BOOK TW TWO O
Solutions Manual
Brian Heimbeck Heimbecker er Igor Nowikow Christopher T. Howes Jacques Mantha Brian P. Smith Henri M. van Bemmel
Physics: Concepts and Connections Book Two Solutions Manual Authors Brian Heimbecker Igor Nowikow Christopher T. Howes Jacques Mantha Brian P. Smith Henri M. van Bemmel
NELSON Director of Publishing David Steele
First Folio Resource Group Project Management Robert Templeton
Publisher Kevin Martindale
Composition Tom Dart
Project Editor Lina Mockus-O’Brien
Proofreading and Copy Editing Christine Szentgyor Szen tgyorgi gi Patricia Trudell
Editor Kevin Linder
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Illustrations Greg Duhaney Claire Milne
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Table of Contents I
Solutions to Applying the Concepts Questions
Chapter 1 Section 1. 1.3 1.4 1.6 1.7 1.8 1.11 1.12 1.13 1.14 1.15 Chapter 2 Section 2. 2.1 2.2 2.3 2.4 2.5 2.6 2.7 2.8
1 1 1 3 3 4 5 6 6 7
8 9 10 11 12 13 14 15
Chapter 3 Section 3.3 3.4 3.5 3.6 3.7 3.8 3.9
17 18 18 20 21 22 22
Chapter 4 Section 4.2 4.3 4.4 4.5 4.6
24 24 25 25 26
Chapter 5 Section 5.2 5.3 5.4 5.5 5.6 5.7
28 28 29 29 30 31
Chapter 6 Section 6.1 6.2 6.3
33 33 34
Chapter 7 Section 7.2 7.3 7.4 7.5 7.6 7.7 7.8 7.9 7.10 7.1 7. 11
36 36 36 37 38 38 38 39 39 40
Chapter 8 Section 8.4 8.5 8.6 8.7 8.8 8.9
41 41 42 43 44 44
Chapter 9 Section 9.5
45
Chapter 10 Sec ecti tion on 10.2 10.3 10.4 10.5
47 47 48 48
Chapter 11 Sec ecti tion on 11. 1.4 4 11.5 11.6 11.8 11.9 11. 1.1 10
49 49 49 49 50 51
Chapter 12 Sec ecti tion on 12. 2.2 2 12.3 12.4 12.5 12.6 12.8
52 52 52 53 53 54
Chapter 13 Sec ecti tion on 13. 3.1 1 13.2 13.3 13.4 13.5 13.6 13.7 13.8
55 55 55 56 56 57 57 58
Chapter 14 Sec ecti tion on 14. 4.1 1 14.2 14.3 14.4 14.5 14.6 14.7 14.8
59 59 59 59 59 60 60 60
Table of Contents
II Answers to End-of-chapter Conceptual Questions Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14
61 63 65 66 67 68 69 71 75 77 79 80 81 83
IIII Solu II Soluti tion onss to to EndEndof-chapter Problems Chapter 1 Chapter 2 Chapter 3 Chapter 4 Chapter 5 Chapter 6 Chapter 7 Chapter 8 Chapter 9 Chapter 10 Chapter 11 Chapter 12 Chapter 13 Chapter 14
87 95 107 120 126 134 140 151 160 165 170 178 183 191
iii
PART 1
Solutions to Applying the Concepts
In this section, solutions have been provided only for problems requiring calculation.
Section 1.3 24 h 60 min 60 s 1. (30 days) 1 day 1h 1 min 6 2.6 10 s (units cancel to give answer in seconds) 1 mile 1 km 2. (7 furlongs) 1.4 km 8 furlong 0.63 mile (units cancel to give answer in kilometres) 20 oz 27.5 mL 3. (1 quart) 1 quart 1 oz 2 5.5 10 mL (units cancel to give answer in millilitres)
Section 1.4 1. Since the question is asking for velocity, the answer must include a direction. Since the direction in which the train travels is constant,
avg
v
d
v
9.0 m [E] 0 m [E] 8.0 s
v
1.1 m/s [E]
avg
avg
avg
t
b) The car’s instantaneous velocity at 5 s can
be approximated by the difference between the distance travelled after 6 s and the distance travelled after 5 s, divided by the time during that interval:
v
d
v
8.0 m [E] 8.0 m [E] 6.0 s 5.0 s
v
0 m/s
avg
avg
avg
t
Section 1.6 1. v22 v12 2ad
t
4
v
avg
velocity, direction is important. Since the direction is constant,
v
v
d
3. a) Since the question asks for the car’s
avg
2.5 10 m [N] 1.8 10 s 3
d
14 m/s [N]
2. a) Since the question is asking for average
speed, direction is not required. vavg
d
vavg
8.0 km 5.0 h
b) Since the question is asking for average
v
avg
avg
v
avg
d
t
3.0 km [W] 5.0 km [E] 5.0 h 3.0 km [E] 5.0 km [E] 5.0 h 2.0 km [E] 5.0 h
v
v
0.40 km/h [E]
avg
avg
2
2
2
vavg 1.6 km/h
v
2
d
d 9.4 103 m 2. 10 cm 1.0 101 m (v v2)t d 1
t
velocity, direction is required.
v22 v12 a
2 (600 m/s) (350 m/s) 2(12.6 m/s )
v1
2d v2 t
v1
2(1.0 10 m) 0.05 m/s 3.0 s
1
v1 1.7 102 m/s 3. a) Igor: dI vI t
1 Brian: dB aBt 2 2 If they meet, dI dB 8.0 m 1 vIt 8.0 m aBt 2 2 1 0 (2.8 m/s2)t 2 (7.0 m/s)t 8.0 m 2 0 (1.4 m/s2)t 2 (7.0 m/s)t 8.0 m Solutions to Applying the Concepts
1
ac b b2 4 t 2a
6. 100 km/h 27.8 m/s
1 2
d v1t at 2
t
7.0 m/s (7.0 m/s) .4 m/s )(8.0 m) 4(1 2(1.4 m/s )
t
7.0 m/s 2.05 m/s 2.8 m/s
2
2
2
2
t 3.2 s or t 1.8 s
We will take the lower value: t 1.8 s. 1 b) dB (2.8 m/s2)(1.8 s)2 2 dB 4.4 m 4. 8.0 cm 8.0 102 m v22 v12 2ad v 2 v12 a 2 2d
1 500 m (27.8 m/s)t (30 m/s2)t 2 2 2 2 0 (15 m/s )t (27.8 m/s)t 500 m t
(0 m/s)2 (350 m/s)2 a 2(8.0 102 m) a 7.7 105 m/s2
5. t total t 1 t 2 t 3 t total 3.0 s 6.0 s 10 s t total 19.0 s
In the first 3.0 s, the truck travels a distance of: 1 d1 (v1 v2)t 1 2 1 d1 (0 m/s 8.0 m/s)(3.0 s) 2 d1 12 m Since the truck travels at a constant speed over the second interval, d2 v2 t 2 d2 (8.0 m/s)(6.0 s) d2 48 m
For the final interval, 1 d3 v1t 3 at 32 2 2
2
2
t 4.9 s
1 2 1 80 m (17 m/s)t (9.8 m/s2)t 2 2 2 2 0 (4.9 m/s )t (17 m/s)t 80 m
7. a) d v1t at 2
(17 m /s) 4(4.9 m/s ) (80 m) 17 m/s 2(4.9 m/s ) 2
2
2
t 2.7 s b) v22 v12 2ad v2 v12 2 ad v2 (17 m/s)2 2(9.8 m/s2)(80 m ) v2 43 m/s v v1 8. a) a 2 t v v1 (eq.1) t 2 a
v v d t 2 2
1
(eq. 2)
Substituting equation 1 into equation 2,
2
v2 v1 a 2 2 2ad v2 v1 v2v1 v1v2 v22 v12 2ad v v1 b) a 2 t (eq. 1) v1 v2 at v2 v1 d t (eq. 2) d
v2 v1
2
1 2
d3 (8.0 m/s)(10 s) (2.5 m/s2)(10 s)2 d3 2.1 10 m dtotal d1 d2 d3 dtotal 12 m 48 m 2.1 102 m dtotal 2.7 102 m dtotal vavg t total
Substituting equation 1 into equation 2, d
2 v2 v2 at
1 2
d v2t at 2
2
2.7 10 m 19.0 s
vavg 14 m/s 2
2
t
vavg
(27.8 m/s) 5 m/s )(50 0 m) 27.8 m/s 4(1 2(15 m/s )
Solutions to Applying the Concepts
t
Section 1.7 1. a)
v22
v12
Section 1.8
2ad
1. a) at7.0s
Assuming up is positive, v22
2
at7.0s
2
1
b) The distance travelled by Puddles from t 5.0 s to t 13 s can be found by
1 2 Assuming down is positive, 1 30.0 m (4.0 m/s)t (9.8 m/s2)t 2 2 2 2 0 (4.9 m/s )t (4.0 m/s)t 30.0 m b b2 4 ac t 2a
2. a) d v1t at 2
4.0 m/s (4.0 m/s)2 4(4.9 m/s2)(30.0 m)
t
