (eq. 2)

Dividing equation 2 by equation 1:

�

v1o � v1f � v2f

v1f � v2f � v1o (eq. 3)Substituting equation 3 into equation 1:

m1(v1o � v1f) � m2v2f

m1(v1o � v2f � v1o) � m2v2f

m1(2v1o � v2f) � m2(v2f)v1o(2m1) � v2f(m1 � m2)

v2f �

57. a) v1f � v1o

v1f � (3 m/s)

v1f � 2 m/s

v2f � v1o

v2f � (3 m/s)

v2f � 5 m/s

b) Ek � �12

�mv2

Ek � �12

�(3 kg)(5 m/s)2

Ek � 37.5 JEk � 38 J

58. pTf � pTo

(m1 � m2)vf � m1v1o � m2v2o

(0.037 kg)vf � (0.035 kg)(8 m�s) �

(0.002 kg)(�12 m/s)vf � 6.9 m/s

2(15 kg)��15 kg � 3 kg

2m1�m1 � m2

15 kg � 3 kg��15 kg � 3 kg

m1 � m2�m1 � m2

2m1v1o�m1 � m2

m2v2f2

�m2v2f

m1(v1o2 � v1f

2)��m1(v1o � v1f)

255.78 J�

1.0 s

1 hp�746 W

19 983 W��

0.46

459 620 J��

23 s

E��t

1 kWh�3600 kJ

E��t

60 s�1 min

60 s�1 min

60 min�

1 h

60 s�1 min

60 min�

1 h24 h�1 d


59. a) pTo � m1v1o�m2v2o

pTo � (3.2 kg)(2.2 m/s) � (3.2 kg)(0)pTo � 7.0 kg·m/s

Ek-To � �12

�mv2

Ek-To � �12

�(3.2 kg)(2.2 m/s)2

Ek-To � 7.7 Jb) Using the conservation of momentum and

m1 � m2:m1v1o � m2v2o � m1v1f � m2v2f

2.2 m/s � 0 � 1.1 m/s � v2f

v2f � 1.1 m/s

c) Ek-Tf � �12

�mv1f2 � �

12

�mv2f2

Ek-Tf � �12

�(3.2 kg)(1.1 m/s)2 �

�12

�(3.2 kg)(1.1 m/s)2

Ek-Tf � 3.8 Jd) The collision is not elastic since there was

a loss of kinetic energy from 7.7 J to 3.8 J.60. pTo � pTf


(0.015 kg)(375 m/s) � 0 � (0.015 kg)(300 m/s) �

(2.5 kg)v2f

v2f � 0.45 m/s61. m1 � 6m

v1o � 5 m/sm2 � 10mv2o � �3 m/sChanging the frame of reference so that v2f � 0:v1o � 8 m/s

v1f � v1o

v1f � (8 m/s)

v1f � �2 m/s

v2f � v1o

v2f � (8 m/s)

v2f � 6 m/sReturning to our original frame of reference(subtract 3 m/s):v1f � �2 m/s � 3 m/s � �5 m/s,v2f � 6 m/s � 3 m/s � 3 m/s

62. a) v1f � v1o

v1f � (5 m/s)

v1f � 2.5 m/s

b) v2f � v1o

v2f � (5 m/s)

v2f � 7.5 m/s63. mw � 0.750 kg

k � 300 N/mmb � 0.03 kgx � 0.102 ma) Ee-gained � Ek-lost

kx2 � mv2

(300 N/m)(0.102 m)2 � (0.78 kg)v2

v � 2 m/sUsing the conservation of momentum:

pTo � pTf

mbvbo � mwvwo � m(b�w)vf

(0.03 kg)vbo � 0 � (0.78 kg)(2.0 m/s)vbo � 52 m/s

b) The collision is inelastic since:

Eko � �12

�(0.03 kg)(52 m/s)2

Eko � 40.56 JandEkf � 0The kinetic energy is not conserved.

64. a) mgh � �12

�mv2

(2.05 kg)(�9.8 m/s)(0.15 m) � �12

�(2.05 kg)v2

v � 1.7 m/sb) m1v1 � v2(m1 � m2)

(0.05 kg)v1 � (1.71 m/s)(2.05 kg)v1 � 70 m/s

65. Using the conservation of momentum and m1 � m2 � m:

pTo � pTf

mv1o � mv2o � mv1f � mv2f

v1o � 0 � v1f � v1f (eq. 1)

1�2

1�2

2(3m2)��3m2 � m2

2m1�m1 � m2

3m2 � m2��3m2 � m2

m1 � m2�m1 � m2

2(6m)��6m � 10m

2m1�m1 � m2

6m � 10m��6m � 10m

m1 � m2�m1 � m2


Using the conservation of kinetic energy:Ek�To � Ek�Tf

mv1o2 � mv2o

2 � mv1f2 � mv2f

2

v1o2 � 0 � v1f

2 � v2f2

v1o2 � v1f

2 � v2f2 (eq. 2)

Equation 1 can be represented by the vectordiagram:

The angle � is the angle between the finalvelocity of the eight ball and the cue ball afterthe collision.Using the cosine law and equation 2:v1o

2 � v1f2 � v2f

2 � 2(v1f)(v2f) cos �v1o

2 � v1o2 � 2(v1f)(v2f) cos �

0 � �2(v1f)(v2f) cos �0 � cos �� 90°

Therefore, the angle between the two ballsafter collision is 90°.

θ

v1o

v2fv1f

1�2

1�2

1�2

1�2


Chapter 613. vi � 4 km/s � 4 � 103 m/s, vf � 80 m/s

�E � �12

�mvi2 � �

12

�mvf2

�E � �12

�(100 000 kg)[(80 m/s)2 � (4000 m/s)2]

�E � �7.9968 � 1011 JIt has released 7.9968 � 1011 J of energy to theatmosphere.The shuttle’s initial height was 100 km, and itlanded on Earth’s surface, therefore its changein height is 100 km.

14. mE � 5.98 � 1024 kg, msat � 920 kg, Ek � 7.0 � 109 Ja) At the start, the height is rE � 6.38 � 106 m.

Therefore, the total energy isET � Eki � Epi

ET � 7.0 � 109 J �

ET � 7.0 � 109 J �

ET � �5.051 672 � 1010 JSince velocity is 0 at maximum height,total energy at maximum height � Epf

ET �

ET �

ET �

The total energy is constant, therefore,

�5.051 672 � 1010 J�

h � 884.1 kmb) Escape velocity from Earth’s surface is

given by:

vesc � ��vesc � ��vesc � 1.12 � 104 m/sTherefore, the initial kinetic energy requiredfor the escape should be greater than:

E � �12

�mv2

E � �12

�(920 kg)(1.12 � 104 m/s)2

E � 5.75 � 1010 Jc) The initial speed needed to keep going

indefinitely should be greater than theescape speed, i.e., greater than 11.2 km/s.

15. ms � 550 kg, mE � 5.98 � 1024 kg, h � 6000 km � 6 � 106 m, rE � 6.38 � 106 m

a) �Ep � � � ��Ep � GMm� � ��Ep � (6.67 � 10�11 N·m2/kg2)

(5.98 � 1024 kg)(550 kg)

� �

��Ep � 1.67 � 1010 J

b) At the maximum height of 6000 km, thekinetic energy is 0 since the velocity iszero. Therefore, the change in Ep is the ini-tial kinetic energy,i.e., Eki � 1.67 � 1010 J.

16. mE � 5.98 � 1024 kg, rE � 6.38 � 106 m, mm � 20 000 kg,vi � 3.0 km/s � 3.0 � 103 m/s, h � 200 km � 2.0 � 105 mSince the meteorite is headed from outer space,

Epi � 0 and Eki � �12

�mvi2

Therefore, ET � �12

�mvi2

At 200 km,ET � Ekf � Epf

�12

�mvi2 � mvf

2 �

�12

�vi2 � vf

2 �

�12

�(3.0 � 103 m/s)2 � �12

�vf2 �

vf � 11 412.1 m/sThe meteorite’s speed 200 km above Earth’ssurface is approximately 11.4 km/s.

(6.67 � 10�11 N·m2/kg2)(5.98 � 1024 kg)��

6.58 � 106 m

GM�h � r

1�2

GMm�h � r

1�2

1��6.38 � 106 m

�1��6.38 � 106 m � 6 � 106 m

1�r

� 1�r � h

�GMm�

r�GMm�

r � h

2(6.67 � 10�11 N·m2/kg2)(5.98 � 1024 kg)��

(6.38 � 106 m)

2GM�

r

� 3.67 � 1017 N·m2

��6.38 � 106 m � h

� 3.67 � 1017 N·m2

��6.38 � 106 m � h

�(6.67 � 10�11 N·m2/kg2)(5.98 � 1024 kg)(920 kg)��

6.38 � 106 m � h

�GMm�

r � h

(6.67 � 10�11 N·m2/kg2)(5.98 � 1024 kg)(920 kg)��

6.38 � 106 m

GMm�

r


17. vesc � c � 2.99 � 108 m/s, mE � 5.98 � 1024 kg

vesc � ��r �

r �

r � 8.92 � 10�3 m18. Given: dE�M � 3.82 � 108 m,

mMoon � 7.35 � 1022 kg, mEarth � 5.98 � 1024 kgEquating the forces of gravity between Earthand the Moon, using the distance from Earthas r,

�

�

MMoonr2 � MEarth(3.82 � 108 m � r)2

0 � MEarth(1.46 � 1017 m �

7.64 � 108r � r2) � MMoonr2

0 � 8.73 � 1041 m � 4.57 � 1033r �

5.98 � 1024r2 � 7.36 � 1022r2

0 � 5.91 � 1024r2 � 4.57 � 1033r �

8.73 � 1041 mr � 4.29 � 108 m, 3.45 � 108 m

The forces of gravity from Earth and theMoon are equal at both 4.43 � 108 m and 3.45 � 108 m from Earth’s centre.

19. mEarth � 5.98 � 1024 kg, mMoon � 7.35 � 1022 kg,rE � 6.38 � 106 m,rM � 1.738 � 106 mLet m be the mass of the payload.

E � ��GM

rMoonm� � �

GMREarthm�

E � (6.67 � 10�11 N·m2/kg2)

� � �E � 5.97 � 107 JThe energy required to move a payload fromEarth’s surface to the Moon’s surface is 5.97 � 107 J/kg.

20. mE � 5.98 � 1024 kg, rE � 6.38 � 106 m, h � 400 km � 4.0 � 105 mOrbital speed is given by:

v � ��v � ��v � 7.67 km/sThe period of the orbit is the time required bythe satellite to complete one rotation aroundEarth. Therefore, the distance travelled, �d, isthe circumference of the circular orbit.Therefore, d � 2π(r � h)d � 2(3.14)(6.38 � 106 m � 4.0 � 105 m)d � 42 599 996 mHence, speed is given by,

v �

T �

T �

T � 5552 sThe period of the orbit is 5552 s or 92.5 min.

21. mE � 5.98 � 1024 kg, rE � 6.37 � 106 mSince the orbit is geostationary, it has a periodof 24 h � 86 400 s. Using Kepler’s third law,

�

r � � ��13

�

r � � ��13

�

r � 4.22 � 107 mSubtracting Earth’s radius,r � 4.22 � 107 m � 6.37 � 106 mr � 3.59 � 107 mThe satellite has an altitude of 3.59 � 104 km.

22. mE � 5.98 � 1024 kg, rE � 6.37 � 106 m, r1 � 320 km � 3.2 � 105 m,r2 � 350 km � 3.5 � 105 m

(6.67 � 10�11 N·m2/kg2)(5.98 � 1024 kg)(86 400 s)2

��4(3.14)2

GMT2

�42

GM�42

r3

�T2

42 599 996 m��

7670 m/s

d�v

d�T

(6.67 � 10�11 N·m2/kg2)(5.98 � 1024 kg)��

6.38 � 106 m � 4.0 � 105 m

GM�r � h

5.98 � 1024 kg��6.38 � 106 m

�7.35 � 1022 kg��

1.738 � 106 m

MEarth�r2

MMoon��(3.82 � 108 m � r)2

GMEarthm��

r2

GMMoonm��(3.82 � 108 m � r)2

2(6.67 � 10�11 N·m2/kg2)(5.98 � 1024 kg)��

(2.99 � 108 m/s)2

2GM�(vesc)2

2GM�

r


Energy added to the station’s orbit is given by:

E � � � �E � �GMm� � �E � �(6.67 � 10�11 N·m2/kg2)

(5.98 � 1024 kg)m

� � �E � 2.65 � 105m JThe shuttle has added 2.65 � 105m J ofenergy to the station’s orbit.

23. a) The total energy of a satellite in an orbit isthe sum of its kinetic and potential ener-gies. In all cases, total energy remains con-stant. Therefore, when r is increased, thegravitational potential energy increases as

Ep � ��G

rMm�. As r increases, the energy

increases as it becomes less negative. Thus,when potential energy increases, kineticenergy decreases to maintain the total

energy a constant. Since Ek � �12

�mv2, if

kinetic energy decreases, v also decreasesand when r increases, v decreases.

b) In Kepler’s third law equation � K,

r is directly proportional to T. Therefore,as r increases, T also increases.

24. mE � 5.98 � 1024 kg, mM � 7.35 � 1022 kg, r � 3.82 � 108 mThe Moon’s total energy in its orbit aroundEarth is given by:

ET � ��12

�Ep

ET � ��12

�

ET � ��12

�

ET � �3.84 � 1028 J

25. mSaturn � 5.7 � 1026 kg, rSaturn � 6.0 � 107 mEquating two equations for kinetic energy,

mv2 �

v � ��v � ��v � 2.5 � 104 m/s

If an object is orbiting Saturn, it must have aminimum speed of 2.5 � 104 m/s.

26. mM � 7.35 � 1022 kg, r � rM � 100 kmr � 1.738 � 106 m � 1 � 105 m r � 1.838 � 106 m

vesc � ��vesc � ��vesc � 2.31 � 103 m/sThe escape speed from the Moon at a heightof 100 km is 2.31 km/s.

27. According to Kepler’s third law,

�

T2 �

T � ��T � 7071 s

It would take the Apollo spacecraft 7071 s or1 h 58 min to complete one orbit around theMoon.

28. dM�S � 2.28 � 1011 m, rM � 3.43 � 106 m, mM � 6.37 � 1023 kg, mS � 2.0 � 1030 kga) Orbital speed is given by:

v � ��v � ��v � 24.2 km/s

(6.67 � 10�11 N·m2/kg2)(2.0 � 1030 kg)��

2.28 � 1011 m

GM�

r

4(3.14)2(1.838 � 106 m)3

��(6.67 � 10�11 N·m2/kg2)(7.35 � 1022 kg)

42r3

�GmMoon

GM�42

r3

�T2

2(6.67 � 10�11 N·m2/kg2)(7.35 � 1022 kg)��

1.838 � 106 m

2GmMoon�r

(6.67 � 10�11 N·m2/kg2)(5.7 � 1026 kg)��

6 � 107 m

GM�

r

GMm�

2r1�2

(6.67 � 10�11 N·m2/kg2)(5.98 � 1024 kg)(7.35 � 1022 kg)��

3.82 � 108 m

GMm�

r

r3

�T2

1��6.70 � 106 m

1��6.73 � 106 m

1�r1 � rE

1�r2 � rE

�GMm�r1 � rE

�GMm�r2 � rE


b) h � 80 km � 8 � 104 m

v � ��v � ��v � 3.48 km/sThe speed required to orbit Mars at an alti-tude of 80 km is 3.48 km/s.

29. mM � 7.35 � 1022 kg, rM � 1.738 � 106 mEscape speed is given by:

vesc � ��vesc � ��vesc � 2.38 km/s

30. Three waves pass in every 12 s, with 2.4 mbetween wave crests.

f �

f � �12

3s

�

f � 0.25 Hz31. k � 12 N/m, m � 230 g � 0.23 kg,

A � 26 cm � 0.26 mAt the maximum distance, i.e., A, v � 0,therefore the total energy is:

E � �12

�kA2

Also, at the equilibrium point, the displace-ment is zero, therefore the total energy is thekinetic energy:

E � �12

�mv2

Hence,

�12

�kA2 � �12

�mv2

v � ��v � ��v � 1.88 m/s

The speed of the mass at the equilibriumpoint is 1.88 m/s.

32. m � 2.0 kg, x � 0.3 m, k � 65 N/m

a) E � �12

�kx2

E � �12

�(65 N/m)(0.3 m)2

E � 2.925 JInitial potential energy of the spring is2.925 J.

b) Maximum speed is achieved when the totalenergy is equal to kinetic energy only. Therefore,

E � �12

�mv2

2.925 J � �12

�(2.0 kg)v2

v � �2.925�v � 1.71 m/s

The mass reaches a maximum speed of1.71 m/s.

c) x � 0.20 mTotal energy of the mass at this location isgiven by:

E � �12

�mv2 � �12

�kx2

2.925 J � �12

�(2.0 kg)v2 �

�12

�(65 N/m)(0.2 m)2

v � �1.625�v � 1.275 m/s

The speed of the mass when the displace-ment is 0.20 m is 1.275 m/s.

33. Given the information in problem 32,a) Maximum acceleration is achieved when

the displacement is maximum sinceF � kx and F � maTherefore, maximum displacement isx � 0.30 mHence,ma � kx

a �

a �

a � 9.75 m/s2

The mass’ maximum acceleration is9.75 m/s2.

(65 N/m)(0.30 m)��

(2.0 kg)

kx�m

(12 N/m)(0.26 m)2

��0.23 m

kA2

�m

number of waves��

time

2(6.67 � 10�11 N·m2/kg2)(7.35 � 1022 kg)��

1.738 � 106 m

2GM�

r

(6.67 � 10�11 N·m2/kg2)(6.37 � 1023 kg)��

3.43 � 106 m � 8 � 104 m

Gm�r � h


b) x � 0.2 m

a �

a �

a � 6.5 m/s2

The mass’ acceleration when the displace-ment is 0.2 m is 6.5 m/s2.

34. dtide � 15 m, mfloats � m, spanfloats � 10 km, Ttide � 12 h 32 min � 45 120 sa) Finding the work done by the upward

movement of the floats,Wup � FgdWup � m(9.8 m/s2)(15 m)Wup � 147m JSince there is a downward movement aswell,Wup, down � 2Wup

Wup, down � 294m JSince the linkages are only 29% efficient,Wactual � 0.29(294m J)Wactual � 85.26m JTo find power:

P �

P �

P � 1.89 � 10�3m WP � 1.89m mWThe power produced would be 1.89m mW.

b) 1.89m mW from the hydroelectric linkagesis not even comparable to 900 MW from areactor at Darlington Nuclear PowerStation. In order for the linkages to pro-duce the same power, the total mass of thefloats would have to be 4.76 � 1011 kg, or476 million tonnes.

35. m � 100 kg, d � 12 m,x � 0.64 cm � 0.0064 mFirst, we must find the speed at which themass first makes contact with the spring.Using kinematics,v2 � vo

2 � 2adv � �vo

2 � 2�ad�v � �0 � 2(�9.8 m/�s2)(12� m)�v � 15.34 m/s

Finding the maximum kinetic energy of themass (instant before compression of spring),

Ekmax � �12

�mv2

Ekmax � �12

�(100 kg)(15.34 m/s)2

Ekmax � 11 760 JSince kinetic energy is fully converted to elas-tic potential energy when the spring is fullycompressed,

Epmax � �12

�kx2

k �

k �

k � 5.7 � 108 N/mThe spring constant is 5.7 � 108 N/m.

36. k � 16 N/m, A � 3.7 cmSince total energy is equal to maximum poten-tial energy, maximum amplitude � x at thepoint of maximum potential energy:Ep � Etotal

Ep � �12

�kx2

Ep � �12

�(16 N/m)(0.037 m)2

Ep � 0.011 JThe total energy of the system is 0.011 J.