finding the area under the curve between those times. We must consider two separate intervals: between 5.0 s and 10 s, and between 10 s and 13 s. The area under the graph in the first interval can be expressed as the sum of the areas of a triangle and a rectangle: d1
2
2(4.9 m/s )
d1
4.0 m/s 24.6 m/s
9.8 m/s 2
1 b) d v1t at 2 2 Assuming down is positive, 1 30.0 m (4.0 m/s)t (9.8 m/s2)t 2 2 2 2 0 (4.9 m/s )t (4.0 m/s)t 30.0 m ac b b2 4 t 2a
2 2 (4.0 m/s) .9 m/s )(30 .0 m) 4(4 t 4.0 m/s 2
2(4.9 m/s ) 4.0 m/s 24.6 m/s 9.8 m/s 2
t 2.9 s
1 2 Assuming down is positive, 1 35 m v1(3.5 s) (9.8 m/s2)(3.5 s)2 2 v1 7.2 m/s or 7.2 m/s [up]
t 1v1
t v 2 1
1
(10 s 5.0 s)(60 m/s 50 m/s) 2 (10 s 5.0 s)(50 m/s)
t 2.1 s
3. d v1t at 2
55.0 m/s 51.0 m/s 8.0 s 6.0 s
t
t1
at7.0s 2.0 m/s2 60 m/s 60 m/s at12s 13 s 11 s at12s 0 m/s2 32.0 m/s 8.0 m/s at3.0s 4.0 s 2.0 s at3.0s 12 m/s2
t
t2
2
v1 a 0 (80.0 m/s)2 d 2(9.8 m/s2) d 330 m b) v2 v1 at v v1 t 2 a 0 80.0 m/s t 9.8 m/s2 t 8.16 s c) 2(8.16 s) 16.3 s d
v v t t
d1 275 m
The area under the graph in the second interval can be expressed as a rectangle: d2 t 2v2 d2 (13 s 10 s)(60 m/s 0 m/s) d2 180 m dT d1 d2 dT 275 m 180 m dT 455 m 2. a) For Super Dave, Sr., d vavg t d t vavg t
50 m 10 m/s
t 5.0 s
Solutions to Ap ply in g the Con ce pts
3
We can find the acceleration of Super Dave, Jr. from the slope of his vt graph:
For segment 2, d2 2.0 m 2.0 m d2 0 m vavg 0 m/s For segment 3, d3 1.0 m 2.0 m d3 1.0 m t 3 1.8 s 1.0 s t 3 0.8 s
a
v t
2
6 m/s a 2s a 3 m/s2
d v1t
at 2
2
But v1 0 m/s, so 2
d
at 2
2d a 2(50 m) t 3 m/s
1.0 m 0.8 s
t 3
vavg 1.25 m/s 3
For segment 4, d4 2.2 m 1.0 m d4 1.2 m t 4 2.6 s 1.8 s t 4 0.8 s
t 6 s
b) Super Dave, Sr. wins the race by 1 s. c) Super Dave, Sr.: d vavg t d t vavg
100 m 10 m/s
vavg 4
vavg 4
d4 t
4
1.2 m 0.8 s
vavg 1.5 m/s dtotal b) vavg t total 4
t 10 s
Super Dave, Jr.:
vavg
2
d
vavg 3
2
d v1t
d3
3
t
t
vavg
at , where v1 0 m/s, so 2
vavg 0.65 m/s
at 2
2
2.2 m 0.5 m 2.2 s 0.0 s
Section 1.11
2d a 2(100 m) t 3 m/s t
1. a)
F n
2
t 8 s
F k
Super Dave, Jr. wins. 3. a) For segment 1, d1 2.0 m 0.5 m d1 1.5 m t 1 0.6 s 0.0 s t 1 0.6 s vavg
d1
vavg
1.5 m 0.6 s
1
1
F g
t 1
vavg 2.5 m/s 1
4
Ball
Solutions to Applying the Concepts
Forces are unbalanced as the force provided by the kicker, F k, will cause the ball to accelerate.
b)
F support
F m
Gun
The forces are balanced. The force he provides on the gun, F m, F B will balance the force of the bullet.
F g
c)
F buoyant
The forces are not balanced, as the penny still accelerates downward, but at a slower rate.
Penny
F c) 1 m1a3
2
F 1 2m1a3 a a3 1
2 5.0 m/s2 a3 2 a3 2.5 m/s2
F ma F g F f ma F f m( g a) F f (90 kg)(9.8 m/s2 6.8 m/s2) F f 270 N 2 3. v2 v12 2ad v 2 v12 a 2 2d (0 m/s)2 (15 m/s)2 a 2(4.5 103 m) 2.
F g
d)
F parachute
These forces are balanced, and the soldier falls downward at a constant speed.
Soldier
a 2.5 104 m/s2 F ma F (8.0 102 kg)(2.5 104 m/s2) F 2000 N 4. For the first kilometre,
1 2
d v1t a1t 2 F g
Section 1.12 1. a) F 1 m1a1 F 1 a1 m1 a1
1 2 2d
d a1t 2
10 N 2.0 kg
a1 5.0 m/s2 b) F 1 2m1a2 F 1 a2 2m1 F 1 a1 m1 a a2 1
2 5.0 m/s2 a2 2 a2 2.5 m/s2
a1 a1
t 2 2(1000 m) (21.0 s) 2
a1 4.54 m/s2 v22 v12 2ad v2 2ad m) v2 2(4.54 m/s2)(1000 v2 95.3 m/s
For the last 1.4 km, the car’s acceleration is: v22 v12 2a2d a2
v22 v12 d
2
(0 m/s)2 (9.53 m/s)2 a2 2(1.40 103 m) a2 3.24 m/s2
F f ma2 F f (600 kg)(3.24 m/s2) F f 1.94 103 N
Solutions to Ap ply in g the Con ce pts
5
5.
During the first kilometre, the forces acting on the car are the force due to the engine and the frictional force:
F 1-2 (m2 m3)a F 1-2 (5000 kg 4000 kg)(1.5 m/s 2) F 1-2 1.35 104 N
F engine F f ma1 F engine ma1 F f F engine (600 kg)(4.54 m/s 2) (1.94 103 N) F engine 4.66 103 N v22 v12 2ad v 2 v12 a 2 2d (0 m/s)2 (28 m/s)2 a
The tension force in the rope between barges 2 and 3 can be found two ways: i) The difference between the force required to accelerate all the barges at a rate of 1.5 m/s2 minus the force required to accelerate the first two barges at the same rate:
F 2-3 F T F 12 F 2-3 2.25 104 N (6000 kg 5000 kg)(1.5 m/s2) F 2-3 6.0 103 N
2(0.35 m)
a 1.12 103 m/s2 F mitt ma F mitt (0.25 kg)(1.12 103 m/s2) F mitt 280 N
Section 1.13 1. a) Action: Foot striking the ball east Reaction: Ball pushing west on the foot b) Action: Paddle pushing backward on the water Reaction: Water pushing forward on the paddle c) Action: Balloon compressing and pushing air out Reaction: Air pushing back the other way on the balloon d) Action: Earth’s gravity pulling down on the apple Reaction: Apple’s gravity pulling up on Earth e) Action: Gravitational force downward of the laptop on the desk Reaction: Normal force upward of the desk on the laptop 3. a) F T mTa F T (6000 kg 5000 kg 4000 kg) 2
(1.5 m/s ) F T 2.25 104 N b) The tension force in the rope between barges 1 and 2 is equal to the force required to accelerate barges 2 and 3 at a rate of 1.5 m/s2.
ii) The force required to accelerate barge 3
at a rate of 1.5 m/s2: F 2-3 m3a F 2-3 (4000 kg)(1.5 m/s 2) F 2-3 6.0 103 N 4. a) F T mTa F sled F T (m1 m2)a F ( F f F f ) a sled m1 m2 1
a 0.83 m/s2 b) To find the tension force in the rope
joining the two toboggans, we consider the forces acting on the second toboggan: F T m2a F rope F f m2a F rope m2a F f F rope (300 kg)(0.83 m/s 2) 100 N F rope 350 N
Section 1.14 1. a) Friction is the only force acting on the truck, so F f ma v v1 a 2 t m(v2 v1) F f t
F f
(4000 kg)(0 m/s 16.7 m/s) 10 s
F f 6.7 103 N 6
2
700 N 200 N a 600 kg
Solutions to Applying the Concepts
b) F f k F n F n mg F f kmg F f k mg 6.7 103 N k
2. F g
(4000 kg)(9.8 m/s )
F g
(6.67 10 Nm kg )(0.013)(5.97 10 kg) (3.82 10 m)
2
11
1 2
1 8
2
b) F g 2
2
G (2m )m (3r ) 1
2
2
2 Gm1m2 9 r 2 2 F g ( F g ) 9 G 4m1m2 c) F g (2r )2 1
2
Gm m r
F g F g 2
1 2
2
1
1 2
( F gEarth) F g2
4.
1 GmyoumEarth GmyoumEarth 2 2 (r Earth r 2)2 r Earth 1 1 2 2 2(r Earth ) r Earth 2r Earthr 2 r 22 r 22 2r Earthr 2 r Earth2 0 2r Earth (2r Earth)2 4(1)(r Earth2) r 2 2 r 2 2.6 106 m
(0 m/s)2 (22.2 m/s)2 d 2(0.60)(9.8 m/s2) d 42 m
5.