37. mbullet � 5 g � 0.005 kg, mmass � 10 kg, k � 150 N/m, vo bullet � 350 m/sTo find the final velocity, use the law of con-servation of linear momentum:

po � pf

(0.005 kg)(350 m/s) � 0 � (10.005 kg)vv � 0.175 m/s

Therefore, the mass and bullet’s kineticenergy is:

Ek � �12

�mv2

Ek � �12

�(10.005 kg)(0.175 m/s)2

Ek � 0.153 J

2(11 760 J)��(0.0064 m)2

2Epmax�x2

85.26m J��45 120 s

W�

t

(65 N/m)(0.2 m)��

2.0 kg

kx�m


Since all of this energy is transferred to elasticpotential energy,

Ep � Ek

�12

�kx2�0.153 J

x � ��x � 0.045 m

38. b � 0.080 kg/s, m � 0.30 kg, xo � 8.5 cm,

x � xoe��

2mbt�

a) t � 0.1 s

x � (8.5 cm)ex � 8.39 cm

b) t � 1.5 s

x � (8.5 cm)ex � 6.96 cm

c) t � 15.5 s

x � (8.5 cm)ex � 1.076 cm

d) t � 3.0 min � 180 s

x � (8.5 cm)ex � 3.21 � 10�10 cm

e) t � 5.2 h � 18 720 s

x � (8.5 cm)ex � 0 cm

39. x � �12

�xo

Hence,

x � xoe��

2mbt�

�12

�xo � xoe��

2mbt�

0.5 � e��(

20(.00.8300

kkgg/)s)t

�

ln 0.5 �

t � 5.2 sTherefore, the time required for the amplitudeto reach one-half its initial value is 5.2 s.

40. k � 100 N/m

E � �12

�kxo2e�

�

mbt�

a) Initial energy:

Eo � �12

�kxo2e�

�

mbt�(t � 0 s)

Eo � �12

�kxo2e0

Eo � �12

�kxo2

One-half of the initial energy is:

�12

��12

�kxo2� � �

14

�kxo2

Therefore, the time required for the energyto reach this value is:

Ef � �12

�kxo2e�

�

mbt�

�14

�kxo2 � �

12

�kxo2e�

�

mbt�

�12

� � e��

mbt�

ln��12

��

mbt�

ln��12

��

t � 2.6 sTherefore, it takes 2.6 s for the mechanicalenergy to drop to one-half of its initialvalue.

b) i) t � 0.1 s

E � �12

�(100 N/m)(0.085 m)2e

E � 0.352 Jii) t � 22.3 s

E � �12

�(100 N/m)(0.085 m)2e

E � 9.45 � 10�4 Jiii) t � 2.5 min � 150 s

E � �12

�(100 N/m)(0.085 m)2e

E � 1.53 � 10�18 Jiv) t � 5.6 a � 176 601 600 s

E � �12

�(100 N/m)(0.085 m)2e

E � 0 J

�(0.080 kg/s)(176 601 600 s)��

0.30 kg

�(0.080 kg/s)(150 s)��

0.30 kg

�(0.080 kg/s)(22.3 s)��

0.30 kg

�(0.080 kg/s)(0.1 s)��

0.30 kg

�(0.080 kg/s)t��

0.3 kg

�(0.080 kg/s)t��

0.30 kg

�(0.080 kg/s)(18 720 s)��

2(0.30 kg)

�(0.080 kg/s)(180 s)��

2(0.30 kg)

�(0.080 kg/s)(15.5 s)��

2(0.30 kg)

�(0.080 kg/s)(1.5 s)��

2(0.30 kg)

�(0.080 kg/s)(0.1 s)��

2(0.30 kg)

2(0.153 J)��150 N/m


Chapter 7

17. a) � 0.0175 rad

b) 90° � �14

�2 rad

90° � �

2� rad

c) � 3.84 rad

d) � 8.01 rad

e) � 20.9 rad

18. a) (15.3 rev)(2 rad/rev) � 96.1 rad

b) turn � � rad

c) 4.4 h � � 2.3 rad

d) 28.5 h � � 7.46 rad

19. a) 0 rad � 0°

b) � rad�(57.3°/rad) � 120°

c) (20 rad)(57.3°/rad) � 3600°d) (466.6 rad)(57.3°/rad) � 2.67 � 104°

20. a) � 0.56 cycles

b) rad � �12

� cycle

c) 50° � � 0.14 cycle

d) 450° � � 1.25 cycles

21. a) s � r�

s � (40 m)(2 rad)s � 80 m

b) s � r�

s � (40 m)(6.7 rad)s � 268 m

c) � 2.16 rad

s � r�

s � (40 m)(2.16 rad)s � 86 m

d) � 9.77 rad

s � r�

s � (40 m)(9.77 rad)s � 3.9 � 102 m

22. a) � � (15 cycles)(2 rad/cycle)� � 30 rad

b) �t � 3.5 s

� �

� �

� � 27 rad/sc) � and � become negative.

23. �t � 26 s� � (4 cycles)(2 rad/cycle)� � 8 rad

� �

� �

� � 0.97 rad/s

24. a) � � � �

� � 178.0 rad/s

b) � �

�� t�� (178.0 rad/s)(0.56 s)�� 1.0 � 102 rad

25. a) �1 � 0�2 � 2.55 rad/s�t � 115 s

� �

� �

� � 0.0222 rad/s2

b) fmax �

fmax � 0.406 Hz

26. �1 � � �

�1 � 4.7 rad/s�2 � 0�t � 22.5 s

60 s�1 min

2 rad�1 rev

45 rev�1 min

2.55 rad/s��2 rad/cycle

2.55 rad/s � 0��

115 s

(�2 � �1)��t

��t

1 min�60 s

2 rad�

rev1700 rev�

1 min

8 rad � 0��

26 s

��t

30 rad � 0��

3.5 s

��t

560°��57.3°/rad

124°��57.3°/rad

1 cycle�360°

1 cycle�360°

3.5 rad��2 rad/cycle

2�3

2 rad�

24 h

2 rad�

12 h

3�2

2 rad�

turn3�4

1200°��57.3°/rad

459°��57.3°/rad

220°��57.3°/rad

1°��57.3°/rad


� �

� �

� � �0.21 rad/s2

27. �1 � 18.0 rad/s�2 � 0�t � 22.0 s

a) � �

� �

� � �0.818 rad/s2

b) �22 � �1

2 � 2��

��

��

�� 198 rad

c) number of cycles �

number of cycles � 31.5d) �2 � �1 � ��t

�2 � 18.0 rad/s � (�0.818 rad/s2)(8.7 s)�2 � 11 rad/s

28. � � 0.95 rad/s2

�1 � �1.2 rad/sa) �t � 0.30 s

�2 � �1 � ��t�2 � �1.2 rad/s � (0.95 rad/s2)(0.30 s)�2 � �0.92 rad/s

b) �t � 1.26 s�2 � �1 � ��t�2 � �1.2 rad/s � (0.95 rad/s2)(1.26 s)�2 � �3.0 � 10�3 rad/s

c) �t � 13.5 s�2 � �1 � ��t�2 � �1.2 rad/s � (0.95 rad/s2)(13.5 s)�2 � 12 rad/s

29. r � 0.028 mv � 0.12 m/sv � r�

� � �vr

�

� �

� � 4.3 rad/s

Therefore, the angular speed of the reel isapproximately 4.3 rad/s.

30. r � 0.50 m� � (3.5 rev/s)(2 rad/rev)� � 22 rad/sac � r�2

ac � (0.50 m)(22 rad/s)2

ac � 2.4 � 102 m/s2

31. ac � 7.98 m/s2

r �

r � 1.25 � 103 m

a) ac �

v � �acr�v � �(7.98�m/s2)(�1.25 �� 103 m�)�v � 99.9 m/s

b) � � 0

c) � � �vr

�

� �

� � 0.0799 rad/snumber of revolutions �

� 24 h �

number of revolutions � 1.10 � 103

d) � � � 45 min �

� � 216 rads � r�

s � (1.25 � 103 m)(216 rad)s � 2.70 � 105 m

32. c � 3.0 � 108 m/sr � 0.80 m�d � 2(10.0 km)�d � 20 000 m

a) �t �

�t �

�t � 6.7 � 10�5 s

��

�� 0.0174 rad

1°��57.3°/rad

20 000 m��3.0 � 108 m/s

�d�

c

60 s�1 min

0.0799 rad��

1 s

3600 s�

1 h0.0799 rad/s��2 rad/rev

99.9 m/s��1.25 � 103 m

v2

�r

2.50 � 103 m��

2

0.12 m/s��0.028 m

198 rad��2 rad/cycle

�(18.0 rad/s)2

��2(�0.818 rad/s2)

��12

�2�

(0 � 18.0 rad/s)��

22.0 s

(�2 � �1)��t

0 � 4.7 rad/s��

22.5 s

(�2 � �1)��t


� �

� �

� � 2.6 � 102 rad/sb) v � r�

v � (0.80 m)(2.6 � 102 rad/s)v � 2.1 � 102 m/s

33. Both people are travelling at the same angularspeed but in the opposite direction.Therefore, they will meet halfway, after each

person has travelled �

2� rad.

� �

�t �

�t �

�t � 1.2 s34. �A � �1.3 rad/s

�B � 1.6(1.3 rad/s)�B � 2.08 rad/s

�A � �B �

�B � � �A

tA � tB

�

�

(2.08 rad/s)�A � (1.3 rad/s) � (1.3 rad/s)�A

�A � 1.208 radt � tA

t �

t �

t � 0.93 s35. �1 � 4.2 rad/s

� � 1.80 rad/s2

�t � 2.8 sa) �2 � �1 � ��t

�2 � 4.2 rad/s � (1.80 rad/s2)(2.8 s)�2 � 9.2 rad/s

b) �� 1�t � �12

��t2

�� (4.2 rad/s)(2.8 s) �

�12

�(1.8 rad/s2)(2.8 s)2

�� 19 rad36. �1 � 190 rad/s

�2 � �80 rad/s�t � 6.4 s

a) � �

� �

� � �42 rad/s2

b) �� t

�� (6.4 s)

�� 3.5 � 102 radc) (3.5 � 102 rad)(57.3°/rad) � 2.0 � 104°d) �2 � 0

� � �42 rad/s2

�t �

�t �

�t � 4.5 s37. � � 3.8 rad/s2

�t � 3.5 s�� 110 rad

a) �� 1�t � ��t2

�1 �

�1 �

�1 � 24.77 rad/s�1 � 25 rad/s

b) �22 � �1

2 � 2��

�2 � �(25 ra�d/s)2 �� 2(3.8�rad/s2)�(110 ra�d)��2 � 38.22 rad/s�2 � 38 rad/s

110 rad � �12

�(3.8 rad/s2)(3.5 s)2

��3.5 s

�� 12

��t2

��t

1�2

0 � 190 rad/s��

�42 rad/s2

�2 � �1��

(�80 rad/s � 190 rad/s)��

2

(�2 � �1)��2

�80 rad/s � 190 rad/s��

6.4 s

�2 � �1��t

1.208 rad��1.3 rad/s

�A��A

� �A��2.08 rad/s

�A��1.3 rad/s

�B��B

�A��A

�

2� rad

��1.3 rad/s

��

��t

0.0174 rad��6.7 � 10�5 s

��t


c) � �

� �

� � 3.8 rad/s2

d) �� t

�� (3.5 s)

�� 110 rad

�� 2�t � �12

��t2

�� (38.22 rad/s)(3.5 s) �

�12

�(3.8 rad/s2)(3.5 s)2

�� 110 rad

38. �1 � � �

�1 � 41.9 rad/s�t � 1.2 sa) �� (10 turns)(2 rad/turn)

�� 20 rad

b) �� t

�2 � � �1

�2 � � 41.9 rad/s

�2 � 62.8 rad/s�2 � 63 rad/s

c) � �

� �

� � 17 rad/s2

39. �� 2.0 � 104 rad�1 � 3.5 � 103 rad/s�2 � 2.5 � 104 rad/s

a) �� t

�t �

�t �

�t � 1.4 s

b) � �

� �

� � 1.5 � 104 rad/s2

40. � � 1.5 � 104 rad/s2 (from 39b)�1 � 0�2 � 3.5 � 104 rad/s

�t �

�t �

�t � 2.3 s41. �2 � 15 rad/s

�t � 3.4 s� � 2.3 rad/s2

a) �� 2�t � �12

��t2

�� (15 rad/s)(3.4 s) � �12

�(2.3 rad/s2)(3.4 s)2

�� 38 rad

b) � �

�1 � �2 � ��t�1 � 15 rad/s � (2.3 rad/s2)(3.4 s)�1 � 7.2 rad/s

42. TM � 5.94 � 107 sTE � 3.16 � 107 sheadstart � 30°

headstart � rad

rad � �M�t � �E�t

�t �

�t �

�t � 5.63 � 106 s43. �A � 0.380 rad/s

�B � 0.400 rad/s�A � 0.080 rad/s2

�B � 0

��

6� rad

��

��5.924

�

ra1d07 s

�� 3.126

�

ra1d07 s

��

��

6� rad

��M � �E

�6

�6

�2 � �1��t

3.5 � 104 rad/s � 0��

1.5 � 104 rad/s2

�2 � �1��

(2.5 � 104 rad/s) � (3.5 � 103 rad/s)��

1.4 s

�2 � �1��t

2(2.0 � 104 rad)��(3.5 � 103 rad/s) � (2.5 � 104 rad/s)

2��1 � �2

(�1 � �2)��2

62.8 rad/s � 41.9 rad/s��

1.2 s

�2 � �1��t

2(20 rad)��

1.2 s

2��t

(�1 � �2)��2

1 min�60 s

2 rad�

rev400 rev�1 min

(38.22 rad/s � 24.77 rad/s)��

2

(�2 � �1)��2

38.22 rad/s � 24.77 rad/s��

3.5 s

�2 � �1��t


headstart �

headstart � 0.436 rad��A � 0.436 rad � ��B

�A�t � �A�t2 � 0.436 rad � �B�t

0 � �A�t2 � �A�t � �B�t � 0.436 rad

0 � (0.080 rad/s2)�t2 � (0.380 rad/s)�t �

(0.400 rad/s)�t � 0.436 rad0 � (0.040 rad/s2)�t2 � (0.020 rad/s)�t �

0.436 radUse the quadratic formula:

�t �

�t � 3.56 s

44. I � �12

�mr2

Ia � 14 kg·m2

Ib � 4.8 kg·m2

Ic � 6.8 kg·m2

The order is a, c, b.45. Ia � mr2

Ib � �12

�mr2

Ic � �12

�mr2

Id � �112�(3m)l2

Id � �112�(3m)(2r)2

Id � mr2

Ie � �25

�mr2

Ie � �25

�(2m)��12

�r�2

Ie � �15

�mr2

The order is a and d, b and c, e.46. m � 4200 kg

r � 0.3 m

I � �12

�mr2

I � �12

�(4200 kg)(0.3 m)2

I � 189 kg·m2

47. m � 3.5 kga) I � mr2

I � (3.5 kg)(0.21 m)2

I � 0.15 kg·m2

b) I � �12

�mr2

I � �12

�(3.5 kg)(0.21 m)2

I � 0.077 kg·m2

c) I � �25

�mr2

I � �25

�(3.5 kg)(0.25 m)2

I � 0.088 kg·m2

d) I � �12

�mr2

I � �12

�(3.5 kg)(0.50 m)2

I � 0.44 kg·m2

48. m � 1.4 kgr � 0.12 m

a) I � �12

�mr2

I � �12

�(1.4 kg)(0.12 m)2

I � 0.010 kg·m2

b) � � (60 rev/s)(2 rad/rev)� � 120 rad/s� � 377 rad/s

49. m � 10.0 kg

ri � �12

�(0.54 m)

ri � 0.27 m

re � �12

�(0.54 m)(1.4)

re � 0.378 m

I � �12

�m(ri2 � re

2)

I � �12

�(10.0 kg)[(0.27 m)2 � (0.378 m)2]

I � 1.08 kg·m2

50. m � 2.0 kgr � 1.5 m

a) I � �23

�mr2

I � �23

�(2.0 kg)(1.5 m)2

I � 3.0 kg·m2

0.020 rad/s � �(�0.0�20 rad�/s)2 �� 4(0.0�40 rad�/s2)(��0.436�rad)��

2(0.040 rad/s2)

1�2

1�2

1�2

25°��57.3°/rad


b) � � � �

� � 20.9 rad/s

c) I � �25

�mr2

I � �25

�(2.0 kg)(1.5 m)2

I � 1.8 kg·m2

51. m � 20 kgr � 0.9 m�2 � 0�1 � (12.3 rev/s)(2 rad/rev)�1 � 77.3 rad/sWR � ��

WR � I��

WR � (mr2)� �� t�WR � �

12

�mr2(0 � �1)(�1 � 0)

WR � ��12

�(20 kg)(0.9 m)2(77.3 rad/s)2

WR � �4.8 � 104 J52. m � 1450 kg

r �

r � 0.675 m� � 1.40 rad/s

a) I � �12

�mr2

I � �12

�(1450 kg)(0.675 m)2

I � 330 kg·m2

b) Erot � �12

�I�2

Erot � �12

�(330 kg·m2)(1.40 rad/s)2

Erot � 3.24 � 102 Jc) v � r�

v � (0.675 m)(1.40 rad/s)v � 0.945 m/s

d) � �

�� t�� (1.40 rad/s)(6.5 s)�� 9.1 rad

number of turns �

number of turns � 1.453. m � 5.98 � 1024 kg

r � 6.38 � 106 m�t � 3.16 � 107 s�� 2 rad

a) Erot � �12

�I�2

Erot � �12

��25

�mr2��2

Erot � �15

�mr2� �2

Erot � �15

�(5.98 � 1024 kg)(6.38 � 106 m)2

� �2

Erot � 1.92 � 1024 Jb) v � r�

v � (6.38 � 106 m)� �v � 1.27 m/s

54. m � 8.30 � 10�25 kgr � 3.5 ma) Ie � mr2

Ie � (8.30 � 10�25 kg)(3.5 m)2

Ie � 1.0 � 10�23 kg·m2

b) � �

� � 6.3 � 103 rad/s

c) Ek � �12

�mv2

Ek � �12

�m(r�)2

Ek � �12

�(8.30 � 10�25 kg)(3.5 m)2

(6.3 � 103 rad/s)2

Ek � 2.0 � 10�16 J55. me � 9.11 � 10�31 kg

mn � 1.67 � 10�27 kgr � 5.0 � 10�11 mL � 1.05 � 10�34 kg·m2/sa) I � mer2

I � (9.11 � 10�31 kg)(5.0 � 10�11 m)2

I � 2.3 � 10�51 kg·m2

(1000 cycles)(2 rad/cycle)��

1.0 s

2 rad��3.16 � 107 s

2 rad��3.16 � 107 s

��t

9.1 rad��2 rad/turn

��t

1.35 m�

2

(�1 � �2)��2

��t

1 min�60 s

2 rad�1 rev

200 rev�1 min


b) L � I�

� �

� �

� � 4.6 � 1016 rad/s

c) Erot � �12

�I�2

Erot � �12

�(2.3 � 10�51 kg·m2)