Section 1.15 Gm m 2
r
31
2
(6.67 10 N)(9.11 10 kg) (0.01 m)
Gm m r Gm m r
F g myou g Jupiter
F g 5.5 1067 N
2
1
F g
F g
24
2
F g ( F g )
2
2
2
8
2
11
2
g2
G (0.013)m r
2
F n mg F app F f F app k F n F app k mg F app a m a k g a (0.15)(9.8 m/s2) a 1.5 m/s2 F f ma k F n ma kmg ma a k g v22 v12 2ad v 2 v12 d 2 2k g
F g
2 Earth
F g
1 Gm m F 8 r
constant velocity, it is not being acted upon by an unbalanced force. Therefore, the forces must have equal magnitudes and opposite directions. b) From a), we know that the applied force, F app, is equal in magnitude to the force due to friction, F f .
1 2
Moon
1
2. a) Since the toy duck is travelling at a
1. F g
Earth 2
F g 2.1 1020 N Gm1m2 3. a) F g r 2
2
k 0.17
3.
Gm m r
you
Jupiter 2 Jupiter you
Jupiter 2 Jupiter
g Jupiter
Gm r
g Jupiter
(6.67 10 Nm kg )(1.9 10 kg) (7.2 10 m)
Jupiter 2 Jupiter
11
2
2
7
27
2
g Jupiter 24 m/s2
Solutions to Ap ply in g the Con ce pts
7
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 dy (50 m) cos 14° [S] dy 49 m [S] In the E-W direction, dx d sin dx (50 m) sin 14° [E] dx 12 m [E] b) In the N-S direction vy v sin v (200 m/s) sin 30° [S] y v 100 m/s [S] y In the E-W direction, vx v cos v (200 m/s) cos 30° [W] x v 173 m/s [W] x c) In the N-S direction, ay a sin ay (15 m/s2) sin 56° [N] ay 12 m/s2 [N] In the E-W direction, ax a cos ax (15 m/s2) cos 56° [E] ax 8.4 m/s2 [E] 3. Horizontally, vx v cos vx (5.0 m/s) cos 25° vx 4.5 m/s Vertically, vy v sin vy (5.0 m/s) sin 25° vy 2.1 m/s
v v b
w tan
53° v 5.0 m/s [up 53° forward] g 5. a) Component Method: v v v f 1 2 For the x components, v v v fx 1x 2x v (50 m/s) cos 36° [W] fx
(70 m/s) cos 20° [E] vfx (50 m/s) cos 36° (70 m/s) cos 20° v 25.3 m/s [E] fx For the y components,
v
fy
v
1y
v
2y
v
(50 m/s) sin 36° [N] (70 m/s) sin 20° [S] vfy (50 m/s) sin 36° (70 m/s) sin 20° v 5.45 m/s [N] fy vf vf 2 vf 2 vf (25.3 m/s)2 (5.45 m/s) 2 vf 26 m/s fy
x
y
v vf
tan f x y
tan1
4. vg vw v b v 4.0 m/s [forward] 3.0 m/s [upward] g Since vw and v b are perpendicular, vg vw2 v b2 /s)2 vg (4.0 m/s)2 (3.0 m vg 5.0 m/s
4.0 m/s 3.0 m/s
tan1
25.3 m/s 5.45 m/s
78° v 26 m/s [N78°E] f Sine/Cosine Method: 54° 90° 54° 20° 16° vf 2 v12 v22 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,
v v sin sin 1
f
50 m/s 25.9 m/s sin sin 16° 32°
8
Solutions to Applying the Concepts
For the y components,
To find , 180° 54° 78° v 26 m/s [N78°E] f
nety F F 1y F 2y F 3y nety (200 N) sin 30° [N] F
θ
20°
v 2
γ
v 1
F net F net 2 F net 2 F net (24.15 N)2 (18.08 N)2 F net 30.1 N F net tan F net x
v f
b) 37° (parallel line theorem) 180° 53° 37° (supplementary
x
y
angles theorem) 90° Sine/cosine Method: df 2 d12 d22 2d1d2 cos df 2 (28 m)2 (40 m)2
1
49 m 28 m sin 90° sin
tan1
28 m 49 m
Section 2.2 1. a) vog vmg vom vmg cos vom
cos
γ
d f
53°
β ϕ 37° θ
The ship’s heading is [S76°E]. d 1
5.0 km/h 20 km/h
76°
c) Component Method: net F F 1 F 2 F 3 For the x components, netx F F 1x F 2x F 3x netx 140 N [W] (200 N) cos 30° [E] F
d 2
25.8 m/s 1.17 m/s
87° v 26 m/s [N87°W]
35° To find , 180° 37° 18° df 49 m [W18°N]
24.15 N 18.08 N
d d sin sin
sin
tan1
53° F net 30.1 N [N53°W] 6. vx v2 sin 40° v1 sin 15° vx 25.8 m/s [W] vy v2 cos 40° (v1 cos 15°) vy 1.17 m/s [N] v (25.8 m/s)2 (1.17 m/s) 2 v 26 m/s
2(28 m)(40 m) cos 90° df 49 m To find direction, f
y
β
36°
(100 N) cos 35° [S] F nety (200 N) sin 30° (100 N) cos 35° nety 18.08 N [N] F
ϕ
b) v2og v2om v2mg vog (20 km/h)2 (5.0 km/h)2 v 19 km/h [E] og d c) t v
(100 N) sin 35° [W] F netx 140 N (200 N) cos 30° (100 N) sin 35° F netx 24.15 N netx 24.15 N [W] F
t
100 km 19 km/h
t 5.2 h
Solutions to Ap ply in g the Con ce pts
9
2. a) sin
b) vom (2.0 m/s)2 (0.50 m/s)2 v 1.9 m/s [E] om d t vom
vmg vom
sin1
0.50 m/s 3.0 m/s
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 2.96 m/s [N] g
d t v
Section 2.3
1 15 m (0 m/s)t (9.8 m/s2)t 2 2 30 m t 2 9.8 m/s2 t 1.7 s 1 b) dx vi t axt 2 2 1 dx (25 m/s)(1.7 s) (0 m/s2)t 2 2 dx 43 m x
500 m 3.0 m/s
t 167 s c) The boy travels an extra distance west of
the girl’s landing point, caused by the horizontal component of his velocity (equal to the river’s current).
v v t v v t a 0 m/s (35 m/s) sin 40° t 9.8 m/s
2. a) ay
d vt d (0.50 m/s)(167 s) d 83 m
extra 83 m at 5.0 m/s is 17 s. The boy’s total time is 167 s 17 s 184 s. The girl’s time was 169 s. She wins the race. 3. vpw vsw vps v2pw v2sw v2ps v2pw (10 km/h)2 (6.0 km/h)2 vpw 12 km/h
tan
10 km/h 6.0 km/h
cos1
2y
1y
t 2.3 s b) Since the curve Blasto travels is
symmetrical (a parabola), the time he takes to reach maximum height is the same as the time he takes to reach the ground. t total 2(2.3 s) t total 4.6 s Solving for horizontal distance, 1 dx vi t axt 2 2 dx (35 m/s) cos 40°(4.6 s) dx 120 m 3. a) To find the time required for the bomb to reach the ground, 1 dy vi t ayt 2 2 200 m (97.2 m/s) cos 25° t 1 (9.8 m/s2)t 2 2 200 m (88.1 m/s)t (4.9 m/s2)t 2 y
0.50 m/s 2.0 m/s
76°
Terry must throw at [S76°E]. 10
1y
x
59° v 12 km/h [N59°E] pw 4. a) vog vom vmg vmg cos vog
2y
2
d) The time required for the boy to run the
1 2
dy viyt ay t 2
d t v
5.0 m 1.9 m/s
t 2.6 s
1. a)
500 m t 2.96 m/s t 169 s The boy:
t
t
Solutions to Applying the Concepts
0 (88.1 m/s)t (4.9 m/s2)t 2 200 m t
88.1 m/s (88.1 m/s) 4(4 .9 m/s )( 20 0 m) 2(4.9 m/s )
2
2
2
t 2.0 s
To calculate the horizontal distance, 1 dx vi t axt 2 2 Since there is no horizontal acceleration, x
dx vixt dx (97.2 m/s) sin 25°(2.0 s) dx 82 m b) The y component of the final velocity, vfy, is vf y2 viy2 2ad vf y2 [(97.2 m/s) cos 25°] 2
2(9.8 m/s2)(200 m) vf 108 m/s vf (97.2 m/s) sin 25° vf 41.1 m/s m/s)2 vf (108 m/s)2 (41.1 vf 115.6 m/s 41.1 m/s tan 108 m/s 21° v 116 m/s inclined at 21° to the vertical f 4. Since the time it takes for the ball to hit the green is not given, we can find two timerelated equations (one for the horizontal component and one for the vertical component), for the golf ball’s velocity, equate both equations, and solve for horizontal velocity. For the vertical component, 1 dy vi t ayt 2 2 Since the change in height is 0 m, 1 0 (vi sin )t (9.8 m/s2)t 2 2 2 (4.9 m/s )t vi sin v sin (eq. 1) t i 4.9 m/s2 For the horizontal component, 1 dx vi t axt 2 2 250 m (vi cos )t 250 m (eq. 2) t vi cos y
x
x
y
g
g
g
x
Equating equations 1 and 2, 250 m vi sin 2 4.9 m/s vi cos (250 m)(4.9 m/s 2) vi 2 sin cos 1225 m2/s2 vi 2 sin 17°cos 17° vi 66 m/s v 66 m/s, 17° above the horizontal i g
g
g
g
g
g
Section 2.4 1. F p F 1 F 2 p 200 N [N] 300 N [W] F F p F 12 F 22 F p (200 N)2 (300 N) 2 F p 361 N F tan 2 F p F 1
F 2
300 N tan 200 N 56°
θ
F 1
F f
p 361 N [N56°W] F
For the frictional force, F f kmg F f 0.23(200 kg)(9.8 m/s2) F f 451 N
This is the maximum force of friction between the stove and the floor. However, friction only f 361 N [S56°E]. acts to oppose motion, so F
net F F p F f net 361 N [N56°W] 361 N [S56°E] F net 361 N[N56°W] 361 N [N56°W] F F net 0 N F net ma F net a m
0N 200 kg a 0 m/s2 Since the frictional force is stronger than the force provided by the people’s pushing, the stove does not move. a
g
g
Solutions to Ap ply in g the Con ce pts
11
2. a) F net F 1 F 2 F 3 net 25 N [S16°E] 35 N [N40°E] F
45 N [W] Adding the x components, netx F 1x F 2x F 3x F netx (25 N) sin 16° [E] F (35 N) sin 40° [E] 45 N [W] F netx (25 N) sin 16° (35 N) sin 40° 45 N F netx 15.6 N netx 15.6 N [W] F Adding the y components, nety F F 1y F 2y F 3y nety (25 N) cos 16° [S] F
(35 N) cos 40° [N] F nety (25 N) cos 16° (35 N) cos 40° nety 2.78 N [N] F
F net F net 2 F net 2 N)2 F net (15.6 N)2 (2.78 F net 15.8 N F net tan F net x
y
x y
tan
15.6 N 2.78 N
80° net 15.8 N [N80°W] F b) F net ma net F a m
a
16 N [N80°W] 80 kg
a
0.20 m/s2 [N80°W]
3. F net ma net (0.250 kg)(200 m/s 2 [W15°S]) F net 50.0 N [W15°S] F net F F 1 F 2 2 F F net F 1 2 50.0 N [W15°S] 100 N [N25°W] F 2 50.0 N [W15°S] 100 N [S25°E] F
Adding the y components, 2y (50.0 N) sin 15° [S] (100 N) cos 25° [S] F 2y 103.6 N [S] F
F 2 F 2 2 F 2 2 N)2 F 2 (6.03 N)2 (103.6 F 2 104 N F 2 tan F 2 x
y
x
103.6 N tan 6.03 N 86.7° 90° 86.7° 3.3° F 2 104 N [S3.3°W] 4. The only two forces in the x direction are F x and F f .
net F F x F f F x F cos 45° F x (250 N) cos 45° F x 177 N F f k F n n F F g F y F n mg F sin 45° F n (20 kg)(9.8 m/s2) 177 N F n 372 N F f (0.40)(372 N) F f 149 N F net 177 N 149 N F net 27.9 N F net ma F net a m
a
y
27.9 N 20 kg
a 1.38 m/s2
Adding the x components, 2x (50.0 N) cos 15° [W] F (100 N) sin 25° [E] 2x 6.03 N [W] F
Section 2.5 1. The only two unbalanced forces are F || and F f. (eq. 1) F net F || F f (eq. 2) F || F g sin 25° F f F n F f F g cos 25°
12
Solutions to Applying the Concepts
(eq. 3)
Substituting equations 2 and 3 into equation 1, F net F g sin 25° F g cos 25° F net F g(sin 25° cos 25°) F net (2.0 kg)(9.8 m/s2)(sin 25° cos 25°) F net (19.6 N)(sin 25° cos 25°) F net 6.51 N F net ma 6.51 N (2.0 kg)a a 3.26 m/s2
1 2
d vit at 2
1 4.0 m (3.26 m/s2)t 2 2 8.0 m t 2 3.26 m/s t 1.6 s 2. Since there is no friction, the only force that prevents the CD case from going upward is the deceleration due to gravity, F ||.
F net F || F net F g sin 20° Since F net ma, ma mg sin 20° a g sin 20° a 3.35 m/s2 v v1 a 2 t v v1 t 2 a
ramp surface direction: net F F || F f F engine m(0.60 g ) mg sin 30° F n F engine m(0.60 g ) mg sin 30° (0.28)mg cos 30° F engine F engine (0.60)mg mg sin 30° (0.28)mg cos 30° F engine mg (0.60 sin 30°
(0.28) cos 30°) F engine 3.36m N
F net m1a T m1 g m1a For m2, F net m2a m2 g T m2a
(eq. 1)
(eq. 2) Adding equations 1 and 2, m2 g m1 g a(m1 m2)
4.0 m/s 3.35 m/s 2
t 1.2 s
3. To find the distance the skateboarder travels
up the ramp, we need to find the velocity of the skateboarder entering the second ramp at v1. Since there is no change in velocity on the horizontal floor, v1 v2. For the acceleration on ramp 1, F net F || ma mg sin 30° a g sin 30° a 4.9 m/s2 v22 v12 2ad v22 0 m/s 2(4.9 m/s2)(10 m) v2 9.9 m/s
F net F || F n ma mg sin 25° (0.1)mg cos 25° a 5.02 m/s2 For d, v32 v22 2ad (0 m/s)2 (9.9 m/s)2 2(5.02 m/s2)d d 9.8 m 4. F net m(0.60 g ) F net also equals the sum of all forces in the
Section 2.6 1. a) For m1,
t
For the deceleration on ramp 2,
a
m2 g m1 g m1 m2
a
(15 kg)(9.8 m/s ) 0.20(10 kg)(9.8 m/s ) 25 kg
2
2
a 5.1
m/s2 [right] Substitute a into equation 2:
T m2 g m2a T 71 N b) For m1, F net m1a T m1 g sin 35° m1 g cos 35° m1a
(eq. 1) For m2, F net m2a m2 g T m2a (eq. 2)
Solutions to Ap ply in g the Con ce pts
13
Adding equations 1 and 2, m2 g m1 g sin 35° m1 g cos 35° a(m1 m2) g (m2 m1 sin 35° m1 cos 35°) a
m m 1
a
2
(9.8 m/s2)[5.0 kg (3.0 kg) sin 35° 0.18(3.0 kg) cos 35°] 8.0 kg
a
3.5 m/s2 [right] Substitute a into equation 2:
T m2 g m2a T (5.0 kg)(9.8 m/s2) (5.0 kg)(3.5 m/s2) T 32 N c) For m1, F net m1a T m1 g sin 40° 1m1 g cos 40° m1a
(eq. 1) For m2,
(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) g (m2 sin 60° 2m2 cos 60° m1 sin 40° 1m1 cos 40°) m1 m2
m/s )[(30 kg) sin 60° 0.30(30 kg) cos 60° (20 kg) sin 40° 0.20(20 kg) cos 40°] a (9.8 50 kg 2
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/s 2) cos 40° T 1.8 102 N d) For m1,
F net m1a m1 g sin 30° T 1 m1a For m2, F net m2a T 1 T 2 m2a For m3, F net m3a T 2 m3 g m3a
14
a
m m m 1
a
2
3
2
2
(30 kg)(9.8 m/s )sin 30° (10 kg)(9.8 m/s ) 60 kg
a 0.82
m/s2 [left] Substitute a into equation 3:
T 2 m3a m3 g T 2 (10 kg)(0.82 m/s2) (10 kg)(9.8 m/s2) T 2 106 N Substitute a into equation 2: T 1 m2a T 2 T 1 106 N (20 kg)(0.82 m/s2) T 1 122 N
Section 2.7
F net m2a m2 g sin 60° T 2m2 g cos 60° m2a
a
Adding equations 1, 2, and 3, m1 g sin 30° m3 g a(m1 m2 m3) m1 g sin 30° m3 g
(eq. 1)
(eq. 2)
(eq. 3)
v2 1. ac r
(25 m/s)2 30 m ac 21 m/s2 ac
d 2. v t
2
v 25
r t
v2 ac r ac
2
2500 r t 2
25002(1.3 m) ac (60 s)2
ac 8.9 m/s2 v2 3. ac r 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. 2r 4. a) v , where T r 3.8 105 km r 3.8 108 m T 27.3 days T 2.36 106 s
Solutions to Applying the Concepts
v2 ac r 42r ac T 2 42(3.8 108 m) ac (2.36 106 s)2 ac 2.7 103 m/s2
b) F c mac
v2 F c (10 kg) r F c 24 N
c) Friction holds the child to the merry-go-
b) The Moon is accelerating toward Earth. c) The centripetal acceleration is caused by
the gravitational attraction between Earth and the Moon. 5. r 60 mm r 0.06 m ac 1.6 m/s2 v2 ac r v acr v 0.31 m/s 6. Since d 500 m, r 250 m 2r v T
1
f T v 2rf ac g v2 ac r g 42rf 2 g f 42r
9.8 m/s f 4 (250 m) 2
2
round and causes the child to undergo circular motion. 2. Tension acts upward and the gravitational force (mg ) acts downward. F c F net and causes Tarzan to accelerate toward the point of rotation (at this instant, the acceleration is straight upward). F c mac mv2 T mg r
2
T
3. Both tension and gravity act downward. F c mac mv2 T mg r When T 0, mv2 mg r v gr 2 m) v (9.8 m/s2)(1. v 3.4 m/s 4. a) °
N
mg
20°
f 2724 rotations/day
20(2r ) v 180 s v 3.5 m/s
N cos 20
N sin 20°
60 s 60 min 24 h 1 min 1 h 1 day
d 1. a) v t
9.8 m/s2
T 9.7 102 N
f 0.0315 rotations/s f (0.0315 rotations/s)
Section 2.8
(4 m/s) (60 kg) 2.5 m
v2 T m g r
b)
F c mg tan 20° mv2 mg tan 20° r 20° v rg tan v (100 m)(9.8 m/s2) tan 20° v 19 m/s
c) The horizontal component of the normal
force provides the centre-seeking force.