(4.6 � 1016 rad/s)2

Erot � 2.4 � 10�18 J56. r � 0.20 m

h1 � 2.5 mh2 � 0v1 � 0�1 � 0

a) mgh1 � �12

�mv12 � �

12

�I�12

� mgh2 � �12

�mv22 � �

12

�I�22

mgh1 � �12

�mv22 � �

12

��12

�mr2��22

gh1 � �12

�v22 � �

14

�r2� �2

gh1 � �12

�v22 � �

14

�v22

gh1 � �34

�v22

v2 � ��gh1�v2 � ��(9.8 m�/s2)(2.�5 m)�v2 � 5.7 m/s

b) �2 �

�2 �

�2 � 29 rad/s57. r � 0.20 m

h1 � 2.5 mh2 � 0v1 � 0�1 � 0

a) mgh1 � �12

�mv12 � �

12

�I�12

� mgh2 � �12

�mv22 � �

12

�I�22

mgh1 � �12

�mv22 � �

12

�(mr2)�22

gh1 � �12

�v22 � �

12

�r2� �2

gh1 � �12

�v22 � �

12

�v22

gh1 � v22

v2 � �gh1�v2 � �(9.8 m�/s2)(2.�5 m)�v2 � 4.9 m/s

b) �2 �

�2 �

�2 � 25 rad/s

58. mgh1 � �12

�mv12 � �

12

�I�12

� mgh2 � �12

�mv22 � �

12

�I�22

mgh1 � �12

�mv22 � �

12

��25

�mr2�� 2

gh1 � �12

�v22 � �

15

�v22

v2 � ��gh1�59. l � 2.8 m

r � 2.8 mh1 � 2.8 mh2 � 0v1 � 0�1 � 0

mgh1 � �12

�mv12 � �

12

�I�12

� mgh2 � �12

�mv22 � �

12

�I�22

mgh1 � �12

�mv22 � �

12

��13

�ml2�� 2

gh1 � �12

�v22 � �

16

�v22

gh1 � �23

�v22

v2 � ��gh1�v2 � ��(9.8 m�/s2)(2.�8 m)�v2 � 6.4 m/s

3�2

3�2

v2�r

10�7

v2�r

4.9 m/s�0.20 m

v2�r

v2�r

5.7 m/s�0.20 m

v2�r

4�3

4�3

v2�r

1.05 � 10�34 kg·m2�s��

2.3 � 10�51 kg·m2

L�I


60. m � 3.9 kgr � 0.13 m� � 150 rad/s

a) I � �25

�mr2

I � �25

�(3.9 kg)(0.13 m)2

I � 0.0264 kg·m2

b) L � I�L � (0.0264 kg·m2)(150 rad/s)L � 4.0 kg·m2/s

61. m � 2.4 kgr � 0.30 m�1 � 0�2 � 250 rad/s�t � 3.5 s

a) I � �12

�mr2

I � �12

�(2.4 kg)(0.30 m)2

I � 0.108 kg·m2

b) �� 2 � �1

�� 250 rad/s � 0�� 250 rad/s

c) �L � L2 � L1

�L � I��

�L � (0.108 kg·m2)(250 rad/s)�L � 27 kg·m2/s

d) � �

� �

� � 71.4 rad/s2

e) � � I�� (0.108 kg·m2)(71.4 rad/s2)� � 7.7 N·m

62. I � 1.50 � 10�3 kg·m2

�d � 4.5 m�� (3.0 turns)(2 rad/turn)�� 6.0 rada) v � 17.0 m/s

�t �

�t �

�t � 0.2647 s�t � 0.26 s

b) � �

� �

� � 71 rad/sc) L � I�

L � (1.50 � 10�3 kg·m2)(71 rad/s)L � 0.11 kg·m2/s

63. l � 2.5 mm � 3.2 kg�t � 13 sr � 0.010 m�� (28 turns)(2 rad/turn)�� 56 rad

a) I � �12

�mr2

I � �12

�(3.2 kg)(0.010 m)2

I � 1.6 � 10�4 kg·m2

b) L � I�

L � (1.6 � 10�4 kg·m2)� �L � 2.2 � 10�3 kg·m2�s

64. l � 2.5 mm � 3.2 kg�t � 13 s�� (28 turns)(2 rad/turn)�� 56 rad

a) I � �112�ml2

I � �112�(3.2 kg)(2.5 m)2

I � 1.667 kg·m2

I � 1.7 kg·m2

b) L � I�

L � (1.667 kg·m2)� �L � 22 kg·m2�s

65. m � 3.2 kgl1 � 2.5 m

l2 � � 0.5 m

l2 � 0.75 m

2.5 m�

2

56 rad�

13 s

56 rad�

13 s

6.0 rad�0.2647 s

��t

4.5 m��17.0 m/s

�d�

v

250 rad/s��

3.5 s

��t


I � Icm � ml22

I � �112�ml1

2 � ml22

I � 1.7 kg·m2 � (3.2 kg)(0.75 m)2

I � 3.5 kg·m2

66. �1 � 6.85 rad/s�2 � 4.40 rad/sI�1 � xI�2 where x is the factor by which

the moment of inertia changes.

x �

x �

x � 1.5667. I1�1 � I2�2

I1�1 � ��12

�I1��2

�1 � �12

��2

2�1 � �2

Therefore, the angular speed will increase by afactor of 2.

68. Im � 1.5 � 10�3 kg·m2

Is � 8.5 kg·m2

�s � 10 rad/sa) Is�s � Im�m

�m �

�m �

�m � 5.7 � 104 rad/sb) (10 rad)(57.3°/rad) � 573°c) (5.7 � 104 rad)(57.3°/rad) � 3.3 � 106°

d) 45° � �

4� rad

�t �

�t �

�t � 0.0785 s��m � �m�t��m � (5.7 � 104 rad/s)(0.0785 s)��m � 4.45 � 103 rad

number of rotations �

number of rotations � 7.1 � 102

69. rp � 4.3 mrt � 4.3 mmp � 600 kg�p � 6.4 rad/smt � 35 kga) Ip�p � (Ip � It)�f

�f �

�f �

�f �

�f � 5.7 rad/sb) �t � 3.1 rad/s

Ip�p � It�t � (Ip � It)�f

�f �

�f �

�f �

�f � 6.0 rad/sc) �t � �6.4 rad/s

�f �

�f �

�f �

�f � 5.0 rad/s70. m1 � 30 kg

r1 � 1.5 mm2 � 20 kgr2 � 1.0 m�1 � �12 rad/s

�12

�(600 kg)(4.3 m)2(6.4 rad/s) � (35 kg)(4.3 m)2(�6.4 rad/s)

��

��12

�(600 kg)(4.3 m)2 � (35 kg)(4.3 m)2�

�12

�mprp2�p �mtrt

2�t

��

��12

�mprp2 � mtrt

2�

Ip�p � It�t��(Ip � It)

�12

�(600 kg)(4.3 m)2(6.4 rad/s) + (35 kg)(4.3 m)2(3.1 rad/s)

��

��12

�(600 kg)(4.3 m)2 � (35 kg)(4.3 m)2�

�12

�mprp2�p � mtrt

2�t

��

��12

�mprp2 + mtrt

2�

Ip�p � It�t��(Ip � It)

�12

�(600 kg)(4.3 m)2(6.4 rad/s)��

��12

�(600 kg)(4.3 m)2 � (35 kg)(4.3 m)2�

�12

�mprp2�p

��

��12

�mprp2 � mtrt

2�

Ip�p�(Ip � It)

4.45 � 103 rad��2 rad/rotation

�

4� rad

�10 rad/s

��s��s

(8.5 kg·m2)(10 rad/s)��

(1.5 � 10�3 kg·m2)

Is�s�Im

6.85 rad/s��4.40 rad/s

�1��2


a) I1�1 � (I1 � I2)�f

�f �

�f �

�f �

�f � �9.2 rad/sb) I1�1-i � I2�2-i � (I1 � I2)�f

�f �

�f �

�f �

�f � �12 rad/s

c) �f �

�f �

�f �

�f � �6.5 rad/sd) I1�1-i � I2�2-i � (I1 � I2)�f

�2-i �

�2-i �

�2-i �

�2-i � 40 rad/s71. I � 250 kg·m2

r1 � 2.5 mr2 � 1.5 m�t-1 � 2.0 rad/smd � 40 kgI1�1 � I2�2

�2 �

�2 �

�2 �

�2 � 2.9 rad/s

73. m � 0.135 kgI � 8.50 � 10�5 kg·m2

r � 0.0030 m�d � 1.10 m�1 � 0v1 � 0

a) a �

a �9.8 m/s2

�1 � �a � 0.138 m/s2

b) �d � v1�t � �12

�a�t2

�t � ��t � ��t � 3.99 s

c) a �

v2 � v1 � a�tv2 � 0 � (0.138 m/s2)(3.99 s)v2 � 0.551 m/s

d) v2 � r�2

�2 �

�2 �

�2 � 184 rad/s

e) Ek(final) � �12

�mv22

Ek(final) � �12

�(0.135 kg)(0.551 m/s)2

Ek(final) � 0.0205 J

f) Erot(final) � �12

�I�22

Erot(final) � �12

�(8.50 � 10�5 kg·m2)(184 rad/s)2

Erot(final) � 1.43 Jg) ETotal(initial) � mgh1

ETotal(initial) � (0.135 kg)(9.80 m/s2)(1.10 m)ETotal(initial) � 1.46 J

0.551 m/s��0.0030 m

v2�r

(v2 � v1)��t

2(1.10 m)��0.138 m/s2

2�d�

a

8.50 � 10�5 kg·m2

��(0.135 kg)(0.0030 m)2

g��

�1 � �m

Ir2��

[250 kg·m2 � (40 kg)(2.5 m)2](2.0 rad/s)��

(250 kg·m2 � (40 kg)(1.5 m)2)

(I � mdr12)�1��

(I � mdr22)

I1�1�I2

�(30 kg)(1.5 m)2(�12 rad/s)��

(20 kg)(1.0 m)2

�m1r12�1-i��

m2r22

�I1�1-i�I2

(30 kg)(1.5 m)2(�12 rad/s) � (20 kg)(1.0 m)2(12 rad/s)��

((30 kg)(1.5 m)2 � (20 kg)(1.0 m)2)

m1r12�1-i � m2r2

2�2-i��(m1r1

2 � m2r22)

I1�1-i � I2�2-i��(I1 � I2)

(30 kg)(1.5 m)2(�12 rad/s) � (20 kg)(1.0 m)2(�12 rad/s)��

((30 kg)(1.5 m)2 � (20 kg)(1.0 m)2)

m1r12�1-i � m2r2

2�2-i��(m1r1

2 � m2r22)

I1�1-i � I2�2-i��(I1 � I2)

(30 kg)(1.5 m)2(�12 rad/s)��((30 kg)(1.5 m)2 � (20 kg)(1.0 m)2)

m1r12�1��

(m1r12 � m2r2

2)

I1�1�(I1 � I2)


74. m � 0.135 kgI � 8.50 � 10�5 kg·m2

r � 0.0030 m�d � 1.10 mv1 � 1.0 m/s

a) a �

a �9.8 m/s2

�1 � �a � 0.138 m/s2

b) �d � v1�t � �12

�a�t2

0 � �12

�a�t2 � v1�t � �d

0 � �12

�(0.138 m/s2)�t2 �

(1.0 m/s)�t � (1.10 m)0 � (0.0690 m/s2)�t2 �

(1.0 m/s)�t � (1.10 m)Use the quadratic formula:

�t �

�t � 1.03 s

c) a �

v2 � v1 � a�tv2 � 1.0 m/s � (0.138 m/s2)(1.03 s)v2 � 1.14 m/s

d) v2 � r�2

�2 �

�2 �

�2 � 380 rad/s

e) Ek(final) � �12

�mv22

Ek(final) � �12

�(0.135 kg)(1.14 m/s)2

Ek(final) � 0.088 J

f) Erot(final) � �12

�I�22

Erot(final) � �12

�(8.50 � 10�5 kg·m2)(380 rad/s)2

Erot(final) � 6.16 J

g) ETotal(initial) � mgh1 � �12

�mv12 � �

12

�I�12

ETotal(initial) � (0.135 kg)(9.80 m/s2)(1.10 m) �

�12

�(0.135 kg)(1.0 m/s)2 �

�12

�(8.50 � 10�5 kg·m2)

� �2

ETotal(initial) � 6.24 J

1.0 m/s��0.0030 m

1.14 m/s��0.0030 m

v2�r

(v2 � v1)��t

�1.0 m/s � �(1.0 m�/s)2 �� 4(0.0�690 m�/s2)( �� 1.10 �m)��

2(0.0690 m/s2)

8.50 � 10�5 kg·m2

��(0.135 kg)(0.0030 m)2

g��

�1 � �m

Ir2��


Chapter 833. Positive signs: protons

Negative signs: electrons34. a) No charge

b) Negativec) Positived) No chargee) Positive

35. a) Negativeb) Positivec) Negatived) Positive

36. a) Negativeb) Electrons

37. a) Glass: positive; silk: negativeb) Since they have opposite charges, they will

be attracted38. a) Insulator (non-metallic)

b) Conductor (conducts lightning to ground)c) Insulator (non-metallic)d) Insulator (non-metallic)e) Insulator (non-metallic)f) Insulator (non-metallic)

39. Dog hair is positive since a silk shirt rubbedwith wool socks would have a negative charge.

40. a) The electroscope becomes positive becauseit gives up some electrons to the glass rodto reduce the rod’s deficit of electrons. Thisis called charging by contact.

b) The leaves become positively charged aswell. In charging by contact, the chargedobject receives the same charge as thecharging rod.

c) Negative charges will enter the leaves if thesystem is grounded.

41. 1 C � 6.25 � 1018 e�, q � 15 Cq � (15 C)(6.25 � 1018 e�/C)q � 9.38 � 1019 e�

42. q � 1.1 �Cq � 1.1 � 10�6 Cq � (1.1 � 10�6 C)(6.25 � 1018 e�/C )q � 6.9 � 1012 e�

43. The electroscope has an overall positivecharge:q � �4.0 � 1011 e�

q � (�4.0 � 1011 e�)(�1.602 � 10�19 C/e�)q � �6.4 � 10�8 C

44. q � (�5.4 � 108 e�)

q � (�5.4 � 108 e�)(�1.602 � 10�19 C/e�)

q � �4.3 � 10�11 C45. qn � �2.4 � 10�12 C

(2.4 � 10�12 C)(6.25 � 1018 e/C) � 1.5 � 107 elementary chargesThis means that there are 1.5 � 107 protonsin the nucleus, so the neutral atom must havean equal number of electrons: 1.5 � 107.

46. Fe �

a) Fe1 �

Fe1 �

Fe1 � Fe

b) Fe2 �

Fe2 �

Fe2 � 4Fe

c) Fe3 � Fe

Fe2 � Fe

47. Each sphere loses half of its charge to balancewith its identical neutral sphere.

q1� � �12

�q1, q2� � �12

�q2

Fe1 �

Fe2 �

Fe2 �

Fe2 �kq1q2�4r2

2

k��12

�q1��12

�q2��

r22

kq�1q�2�r2

2

kq1q2�r2

1

1�4

4�16

4kqq�

r2

k(2q)(2q)��

r2

1�16

kqq�16r2

kqq�(4r)2

kqq�

r2

1�2

1�2


But Fe2 � Fe1

�

�

r22 �

Therefore, r2 � �12

�r1

The spheres should be placed one-half theiroriginal distance apart to regain their originalrepulsion.

48. r � 100 pm � 100 � 10�12 m � 1.00 � 10�10 m,q1 � q2 � 1.602 � 10�19 C

Fe �

Fe �

Fe � 2.3 � 10�8 N

49. r � 25.0 cm � 0.250 m, Fe � 1.29 � 10�4 N,q1 � q2 � q � qo

a) Fe �

q � ��q � ��q � 3.00 � 10�8 C

b) q is �23

� the original charge on each sphere.

qo � �32

�q

qo � �32

�(3.00 � 10�8 C)

qo � 4.5 � 10�8 CThe type of charge, positive or negative,does not matter as long as they are boththe same. (Like charges repel.)

50. q1 � �q, q2 � �3qqT � q � (�3q) � �2q

So q1� � q2� � � �q

�

�

�

� ��13

�

The magnitude of Fe2 is �13

�Fe1, and in the

opposite direction of Fe1.51. a)

b) q1 � �1.602 � 10�19 C, q2 � �1.602 � 10�19 C, m � 9.1 � 10�31 kg, g � �9.8 m/s2

Fg � Fe

mg �

r � ��r � ��r � 5.1 m

52. q1 � �2.0 � 10�6 C, q2 � �3.8 � 10�6 C, q3 � �2.3 � 10�6 Ca) r1 � 0.10 m, r2 � 0.30 m

1Fe3 �

1Fe3 �

1Fe3 � �4.14 N (attraction)

1F��e3 � 4.14 N [right]

2Fe3 �

2Fe3 �

2Fe3 � �0.87 N (repulsion)

2F��e3 � 0.87 N [left]

(9.0 � 109 N·m2/C2)(3.8 � 10�6 C)(2.3 � 10�6 C)��

(0.30 m)2

kq2q3�r2

2

(9.0 � 109 N·m2/C2)( � 2.0 � 10�6 C)(2.3 � 10�6 C)��

(0.10 m)2

kq1q3�r2

1

�(9.0 � 109 N·m2/C2)(1.602 � 10�19 C)2

��(9.1 � 10�31 kg)(�9.8 m/s2)

kq1q2�mg

kq1q2�r2

e

p

Fe

Fg

Fe2�Fe1

q2

��3q2

Fe2�Fe1

(�q)(�q)��(�q)(�3q)

Fe2�Fe1

��kq�

r12

q�2��

��kqr12

q2��Fe2�Fe1

� 2q�

2

(1.29 � 10�4 N)(0.25 m)2

��(9.0 � 109 N·m2/C2)

Fer2

�k

kqq�

r2

2�3

(9.0 � 109 N·m2/C2)(1.602 � 10�19 C)2

��(1.00 � 10�10 m)2

kq1q2�r2

r21�

4

1�r2

1

1�4r2

2

kq1q2�r2

1

kq1q2�4r2

2


F��eT � 4.14 N [right] � 0.87 N [left]F��eT � 3.3 N [right]

b) r1 � 0.30 m, r2 � 0.10 m

1Fe3 �

1Fe3 �

1Fe3 � �0.46 N (attraction)

1F��e3 � 0.46 N [left]

2Fe3 �

2Fe3 �

2Fe3 � � 7.86 N (repulsion)

2F��e3 � 7.86 N [right]F��eT � 0.46 N [left] � 7.86 N [right]F��eT � 7.4 N [right]

c) 1Fe3 � �4.14 N (attraction)

1F��e3 � 4.14 N [left]

2Fe3 � 7.86 N (repulsion)

2F��e3 � 7.86 N [left]F��eT � 4.14 N [left] � 7.86 N [left]F��eT � 12 N [left]

d) The third charge could only be placed tothe left or to the right of the two basiccharges for the forces to balance and give aforce of 0.For the charge to be placed a distance of rx

metres to the left of the first charge:�1Fe3 � 2Fe3

� �

� �

(3.8 � 10�6)rx2 � (2.0 � 10�6)(4.0 � 10�2

� 4.0 � 10�1rx � rx2)

(3.8 � 10�6)rx2 � 8.0 � 10�8 � 8.0 �

10�7rx � 2.0 � 10�6rx2

Rearranging:1.8 � 10�6rx

2 � 8.0 � 10�7rx � 8.0 � 10�8

� 0Solve for rx using the quadratic formula.

rx �

So rx � 0.53 m or �0.084 m.

Therefore, the charge must be placed0.53 m to the left of the first charge. Theother answer, �0.084 m, would place thecharge between the two base charges andtherefore is an inappropriate answer. For acharge placement to the right of the twocharges, two inappropriate answers are cal-culated, meaning that the only possibleplacement for the charge is at 0.53 m to theleft of the first charge.

53. The forces on the test charge from the repul-sion by the other two charges must equal oneanother for the test charge to come to restthere. The force of charge 1 on the test charge(1Fqt) must equal the force of charge 2 on thetest charge (2Fqt).

1Fqt � 2Fqt

�

�

r2 � r2

Therefore, the net force on the charge would

be 0 if it was placed �13

� of the distance

between the two charges.54. q2 � q1 � q3 � �1.0 � 10�4 C,

r1 � r2 � r3 � 0.40 m

For q1: The force is the vector sum of twoforces, 2F��e1 and 3F��e1. These two magnitudesmust have the same value.

2Fe1 �

2Fe1 �

2Fe1 � 5.6 � 102 N � 3Fe1

F2eT � 2F2

e1 � 3F2e1 � 2(2Fe1)(2Fe1)(cos 120°)

FeT � �2(5.6�� 102�C)2 ��2(5.6�� 102�C)2(co�s 120°�)�

FeT � 9.7 � 102 N

(9.0 � 109 N·m2/C2)(1.0 � 10�4 C)2

��(0.40 m)2

kqq�

r2

q

q1

q1

q32

40 cm

30°

120°

30°

4r2

�4(9)

r2

�9

k4qqt�

��23

�r�2

kqqt�

��13

�r�2

�(�8.0 � 10�7) � �(�8.0� � 10��7)2 � 4�(1.8 �� 10�6)�(�8.0� � 10��8)��

2(1.8 � 10�6)

(3.8 � 10�6 C)��(2.0 � 10�1 m � rx)2

( � 2.0 � 10�6 C)��

rx2

kq2q3��(0.20 m � rx)2

kq1q3�rx

2

(9.0 � 109 N·m2/C2)(3.8 � 10�6 C)(2.3 � 10�6 C)��

(0.10 m)2

kq2q3�r2

2

(9.0 � 109 N·m2/C2)( � 2.0 � 10�6 C)(2.3 � 10�6 C)��

(0.30 m)2

kq1q3�r2

1


From the isosceles triangle with angles of 30°,the total angle is 30° � 60° � 90°.F��eT1 � 9.7 � 102 N [up]F��eT2 � 9.7 � 102 N [left 30° down]F��eT3 � 9.7 � 102 N [right 30° down]Each force is 9.7 � 102 N [at 90° from theline connecting the other two charges].

55. a) l � 2.0 � 10�2 m, q1 � q2 � q3 � q4 � �1.0 � 10�6 C

2Fe1 �

2Fe1 �

2F��e1 � 22.5 N [left]

4F��e1 � 22.5 N [up]

3Fe1 �

3Fe1 �

3F��e1 � 11.25 N [left 45° up]From Pythagoras’ theorem:

2Fe1 � 4Fe1 � �2(22.5� N)2�2F��e1 � 4F��e1 � 31.82 N [left 45° up]Therefore,F��eT1 � (31.82 N � 11.25 N) [left 45° up]F��eT1 � 43.1 N [left 45° up]F��eT2 � 43.1 N [right 45° up]F��eT3 � 43.1 N [right 45° down]F��eT4 � 43.1 N [left 45° down]Each force is 43.1 N [symmetrically out-ward from the centre of the square].

b) The force on the fifth charge is 0 Nbecause the forces from each charge arebalanced.

c) Sign has no effect. If the new fifth chargewere either positive or negative, the attrac-tive/repulsive forces would still balanceone another.

56.

57. The field is similar to the one above, but isnow asymmetrical and has its inflectionpoints pushed farther to the right.

58. Parallel plates: Coaxial cable:

59. q � �2.2 � 10�6 C, Fe � 0.40 N

� �

� �

� � 1.8 � 105 N/C60. Fe � 3.71 N, � � 170 N/C

q �

q �

q � 2.2 � 10�2 C

3.71 N�170 N/C

Fe��

0.40 N��2.2 � 10�6 C

Fe�q

+ –

(9.0 � 109 N·m2/C2)( � 1.0 � 10�6 C)2

��2(2.0�� 10��2 m)2�

kqq�

r23

(9.0 � 109 N·m2/C2)( � 1.0 � 10�6 C)2

��(2.0 � 10�2 m)2

kqq�

r22

q

q

q

q

2.0 cm

2.0 cm


61. q1 � �4.0 � 10�6 C, q2 � �8.0 � 10�6 C, r � 2.0 m

� � �

� � �

� � 3.6 � 104 N/CTherefore, the field strength is 3.6 � 104 N/Ctowards the smaller charge.

62. a) q � 2.0 � 10�6 C, F��e � 7.5 N [left]

��

��

�� 3.8 � 106 N/C [left]b) q2 � �4.9 � 10�5 C

Take right to be positive.F��e � q��

Fe � (�4.9 � 10�5 C)(3.8 � 106 N/C)Fe � �1.86 � 102 NThe force would be 1.86 � 102 N [left].

63. r � 0.5 m, q � 1.0 � 10�2 C

� �

� �

�� 3.6 � 108 N/C [left]64. q1 � 4.0 � 10�6 C, q2 � �1.0 � 10�6 C

Take right to be positive.�� 2 � ��1

� � �

�� 3.25 � 105 N/C [right]

65. r � 5.3 � 10�11 m, q � 1.602 � 10�19 C

� �

� �

� � 5.1 � 1011 N/C

66. rT � 0.20 m, q1 � �1.5 � 10�6 C, q2 � �3.0 � 10�6 Cr2

2 � (0.20 � r1)2

�1 � �2

�

�

r22 � 2r2

1

Substitute for r22 and rearrange:

0 � r12 � 0.4r1 � 4.0 � 10�2

r1 �

r1 � 8.3 � 10�2 m, therefore,r2 � 1.17 � 10�1 m � 1.2 � 10�1 m� � 0 at 1.2 � 10�1 m from the larger charge,or 8.3 � 10�2 m from the smaller charge.

67. q1 � q2 � q3 � q4 � �1.0 � 10�6 C, r � 0.5 m

Since the magnitudes of all four forces areequal, and they are paired with forces in theopposite direction (F��e2 � �F��e4 and F��e1 � �F��e3), there is no net force. Therefore,there is no net field strength.� � 0 N/C

68. q1 � q2 � �2.0 � 10�5 C, r � 0.50 m

�1 �

�1 �

�1 � 7.2 � 105 N/C�1 � �2 and ��T � ��1 � ��2

Therefore, �T � �2(�1)2�� 2(�1�)2(cos�120°)��T � 1.2 � 106 N/C [at 90° from the line con-necting the other two charges]

(9.0 � 109 N·m2/C2)(2.0 � 10�5 C)��

(0.50 m)2

kq�r2

q

q

0.50 m

1

2P

P

q

q

q

q

0.5 m

0.5 m

�0.4 � �(0.4)2�� 4(��4.0 ��10�2)��

2

3.0 � 10�6 C��

r22

1.5 � 10�6 C��

r21

kq2�r2

2

kq1�r2

1

(9.0 � 109 N·m2/C2)(1.602 � 10�19 C)��

(5.3 � 10�11 m)2

kq�r2

(9.0 � 109 N·m2/C2)( � 1.0 � 10�6 C)��

(0.30 m)2

(9.0 � 109 N·m2/C2)(4.0 � 10�6 C)��

(0.40 m)2

(9.0 � 109 N·m2/C2)(1.0 � 10�2 C)��

(0.5 m)2

kq�r2

7.5 N [left]��2.0 � 10�6 C

F��e�q

(9.0 � 109 N·m2/C2)(4.0 � 10�6 C)��

��2.02

m��

2

(9.0 � 109 N·m2/C2)(8.0 � 10�6 C)��

��2.02

m��

2

kq1�

��12

�r�2

kq2�

��12

�r�2


69. q � 0.50 C, ∆V � 12 VW � q∆VW � (0.50 C)(12 V)W � 6.0 J

70. W � 7.0 � 102 J, ∆V � 6.0 V

q �

q �

q � 1.2 � 102 C71. q � 1.5 � 10�2 C, Fe � 7.5 � 103 N,

∆d � 4.50 cm � 4.50 � 10�2 m

�V �

�V �

�V �

�V � 2.3 � 104 V72. � � 130 N/C, Fe � 65 N, ∆V � 450 V

W �

W �

W �

W � 2.3 � 102 J73. d � 0.30 m, q � �6.4 � 10�6 C

V �

V �

V � 1.9 � 105 V74. a) q1 � �1.0 � 10�6 C, q2 � �5.0 � 10�6 C,

r � 0.25 m

Ee �

Ee �

Ee � 0.18 J (repulsion)

b) Ee1 �

Ee1 �

Ee1 � 0.045 J (repulsion)W � ∆Ee

W � Ee2 � Ee1

W � 0.18 J � 0.045 JW � 0.14 J

75. Position in the field has no bearing on thefield strength.� � 5.0 � 103 N/C, d � 5.0 cm � 5.0 � 10�2 mV � d�

V � (5.0 � 10�2 m)(5.0 � 103 N/C)V � 2.5 � 102 V

76. a) q � 1 � 10�5 C, � � 50 N/CFe � q�

Fe � (1 � 10�5 C)(50 N/C)Fe � 5.0 � 10�4 N

b) ∆d � 1.0 m∆Ek � W∆Ek � Fe∆d∆Ek � (5.0 � 10�4 N)(1.0 m)∆Ek � 5.0 � 10�4 J

c) v � 2.5 � 104 m/s

Ek � mv2

m �

m �

m � 1.6 � 10�12 kg77. d1 � 1.0 � 10�9 m, d2 � 1.0 � 10�8 m,

q1 � q2 � �1.602 � 10�19 C, m1 � m2 � 9.11 � 10�31 kg�Ee � E2 � E1

�Ee � �

�Ee � kq1q2� � ��Ee � �2.08 � 10�19 JTherefore, the electric potential energy wasreduced by 2.08 � 10�19 J, which was trans-ferred to kinetic energy. The energy is spreadover both electrons, so the energy for eachelectron is 1.04 � 10�19 J.

1�d1

1�d2

kq1q2�d1

kq1q2�d2

2(5.0 � 10�4 J)��(2.5 � 104 m/s)2

2Ek�v2

1�2

(9.0 � 109 N·m2/C2)( � 1.0 � 10�6 C)( � 5.0 � 10�6 C)��

1.00 m

kq1q2�r

(9.0 � 109 N·m2/C2)( � 1.0 � 10�6 C)( � 5.0 � 10�6 C)��

0.25 m

kq1q2�r

(9.0 � 109 N·m2/C2)(6.4 � 10�6 C)��

0.30 m

kq��d

(450 V)(65 N)��

(130 N/C)

�VFe��

�V�

q

(7.5 � 103 N)(4.5 � 10�2 m)��

1.5 � 10�2 C

Fe�d�

q

W�q

7.0 � 102 J��

6.0 V

W��V


For one electron:

Ek � �12

�mv2

v � ��v � ��v � 4.78 � 105 m/s

78. V2 � 2V1 and Ek � ∆Ee � qVWith the same charge on each electron, thekinetic energy is also doubled, i.e., Ek2 � 2Ek1

�

� 2

v22 � 2v2

1

v2 � �2�v1

Therefore, the speed is 1.41 times greater.79. a) V � 15 kV � 1.5 � 104 V, P � 27 W,

1 C � 6.25 � 1018 e

number of electrons/s �

number of electrons/s � (27 J/s)� �(6.25 � 1018 e/C)

number of electrons/s � 1.1 � 1016

b) q � 1.602 � 10�19 C, m � 9.11 � 10�31 kgAccelerating each electron from rest,

Ek � �Ee

�12

�mv2 � Vq

v � ��v � ��v � 7.3 � 107 m/s

80. a) d � 1.2 m, V � 7.5 � 103 V, m � 3.3 � 10�26 kgFe � q�

ma �

a �

a �

a � 3.0 � 1010 m/s2

b) E � VqE � (1.602 � 10�19 C)(7.5 � 103 V)E � 1.202 � 10�15 J

c) At this speed and energy, relativistic effectsmay be witnessed. Although the speed maynot be what is predicted by simple mechan-ics, the total energy should be the same butmay be partly contributing to a massincrease of the ion.

81. q1 � q2 � �1.602 � 10�19 C, m1 � m2 � 1.67 � 10�27 kg, v1 � v2 � 2.7 � 106 m/s∆Ek � ∆Ee

The total energy for both ions is:

(2)�12

�mv2 �

r �

r �

r � 1.9 � 10�14 m82. a) q� � �2e, m� � 6.696 � 10�27 kg,

v1v � 0 m/s, v1h � 6.0 � 106 m/s, V � 500 V, dv � 0.03 m, dh � 0.15 mAcceleration is toward the negative plate:

a �

a �

a �

a �

a � 7.97 � 1011 m/s2

2(1.602 � 10�19 C)(500 V)��(6.696 � 10�27 kg)(3.0 � 10�2 m)

qV�mdv

q��m

Fe�m

(9.0 � 109 N·m2/C2)(1.602 � 10�19 C)2

��(1.67 � 10�27 kg)(2.7 � 106 m/s)2

kq1q2�mv2

kq1q2�r

(1.602 � 10�19 C)(7.5 � 103 V)��

(3.3 � 10�26 kg)(1.2 m)

qV�md

qV�

d

2(1.5 � 104 V)(1.602 � 10�19 C)��

9.11 � 10�31 kg

2Vq�

m

1 C��1.5 � 104 J

P(6.25 � 1018 e/C)��

V

�12

�mv22

��12

�mv21

2Ek1�Ek1

Ek2�Ek1

2(1.04 � 10�19 J)��9.11 � 10�31 kg

2Ek�m


Time between the plates is:

t �

t �

t � 2.5 � 10�8 sTherefore,

�dh � �12

�at2

�dh � �12

�(7.97 � 1011 m/s2)(2.5 � 10�8 s)2

�dh � 2.5 � 10�4 m�dh � 0.025 cmThe alpha particle is 3.0 cm � 0.025 cm � 2.975 cm from thenegative plate if it enters at the positiveplate or 1.475 cm from the negative plateif it enters directly between the two plates.

b) v2v � v1v � atv2v � 0 � (7.97 � 1011 m/s2)(2.5 � 10�8 s)v2v � 2.0 � 104 m/sFrom Pythagoras’ theorem,v2 � �(6.0 �� 106 m�/s)2 �� (2.0 �� 104 m�/s)2�v2 � 6.0 � 106 m/s

83. d � 0.050 m, V � 39.0 V

� �

� �

� � 7.80 � 102 N/C84. � � 2.85 � 104 N/C,

d � 6.35 cm � 6.35 � 10�2 mV � d�

V � (6.35 � 10�2 m)(2.85 � 104 N/C)V � 1.81 � 103 V

85. a) m � 2mP � 2mn � 4(1.67 � 10�27 kg), g � 9.80 N/kg, q � �2eFe � Fg

� �

� �

� � 2.04 � 10�7 N/C

b) d � 3.0 cm � 3.0 � 10�2 mV � d�

V � (3.0 � 10�2 m)(2.04 � 10�7 N/C)V � 6.1 � 10�9 V

86. d � 0.12 m, V � 92 V

� �

� �

� � 7.7 � 102 N/C87. � � 3 � 106 N/C, d � 1.0 � 10�3 m

V � d�

V � (1.0 � 10�3 m)(3 � 106 N/C)V � 3 � 103 VTherefore, 3.0 � 103 V is the maximum poten-tial difference that can be applied. Exceedingit would cause a spark to occur between theplates.

88. V � 50 V, � � 1 � 104 N/C

d �

d �

d � 5.0 � 10�3 m89. V � 120 V, � � 450 N/C

d �

d �

d � 2.67 � 10�1 m90. a) m � 2.2 � 10�15 kg, d � 5.5 � 10�3 m,

V � 280 V, g � 9.80 N/kgFe � Fg

q� � mg

� mg

q �

q �

q � 4.2 � 10�19 C

b) N �

N � 2.63 e � 3 eThe droplet has three excess electrons.

4.2 � 10�19 C��1.602 � 10�19 e/C

(2.2 � 10�15 kg)(9.80 N/kg)(5.5 � 10�3 m)��

280 V

mgd�

V

qV�

d

120 V�450 N/C

V��

50 V��1 � 104 N/C

V��

92 V�0.12 m

V�d

4(1.67 � 10�27 kg)(9.80 N/kg)��

2(1.602 � 10�19 C)

mg�

q

39.0 V�0.050 m

V�d

0.15 m��6.0 � 106 m/s

dh�vh


91. V � 450 V, me � 9.11 � 10�31 kg, e � 1.602 � 10�19 Ca) ∆Ee � qV

∆Ee � (1.602 � 10�19 C)(450 V)∆Ee � 7.21 � 10�17 JTherefore,

Ek � �Ee

�12

�mv2 � �Ee

v � ��v � ��v � 1.26 � 107 m/s

b) �12

�mv2 � �13

��Ee

v � ��v � ��v � 7.26 � 106 m/s

92. k � 6.0 � 10�3 N/m, d � 0.10 m, V � 450 V,x � 0.01 m

a) � �

� �

� � 4.5 � 103 N/Cb) The force to deform one spring is:

F � kxF � (6.0 � 10�3 N/m)(0.01 m)F � 6.0 � 10�5 NThe force to deform both springs is: 2(6.0 � 10�5 N) � 1.2 � 10�4 N

c) The force on the pith ball must also be1.2 � 10�4 N

d) Fspring � Fe

Fspring � q�

q �

q �

q � 2.7 � 10�8 C

1.2 � 10�4 N��4.5 � 103 N/C

Fspring��

450 V�0.10 m

V�d

2(7.21 � 10�17 J)��3(9.11 � 10�31 kg)

2�Ee�3m

2(7.21 � 10�17 J)��9.11 � 10�31 kg

2�Ee�m


Chapter 922. I � 12.5 A

B � 3.1 � 10�5 T

B �

r �

r �

r � 8.1 � 10�2 m23. r � 12 m

I � 4.50 � 103 A

B �

B �

B � 7.5 � 10�5 T24. I � 8.0 A

B � 1.2 � 10�3 TN � 1

B �

r �

r �

r � 4.2 � 10�3 m25. N � 12

r � 0.025 mI � 0.52 A

B �

B �

B � 1.6 � 10�4 T

26. � �

� 3500 turns/m

I � 4.0 A

B �

B � � I

B � (4 � 10�7 T·m/A)� �(4.0 A)

B � 1.8 � 10�2 TNOTE: The solutions to problem 27 are basedon a distance between the two conductors of1 cm.

27. a)

Referring to the above diagram, the mag-netic fields will cancel each other outbecause the field from each wire is of thesame magnitude but is in the oppositedirection.

b)

I � 10 Ar � 1.0 � 10�2 m

B �

B �

B � 2.0 � 10�4 TBut this field strength (2.0 � 10�4 T) is foreach of the two wires. Referring to theabove diagram, the two fields flow in thesame direction when the current in thetwo wires moves in the opposite direction.The result is that the two fields will add toproduce one field with double the strength(4.0 � 10�4 T).

28. Coil 1:N � 400L � 0.1 mI � 0.1 ACoil 2:N � 200L � 0.1 mI � 0.1 A

(4 � 10�7 T·m/A)(10 A)��

2(1.0 � 10�2 m)

�I�2r

F F

Currents in opposite directions—wires forced apart

x

F F

Currents in the same direction—wires forced together

3500 turns��

1 m

N�L

�NI�

L

N�L

100 cm�

1 m35 turns�

1 cmN�L

(4 � 10�7 T·m/A)(12)(0.52 A)��

2(0.025 m)

�NI�

2r

(4 � 10�7 T·m/A)(1)(8.0 A)��

2(1.2 � 10�3 T)

�NI�2B

�NI�

2r

(4 � 10�7 T·m/A)(4.50 � 103 A)��

2(12 m)

�I�2r

(4 � 10�7 T·m/A)(12.5 A)��

2(3.1 � 10�5 T)

�I�2B

�I�2r


BTotal � Bcoil1�Bcoil2

BTotal � �

BTotal � �

BTotal � 2.5 � 10�4 T29. The single loop:

Bsingle �

Bsingle �

Solenoid:L � 2rsingle loop

L � 2(0.02 m)L � 0.04

N � � � L

N � (1500 turns/m)(0.04 m)N � 188

Bsol �

Bsol �

Bsol � 7.5 � 10�4 TTo cancel the field, the magnitude of the twofields must be equal but opposite in direction.

Bsol � Bsingle

7.5 � 10�4 T �

I �

I � 24 A30. a) � � 45°

L � 6.0 mB � 0.03 TI � 4.5 AF � BIL sin �F � (0.03 T)(4.5 A)(6.0 m) sin 45°F � 0.57 NThe direction of this force is at 90° to theplane described by the direction of the cur-rent vector and that of the magnetic field,i.e., upwards.

b) If the current through the wire was to bereversed, the magnitude and direction ofthe resultant force would be 0.57 N [down-wards].