Solutions to Ap ply in g the Con ce pts
15
d) If the velocity were greater (and the radius
remained the same), the car would slide up the bank unless there was a frictional force to provide an extra centre-seeking force. The normal force would not be sufficient to hold the car along its path. e) Friction also provides a centre-seeking force. 5. G 6.67 1011 N·m2/kg2, mE 5.98 1024 kg
GmM
4002(1.93 106 m)3 (6.67 1011 N m2/kg2)(7.77 1022 kg)
T 7.4 104 s
42r 3 E
42(3.4 108 m)3 (6.67 10 N m2/kg2)(5.98 1024 kg) 11
T 1.97 106 s T 22.8 days 6. G 6.67 1011 N·m2/kg2, mE 5.98 1024 kg, r E 6.37 106 m F c mHac GmEmH mHv2 r 2 r 2 GmE v r GmE v r r height of orbit r E r 6.00 105 m 6.37 106 m r 6.97 106 m GmE v r
v (6.67 1011 N m2/kg2)(5.98 1024 kg) 6.97 106 m
v 7.57 103 m/s 7. G 6.67 1011 N·m2/kg2 mM (0.013)mE mM 7.77 1022 kg r M 1.74 106 m F c mApolloac GmMmApollo mApollov2 r 2 r 2 GmM v r
16
2 3
M
Gm T
2
400 r Gm T T
T
10(2r ) , where v T T
r height of orbit r M r 1.9 105 m 1.74 106 m r 1.93 106 m
F c mMac GmEmM mMv2 r 2 r 2 GmE v r 42r 3 2r GmE 2 , where v T T
4002r 3
Solutions to Applying the Concepts
c) F net net 2T v F g F net net ma F net net 0 F g 2T v 2T sin m g 2(85 N) sin 1.5° m 9.8 N/kg
Section 3.3 1.
T h = 1.0 × 104 N
T v
60°
T h
m 0.45 kg
Horizontal: T h T cos 60° T h (1.0 104 N) cos 60° T h 5.0 103 N Vertical: T v T sin 60° T v (1.0 104 N) sin 60° T v 8.7 103 N
4. a) F B
0.65 m
F g = mg
tan
T a = 100.0 N
F net net T v T A T A F net net ma F net net 0 T v T A T A T v 2T A T v 2(100.0 N) cos 70° T v 68.4 N 3. a) 5° 5° T = 85 N
2 sin 71.1°
F B 20.7 N b) F h F B cos F h (20.7 N) cos 71.1° F h 6.71 N c) F v F B sin F v (20.7 N) sin 71.1° v 19.6 N [down] (not including the F
T = 85 N
bag
1.90 m 0.650 m
71.1° F net net mg 2FBv F net net ma F net net 0 0 mg 2 F B sin mg F B 2 sin (4.0 kg)(9.8 N/kg) F B
+
70°
+ θ
T v
70°
1.90 m
pail
2.
T a = 100.0 N
F B θ
+
weight of the beams) 6.
F n
F f
F g = mg
b) dv (1.5 m) sin 1.5° dv 0.039 m dv 3.9 cm
θ
T
boat F F ||
F |||| mg sin F f f mg cos F net net T F f f F || || F net net ma
Solutions to Applying the Concepts
17
w 1000 kg/m3
F net net 0 T F |||| F f f T mg sin mg cos (sin cos ) T mg (sin T (400.0 kg)(9.8 N/kg) (sin (s in 30 30° ° (0.25) cos 30°) T 1.11 103 N
1m 1.00 10 cm 3
6
1. a)
F g = mg
b) Position B provides the greatest torque
because the weight weight is directed directed at 90° to the wheel’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
45.0 kg
50°
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·m r 1.5 m 90° F ? rF sin F r sin 2.0 103 N·m F
3
V w 0.0100 m3 mw w·V w mw (1000 kg/m3)(0.0100 m3) mw 10.0 kg F g mg F g (10.0 kg)(9.8 N/kg) F g 98.0 N
Section 3.4
1.50 m
1000 cm 1L 3
V w (10.0 L)
C A C 1.7 102 N·m
d) A larger-radius wheel or more and larger
compartments would increase the torque torque..
Section 3.5 1. 20.0 kg
P
(1.5 m) sin 90°
F 1.3 103 N
0.75 m
3.
3.0 m
A
10.0 L
2.5 m
90° r 1 ? m1 45.0 kg
m2 20.0 kg
B
0.75 3.0
m2 5.0 kg C
a) V w 10.0 L
18
r 2
0.75 m 2
r 2 0.375 m m3 20.0 kg m2 m3 15.0 kg
Solutions to Applying the Concepts
r 3
r 3 1.12 m 0 1 2 3 0 r 1 F 1 sin 1 r 2 F 2 sin 2 r 3 F 3 sin 3 r F r 2 F 2 r 1 3 3 F 1
r 1
At maximum height: H (1.75 m)(45 kg)(9.8 N/kg) sin 75.5° H 7.5 102 N·m % (7.7 102 N·m 27.5 102 N·m) 100
3.0 m 0.75 m 2
7.7 10 N·m
% 2.6% 3.
P
(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)
40°
r 1 0.332 m 2. a) 1.7 m
40°
4.0 m
a)
t-t rF sin t-t
(1.0 m)(30.0 kg)(9.8 N/kg) 2
2 0 r 1 F 1 sin r cm cm F g sin r cm F g sin F 1 cm r 1 sin
1
t-t 147 N·m
F 1
This torque applies to both sides of the teeter-totter,, so the torques balance each teeter-totter other. b)
r h = 1.75 m
F 1
50°
F g
(0.375 m)(5.00 kg)(9.8 N/kg) 0.75 m
1 24.5 N F b) F rv rv F v2 v2 0 F rv rv F v2 v2 F rv rv (5.00 kg)( 9.8 N/kg) [up] p] F rv 49 N [u F rh rh F h1 0 F rh rh F h1 F rh rh 24.5 N rh 24. 24.5 5 N [le [left] ft] F
r l = ?
H L 0 L H r m g r L H H mL g
The vertical reaction force is 49 N [up] and the horizontal reaction force force is 24.5 N [left].
r L
(1.75 m)(45.0 kg) 30.0 kg
r L 2.63 m
4.
F 4
F 3
P
c)
F 1
2.0 m
F 2 1.6 m
θ
0.50 m
2 3 0 r 1 F 1 r 2 F 2 r 3 F R3 R3 0
cos
0.5 m 2.0 m
75.5°
At the horizontal position: H (1.75 m)(45 kg)(9.8 N/kg) H 7.7 102 N·m
0.4 m
1
r 1
0.75 m 2
r 1 0.375 m
Solutions to Applying the Concepts
19
2.
2.0 m 2
r 2
r 2 1.0 m r3 1.60 m 90° sin 1 r 1 F 1 r 2 F 2 F 3 r 3
F 2
θ2 F 1 = F n
4 cm
P
+ 8 cm
(0.375 m)(120.0 kg)(9.8 N/kg) (1.0 m)(5.0 kg)(9.8 N/kg)
F 3
θ1
1.60 m
3 306 N [up [up]] F F 4 F 1 F 2 F 3 F 4 (120.0 kg)(9.8 N/kg) (5.0 kg)(9.8 N/kg) 306 N RP 919 N [up] F
2 0 2 1 2 1 r 2 F 2 sin 2 r 1 F 1 sin 1 r F sin 1 F 2 1 1 r 2 sin 2
Left saw horse: 919 N [up] Right saw horse: 306 N [up]
1
F 2
Section 3.6 1.
τ
48 m
2
(8.0 10 m)(27 kg)(9.8 N/kg) sin (4.0 10 m) sin 2
F 2 529.2 N F 2 5.3 102 N
The angle makes no difference — it cancels out.
45° +
3.