31. a) d(linear density) � 0.010 kg/mB � 2.0 � 10�5 T� � 90°

� dg

� (0.010 kg/m)(9.8 N/kg)

� 9.8 � 10�2 N/m (linear weight)

F � BIL sin �

I �

I �

I �

I � 4900 Ab) This current would most likely melt the

wire.32. a) N � 60

I � 2.2 AB � 0.12 T

B � ��

LNI�

L � ��

BNI�

L �

L � 1.38 � 10�3 mF � BIL sin �F � (0.12 T)(2.2 A)(1.38 � 10�3 m)

(sin 90°)F � 3.64 � 10�4 N

b) F � ma

a �

a �

a � 1.46 � 10�2 m/s2

33. B � 0.02 Tv � 1.5 � 107 m/s� � 90°

3.64 � 10�4 N��

0.025 kg

F�m

(4 � 10�7 T·m/A)(60)(2.2 A)��

0.12 T

9.8 � 10�2 N/m��(2.0 � 10�5 T) sin 90°

�FL

�

�B sin �

F�BL sin �

F�L

F�L

F�L

(7.5 � 10�4 T)2(0.02 m)��

4 � 10�7 T·m/A

(4 � 10�7 T·m/A)(1)I��

2(0.02 m)

(4 � 10�7 T·m/A)(188)(0.4 A)��

0.04

�NI�

L

100 cm�

1 m15 turns�

1 cm

(4 � 10�7 T·m/A)(1)I��

2(0.02 m)

�NI�

2r

(4 � 10�7 T·m/A)(200)(0.1 A)��

(0.1 m)

(4 � 10�7 T·m/A)(400)(0.1 A)��

(0.1 m)

�NI�

L�NI�

L


q � 1.602 � 10�19 Cm � 9.11 � 10�31 kg

Fc � FB

� qvB sin �

r �

r �

r � 4.3 � 10�3 m34. qalpha � 2(1.602 � 10�19 C)

qalpha � 3.204 � 10�19 Cv � 2 � 106 m/sB � 2.9 � 10�5 Tmalpha � 2(protons) � 2(neutrons)malpha � 4(1.67 � 10�27 kg)malpha � 6.68 � 10�27 kg

r �

r �

r � 1.4 � 103 m35. Fg � mg

Fg � (9.11 � 10�31 kg)(9.8 N/kg)Fg � 8.9 � 10�30 NFmag � Bqv sin �Fmag � (5.0 � 10�5 T)(1.602 � 10�19 C)

(2.8 � 107 m/s)Fmag � 2.24� 10�16 NThe magnetic force has considerably moreinfluence on the electron.

36. q � 1.5 � 10�6 Cv � 450 m/sr � 0.15 mI � 1.5 A� � 90°F � Bqv sin �

B �

F �

F �

F � 1.3 � 10�9 N

According to the right-hand rules #1 and #3,this charge would always be forced towardsthe wire.

37. a) v � 5 � 107 m/sr � 0.05 mI � 35 Aq � �1.602 � 10�19 CF � Bqv sin �

B �

F �

F �

F � 1.12 � 10�15 NAccording to the right-hand rules #1 and#3, this charge would always be forcedtoward the wire.

b) If the electron moved in the same directionas the current, then it would be forcedaway from the wire.

38. a) v � 2.2 � 106 m/sr � 5.3 � 10�11 mq � �1.602 � 10�19 Cm � 9.11 � 10�31 kgAt any given instant, the electron can beconsidered to be moving in a straight linetangentially around the proton.

Fmag � Fc

qvB sin � �

B �

B �

B � 2.36 � 105 TBut this field would always be met by afield of the same magnitude but oppositedirection when the electron was on theother side of its orbit. Therefore, the netfield strength at the proton is zero.

b) To keep an electron moving in a circularartificially simulated orbit, the scientistmust apply a field strength of 2.36 � 105 T.

(9.11 � 10�31 kg)(2.2 � 106 m/s)��(�1.602 � 10�19 C)(5.3 � 1011 m)

mv�qr

mv2

�r

(4 � 10�7 T·m/A)(35 A)(�1.602 � 10�19 C)(5 � 107 m/s) sin 90°��

2(0.05)

�Iqv sin ��

2r

�I�2r

(4 � 10�7 T·m/A)(1.5 A)(1.5 � 10�6 C)(450 m/s)sin 90°��

2(0.15 m)

�Iqv sin ��

2r

�I�2r

(6.68 � 10�27 kg)(2 � 106 m/s)��(3.204 � 10�19 C)(2.9 � 10�5 T)

mv�qB

(9.11 � 10�31 kg)(1.5 � 107 m/s)��

(1.602 � 10�19 C)(0.02 T)

mv�qB

mv2

�r


39. � � 475 V/mB � 0.1 TThe electron experiences no net force becausethe forces from both the electric and magneticfields are equal in magnitude but opposite indirection.If all the directions are mutually perpendicu-lar, both the electric and magnetic fields willmove the electron in the same direction (basedon the right-hand rule #3). Therefore,Fmag � Fe

qvB � q�

v �

v �

v � 4750 m/s40. B � 5.0 � 10�2 T

d � 0.01 mv � 5 � 106 m/sq � 1.602 � 10�19 CFmag � Fe

qvB � q�

qvB � q

V � dvBV � (0.01 m)(5 � 106 m/s)(5.0 � 10�2 T)V � 2500 V

41. r � 3.5 mI � 1.5 � 104 A

F �

F �

F � 2.44 � 103 N42. L � 0.65 m

I � 12 AB � 0.20 TF � BIL sin �F � (0.20 T)(12 A)(0.65 m)(sin 90°)F � 1.56 N [perpendicular to wire]At the angle shown, the force is:(1.56 N)sin 30° � 0.78 N

43. a) v � 5.0 � 106 m/sr � 0.001 mq � �1.602 � 10�19 Cm � 9.11 � 10�31 kg

Fc � Fmag

� qvB

B �

B �

B � �2.8 � 10�2 Tb) Fc � mac

Fc � qvBmac � qvB

ac �

ac �

ac � 2.5 � 1016 m/s2

44. a) r � 0.22 mB � 0.35 Tq � 1.602 � 10�19 Cm �1.67 � 10�27 kg

�

v �

v �

v � 7.4 � 106 m/s

b) Fc � �m

rv2

�

Fc �

Fc � 4.2 � 10�13 N

(1.67 � 10�27 kg)(7.4 � 106 m/s)2

��0.22 m

(1.602 � 10�19 C)(0.35 T)(0.22 m)��

(1.67 � 10�27 kg)

qBr�

m

v�Br

q�m

(1.602 � 10�19 C)(5.0 � 106 m/s)(0.028 T)��

9.11 � 10�31 kg

qvB�

m

(5.0 � 106 m/s)(9.11 � 10�31 kg)��

( � 1.602 � 10�19 C)(0.001 m)

vm�qr

mv2

�r

(4 � 10�7 T·m/A)(1.5 � 104 A)2(190 m)��

2(3.5 m)

�I1I2L�

2r

V�d

(475 V/m)��

(0.1 T)

��B


45. � 5.7 � 108 C/kg

B � 0.75 T

�

r �

v �

v �

T �

T �

T �

T �

T � 1.5 � 10�8 s46. m � 6.0 � 10�8 kg

q � 7.2 � 10�6 CB � 3.0 T

t � �12

�T

t � � �� t �

t � 8.7 � 10�3 s47. Falling through the top of the loop, the cur-

rent is clockwise.Falling out of the bottom, the current is counterclockwise.

48. a) The conventional current flow is clockwise(looking down from top).

b) The induced magnetic field is linear (downat the south end).

c) Yes, the falling magnet would experience amagnetic force that is opposing its motion,as described by Lenz’s law.

d) No, the results would be exactly the same.The induced current flow would be in theopposite direction if the poles of the mag-net were reversed, but the reduction inspeed would be the same.

49. The copper conductor is cutting through themagnetic field lines as it moves, and thereforeexperiences a force that opposes its motion.The induced and external magnetic fields arein opposite directions, which causes the oppo-sition. Aluminum wire would make no differ-ence as long as it conducts electricity.

(6.0 � 10�8 kg)��(3.0 T)(7.2 � 10�6 C)

2m�

Bq1�2

2��(0.75 T)(5.7 � 108 C/kg)

2�B��

mq��

2��mBq

v��

�v

2r�

v

2r�

T

d�t

mv�Bq

v�Br

q�m

e�m


Chapter 1021. a) � � 4 m

b) A � 5 cmc) T � 8 s

d) f �

f �

f � 0.1 s�1

e) v � �fv � (4 m)(0.1 s�1)v � 0.4 m/s

22. f �

f �

f � 3.125 cycles/s

T �

T �

T � 0.32 s/cycle

23. f �

f �

f � 1.2 cycles/s

T �

T �

T � 0.83 s/cycle24. f � 60 Hz

T �

T � 0.017 s/cycle

25. a) f �

f � 2.5 Hz

b) T �

T � 0.4 s/cycle26. For 78 rpm:

f �

f � 1.3 Hz

T � �1f�

T � 0.77 s/cycleFor 45 rpm:

f �

f � 0.75 Hz

T �

T � 1.33 s/cycleFor 33�

13

� rpm:

f � rpm

f �

f � 0.56 Hz

T �

T � 1.8 s/cycle27. a) x � A cos �

x � 1 cos (10°)x � 0.98 m

b) x � A cos �x � 1 cos (95°)x � �0.087 m

c) x � A cos �

x � 1 cos � rad�x � �0.71 m

d) x � A cos �x � 1 cos (2π rad)x � 1 m

28.

At equilibrium (x � 0), v is a maximum (sin 90° � 1). When x � A, v is a minimum(sin 0° � 0).

29.

The object always accelerates toward equilib-rium and slows down as it moves away fromequilibrium.

180° 360°θ

a(m

/s2 )

A

180° 360°θ

θ180° 360°

x(m

)v

(m/s

)

–A

3�4

1�f

100 cycles��

180 s

100�

3

1�f

45 cycles�

60 s

78 cycles��

60 s

1�f

150 cycles��

60 s

1�f

1��1.2 cycles/s

1�f

72 cycles��

60 s

cycles�

s

1��3.125 cycles/s

1�f

10 cycles��

3.2 s

cycles�

s

1�8 s

1�T


a is a maximum when x � A (cos 0°); a is aminimum at equilibrium (cos 90°).The a vector is always directed toward theequilibrium position.

30. a) T � 2��T � 2��T � 2.9 s/cycle

b) T � 2��T � 2��T � 18 s/cycle

c) T � 2��T � 2��T � 0.78 s/cycle

31. a) i) T � 2��T � 2��T � 7.2 s/cycle

ii) T � 2��T � 2��T � 44 s/cycle

iii) T � 2��T � 2��T � 1.9 s/cycle

b) i) T � 2��T � 2��T � 1.8 s/cycle

ii) T � 2��T � 2��T � 11 s/cycle

iii) T � 2��T � 2��T � 0.49 s/cycle

32. a) T � 2��T � 2��T � 0.711 s/cycle

b) T � 2��T � 2��T � 0.889 s/cycle

c) T � 2��T � 2��T � 0.204 s/cycle

33. a) f � ��k � 42f 2mk � 42(12 Hz)2(0.402 kg)k � 2.3 � 103 N/m

b) F � kxF � (2.3 � 103 N/m)(0.35 m)F � 8.0 � 102 N

34. a) v � �f

f �

f �

f � 4.62 � 1014 Hz

b) f �

f �

f � 5.00 � 1014 Hz

c) f �

f �

f � 5.17 � 1014 Hz

3.00 � 108 m/s��5.80 � 10�7 m

v��

3.00 � 108 m/s��6.00 � 10�7 m

v��

3.00 � 108 m/s��6.50 � 10�7 m

v��

k�m

1�2

0.21 kg��200 N/m

m�k

0.40 kg�20 N/m

m�k

0.30 kg��23.4 N/m

m�k

0.15 m��24.6 m/s2

L�gJupiter

80 m��24.6 m/s2

L�gJupiter

2.1 m��24.6 m/s2

L�gJupiter

0.15 m�1.6 m/s2

L�gMoon

80 m�1.6 m/s2

L�gMoon

2.1 m�1.6 m/s2

L�gMoon

0.15 m�9.8 m/s2

L�g

80 m�9.8 m/s2

L�g

2.1 m�9.8 m/s2

L�g


d) f �

f �

f � 5.77 � 1014 Hz

e) f �

f �

f � 6.32 � 1014 Hz

f) f �

f �

f � 7.50 � 1014 Hz

35. a) �t �

�t �

�t � 497 s�t � 8.28 min�t � 0.138 h

b) �t �

�t �

�t � 1.3 s�t � 2.1 � 10�2 min�t � 3.5 � 10�4 h

c) �t �

�t �

�t � 1.9 � 104 s�t � 3.2 � 102 min�t � 5.4 h

d) �t �

�t �

�t � 3.03 � 102 s�t � 5.1 min�t � 8.4 � 10�2 h

36. 1 a � (3600 s/h)(24 h/d)(365 d/a)� 3.1536 � 107 s

d � v∆t

d � (3.00 � 108 m/s)(3.1536 � 107 s)d � 9.46 � 1015 m

37. d � 100 light years � 9.46 � 1017 m, v � 3.00 � 108 m/s

�t �

�t � 3.15 � 109 s�t � 100 a

38. �t �

�t �

�t � 5.33 � 10�7 s39. rEarth � 6.38 � 106 m, cEarth � 2π(6.38 � 106 m)

� 4.01 � 107 m

�t �

�t �

�t � 0.134 s40. For the minimum frequency, � � 4 � 10�7 m

f �

f �

f � 8 � 1014 HzFor the maximum frequency, � � 8 � 10�8 m

f �

f �

f � 4 � 1015 HzThus, the range is 8 � 1014 Hz to 4 � 1015 Hz.

41. For the car:∆tcar � (50 h)(3600 s/h)∆tcar � 180 000 sFor light:

�tlight �

�tlight �

�tlight � 0.01 sComparing the two,

� � 1.8 � 107 times180 000 s��

0.01 s�tcar��tlight

4.00 � 106 m��3.00 � 108 m/s

d�v

3.00 � 108 m/s��

8 � 10�8 m

v��

3.00 � 108 m/s��

4 � 10�7 m

v��

4.01 � 107 m��3.00 � 108 m/s

d�v

160 m��3.00 � 108 m/s

d�v

d�v

9.1 � 1010 m��3.00 � 108 m/s

d�v

5.8 � 1012 m��3.00 � 108 m/s

d�v

3.8 � 108 m��3.00 � 108 m/s

d�v

1.49 � 1011 m��3.00 � 108 m/s

d�v

3.00 � 108 m/s��4.00 � 10�7 m

v��

3.00 � 108 m/s��4.75 � 10�7 m

v��

3.00 � 108 m/s��5.20 � 10�7 m

v��


42. a) sin 30° � 0.5b) sin 60° � 0.866c) sin 45° � 0.707d) sin 12.6° � 0.218e) sin 74.4° � 0.963f) sin 0° � 0g) sin 90° � 1

43. a) sin�1 (0.342) � 20°b) sin�1 (0.643) � 40°c) sin�1 (0.700) � 44.4°d) sin�1 (0.333) � 19.5°e) sin�1 (1.00) � 90°

44. v �

v �

v � 3.3 � 108 m/sThis speed is impossible, since it is greaterthan the speed of light.

45. n1 � 1.00, n2 � 1.98, �1 � 2�2

n1 sin �1 � n2 sin �2

sin 2�2 � 1.98 sin �2

2 sin �2 cos �2 � 1.98 sin �2

cos �2 �

�2 � 8.1°46. 1.5 sin 30° � n2 sin 50°

n2 � 0.98As in problem 44, this value is impossible.

47.

48. a) v �

v �

v � 1.24 � 108 m/s

b) v �

v �

v � 1.97 � 108 m/s

c) v �

v �

v � 2.26 � 108 m/s

d) v �

v �

v � 2.31 � 108 m/s

49. Use with n1 � 1.00:

a) 0.413b) 0.658c) 0.752d) 0.769

50. v �

v �

v � 2.26 � 108 m/s

�t �

�t �

�t � 5.31 � 10�5 s51. a) tan �1 � tan �B

tan �1 �

tan �1 � 1.42�1 � 54.8°

b) n1 sin �1 � n2 sin �2

�2 � sin�1 �1.00· ��2 � 35.2°

c) 54.8° (�i � �r)52. Polaroid glasses are most effective when the

light is most polarized. The light is 100%polarized at Brewster’s angle, �B.

tan �B �

tan �B �

�B � 53.1°�elevation � 90° � 53.1°�elevation � 36.9°

1.33�1.00

n2�n1

sin 54.8°��

1.42

n2�n1

1200 m��2.26 � 108 m/s

d�v

3.00 � 108 m/s��

1.33

c�n

n1�n2

3.00 � 108 m/s��

1.30

c�n

3.00 � 108 m/s��

1.33

c�n

3.00 � 108 m/s��

1.52

c�n

3.00 � 108 m/s��

2.42

c�n

wavefrontslight ray

glass

1.98�

2

3.00 � 108 m/s��

0.90

c�n


53. a) I2 � 0.5Io cos2 �

I2 � 0.5Io cos2 30°I2 � 0.375Io

� 37.5%

b) � 20.7%

c) � 5.85%

54. When viewing something through a doublyrefracting crystal, two images are seen, sincetwo rays of polarized light are produced. Ifanother crystal was laid over-top and rotated,nothing would be seen, since the two crystalsnow act as a polarizer�analyzer pair, with anangle of 90° between their axes.

55. Use an analyzer (another polarizer, rotated).

56. tan �B �

�B � tan�1 � ��B � 53°

57. a) tan �B �

�B � tan�1 � ��B � 53.1°

b) tan �B �

�B � tan�1 � ��B � 56.3°

c) tan �B �

�B � tan�1 � ��B � 41.6°

d) tan �B �

�B � tan�1 � ��B � 45.7°

58. tan �B �

tan 60° �

n2 � tan 60°n2 � 1.73

59. The first Polaroid will remove exactly onecomponent (50%) of the incident light. Thethird Polaroid, having been placed at anyangle but 90° to the first one, will remove afraction of the remaining light, allowing onecomponent of the light to pass through. Thesecond Polaroid will then remove only a singlecomponent of the residual light. Thus, a frac-tion of the incident light passes through allthree Polaroids.

60. a) I2 � 0.5Io cos2 �

I2 � 0.5Io cos2 10°I2 � 0.485Io

� 48.5%

b) � 37.5%

c) � 5.85%

d) � 0.380%

61. I2 � 0.4Io

I2 � 0.5Io cos2 �

0.4Io � 0.5Io cos2 �

� � cos�1 �� 26.6°

62. I2 � 0.5Io cos2 �1 and I3 � I2 cos2 �2

I3 � 0.5Io cos2 �1 cos2 �2

I3 � 0.5Io cos2 60° cos2 10°I3 � 0.121Io

� 12.1%I3�Io

0.4�0.5

I2�Io

I2�Io

I2�Io

I2�Io

n2�1.00

n2�n1

1.33�1.30

n2�n1

1.33�1.50

n2�n1

1.50�1.00

n2�n1

1.33�1.00

n2�n1

1.33�1.00

n2�n1

I2�Io

I2�Io

I2�Io


Chapter 1123. a) constructive

b) constructivec) partiald) destructive

24.

25.

26. m � 2� � 550 nma) d � 2.0 � 10�6 m

sin �m �

sin �2 �

sin �2 � 0.55�2 � 33.4°

b) d �

d � 9.52 � 10�5 m

sin �m �

sin �2 �

sin �2 � 0.01155�2 � 0.662°

27. L � 1.0 m

m� �

xm �

x2 �

x2 � 0.55 m28.

29. � � 560 nmd � 4.5 � 10�6 ma) m � 1

sin �m �

sin �1 �

sin �1 � 0.12444�1 � 7.14°

(1)(5.60 � 10�7 m)��

4.5 � 10�6 m

m��

d

1 10

2 2

33

2

11

2

Maximanumbers

Minimanumbers

2(5.50 � 10�7 m)(1.0 m)��

2.0 � 10�6 m

m�L�

d

dxm�L

(2)(5.50 � 10�7m)��

9.52 � 10�5 m

m��

d

1.0 m��10 500 slits

(2)(5.50 � 10�7 m)��

2.0 � 10�6 m

m��

d

2

2

1

1S2

S1

0

1

1

S1

S1


b) sin � �

sin � �

sin � � 0.18667� � 10.8°

c) sin �m �

sin �3 �

sin �3 � 0.37333�3 � 21.9°

d) sin �m �

sin �3 �

sin �3 � 0.43556�3 � 25.8°

30. � � 610 nmm � 2�2 � 23o

d �

d �

d � 3.12 � 10�6 md � 3.12 �m

31. d � 0.15 mmm � 2x2 � 7.7 mL � 1.2 m

� �

� �

� � 4.81 � 10�4 m� � 481 �m

32. � � 585 nmL � 1.25 mm � 9x � 3.0 cm

d �

d �

d � 2.3 � 10�4 md � 0.23 mm

33. � � 630 nmd � 3.0 � 105 m

m �

For maximization, sin � � 1.

m �

m �

m � 4.76 � 1011

34. a) The light now travels an extra ��

4� twice

between the original and the second posi-tions. This produces an extra shift

of ��

2�. The observer therefore sees a dark

band and the fringe pattern moves by halfa band.

b) The light now travels an extra ��

2� twice

between the original and the second posi-tions. This produces an extra shift of �.The observer therefore sees a bright bandand the fringe pattern moves by a fullband.

c) The light now travels an extra �34�� twice

between the original and the second posi-tions. This produces an extra shift

of �32��. The observer therefore sees neither

a bright nor a dark band and the fringe

pattern moves by �32

� of a band.