30 cm 45 cm
F L
F m 11°
P
F w
45° r w 48.0 cm r w 0.480 m mw 10.0 kg
r L
a)
48.0 cm 2
20
b s 0 m b s 0 m b s r m F m sin m r b F b sin b r s F s sin s r F sin b r s F s sin s F m b b r m sin m
F g–s
m
r L 24.0 cm r L 0.240 m mL 5.00 kg 0 w L w L (r w F w sin 45°) (r L F L sin 45°) (0.480 m)(10.0 kg)(9.8 N/kg) sin 45° (0.240 m)(5.00 kg)
F g–b
15°
(9.8 N/kg) sin 45° 41.6 .6 N·m [cloc [clockwise kwise]] 41
Solutions to Applying the Concepts
F m
sin sin sin
F m
sin ( ) sin
F m
r bm b g
rg
b r sms g r m m
s
m b ms r m m
–2
(75 10 m)(9.8 N/kg) sin 75°[(0.57)85 kg 19.0 kg] (45 10 m) sin 11°
–2
(tensi nsion) on) F m 5.57 103 N (te
Reaction forces: 0 F py F my F by F sy
Three-wheeled ATV:
F py F my F by F sy F py (5.57 103 N)(sin 4°) (19.0 kg)(9.8 N/kg) (0.57)(85 kg)( 9.8 N/kg) F py 1049.6 N py 1.05 103 N [up] F 0 F px F mx F bx F sx F px F mx F bx F sx F px (5.57 103 N)(cos 4°) 0 0 px 5.55 103 N [right] F Horizontal force: 1.49 103 N [right]; vertical force: 7.65 102 N [up]
θ
Top View
1.25 m
0.6 m
τ
0.55 m
θ
Section 3.7
x
34.0 cm h 34.0 cm sin 43°
1. a) sin 43°
tipped
htipped
0.7 m
htipped 49.8 cm
34.0 cm h 34.0 cm tan 43°
0.6 m
b) tan 43°
straight
hstraight
Back View
hstraight 36.5 cm 2. Four-wheeled ATV:
θ
T
1.0 m m 0 . 1
θ x θT
θ 0.6 m
tan T
0.60 m 1.0 m
T 31.0°
tan
0.60 m 1.25 m
25.64° x sin
0.55 m
x (0.55 m)(sin 25.64°) x 0.237 m
tan T
0.237 m 1.00 m
tan T 13.3°
Solutions to Applying the Concepts
21
Section 3.8 1. k 16.0 N/m ∆ x 30.0 cm ∆ x 30.0 102 m
A
a) F k∆ x F (16.0 N/m)(30.0 102 m) F 4.80 N b) F ma F a m
4.80 N a 2.7 103 kg a 1.78 103 m/s2 2. F g (67.5 kg)(9.8 N/m) F g 661.5 N
E
FL A L
F
AE L L
F
2 d
2
E L
4 L
0.29 10 m (200 10 N/m )(0.22 10 m) 2 F 4(0.90 m) 3
2
9
2
3
F 0.807 N
661.5 N k 1.0 102 m
k 66150 N/m k 6.61 104 N/m F g-truck mg F g-truck (2.15 103 kg)(9.8 N/kg) F g-truck 2.1 104 N
This weight is distributed equally over four springs. 2.1 104 N F s 4 springs F s 5267.5 N/spring
F x s k
5267.5 N/spring 6.6150 10 N/m
For nylon, Enylon 5 109 N/m2 4 FL d2
4(0.807 N)(0.90 m) 2 (5 10 N/m )(0.22 10 m) E L
d
9
2
3
d 1.83 mm d 1.83 103 m 2. Emarble 50 109 N/m2 A 3.0 m2 m 3.0 104 kg F a) Stress A (3.0 104 kg)(9.8 N/kg) Stress 2
3.0 m
4
x 0.0796 m x 8.0 102 m 3. F kx F (120 N/m)(30.0 102 m) F 36 N
22
F A L L
F kx F k x
Section 3.9 1. d 0.29 mm L 0.90 m ∆ L 0.22 mm Esteel 200 109 N/m2
2
d
E
x
2 4
Stress 9.8 104 N/m2 Stress b) E Strain Stress Strain
E
9.8 104 N/m2 Strain 50 109 N/m2 Strain 2.0 106
Solutions to Applying the Concepts
c) L 15 m ∆ L ? L Strain L L L(Strain) L (15 m)(2.0 106) L 3.0 105 m 3. a) Compressive strength of bone 17 107 N/m2 d bone 4.0 102 m
Bone cross-sectional area is: A r2 A (2.0 102 m)2 A 1.26 103 m2
200 kg
F g — 2
F g — 2
F F b g
2
F b
mg
2
F A
b Strength Breakage occurs if
m g
F b A m g
2
2 A
A
m
Strength
2(Strength) A g
2(17 107 N/m2)(1.26 103 m2) m 9.8 N/kg m 4.4 104 kg
Solutions to Applying the Concepts
23
Section 4.2
Section 4.3
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]
1. a) J F t J (3257 N [forward])(1.3 s) J 4234.1 N·s [forward] J 4.2 103 N·s [forward] b) J F t t J ma v v 1 J m 2 t t
v
(72 km/h [E])
1000 m 1h 1 km 3600 s
v
20 m/s [E] p m v 9.0 104 kg·m/s m
J m( v2 v1) J (0.030 kg)(200 m/s 0 m/s) J 6.0 N·s [out of gun] c) J F t t J ma J (0.500 kg)(9.8 N/kg [down])(3.0 s) J 14.7 N·s [down] J 15 N·s [down] 2. p p2 p1 mv p mv 2 1 p m(v2 v1) 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] J 3. a) F t 2.5 103 N·s F
20 m/s
3
m 4.5 10 kg 3. 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,
p = 16 kg·m/s [S]
b) p mv p (0.5 kg)(45 m/s [N]) p 22.5 kg·m/s [N]
c) p p2 p1 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]
0.2 s
p = 22.5 kg·m/s [N]
F 1.3 104 N b) v1 0 v2 120 km/h v2 33.3 m/s v v1 a 2 t 33.3 m/s 0 m/s a
0.2 s
a 166.7 m/s2 p = 38.5 kg·m/s [N]
1 2
d v1t at 2
1 2
d (0 m/s)(0.2 s) (166.7 m/s2)(0.2 s)2 d 3.3 m
24
Solutions to Applying the Concepts
1 2 1 J (5 s)(25 N [S]) 2 J 62.5 N·s [S] b) J Area under triangle rectangle 1 J (500 250 N [W])(3 s) 2 (250 N [W])(6 s) J 1875 N·s [W] c) J Area above area below (counting the squares: approximately) J (13 squares above) (4 squares below) J 9 squares Multiplying 9 by the length and width of each square, J 9(0.05 s)(100 N [E]) J 45 N·s [E]
4. a) J bh
v
2.0 m/s [forward] 4. m1 m, m2 80m, m(12) 81m, v(12)o ?, v1f 1.5 106 m/s, v2f 4.5 103 m/s, po 7.9 1017 kg·m/s 2f
po pf po m1v1f m2v2f 7.9 1017 kg·m/s m(1.5 106 m/s) 80m(4.5 103 m/s) 7.9 1017 kg·m/s m[1.5 106 m/s 80(4.5 103 m/s)] 7.9 1017 kg·m/s m 1.5 106 m/s 80(4.5 103 m/s) m 6.9 1023 kg 5. m1 5m, v1o v, v(12)f ?, m2 4m, v2o 0 po pf m1v1 m2v2 (m1 m2)v(12)f (5m)(v) (4m)(0) (5m 4m)v(12)f 5mv 9mv(12)f
5 9
v(12)f v
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 2.5 m/s [forward] 2f 2. m1 30 g 0.03 kg, v1o 0, v1f 750 m/s, m2 1.9 kg, v2o 0, v2f ? po pf m1v1o m2v2o m1v1f m2v2f m1v1o m2v2o m1v1f v2f m2
Section 4.5 1. m1 m2 2.0 kg, v1o 5.0 m/s [W], v2o 0, v 3.0 m/s [N35°W], v ? 1f 2f p1o (2.0 kg)(5.0 m/s [W]) p1o 10 kg·m/s [W] p1f (2.0 kg)(3.0 m/s [N35°W]) p1f 6.0 kg·m/s [N35°W]
po
pf p1o p2o p1f p2f , where p2o 0 p1o p1f p2f
θ
v2f
p 1f = 6.0 kg·m/s
p 2f
35°
(0.03 kg)(0) (1.9 kg)(0) (0.03 kg)(750 m/s) 1.9 kg
θ
v
2f 11.8 m/s [back]
3. m1 400 g 0.400 kg, v 3.0 m/s [forward], 1o v 1.0 m/s [forward], 1f m2 0.400 kg, v2o 0, v2f ? po pf m1 v1o m2 v2o m1 v1f m2 v2f m1 v1o m2 v2o m1 v1f v2f m2
2f
v
(0.400 kg)(3.0 m/s [forward]) (0.400 kg)(0) (0.400 kg)(1.0 m/s [forward]) 0.400 kg
p 1o = 10 kg·m/s
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
8.2 kg·m/s 2 kg
v2f 4.1 m/s
Solutions to Applying the Concepts
25
Using the sine law to find direction, sin sin 55° 6.0 kg·m/s 8.2 kg·m/s 37° v 4.1 m/s [W37°S] 2f 2. m1 85 kg, v1o 15 m/s [N], p1o 1275 kg·m/s [N], m2 70 kg, v 5 m/s [E], p 350 kg·m/s [E] 2o 2o
po
pf p1o p2o pf
v3f
0.17 kg·m/s 0.2 kg
v3f 0.87 m/s
Using the sine law to find direction, sin sin 10° 1.0 kg·m/s 0.1743 kg·m/s 85° v 0.87 m/s [S85°W] or 0.87 m/s [W5°S] 3f 4. m1 0.5 kg, v1o 2.0 m/s [R], p1o 1.0 kg·m/s [R], m2 0.30 kg, v2o 0, p 0, v 1.5 m/s [R30°U], 2o 1f p1f 0.75 kg·m/s [R30°U], v2f ?, p2f ?