3.0 � 105 m��6.30 � 10�7 m

d��

d sin ��

�

��129��(5.85 � 10�7 m)(1.25 m)

��3.0 � 10�2 m

�m �12

��L��

x

(1.5 � 10�4 m)(7.7 m)��

(2)(1.2 m)

dxm�mL

(2)(6.10 � 10�7 m)��

sin 23°

m��sin �m

��72

��(5.60 � 10�7 m)��

4.5 � 10�6 m

�m �12

��

��d

(3)(5.60 � 10�7 m)��

4.5 � 10�6 m

m��

d

��32

��(5.60 � 10�7 m)��

4.5 � 10�6 m

�m �12

��

��d


d) The light now travels an extra � twicebetween the original and the second posi-tions. This produces an extra shift of 2�.The observer therefore sees a bright bandand the fringe pattern moves by two fullbands.

35. ∆PD � 4n � 1.42� � 600 nm

t �

t �

t � 2.857 � 10�6 mt � 2.86 �m

36. ∆PD � 12t � 3.60 microns� � 640 nm

n � 1

n � 1

n � 2.0737. ∆PD � 10

vm � 1.54 � 108 m/st � 2.80 microns

nm �

nm �

nm � 1.948nm � 1.95

� �

� �

� � 5.309 � 10�7 m� � 531 nm

38. t � 364 nm� � 510 nmng � 1.40

�g �

�g �

�g � 3.64 � 10�7 m

Because there is a half-phase shift between airand gas,

m �

m �

m � 2.5The interference is destructive.

39. a) 2� � � � �

destructive

b) 2� � � � �

constructive

c) 2� � � � 3�

constructive

d) 2� � � � �

destructive40. ng � 1.40

� � 560 nmt � 4.80 � 10�6 m

�g �

�g �

�g � 4.00 � 10�7 mBecause there is a half-phase shift between airand gas,

m �

m �

m � 24.5The interference is destructive and a darkarea will result.

2(4.80 � 10�6 m) �12

�(4.00 � 10�7 m)��

4.00 � 10�7 m

2t �12

��g

��g

5.60 � 10�7 m��

1.40

��ng

15�2

1�2

7��2

1�2

5��4

1�2

��4

3�2

1�2

��2

2(3.64 � 10�7 m) �12

�(3.64 � 10�7 m)��

3.64 � 10�7 m

2t �12

��g

��g

5.10 � 10�7 m��

1.40

��ng

2(2.80 � 10�6 m)(0.948)��

10

2t(n � 1)��

PD

3.0 � 108 m/s��1.54 � 108 m/s

c�vm

(12)(6.40 � 10�7 m)��

2(3.60 � 10�6 m)

PD��

2t

(4)(6.00 � 10�7 m)��

2(0.42)

PD��2(n � 1)


41. � � 500 nma) nf � 1.44

�f �

�f �

�f � 3.47 � 10�7 mBecause there is a half-phase shift,

m �

t �

t �

t � 8.675 � 10�7 mt � 86.8 nm

b) nf � 1.23

�f � �n�

f�

�f �

�f � 4.07 � 10�7 mBecause the shifts cancel,

t � �m

2�f�

t �

t � 2.03 � 10�7 mt � 203 nm

42. � � 580 nmns � 1.33

�s � �n�

s�

�s �

�s � 4.36 � 10�7 ma) Because of the phase shift and destructive

interference,

t � �m

2�s�

t �

t � 2.18 � 10�7 mt � 218 nm

b) Because of the phase shift and constructiveinterference,

t �

t �

t � 1.09 � 10�7 mt � 109 nm

43. a) � � �vf�

� �

� � 1.40 m

b) � � �vf�

� �

� � 5.20 � 10�7 m� � 520 nm

c) � � �vf�

� �

� � 2.5 m

44. a) f � ��

v�

f �

f � 1.5 � 1020 Hz

b) v � � �

v � 3.889 m/s

f � ��

v�

f �

f � 3.2 Hz

3.889 m/s��

1.2 m

1000 m�

1 km1 h

�3600 s

14 km�

1 h

3.0 � 108 m/s��2.0 � 10�12 m

3.0 � 108 m/s��1.2 � 108 Hz

2.50 � 108 m/s��4.81 � 1014 Hz

350 m/s�250 Hz

�1 � �12

��(4.36 � 10�7 m)��

2

�m � �12

��s

��2

(1)(4.36 � 10�7 m)��

2

5.80 � 10�7 m��

1.33

(1)(4.07 � 10�7 m)��

2

5.00 � 10�7 m��

1.23

�1 � �12

��(3.47 � 10�7 m)��

2

�m � �12

��f

��2

2t �12

��f

��f

5.00 � 10�7 m��

1.44

��nf


46. a) m � 2� � 580 nmw � 2.2 � 10�5 m

sin � �

sin � �

sin � � 0.0659� � 3.78°

b) m � 2� � 550 nmw � 2.2 � 10�5 m

sin �m �

sin �2 �

sin �2 � 0.05�2 � 2.87°

47. w � 1.2 � 10�2 mmm � 1�1 � 4°

� �

� �

� � 8.37 � 10�7 m� � 837 nm

48. L � 1.0 mm � 2� � 837 nmw � 1.2 � 10�2 mmMinimum:

xm �

x2 �

x2 � 0.1395 mx2 � 140 mm

Maximum:

x �

x �

x � 0.174 mx � 174 mm

49. w � 1.1 � 10�5 m� � 620 nmm � 2a) Minimum:

sin �m �

sin �2 �

sin �2 � 0.113�2 � 6.47°

b) Maximum:

sin � �

sin � �

sin � � 0.141� � 8.10°

50. Width of central maximum � 6.6°� � 400 nm

w �

w �

w � 6.949 � 10�6 mw � 6.95 �m

51. � � 585 nmw � 1.23 � 10�3 cmL � 1.2 ma) m � 3

xm �

x3 �

x3 � 0.171 mx3 � 171 mm

(3)(5.85 � 10�7 m)(1.2 m)��

1.23 � 10�5 m

m�L�

w

4.00 � 10�7 m��

sin 3.3°

��sin �

�2 �12

��(6.20 � 10�7 m)��

1.1 � 10�5 m

�m �12

��

��w

(2)(6.20 � 10�7 m)��

1.1 � 10�5 m

m��

w

�2 �12

��(8.37 � 10�7 m)(1.0 m)��

1.2 � 10�5 m

�m �12

��L

��w

(2)(8.37 � 10�7 m)(1.0 m)��

1.2 � 10�5 m

m�L�

w

(1.2 � 10�5 m) sin 4°��

1

w sin �m�m

(2)(5.50 � 10�7 m)��

(2.2 � 10�5 m)

m��

w

�2 �12

��(5.80 � 10�7 m)��

2.2 � 10�5 m

�m �12

��

��w


b) m � 2

x �

x �

x � 0.1426 mx � 143 mm

52. w � 1.10 � 10�3 cm� � 470 nmm � 1

sin � � �

sin � � �

sin � � � 0.0427

� � 4.90°53. � � 493 nm

w � 5.65 � 10�4 mL � 3.5 mm � 1

a) xm �

x � 2

x � 6.1 � 10�3 mx � 6.1 mm

b) sin � � �

sin � � �

sin � � � 8.73 � 10�4

� � 0.10°54. 450 nm55. � � 530 nm

N � 10 000 slitsw � 1 cmm � 1

d � �Nw

�

d �

d � 1 � 10�6 m

sin �m �

sin �1 �

sin �1 � 0.53�1 � 32°

56. � � 650 nmN � 2000 slitsw � 1 cm�m � 11.25°

d �

d �

d � 5.00 � 10�6 m

m � �

m � �

m � 157. m � 2

d �

d � 4.35 � 10�5 mmL � 0.95 m� � 610 nm

xm �

x2 �

x2 � 27 m58. N � 10 000 slits

w � 1.2 cm

d �

d �

d � 1.2 � 10�6 m

1.2 � 10�2 m��

10 000

w�N

(2)(6.10 � 10�7 m)(0.95 m)��

(4.35 � 10�8 m)

m�L�

d

1��2.3 � 104 slits/mm

1�2

(5.00 � 10�6 m) sin 11.25°��

(6.50 � 10�7 m)

1�2

d sin �m��

1 � 10�2 m��

2000

w�N

(1)(5.30 � 10�7 m)��

1 � 10�6 m

m��

d

1 � 10�2 m��

10 000

��2

(1)(4.93 � 10�7 m)��

(5.65 � 10�4 m)��2

m��

w��2

(1)(4.93 � 10�7 m)(3.5 m)��

(5.65 � 10�4 m)

m�L�

w1�2

��2

(1)(4.70 � 10�7 m)��

1.10 � 10�5 m��2

m��

w��2

�2 �12

��(5.85 � 10�7 m)(1.2 m)��

1.23 � 10�5 m

�m �12

��L

��w


a) � � 600 nm

m �

m �

m � 2b) � � 440 nm

m �

m �

m � 2.7m � 2

59. d �

d � 1 � 10�5 m

m400 �

m400 �

m400 � 25

m700 �

m700 �

m700 � 14.3m700 � 14Orders needed:25 � 14 � 11

60. d � 1.0 micronsa) � � 610 nm

m �

m �

m � 1.6m � 1

b) � � 575 nm

m �

m �

m � 1.7m � 1

c) � � 430 nm

m �

m �

m � 2.3m � 2

61. �1 � 589 nm�2 � 589.59 nmw � 2.5 cmN � 104

d �

d �

d � 2.5 � 10�6 m

� � sin�1 � � � sin�1 � �� sin�1 � � �

sin�1 � �� 13.641° � 13.627°� � 1.39 � 10�2°

62. ∆� � �2 � �1

∆� � 5.8959 � 10�7 m � 5.89 � 10�7 m∆� � 5.9 � 10�10 m

�avg �

�avg �

�avg � 5.8930 � 10�7 mm � 2

N �

N �

N � 50063. N � 106

w � 2.5 cm� � 520 nm

d �

d �

d � 2.5 � 10�8 m

2.5 � 10�2 m��

106

w�N

(5.8930 � 10�7 m)��(5.9 � 10�10 m)(2)

�avg��m

(5.89 � 10�7 m) (5.8959 � 10�7 m)��

2

�1 �2�2

5.89 � 10�7 m��2.5 � 10�6 m

5.8959 � 10�7 m��

2.5 � 10�6 m

m�1�d

m�2�d

2.5 � 10�2 m��

104

w�N

1.0 � 10�6 m��4.30 � 10�7 m

d��

1.0 � 10�6 m��5.75 � 10�7 m

d��

1.0 � 10�6 m��6.10 � 10�7 m

d��

1 � 10�5 m��7.00 � 10�7 m

d��

1 � 10�5 m��4.00 � 10�7 m

d��

1 � 10�2 m��

1000

1.2 � 10�6 m��4.40 � 10�7 m

d��

1.2 � 10�6 m��6.00 � 10�7 m

d��


m �

m �

m � 0.0481m � 0

64. N � 4000m � 1∆� � (6.5648 � 10�7 m) � (6.5630 � 10�7 m)∆� � 1.8 � 10�10 m

�avg �

�avg � 6.5639 � 10�7 m

R �

R �

R � 3647Nm � (4000)(1)Nm � 4000R � Nm, therefore it will not be resolved.

65. � � 0.55 nm

d �

d � 4.0 � 10�7 mm � 1

sin � �

sin � �

sin � � 1.375 � 10�3

� � 7.9 � 10�2°Diffraction is not apparent.

66. d � 0.40 nm� � 0.20 nmm � 3

sin � �

sin � �

sin � � 0.75� � 49°

(3)(2.0 � 10�10 m)��2(4.0 � 10�10 m)

m��2d

(1)(5.5 � 10�10 m)��

4.0 � 10�7 m

m��

d

1 m��2.5 � 106

6.5639 � 10�7 m��

1.8 � 10�10 m

�avg��

(6.5648 � 10�7 m) (6.5630 � 10�7 m)��

2

2.5 � 10�8 m��5.20 � 10�7 m

d��


Chapter 1219. �max � 597 nm � 5.97 � 10�7 m

The temperature can be found using Wien’s law:

�max �

T �

T �

T � 4854.27 KT � 4854.27 � 273°CT � 4581.27°C

20. T � 2.7 K�max can be found using Wien’s law:

�max �

�max �

�max � 1.07 � 10�3 m21. T � 125 K

�max can be found using Wien’s law:

�max �

�max �

�max � 2.32 � 10�5 mThe peak wavelength of Jupiter’s cloud is2.32 � 10�5 m. It belongs to the infraredpart of the electromagnetic spectrum.

22. P � 2 W, � � 632.4 nm � 6.324 � 10�7 mWe are to find the number of photons leavingthe laser tube per second. Let us symbolizethis quantity by N�.Using Planck’s equation, we can express theenergy for a single photon:

E� � �h�

c�

The number of photons leaving the tube canbe found as follows:

N� � �EP

�

�

N� � �Ph�

c�

N� �

N� � 6.36 � 1018 photons/s

23. E� � 4.5 eV, W0(gold) � 5.37 eVE� W0. The gold will absorb all of theenergy of the incident photons, hence therewill be no photoelectric effect observed (seeFigure 12.13).

24. � � 440 nm � 4.4 � 10�7 m, W0(nickel) � 5.15 eVFirst, we shall calculate the energy of the inci-dent photons. Using Planck’s equation:

E� �

E� �

E� � 4.52 � 10�19 J

E� �

E� � 2.82 eVSince E� W0, the photoelectric effect willnot be exhibited (see Figure 12.13).

25. P � 30 W, � � 540 nm � 5.4 � 10�7 mWe are to find the number of photons radiatedby the headlight per second. Let us symbolizethis quantity by N�.Using Planck’s equation, we can express theenergy for a single photon:

E� � �h�

c�

The number of photons radiated by the head-light can be found as follows:

N� � �EP

�

�

N� � �Ph�

c�

N� �

N� � 8.15 � 1019 photons/s26. W0 � 3 eV � 4.8 � 10�19 J,

� � 219 nm � 2.19 � 10�7 ma) The energy of photons with cut-off fre-

quency is equal to the work function of themetal. Hence,E� � W0 � 4.8 � 10�19 J

(30 W)(5.4 � 10�7 m)��(6.626 � 10�34 J·s)(3.0 � 108 m/s)

4.52 � 10�19 J��1.6 � 10�19 C

(6.626 � 10�34 J·s)(3.0 � 108 m/s)��

4.4 � 10�7 m

hc��

(2 W)(6.324 � 10�7 m)��(6.626 � 10�34 J·s)(3.0 � 108 m/s)

2.898 � 10�3

��125 K

2.898 � 10�3

��T

2.898 � 10�3

��2.7 K

2.898 � 10�3

��T

2.898 � 10�3

��5.97 � 10�7 m

2.898 � 10�3

��max

2.898 � 10�3

��T


The frequency can be found using Planck’sequation:E� � hf

f �

f �

f � 7.24 � 1014 Hzb) The maximum energy of the ejected pho-

tons can be found using the equation:Ekmax � E� � W0

Ekmax � � W0

Ekmax � �

4.8 � 10�19 JEkmax � 4.28 � 10�19 J

27. a) To avoid unwanted electrical currents andchange in bonding structure of the materialof the satellite, the number of electronsejected from the material should be mini-mal. The greater the work function of themetal, the more photon energy it willabsorb and the fewer electrons will beejected. Hence, the material selected shouldhave a relatively high work function.

b) The longest wavelength of the photons thatcould affect this satellite would have anenergy equal to the work function of thematerial, i.e., E� � W0

Using Planck’s equation E� � �h�

c�,

�max � (if W0 is in Joules)

�max � (if W0 is in eV)

28. W0(platinum) � 5.65 eV � 9.04 � 10�19 JFrom problem 27, we know that:

�max �

�max �

�max � 2.2 � 10�7 mThe maximum wavelength of the photon thatcould generate the photoelectric effect on theplatinum surface is 2.2 � 10�7 m.

29. a) For a material with a work functiongreater than zero, the typical photoelectriceffect graph has a positive x intercept. Ifthe graph passes through the origin, thework function of the material is zero,which means that the photoelectric effectwould be observed with incident photonshaving any wavelength.

b) If the graph has a positive y intercept, wewould observe the photoelectric effectwithout the presence of incident photons.

30. � � 400 pm � 4.0 � 10�10 ma) The frequency of the photon can be found

using the wave equation:

f �

f �

f � 7.5 � 1017 Hzb) The momentum of the photon can be com-

puted using de Broglie’s equation:

p �

p �

p � 1.66 � 10�24 N·sc) The mass equivalence can be found using

de Broglie’s equation:p � mv

m � �pc

�

m �

m � 5.53 � 10�33 kg31. mproton � 1.673 � 10�27 kg

First, we have to express the rest energy of theproton. It can be found using:Eproton � mc2

The energy of the photon, which is equal tothe rest energy of the proton, can be expressedusing Planck’s equation:

E� � �h�

c�

1.66 � 10�24 N·s��

3.0 � 108 m/s

6.626 � 10�34 J·s��

4.0 � 10�10 m

h��

3.0 � 108 m/s��4.0 � 10�10 m

c��

(6.626 � 10�34 J·s)(3.0 � 108 m/s)��

9.04 � 10�19 J

hc�W0

hc�W0e

hc�W0

(6.626 � 10�34 J·s)(3.0 � 108 m/s)��

2.19 � 10�7 m

hc��

4.8 � 10�19 J��6.626 � 10�34 J·s

E��h


Then,Eproton � E�

mc 2 �

mc �

Using de Broglie’s equation:

p �

Hence,p � mcp � (1.673 � 10�27 kg)(3.0 � 108 m/s)p � 5.02 � 10�19 N·s

32. � � 10 �m � 1 � 10�5 mUsing de Broglie’s equation:

p �

p �

p � 6.63 � 10�29 N·s33. �f � 1 nm � 1 � 10�9 m

Consider the following diagram:

From the conservation of energy,Ei � Ef Ek

� mvf2 (eq. 1)

From the conservation of momentum,pi � pf pe

In the direction of the x axis:

cos 43° � � mvf cos � (eq. 2)

In the direction of the y axis:

sin 43° � mvf sin � (eq. 3)

Using math software to solve the system ofequations that consists of equations 1, 2, and3, the value for �i � 9.9552 � 10�10 m.

To find the Compton shift,� � �f � �i

� � 1 � 10�9 m � 9.9552 � 10�10 m� � 4.48 � 10�12 mThe Compton shift is 4.48 � 10�12 m.

34. � � 180°, vf � 7.12 � 105 m/sFrom the conservation of energy,

Ei � Ef Ek

� mvf2 (eq. 1)

From the conservation of momentum,pi � pf pe

� � mvf (eq. 2)

(The negative sign signifies a scatter angle �equal to 180°.)Multiplying equation 2 by c and adding theresult to equation 1,

� mvf2 cmvf

�i �

�i �

�i � 2.04 � 10�9 m35. �i � 18 pm � 1.8 � 10�11 m, energy loss is 67%

The initial energy of the photon can be com-puted using Planck’s equation:

Ei �

Ei �

Ei � 1.1 � 10�14 JSince 67% of the energy is lost, the finalenergy of the photon is:Ef � 0.33Ei

Ef � 0.33(1.1 � 10�14 J)Ef � 3.64 � 10�15 JThe final wavelength can be calculated usingPlanck’s equation:

�f �

�f �

�f � 5.45 � 10�11 m

(6.626 � 10�34 J·s)(3.0 � 108 m/s)��

3.64 � 10�15 J

hc�Ef

(6.626 � 10�34 J·s)(3.0 � 108 m/s)��

1.8 � 10�11 m

hc��i

2(6.626 � 10�34 J·s)(3.0 � 108 m/s)��(9.11 � 10�31 kg)��

12

�(7.12 � 105 m/s)2 (3.0 � 108 m/s)(7.12 � 105 m/s)�

2hc��m��

12

�vf2 cvf�

1�2

2hc��i

h��f

h��i

1�2

hc��f

hc��i

h��i

h��f

h��i

1�2

hc��f

hc��i

y

x

xf

e

xi

C

θ43°

6.626 � 10�34 J·s��

1 � 10�5 m

h��

h��

h��

hc��


The Compton shift as a percentage is:

� � 100%

� 302%

The wavelength of a photon increases by 302%.36. m � 45 g � 0.045 kg, v � 50 m/s

Using de Broglie’s equation:

� �

� �

� � 2.9 � 10�34 mThe wavelength associated with this ball is2.9 � 10�34 m.

37. mn � 1.68 � 10�27 kg, � � 0.117 nm � 1.17 � 10�10 mUsing de Broglie’s equation:

� �

v �

v �

v � 3371 m/sThe velocity of the neutron is 3371 m/s.

38. mp � 1.67 � 10�27 kg, � � 2.9 � 10�34 mUsing de Broglie’s equation:

� �

v �

v �

v � 1.37 � 1027 m/sThe speed of the proton would have to be 1.37 � 1027 m/s. Since v is much greater thanc, this speed is impossible.

39. Ek � 50 eV � 8 � 10�18 J, me � 9.11 � 10�31 kga) We shall first compute the velocity using

the kinetic energy value:

Ek � �12

�mv2

v � ��v � ��v � 4.19 � 106 m/sNow � can be found using de Broglie’sequation:

� �

� �

� � 1.73 � 10�10 mb) The Bohr radius is 5.29 � 10�11 m. The

wavelength associated with an electron islonger than a hydrogen atom.

40. The photon transfers from n � 5 to n � 2.The energy at level n is given by:

En �

The energy released when the photon transfers from n � 5 to n � 2 is:E � E5 � E2

E �

E � 2.86 eVE � 4.58 � 10�19 JTo compute the wavelength:

� �

� �

� � 4.34 � 10�7 mThe wavelength released when the photontransfers from n � 5 to n � 2 is 4.34 � 10�7 m.It is in the visual spectrum and it wouldappear as violet.

41. a) The electron transfers from n � 1 to n � 4.The energy of the electron is given by:

En �

The energy needed to transfer the electronfrom n � 1 to n � 4 is:E � E4 � E1

E �

E � 12.75 eV

13.6 eV�

12

�13.6 eV��

42

�13.6 eV��

n2

(6.626 � 10�34 J·s)(3.0 � 108 m/s)��

4.58 � 10�19 J

hc�E�

13.6 eV�

22

�13.6 eV��

52

�13.6 eV��

n2

6.626 � 10�34 J·s��(9.11 � 10�31 kg)(4.19 � 106 m/s)

h�mv

2(8 � 10�18 J)��9.11 � 10�31 kg

2Ek�m

6.626 � 10�34 J·s��(1.67 � 10�27 kg)(2.9 � 10�34 m)

h�m�

h�mv

6.626 � 10�34 J·s��(1.68 � 10�27 kg)(1.17 � 10�10 m)

h�m�

h�mv

6.626 � 10�34 J·s��(0.045 kg)(50 m/s)

h�mv

�f��i

5.45 � 10�11 m��1.8 � 10�11 m

�f��i


b) The electron transfers from n � 2 to n � 4.Similarly, the energy needed to transfer theelectron from n � 2 to n � 4 is:E � E4 � E2

E �

E � 2.55 eV42. We need to find the difference in the radius

between the second and third energy levels.The radius at a level n is given by rn � (5.29 � 10�11 m)n2

The difference in radii is:∆r � r3 � r2

∆r � (5.29 � 10�11 m)(3)2 �

(5.29 � 10�11 m)(2)2

∆r � 2.64 � 10�10 m43. n � 1

The radius of the first energy level can befound using:rn � (5.29 � 10�11 m)n2

rn � (5.29 � 10�11 m)(1)2

rn � 5.29 � 10�11 mThe centripetal force is equal to the electro-static force of attraction:

F �

F �

F � 8.22 � 10�8 NThe centripetal force acting on the electron tokeep it in the first energy level is 8.22 � 10�8 N.

44. F � 8.22 � 10�8 N, r � 5.29 � 10�11 mF � m4�2rf 2

f � ��f � ��f � 6.56 � 1015 HzThe electron is orbiting the nucleus 6.56 � 1015 times per second.

45. Consider an electron transferring from n � 4 to n � 1. As computed in problem 41,the energy released is equal to 12.75 eV � 2.04 � 10�18 J. The frequencyis then equal to:

f �

f �

f � 3.08 � 1015 HzThe frequency of the photon is 3.08 � 1015 Hz,or one-half the number of cycles per secondcompleted by the electron in problem 44.

46. Bohr predicted a certain value for energy at agiven energy level. From the quantization ofenergy, there can be only specific values forvelocity, v, and radius, r. Thus, the path of theorbiting electron can attain a specific path(orbit) around the nucleus, which is an orbital.

48. v � 1000 m/s, m � 9.11 � 10�31 kg∆py∆y ≥ h–

∆p � m∆v

y �

y �

y � 1.16 � 10�7 mHence, the position is uncertain to 1.16 � 10�7 m.

49. ∆y � 1 � 10�4 mThe molecular mass of oxygen is 32 mol.The mass of one oxygen molecule is

� 5.32 � 10�26 kg

From ∆py∆y ≥ h– and ∆p � m∆v, the maximumspeed is:

v �

v �

v � 1.98 � 10�5 m/s

1.0546 � 10�34 J·s��(5.32 � 10�26 kg)(1 � 10�4 m)

h–�mv

32 mol��6.02 � 1023 mol/g

1.0546 � 10�34 J·s��(9.11 � 10�31 kg)(1000 m/s)

h–�mv

2.04 � 10�18 J��6.626 � 10�34 J·s

E��h

8.22 � 10�8 N��(9.11 � 10�31 kg)(5.29 � 10�11 m)

1�2�

F�mr

1�2�

(8.99 � 109 N·m2/C2)(1.6 � 10�19 C)2

��(5.29 � 10�11 m)2

ke2

�r2

13.6 eV�

22

�13.6 eV��

42


Chapter 13

28. a) 3 cm/a � � � �

� 9.5 � 10�10 m/s

� 3.16 � 10�18

b) 0.1 mm/s � �

� 1.0 � 10�4 m/s

� 3.3 � 10�13

c) � 3.6 � 10�8

d) Mach 6.54 � 6.54 � 332 m/sMach 6.54 � 2171.28 m/s

� 7.24 � 10�7

e) � 7.33 � 10�3

29. a) Snoopy must fly 50 km/h [N then E].Let y � resultant ground speedy � �(130 km/h)2 � (50 km/h)2y � 120 km/h

b) The Baron going west has a ground speedof:

bv��g � bv��w wv��g

bvg � 130 km/h 50 km/h

bv��g � 180 km/h [W]While going east,

bvg � 130 km/h � 50 km/h

bv��g � 80 km/h [E]c) The time for Snoopy:

� � 6000 s

Time for the Baron:

� � �

� 6500 sTherefore, Snoopy wins the race by 500 sor 0.139 h.

d) �ttB

S� �

�ttB

S� �

�

�

�

� ��ttB

S� � �1 ��30. In our rest frame, we observe the contracted

length:

L � L0�1 � ��L � (1.0 m)�1 � (0.080)2L � 0.6 m

31. L � L0

L � L0�1 ��13

� ��1 �� 1 �

�

v � c��v � 0.943cv � 2.83 � 108 m/s

8�9

8�9

v2

�c2

v2

�c2

1�9

v2

�c2

v2

�c2

1�3

v2

�c2

w2

�v2

v2 � w2

�v2

tS�tB

�v2 � w2��

vtS�tB

�(v � w)(v w)��

vtS�tB

(v � w)(v w)��v�(v � w)(v w)

tS�tB

��(v �

2w0)0(v w)��

��

��(v � w20

)(0vv

w)��

��v22

0�

0w2

��

��v1

00w

� �v1�

00w

��

3600 s�

1 h100 km�80 km/h

100 km��180 km/h

3600 s�

1 h200 km

��120 km/h

2.2 � 106 m/s��3 � 108 m/s

2171.28 m/s��3.0 � 108 m/s

10.8 m/s��3.0 � 108 m/s

1.0 � 10�4 m/s��3.0 � 108 m/s

1 m��1000 mm

0.1 mm�

1 s

9.5 � 10�10 m/s��3.0 � 108 m/s

1 d�86400 s

1 a��365.25 d

1 m�100 cm

3 cm�1 a


32. Length is contracted for the moving stop-watch. The time it measures is:

t �

t �

t �

t � 6.12 � 10�7 s33. We observe the dilated half-life of the muon:

t �

t �

t � 4.11 � 10�7 sThe distance travelled is:d � vtd � (0.998c)(4.11 � 10�7 s) d � 123 m

34. Katrina measures a contracted distance:

L � L0�1 ��L � (7.83 � 1010 m)�1 � (0.25)2L � 7.58 � 1010 m

35. The time the girlfriend measures is:

tf �

The time Henry measures is:

th �

th �

Their time difference is:

t � �

Using the low-speed approximation when v c:

�1 �� 1 �

t �

t �

t � 6.81 � 10�12 s36. Given the muon’s dilated half-life:

t � 2.8 � 10�6 sand its rest half-life:t0 � 2.2 � 10�6 s

t �

�22

.

.28� � �1 ��

1 � � �2

�

c�1 � ��22�.

.28��

2

� � v

1.856 � 108 m/s � vd � circumferenced � 2�rd � vt

r �

r �

r � 82.7 m37. Only the component of L0 in the direction of

travel is contracted:Lx � L0 cos 30°The contracted length seen by Tanya in thedirection of travel (x) is:

Lx� � Lx�1 ��Lx� � L0 cos 30°�1 ��The perpendicular length, Ly, is L0 sin 30° forboth Katrina and Tanya.

v2

�c2

v2

�c2

(1.856 � 108 m/s)(2.8 � 10�6 s)��

2�

vt�2�

v2

�c2

2.2�2.8

v2

�c2

t0��

�1 � �vc2

2

��

35 000 m��2(3.0(3

�

5 m10

/8s)

m

2

/s)2��

35 m/s

L0�1 � �1 � �2vc

2

2��

35 m/s

v2

�2c2

v2

�c2

L0�1 � �vc2

2

��

35 m/sL0�

35 m/s

L0�1 � �vc2

2

��

35 m/s

d�v

L0�35 m/s

v2

�c2

2.6 � 10�8 s��1 � (0.998)2

t0��

�1 � �vc2

2

��

(180 m)�1 � (0.7)2��

0.7(3 � 108 m/s)

L0�1 � �vc2

2

��

v

L�v


� tan 45°

� 1

�

Therefore:

1 �

�1 �� tan 30°

�1 ��

1 � �

�

v � 0.816cv � 2.45 � 108 m/s

38. The time to travel a circumference is:

t �

t0 �

t0 � 1.336 � 105 sFor the clocks on Earth, use the low-speedapproximation for v c:

t �

t t0�1 �The difference in the flying clocks comparedto the ones on Earth is:t � t � t0

t � t0�1 � � t0

t � �1 � 1�t � (1.336 � 105 s)� �t � 6.68 � 10�8 s

39. 1 ca � vt1 ca � (3.0 � 108 m/s)

(365.25 � 24 � 60 � 60 s)1 ca � 9.47 � 1015 m

40. Using spacetime invariance:(∆s2) � c2(∆tJ)2 � (∆xJ)2

and:(∆s2) � c2(∆tT)2 (∆xT)2

For Ted, the distance between events is:c2(1.0 � 10�6 s)2 � (600 m)2 � 0 � (xT)2

9 � 104 m2 � 3.6 � 105 m2 � �(xT)2

(xT)2 � 2.7 � 105 m2

xT � 5.20 � 102 m41. Ted’s length, L, has contracted relative to

Jane’s length, L0:

L � L0�1 ��520 m � (600 m)�1 ��

� 1 �

�

v � 0.499cv � 1.50 � 108 m/s

42. The dilated time of the stationary observer is:

t �

�45

.

.00

ss

� � �1 ��12

65� � 1 �

v � c

The distance travelled in the 5.0 s is:d � vt

d � (3.0 � 108 m/s)(5.0 s)

d � 9.0 � 108 m43. See problem 42:

v � c

v � 1.8 � 108 m/s

3�5

3�5

3�5

v2

�c2

v2

�c2

t0��

�1 � �vc2

2

��

56�225

v2

�c2

v2

�c2

169�225

v2

�c2

v2

�c2

(300 m/s)2

��2(3.0 � 108 m/s)2

v2

�2c2

2�r�

v

v2

�c2

v2

�2c2

t0��

�1 � �vc2

2

��

2�(6.38 � 106 m)��

300 m/s

2�r�

v

2�3

v2

�c2

1�3

v2

�c2

1��3

v2

�c2

v2

�c2

L0 sin 30°��

L0 cos 30°�1 � �vc2

2

��

L0 sin 30°��

L0 cos 30°�1 � �vc2

2

��Ly��Lx�

Ly��Lx�

Ly��Lx�


44. Trevor’s time is:

t0 �

t0 �

His sister’s time is:

t �

The time difference is:t � 1 a

t � �

t � �2

vL0��1 � �1 ��

L0 �

L0 �

L0 � 0.691 ca45. q � 1.6 � 10�19 C

v � 0.8cB � 1.5 T

m �

r � �mqB

v�

r �

r �

r � 1.52 � 10�3 m

46. The centripetal force is provided by the elec-trical coulomb force:

�

r �

r �

r �

r � 6.26 � 10�15 m47. The difference between the dilated and rest

masses is:∆m � m � m0

Use the low-speed binomial approximationwhen v c:

� 1

m � m0�1 � � m0

m � m0�1 � 1�m � � �m � � �m � 3.0 � 10�7 kg

48. Use the high-speed approximation:

1 � 2�1 � �m �

m �

m �

m �

m � 3.55 � 10�26 kg

9.11 � 10�31 kg��2(3.3� 10�10)

9.11 � 10�31 kg��2(1 � 0.999 999 999 67)

m0��

�2�1 �� vc

��

m0��

�1 � �vc2

2

��

v�c

v2

�c2

(3.0 � 104 m/s)2

��(3.0 � 108 m/s)2

(60 kg)�

2

v2

�c2

m0�2

v2

�2c2

v2

�2c2

v2

�2c2

1��

�1 � �vc2

2

��

(9.0 � 109 N·m2�C2)(1.602 � 10�19 C)2�1 � (0.6)2��

(9.11 � 10�31 kg)[0.6(3.0 � 108 m/s)]2

ke2�1 � �vc2

2

��

m0v2

ke2

�mv2

kQq�

r2

mv2

�r

[9.11 � 10�31 kg][0.8(3.0 � 108 m/s)]��(1.602 � 10�19 C)(1.5 T)�1 � (0.8)2

m0v��

qB�1 � �vc2

2

��

m0��

�1 � �vc2

2

��

(1 a)(0.95c)��2(1 � �1 � (0.95c)2)

tv��

2�1 � �1 � �vc2

2

��

v2

�c2

2L0�1 � �vc2

2

��

v2L0�

v

2L0�v

2L0�1 � �vc2

2

��

v

d�v


49. For a charge moving perpendicular to a mag-netic field, the centripetal force equals themagnetic force:

� Bqv

Due to mass dilation, the magnetic field is:

B �

B �

B � 2.26 � 10�2 T

50. density �

density �

where x, y, and z are the rectangular dimen-sions. Contraction occurs only in the directionof motion, so density is:

� �

� �x0�1 � �

vc2�

2

��yz

� �

When the density of an object is dilated twiceas much as its density at rest, 2�0 � �:

2�0 �

1 � �

�

v � 0.7071cv � 2.1 � 108 m/s

51. Using the relativistic equation of velocity addi-tion, the velocity of the light relative to theduck is:

lvd �

lvd �

lvd �

lvd � c52. Using the relativistic equation of velocity addi-

tion, the velocity of star A relative to star B is:

avb �

avb �

avb � 0.472c

avb � 1.42 � 108 m/s53. The speed of rocket A relative to Earth is:

avE �

avE �

avE � 0.962c

avE � 2.88 � 108 m/s54. The speed of the positron relative to the elec-

tron is:

pve �

pve �

pve � 0.996c

pve � 2.988 � 108 m/s55. Bob’s velocity relative to Earth, bvE � 0.3c;

Nicole’s velocity relative to Earth,

nvE � 0.9c � pvE, the phaser bullet’s velocityrelative to Earth.

0.95c 0.85c��1 (0.95)(0.85)

pvg gve��1 �p

vg

c�2gve�

0.8c 0.7c��1 (0.8)(0.7)

avb bvE��1 �a

vb

c�2bvE�

0.2c 0.3c��1 (0.2)(0.3)

avE Evb��1 �a

vE

c�2Evb�

1.2c�1.2

c 0.2c��

1 �(c)(

c02.2c)�

lvc cvd��1 �l

vc

c�2cvd�

1�2

v2

�c2

1�2

v2

�c2

�0�

1 � �vc2

2

�

�0�1 � �

vc2

2

�

m0��

�1 � �vc2�

2

��

m�xyz

mass�volume

(9.1 � 10�31 kg)(3.0 � 108 m/s)(0.999 999 986)��(1.602 � 10�19 C)(450 m)�1 � (0.999 999 986)2

m0v��

qr�1 � �vc2

2

��

mv2

�r


The velocity of the phaser bullet relative toBob, pvb, is:

pvb �

pvb �

pvb � 0.822c

pvb � 2.47 � 108 m/s56. Kirk’s velocity relative to Earth: kvE � X

the module’s velocity relative to Kirk: mvk � Xthe module’s velocity relative to Earth:

mvE � 0.8c

mvE �

0.8c �

0.8c � 2X

0.8X2 � 2cX 0.8c2 � 02X2 � 5cX 2c2 � 0

(2X � c)(X � 2c) � 0The speed of the Enterprise is:

�2c� � 1.5 � 108 m/s

57. The mass, m, equivalent to the chemicalenergy released is:E � mc2

m �

m � 3.56 � 10�13 kg58. The mass, m, equivalent to the chemical

energy released is:E � mc2

m �

m � 1.02 � 10�6 kg59.To find the energy equivalent of 1.0 kg of

bananas:E � mc2

E � (1.0 kg)(3.0 � 108 m/s)2

E � 9.0 � 1016 J

E �

E � 2.5 � 1010 kWh

At a typical consumer rate of $0.08/kWh, 1.0 kg of bananas is equivalent to:(2.5 � 1010 kWh)($0.08) � $2 � 109 or

$2 billionConversely, the rate of relativistic “banana”power is:

� $0.000 000 000 052/kWh

60. E � mc2

E � (m0c2 Ek)The work done in increasing an electron’sspeed is:Ek � Ek� � Ek

Ek � (mc2 � m0c2)� � (mc2 � m0c2)Ek � (mc2)� � mc2

Ek � m0c2� � �For v � 0.5c to v� � 0.9c:

Ek � m0c 2� � �Ek � 1.139m0c2

For v � 0.9c to v� � 0.95c:

Ek � m0c2� � �Ek � 0.908m0c2

It takes more work to increase from 0.5c to0.9c.