p 2o = 350 kg·m/s
pTo pTf
p 1o = 1275 kg·m/s
θ
p1o
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 350 kg·m/s tan 1275 kg·m/s 15.4°
p2o p1f p2f , where p2o 0 p1o p1f p2f
pf
mf vf
v
1322 kg·m/s [N15°E] 85 kg 70 kg
f
v
8.5 m/s [N15°E] 3. m1 0.10 kg, v1f 10 m/s [N], p1f 1.0 kg·m/s [N], m2 0.20 kg, v 5.0 m/s [S10°E], 2f p2f 1.0 kg·m/s [S10°E], m3 0.20 kg, v ? 3f pTo 0 f
p 1f = 0.75 kg·m/s
p 1o = 1.0 kg·m/s
p2f 2 (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/s p mv v2f
0 p1f p2f p3f Using the cosine law,
0.513 kg·m/s 0.30 kg
v2f 1.7 m/s
Using the sine law to find direction, sin sin 30° 0.75 kg·m/s 0.513 kg·m/s 47° v 1.7 m/s [R47°D] or 1.7 m/s [D43°R] 2f
pTo pTf
Section 4.6 3.0 m 1. a) 1.5 m from both objects 2 2.0 kg (60 cm) b) 5.0 kg 2.0 kg 17.1 cm from the larger mass 200 c) (20 km) 600 6.67 km from the larger satellite
10°
p 1f = 1.0 kg·m/s p 2f = 1.0 kg·m/s
θ p 3f
p3f 2 (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 26
p 2f
θ
30°
Using the cosine law,
Solutions to Applying the Concepts
2. a) p1o (2.0 kg) p1o
0.011 m 0.1 s
kg·m/s [S20° [S20°E] E] 0.22 kg·m/s 0.017 m 0.1 s 0.17 kg·m/s kg·m/s [S1 [S10°W] 0°W] 0.013 m (2.0 kg) 0.1 s
p2o (1.0 kg) p2o
p1f p1f
0.26 kg·m kg·m/s /s [S5°W [S5°W]] 0.015 m p2f (1.0 kg) 0.1 s p2f 0.15 kg·m/s kg·m/s [S30° [S30°E] E] 0.013 m pcm (3.0 kg) 0.1 s pcm 0.39 kg·m/ kg·m/ss [S8°E [S8°E]]
b) i) 70°
p 1o
10° p To p 2o
ii) 5°
p 1f
p Tf
30° p 2f
c) The total momentum before and after
collision is the same as the momentum of the centre of mass. The total momentum vectors have the same length and direction as the momentum of the centre of mass. Solutions to Applying the Concepts
27
b) The triangular areas above and below the
Section 5.2 1. a) W F d W (40 N)(0.15 m) W 6.0 J b) W F d W mg d W (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/s To find d, v22 v12 2ad (v 2 v12) d 2 2a (12.5 m/s)2 0 d 2
2(2.5 m/s )
d 31.25 m W F d W (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 (v2 v1) 4. a t (14 m/s 25 m/s) a
5.0 s
a 2.2 m/s2 F ma F (52 000 kg)(2.2 m/s2) F 114 400 N (v22 v12) d 2a [(14 m/s)2 (25 m/s)2] d 2(2.2 m/s2) d 97.5 m W F d W (114 400 N)(97.5 m) W 1.1 107 J 5. a) W F d W (175 N)(55 m) W 9625 J
axis are identical and cancel out, therefore, therefore, W (0.040 m)(20 N) W 0.80 J 6. F ma F (3 kg)(9.8 N/kg) F 29.4 N W d F d
480 J 29.4 N
d 16 m
Section 5.3 1 1. a) Ek mv2 2 1 Ek (20 000 kg)(7500 m/s) 2 2 Ek 5.6 1011 J b) 20 km/h 5.6 m/s 1 Ek mv2 2 1 Ek (1.0 kg)(5.6 m/s)2 2 Ek 15.4 J 1 c) Ek mv2 2 1 Ek (0.030 kg)(400 m/s)2 2 Ek 2.4 103 J 1 2. Ek mv2 2 1 3900 J (245 kg)v2 2 v (3900 J)(2) 245 kg v 5.6 m/s 1 3. Ek mv2 2 2 Ek
m
2
m
2(729 J) (15 m/s)
v
2
m 6.5 kg 4. p 2mEk p 2(9.11 1031 kg)(6000 eV )(1.6 )(1.6 1019 J/eV ) p 4.2 1023 N·s
28
Solutions to Applying the Concepts
5. Ek Ek2 Ek1
1 2
1 2
(60 0 kg) kg)(5. (5.0 0 m/s m/s))2 (60 kg)(14 m/s)2 Ek (6
3. a) Using the law of conservation of energy, Etotal 5460 J
1 2
3
Ek 5.1 10 J
1 2
1 2 1 Ek (0.350 kg)(25.0 m/s)2 2 Ek 1.1 102 J (v22 v12) b) a 2d 0 (25.0 m/s)2 a 2(0.024 m) a 1.3 104 m/s2
(3.0
6. a) Ek mv2
Eg mgh 5460 J (3.0 kg)(9.8 N/kg)h h 185.7 m from the ground h 180.7 m from the pad c) v2 v1 at v2 (60 m/s) (9.8 N/kg)(2.0 s) v2 40.4 m/s
F ma F (0.350 kg)(1.3 104 m/s2) F 4557 N W F d W (4557 N)(0.024 m) W 1.1 102 J c) F avg avg ma F avg avg 4557 N 3 F avg avg 4.6 10 N
1 2 1 Ek (3.0 kg)(40.4 m/s)2 2 Ek 2.4 103 J Ek mv2
Section 5.4 1. a) Eg mgh Eg (3.5 kg)(9.8 N/kg)(1.2 m) Eg 4.1 101 J b) Eg mgh Eg (2000 kg)(9.8 N/kg)(0) Eg 0 J c) Eg mgh Eg (2000 kg)(9.8 N/kg)(1.9 m) Eg 3.7 104 J 2. a) v22 v12 2ad (or use the conservation
of energy) 2 v2 (0) 2(9.8 m/s2)(27 m) v2 23 m/s Efinal Einitial Ekf Ego Eko 1 2 (65 kg)vf (65 kg)(9.8 N/kg)(27 m)
1 kg)(3.0 m/s)2 2 vf 23 m/s
kg)v2 (3.0 kg)(9.8 N/kg)(5.0 m) 5460 J
b)
2
mgh 5460 J
v 60 m/s
b)
2
mv
Ep Etotal Ek Ep 5460 J 2448.24 J Ep 3.0 103 J 4. F kx F k x mg k x k
(5000 kg)(9.8 N/kg) 0.04 m
k 1.2 106 N/m
For only one spring: 1 225 000 N/m k 4 k 3.0 105 N/m
Section 5.5 rise 1. a) k run 20 N k 0.1 m k 200 N/m k 2.0 102 N/m
(65
Solu So luti tion onss to Ap ply in g the Con ce pts
29
b) Maximum elastic potential energy occurs x 0.1 m. at x
1 2 1 Ep (200 N/m)(0.1 m)2 2 Ep 1.0 J Ep kx2
c) Ee Ee2 Ee1
1 2 1 2 (200 N/m)(0.03 m) 2 Ee 7.0 102 J
E (200 N/m)(0.04 m)2
F g F e mg kx (0.500 kg)(9.8 N/kg) k(0.04 m) k 122.5 N/m 3. a) W E W E2 E1 where E1 0 W E2
2.
1 2 1 W (55 N/m)(0.04 m)2 2 W 4.4 102 J W kx2
b) W E W E2 E1 where E1 0 W E2
1 2 1 W (85 N/m)(0.08 m)2 2 W 2.7 101 J W kx2
Ee Ek
4.
1 2 1 2 mv 2 2 (200 N/m)(0.08 m)2 (0.02 kg)v2 v 8.0 m/s kx
5.
Ee Ek
1 2 1 2 mv 2 2 6 2 (5 10 N/m) x (2000 kg)(4.5 m/s)2 x 9 cm kx
30
6. The loss in elastic potential energy is equal to
the gain in kinetic energy. Ee Ek
Let the subscript 1 represent the initial compressed spring and subscript 2 represen represent t the moment after the spring has been released when the cart has a velocity of 0.42 m/s. ( Ee2 Ee1) Ek2 Ek1 1 1 1 kx12 kx22 mv22 0 2 2 2
x2
x2
kx12 mv22 k
(65 N/m)(0 N/m)(0.08 .08 m)2 (1. (1.2 2 kg) kg)(0. (0.42 42 m/s m/s))2 65 N/ N/m m
x2 0.056 m x2 5.6 cm
Section 5.6 1. The energy required to heat the water is Ew (4.2 103 J/°C/L)(65°C 10°C)(2.3 L) Ew 5.31 105 J The energy expended by the stove is E P t Es P t Es (1000 W)(600 s) Es 6.0 105 J
The energy lost to the environment is E Es Ew E 6.9 104 J 2. a) Ep mgh Ep (83.0 kg)(9.8 N/m)(13.0 m) Ep 1.057 104 J E P t 1.057 104 J P
18.0 s
P 590 W b) Ep 1.057 104 J Ep 10 600 J 3. Once the radiation of the Sun reaches Earth,
it has spread out into a sphere surrounding the Sun. This sphere has a surface area of: SA 4r 2 SA 4(1.49 1011 m)2 SA 2.79 1023 m2
Solutions to Applying the Concepts
The ratio of this area to the area of Earth exposed to the radiation will be equal to the ratio of the power radiated by the Sun to the power absorbed by Earth. SASun Sun’s radiation AEarth absorbed radiation 23 2 2.79 10 m 3.9 1026 W
d x ( ) Earth 2
2 2.79 1023 m2 3.9 1026 W x (6.87 106 m)2 (3.9 1026 W)(1.48 1014 m2) x 2.79 1023 m2 x 2 1017 W Therefore, Earth intercepts 2 1017 J of energy from the Sun each second. 4. The total time is 3(20 min)(60 s/min) 3600 s The time the player spends on ice is (3600 s)(0.25) 900 s
E P t E P t E (215 W)(900 s) E 1.935 105 J
b) Since Eko Ekf , the collision is elastic ( EkTotal 6 105 J). c) W Ektruck
1 2 1 2 (3000 kg)(20 m/s) 2 W 4.5 105 J 4. mp 0.5 kg mg 75 kg dp 0.03 m vpo 33.0 m/s vgo 0 vgf 0.30 m/s
W (3000 kg)(10 m/s)2
a) pgo mv pgo (75 kg)(0) pgo 0 Ekgo 0 ppo mv ppo (0.5 kg)(33.0 m/s) ppo 16.5 kg·m/s
1 2 Ekpo 272.25 J
Ekpo (0.5 kg)(33.0 m/s)2
While sitting on the bench, the player expends 100 W of power. He spends 3600 s 900 s 2700 s on the bench. E (100 W)(2700 s) E 2.7 105 J ET (1.935 105 J) (2.7 105 J) ET 4.6 105 J
Section 5.7 3. a) m1 3000 kg v 20 m/s [W] 1o v 10 m/s [W] 1f m2 1000 kg v2o 0 v2f ?