61. To find the equivalent mass of the particle:E � mc2

m �

m � 9.1 � 10�31 kgm � the mass of an electron

62. To find the difference between the dilated relativistic and the classical momentum,p � p � p0

p � mv � m0v

p � m0v� � 1�1��

�1 � �vc2

2

��

8.19 � 10�14 J��(3.0 � 108 m/s)2

1��1 � (0.9)2

1��1 � (0.95)2

1��1 � (0.5)2

1��1 � (0.9)2

1��

�1 � �vc2

2

��1

��

�1 � �vc2

2

��

$1.29��2.5 � 1010 kWh

9.0 � 1016 J��3.6 � 106 J/kWh

9.2 � 1010 J��

c2

3.2 � 104 J��

c2

0.8X2

�c

X X�

1 �Xc2

2

�

mvk kvE��1 �m

vk

c�2kvE�

0.9c 0.3c��1 (0.9)(0.3)

pvE Evb��1 �p

vE

c�2Evb�


Since v � 75 � 103 m/s, or v c, use thelow-speed binomial approximation:

1

p � m0v�1 � 1�p �

p �

p � 0.29 kg·m/s63. Since v c, use the low-speed approxima-

tion:

1

The work done, ∆Ek, in speeding Mercuryfrom rest is given by:Ek � mc2 � m0c2

Ek � m0c2� � 1�Ek � m0c2�1 � 1�Ek �

Ek �

Ek � 3.75 � 1032 JThe mass equivalent, ∆m, to this amount ofenergy is:

m �

m �

m � 4.16 � 1015 kg

64. m �

m �

m �

m � 2.4 � 10�28 kg

65. Accelerating the electron of mass,m0 � 9.1 � 10�31 kg, and charge,q � 1.6 � 10�19 C, from rest, through apotential of V results in a new total energy:

E � m0c2 VqE � mpc2

mpc2 � m0c2 Vq

V �

V �

V � 9.38 � 108 VV � 938 MV

66. Using the energy triangle,E2 � (mvc)2 (m0c2)2

E2 � (m0c2 Ek)2

For particle A:(21 J 8 J)2 � (21 J)2 (mvc)2

(mvc)2 � 841 J2 � 441 J2

(mvc)2 � 400 J2

mvc � 20 JTo find the velocity of A,

� (where E � mc2 � m0c2 + Ek)

�

v � 0.69cFor particle B:(22 J 7 J)2 � (22 J)2 (mvc)2

(mvc)2 � 841 J2 � 484 J2

(mvc)2 � 357 J2

mvc � 18.9 JTo find the velocity of B,

�

�

v � 0.65cParticle A has the greater speed.

67. E2 � (mvc)2 (m0c2)2

E2 � (mvc)2 (938.3 MeV)2

E2 � (0.996mc2)2 � (938.3 MeV)2

E2(1 � 0.9962) � 8.804 � 105 MeV2

E � 1.05 � 104 MeV

18.9 J�29 J

v�c

mvc�mc2

v�c

20 J�29 J

v�c

mvc�mc2

v�c

(3.0 � 108 m/s)2[(1.67 � 10�27 kg) � (9.1 � 10�31 kg)]��

1.60 � 10�19 C

mpc2 � m0c2

��q

(1.602 � 10�19 C)(1.35 � 108 V)��

(3.0 � 10�8 m/s)2

qV�c2

E�c2

3.75 � 1032 J��(3.0 � 108 m/s)2

Ek�c2

(3.28 � 1023 kg)(4.78 � 104 m/s)2

��2

m0v2

�2

v2

�2c2

1��

�1 � �vc2

2

��

v2

�2c2

1��

�1 � �vc2

2

��

(125 kg)(75 000 m/s)3

��2(3.0 � 108 m/s)2

m0v3

�2c2

v2

�2c2

v2

�2c2

1��

�1 � �vc2

2

��


68. E � mc2 � m0c2 Ek

mc2 � (0.511 MeV) (3.1 � 103 MeV)mc2 � 3100.511 MeVFrom the energy triangle:

cos � �

cos � �

� � 89.990557°

sin � �

sin � �

v � c sin �v � (3.0 � 108 m/s) sin 89.990557°v � 2.999 999 96 � 108 m/s

69. Using the energy triangle:

� sin �

�

� tan �

For particle A:

�

� 0.60

tan � � 0.60� � 30.96°

� sin (30.96°)

v � 0.514cFor particle B:

�

� 0.50

v � 0.50cParticle A is faster.

mvc�mc2

(5 � 10�8 N·s)(3 � 108 m/s)��

30 Jmvc�mc2

v�c

mvc�m0c2

(4 � 10�8 N·s)(3 � 108 m/s)��

20 Jmvc�m0c2

mvc�m0c2

v�c

mvc�mc2

mvc�mc2

v�c

mvc�mc2

0.511 MeV��3100.5 MeV

m0c�

E


Chapter 1443. a) Cl

b) Rnc) Bed) Ue) Md

44. For AZ X, Z is the number of protons and A � Z

is the number of neutrons:a) 17 protons, 18 neutronsb) 86 protons, 136 neutronsc) 4 protons, 5 neutronsd) 92 protons, 146 neutronse) 101 protons, 155 neutrons

45. Since 1 u � 931.5 MeV/c2, then 18.998 u � 931.5 MeV/c2/u � 17 697 MeV/c2.

46. Conversely, � 0.114 u.

47. To find the weighted average of the two iso-topes:0.69(62.9296 u) 0.31(64.9278 u) � 63.55 uThis is closest to the mean atomic mass of Cu.

48. B � [Zm(1H) Nmn � m(146C)]c2

B � [6(938.78) 8(939.57) �(14.003 242 u)(931.5)] MeV

B � 105.22 MeV

� � 7.5 MeV/nucleon

49. Since 146C → 14

7N �10 e v, the ratio

changes from �86

� to �77

� or from �43

� to the

more stable �11

�.

50. The binding energy is:B � [m(3He) mn � m(4He)]c2

B � [3.0160 u 1.008 665 u � 4.002 60 u]c2

� 931.5 MeV/c2/uB � 20.55 MeV

51. Since 23292U → 228

90Th 42He Ek,

Ek � [mU � mTh � m�]c2

Ek � [232.037 131 u � 228.028 716 u �4.002 603 u]c2 � 931.5 MeV/c2/u

Ek � 5.41 MeV

52. Assuming the uranium nucleus is fixed at restand the kinetic energy of the alpha particlebecomes electrical potential,

Ek �

r �

r �

r � 5.0 � 10�14 m53. 231

90Th → 23191Pa �1

0e v235

92U → 23190Th 4

2He54. The mass difference is:

∆m � mn � (mp me)∆m � [939.57 � 938.27 � 0.511] MeV/c2

∆m � 0.789 MeV/c2

55. From problem 54, the energy equivalent of0.789 MeV/c2 is 0.789 MeV.

Thus �23

�(0.789 MeV) � 0.526 MeV.

56. Since the total momentum before decay isequal to the total momentum after decay, andp � 0 � p�, the three momentum vectors mustform a right-angle triangle. From Pythagoras’theorem:pC

2 � pe2 p�

2

pC � �(2.64� 10�21)2 (4.76� 10�21)2pC � 5.44 � 10�21 N·s

57. Using Ek � , the recoiling carbon nucleus

will have

Ek �

Ek � 7.42 � 10�16 J58. For a fixed gold nucleus at rest, the kinetic

energy of the 449-MeV alpha particle is converted to electrical potential. Thus, forthe radius,

Ek �

r �

r �

r � 5.07 � 10�16 m

(8.99 � 109 J·m/C2)(1.6 � 10�19 C)2(2)(79)��

(449 � 106 eV)(1.6 � 10�19 J/eV)

kq1q2�Ek

kq1q2�r

(5.44 � 10�21 N·s)2

��2(12.011 u)(1.6605 � 10�27 kg/u)

p2

�2m

(8.99 � 109 J·m/C2)(1.6 � 10�19 C)2(2)(92)��

(5.3 � 106 eV)(1.6 � 10�19 J/eV)

kq1q2�Ek

kq1q2�r

N�Z

105.22 MeV��14 nucleons

B�A

106 MeV/c2

��931.5 MeV/c2/u


59.

When t � 8 h, 39.7% of the original dose isstill radioactive.

60. For carbon-14, T�12

� � 5730 a. Comparing therelative amount, NR, of a 2000-a relic with theamount, NS, in a shroud suspected of being2002 a � 1350 a 650 a, yields:

�NN

R

S� �

�NN

R

S� � ��

12

��2000

57a

3�

06a

50 a�

�NN

R

S� � 0.85

61. The half-life of Po-210 is:T�

12

� � 138 d � 198 720 minThe half-life of Po-218 is T�

12

� � 3.1 minAfter 7.0 min, there will be:

210Po: N � N0��12

�� 12

��1987.0

72m0

imn

in�

log N � (3.5 � 10�5)log �12

�

N � 100%

218Po: N � N0��12

�� 12

��73

.

.01

mm

iinn

�

log N � (2.26)log �12

�

N � 20.9%There will be a total of:1(1 �g) 0.209(1 �g) � 1.21 �gTherefore, 1.21 � 10�6 g of radioactive Poremains.

62. If the amount of radioactive material is 23%of the original amount after 30 d, then,

N � N0��12

��0.23N0 � N0��

12

��log (0.23) � � � log ��

12

��

T�12

� �

T�12

� �14 d

63. The molar amount of 235U is �52.3152

� � 0.0218

and of 207Pb is �32.0472

� � 0.0165. The

original molar amount of 235U was 0.0218 0.0165 � 0.0383. Using thedecay formula where T�

12

� � 7.1 � 108 a,

N � N0��12

��0.0218 � 0.0383��

12

��log � � � � � log � �

t �

t � 5.78 � 108 a

64. Using the activity decay formula where T�

12

� � 5730 a for 14C decay,

N � N0��12

��750 � 900��

12

��573

t0 a�

log � �� log � �

t �

t � 1507 a

log ��56

��(5730 a)��

log ��12

��

1�2

t�5730 a

750�900

t�T�

1

2�

log ��00..00

32

818

3��(7.1 � 108 a)

��log ��

12

��

1�2

t��7.1 � 108 a

0.0218�0.0383

t�T�

1

2�

t�T�

1

2�

(30 d) log ��12

��

log (0.23)

30 d�T�

12

�

30 d�

T�1

2�

t�T�

1

2�

t�T�

1

2�

t�T�

1

2�

��12

��tR�T�

1

2�

�

��12

��tR�T�

1

2�

40

20

4 8 12 16 200

60

80

100

t (h)

% r

adio

activ

e

% of Original Dose still Radioactive vs. Time


65. For an isotope to be doubly stable, its valuesfor both Z and N � A � Z must be “magic”nuclear shell numbers, where the numbersare 2, 8, 20, 28, 50, 82, and 126. The otherdoubly stable isotopes are 4

2He, 168O, 40

20Ca, 4820Ca,

7828Ni, and 132

50 Sn.66. 137

55Cs → 13756Ba �1

0 ev Ek. To determine themaximum Ek available per disintegration, findthe mass difference of the parent nucleon andthe daughter plus the electron.Ek � [136.9071 u � (136.9058 u

0.000 549 u)]c2 � 931.5 MeV/c2/uEk � 0.6996 MeV

67. Dose �

Dose �

Dose � 0.013 mGy68. The pilots fly for 52 weeks � 20 h/week �

1040 h per year. Thus, their exposure is:(7.0 � 10�6 Sv/h)(1040 h/a) � 7.28 � 10�3 Sv/a.Compared with the average of 2 mSv/a, this

value is about �7.

228� � 3.64 times greater.

69. Since 23892U → 206

82Pb, 238 � 206 � 32 nucleons are lost through alpha decay in groups of 4nucleons per decay.

Thus, there are �342� � 8 alpha particles

emitted. The number of beta decays is equal tothe number of neutrons changed into protons.N � protons in Pb �

protons left after alpha decayN � 82 � (92 � 8 � 2)N � 6 beta particles emitted

70. Four beta decays means that four neutronswere changed into protons, or 208

82Pb � 82�4208 X.

Six alpha decays means that 20878X came from

2086�4786�2Y � 232

90Y. From the periodic table, thiselement is thorium-232 or 232Th.

71. The separation distance of an alpha particle(A� � 4) and a nitrogen nucleus (AN � 14)is given by:rs � r� rN

rs � 1.2�3

A� 1.2�3

ANrs � 1.2�

34 1.2�

314

rs � 4.8 fm72. Considering the nitrogen nuclei to be fixed

at rest, the Ek of the incoming alpha particleis converted to electrical potential, or

Ek � , where q1 � 2e and q2 � 7e

Ek �

Ek � 4.2 MeV73. The half-life of hassium-269 is T�

12

� � 9.3 s. The original amount of hassium is

N �

Using the activity equation:

Activity �

Activity �

0.693� �9.3 s

Activity � 1.67 � 1017 BqUsing the decay formula for a time of 1 s:

N � N0��12

��N � ��

12

��91.3

ss

�

N � 92.82%If 92.82% remains after 1 s, then 100% � 92.82% � 7.18% has decayed. This activity equals:

Activity � (7.18%)� �Activity � 1.61 � 1017 Bq

74. The energy released is equivalent to theenergy of the mass difference:E � [m(1H) m(2H) � m(3He)]c2

E � [1.007 825 u 2.014 102 u �3.016 029 u]c2 � 931.5 MeV/c2/u

E � 5.49 MeV

(1.0 � 10�3 g)(6.022 � 1023 mol�1)��

269 g/mol

t�T�

1

2�

(1.0 � 10�3 g)(6.022 � 1023 mol�1)��

269 g/mol

0.693N�

T�12

�

mass of hassium � Avogadro’s number��

mass per mole

(9.0 � 109 J·m/C2)(1.6 � 10�19 C)2(2)(7)��

(4.8 � 10�15 m)(1.6 � 10�13 J/MeV)

kq1q2�r

(3700 Bq)(365 � 24 � 60 � 60 s)(1.0 � 106 eV)(1.6 � 10�19 J/eV)(5%)��

70 kg

activity � time � energy � percentage absorbed��

mass


75. One mole of 235U releases 23 500 GJ of energy.mo � 2(fuel used)mo � 2(moles of U used)(mass/mol)

mo � 2� �(0.235 kg/mol)

mo � 2� �(0.235 kg/mol)

mo � 883 kg

76. %E �

%E �

%E �

%E � 0.242About 24.2% of the fission energy is trans-formed into electrical energy.

77. Since a mole of 235U releases 23 500 GJ ofenergy, the 50 kg releases

� 5 � 106 GJ

� 5 � 1015 J78. Since the electron keeps only 10% of its

kinetic energy with each collision, the energyremaining after x collisions is given by:

Ex � Eo(0.1)x

0.05 eV � (5.0 � 106 eV)(0.1)x

log � � � x log (0.1)

x �

x � 8 collisions79. The incoming speed of a neutron with

3.5 MeV of kinetic energy is:

v � ��v � ��v � 2.5876 � 107 m/sFor head-on elastic collisions,

v� � � �v, where v� is the recoil

velocity of the neutron.

v1� � � �(2.5876 � 107 m/s)

v1�� 1.782 � 104 m/s80. For the reaction 235

92U 10n → 141

56Ba ZAY 31

0n,conservation of atomic mass number for thereaction yields 235 1 � 141 A 3(1), orA � 92. Conservation of atomic numberyields 92 0 � 56 Z 3(0), or Z � 36.The daughter isotope, from the periodic table,is 92

36Kr.81. Working in MeVs, assume the rest mass of

lead-207 is:m0 � (207 u)(931.5 MeV/c2/u)m0 � 1.928 � 105 MeV/c2

Its total energy is:E � m0c2 Ek

E � 1.928 � 105 MeV 7.000 � 106 MeVE � 7.1928 TeVAt relativistic speeds, use Einstein’s energytriangle:(mvc)2 � E2 � (m0c2)2

(mvc)2 � (7.1928 � 1012 eV)2 �

(1.928 � 1011 eV)2

mvc � 7.1902 � 1012 eVRearranging for v,

�

�

v � 0.999639cv � 2.9989 � 108 m/s

82. The de Broglie wavelength is:

� �

� �

� �

� � 1.73 � 10�19 m

(6.626 � 10�34 J·s)(3.0 � 108 m/s)��(7.19 � 1012 eV)(1.6 � 10�19 J/eV)

hc�mvc

h�mv

7.1902 � 1012 eV��7.1928 � 1012 eV

v�c

7.1902 � 1012 eV��

mc2

v�c

1.008 665 u � 1.007 276 u��1.008 665 u 1.007 276 u

mn � mx�mn mx

2(3.5 � 106 eV)(1.602 � 10�19 J/eV)��(1.008 665 u)(1.6605 � 10�27 kg/u)

2Ek�m

log (10�8)��log (10�1)

0.05 eV��5.0 � 106 eV

(50 kg)(23 500 GJ/mol)��

0.235 kg/mol

(0.7 GW)(86400 s)��

��0.2325.5

kkg/gmol

��(23 500 GJ/mol)

(electrical power) � time��

��mm

oalsasr

omfaUss

��(energy/mol)

electrical energy produced��

fission energy released

(0.7 GW)(2)(3.1536 � 107 s)��

23 500 GJ/mol

power � time��

energy�mol


83. At relativistic speeds, the mass becomesdilated:

m �

m �

m � 2.53 � 10�27 kgThe de Broglie wavelength is:

� �

� �

� � 1.16 � 10�15 m� � 1.16 fm

84. f �

B �

B �

B � 2.28 T85. Electrons and protons with the same

de Broglie wavelength have the same

momentum �� . Using Einstein’s

energy triangle and MeV units, for the electron:(mvc)2 � (m0c2 Ek)2 � (m0c2)2

(mvc)2 � (0.511 MeV 9 � 103 MeV)2 �

(0.511 MeV)2

mvc � 9.0005 GeVThe proton has an equal mvc, so(9000.5 MeV)2 � (938.27 MeV Ek)2 �

(938.27 MeV)2

938.27 MeV Ek �

�(9000.5 MeV)2 (938.27MeV)2Ek � 8951.47 MeV � 938.27 MeVEk � 8.1 GeV

86. Using the energy equation mc2 � m0c2 Ek tofind the dilated mass of the proton,

m � m0

m � 938.27 MeV/c2

m � 1338.27 MeV/c2

m � 2.3856 � 10�27 kg

The cyclotron frequency, f � , yields:

B �

B �

B � 1.87 T

87. a) uds � � � � 0

b) ud � � 1

c) db � � � 0

d) cc � � � 0

88. a) lambda (baryon)b) pion or rho (mesons)c) b-zero (meson)d) eta-c (meson)

89. A neutron consists of udd, therefore an anti-neutron is u dd.

90. The mass of the top quark is

� 188.94 u. The element

with the closest atomic mass is osmium (Os),with an atomic mass of 190.2 u.

91. The � pion has a quark combination of udand a charge of e. Conversely, a �� pion hasthe combination u d, and its charge is

�� e� �� e� � �e.

92. t �

t �

t � 8 � 10�24 s93. a) Two protons approach and exchange a vir-

tual meson, then recoil from each other.b) An atom sits at rest, then one of its elec-

trons drops to a lower energy level andemits a photon, so the atom is pushed inthe opposite direction.

c) A pion decays into a muon and a muonneutrino.

2.4 � 10�15 m��3 � 108 m/s

d�

v

1�3

2�3

176 � 103 MeV/c2

��931.5 MeV/c2/u

2�3

2�3

1�3

1�3

1�3

2�3

1�3

1�3

2�3

2�(2.3856 � 10�27 kg)(20 � 106 Hz)��

1.6 � 10�19 C

2�mf�

q

qB�2�m

400 MeV��

c2

Ek�c2

h�mv

2�(2.53 � 10�27 kg)(23 � 106 Hz)��

(1.6 � 10�19 C)

2�mf�

q

qB�2�m

6.626 � 10�34 J·s��(2.53 � 10�27 kg)(0.75c)

h�mv

1.673 53 � 10�27 kg��

�1 � 0.752

m0��

�1 � �vc2

2

��


94. Antiproton decay: p → n W� → n �10e v

95. For a neutron and a proton, the interaction is:p n → p n

96. In the reaction p → n π, the energy associ-ated with the mass difference is: E � (mp � mn)c2

From Einstein’s energy relationship,E2 � [(mp � mn)c2]2

E2 � p2c2 m02c4

p2c2 � E2 � (m0c2)2

p2c2 � (939.6 MeV � 938.3 MeV)2 �

(139.6 MeV)2

p2c2 � �19 486.5 MeVp2 � �2.165� 10�13 N2·s2

This result does not have a solution in the realnumbers, so the momentum, p, is imaginary(or virtual).

97. For a strange, s, quark and an antistrange, s,quark, the two new quarks created at the bro-ken ends could be u and u or d and d, accord-ing to ss → su su or sd sd. These particlesare known as � mesons.

98. The charge of the strange quark, s, is ��13

� and

the charge of the anticharm quark, c, is ��23

�.

The charge of the meson is: � � � �1

99. The charge of the baryon is

ttb � � � 1

100. The blue quark could either emit a blue-antigreen gluon or absorb a green-antiblue gluon.

t

x

Green Green

Blue

Blue-antigreen

gluon

Blue

1�3

2�3

2�3

2�3

1�3

t

x

u

udu ddu

udu ddu

uπ0

NeutronProton

Proton Neutron

e

n

t

x

p

v

W–


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Physics Concept and Connections Book 2 Solution Manual