pTo pTf m1v1o m2v2o m1v1f m2v2f (3000 kg)(20 m/s) 0 (3000 kg)(10 m/s) (1000 kg)v2f v2f 30 m/s
b)
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 1 2 1 Ekp (0.500 kg)(12 m/s)2 2 Ekp 36 J 1 Ekg mgvgf 2 2 1 Ekg (75 kg)(0.30 m/s)2 2 Ekg 3.4 J d) The collision is inelastic due to the loss of kinetic energy. c) Ekp mpvpf 2
Solutions to Ap ply in g the Con ce pts
31
9. a) Counting the squares below the top curve,
5. m1 10 g m2 50 g v1o 5 m/s v2o 0 m m2 v1f v1o 1 m1 m2
v1f (5 m/s)
10 g 50 g 10 g 50 g
v1f 3.3 m/s 2m1 v2f v1o m1 m2
v2f (5 m/s)
there are about 16.5 squares, each with an area of (0.01 m)(166.7 N) 1.6667 J. The amount of energy going into the shock absorber is (16.5)(1.6667 J) 27.5 J. b) There are roughly 6 squares below the lower curve. The energy returned to the shock absorber is (6)(1.6667 J) 10 J 27.5 J 10 J c) % energy lost 100 27.5 J % energy lost 64%
2(10 g) 10 g 50 g
v2f 1.7 m/s 6. m1 0.2 kg m2 0.3 kg v1o 0.32 m/s v2o 0.52 m/s
Changing the frame of reference, v1o 0.84 m/s v2o 0 m/s 0.2 kg 0.3 kg v1f (0.84 m/s) 0.2 kg 0.3 kg v1f 0.168 m/s 2(0.2 kg) v2f (0.84 m/s) 0.2 kg 0.3 kg v2f 0.672 m/s Returning to the original frame of reference, v1f 0.168 m/s 0.52 m/s v1f 0.69 m/s v2f 0.672 m/s 0.52 m/s v2f 0.15 m/s 1 8. a) Estored bh 2 1 Estored (0.06 m 0.02 m)(50 N) 2 Estored 1.0 J 1 b) E lost 1.0 J (0.005 m)(30 N) 2 (0.005 m)(20 N) 1 (0.035 m)(20 N) 2 E lost 1.0 J 0.075 J 0.1 J 0.35 J E lost 0.475 J
32
Solutions to Applying the Concepts
Section 6.1 1. mE 5.98 1024 kg, mS 1.99 1030 kg, r 1.50 1011 m
r 2
1.8 107 J r 1
GMm a) Ek 2r
1.8 107 J 11
2
2
30
24
(6.67 10 N·m /kg )(1.99 10 kg)(5.98 10 kg) 2(1.50 10 m) 11
Ek 2.65 1033 J GMm b) Ep r
Ep
11
2
2
30
24
(6.67 10 N·m /kg )(1.99 10 kg)(5.98 10 kg) (1.50 10 m) 11
Ep 5.29 1033 J c) ET Ek Ep ET 2.65 1033 J (5.29 1033 J) ET 2.65 1033 J GM 2. ag r 2 (6.67 1011 N·m2/kg2)(5.98 1024 kg) ag (6.38 106 m 1 106 m)2 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
(6.67 1011 N·m2/kg2)(5.98 1024 kg)
r 2
Ek
GM G M
2GM r
2(6.67 1011 N·m2/kg2)(5.98 1024 kg) (6.38 106 m 1 106 m)
vesc 10 397 m/s
Since the rocket has only achieved 6000 m/s, it will not escape Earth. 1 b) Ek 1000 km mv2 2 1 Ek 1000 kmm(6000 m/s)2 2 Ek 1000 km1.8 107m J Since all kinetic energy is converted to gravitational potential energy at maximum height, Ek Ep Ek E2 E1 GMm 1.8 107m J r 2
GMm r 1
11
2
2
24
(6.67 10 N·m /kg )(5.98 10 kg) 6.38 10 m 1 10 m 6
6
r 2 1.1 107 m hmax r 2 r E hmax 1.1 107 m 6.38 106 m hmax 4.7 106 m
Section 6.2 1. a) M Sun 1.99 1030 kg, T 76.1 a 2.4 109 s T 2 ka3 a
(2.4 109 s)2 42
1 3
G M
a
(2.4 109 s)2 42 11 (6.67 10 N·m2/kg2)(1.99 1030 kg)
1 3
a 2.7 1012 m b) 0.97 d c) v t 2(2.69 1012 m) v 2.4 109 s v 7031 m/s 2. r altitude 10 000 km 1 107 m, r Jupiter 7.15 107 m, m Jupiter 1.9 1027 kg
vesc vesc
2 GM r
2(6.67 1011 N·m2/kg2)(1.9 1027 kg) 7.15 107 m 1 107 m
vesc 56 000 m/s 3. mMoon 7.36 1022 kg, mEarth 5.98 1024 kg, r 3.82 108 m a) vesc vesc
2 GM r
2(6.67 1011 N·m2/kg2)(5.98 1024 kg) 3.82 108 m
vesc 1445 m/s
Solutions to Ap ply in g the Con ce pts
33
To find the current speed of the Moon, GMm 1 mv2 2 2r
To find the period, 2
T ka , where k
v v
GM r
(6.67 1011 N·m2/kg2)(5.98 1024 kg) 3.82 108 m
v 1022 m/s
To find the additional speed required for escape, vadd esc 1445 m/s 1022 m/s vadd esc 423 m/s 1 1 b) Ek mvesc2 mv 2 2 2 1 Ek (7.36 1022 kg)[(1445 m/s)2 2 (1022 m/s)2] Ek 3.84 1028 J c) This value is comparable to a 900-MW nuclear power plant (e.g., Darlington) running for 2.35 1011 years! 4. Geostationary Earth satellites orbit constantly above the same point on Earth because their period is the same as that of Earth. 5. M 5.98 1024 kg, r 6.378 106 m, v 25 m/s To find the semimajor axis,
2
2
2
24
a 3.19 106 m
2
24
T 1792 s
Section 6.3 1. a) At the equilibrium point, the bob’s kinetic energy accounts for all the energy in the system. This total energy is the same as the maximum elastic potential energy. Ek equil ET Ek equil Epmax
1 2 1 Ek equil(33 N/m)(0.23 m)2 2 Ek equil0.87 J b) 0 1 c) Ek mv2 2 v 2 Ek Ek equilkx2
2(0.87 J) 0.485 kg m
v
v 1.9 m/s
m k
b) At 0.16 m, the elastic potential energy of
2
6
6
the bob is 1 Ep 0.16m kx2 2 1 Ep 0.16m (33 N/m)(0.16 m)2 2 Ep 0.16m 0.42 J ET Ek Ep Ek ET Ep Ek 0.87 J 0.42 J Ek 0.45 J
34
T 0.76 s
24
2
3
T
(6.67 10 N·m /kg )(5.98 10 kg)(6.378 10 m) 2(6.67 10 N·m /kg )(5.98 10 kg) (25 m/s) (6.378 10 m) 2
11
6
0.485 kg 2 33 N/m
11
2
T 2
4 (3.19 10 m) (6.67 10 N·m /kg )(5.98 10 kg)
harmonic motion,
11
T
GM
2. a) To find the period of an object in simple
ET Ep Ek 1 GMm GMm mv2 r 2a 2 GM 2GM v2 a r v2 1 2 a r GM 1 2GM v2r a GMr GMr a 2GM v2r
a
42
3
Solutions to Applying the Concepts
1 2
Ek mv2 v v
2 2(0.45 J) 0.485 kg Ek m
v 1.36 m/s c) Ek 0.45 J, from part b 3.
Position vs. Time ) m 0.4 ( t n 0.2 e m 0 e c –0.2 a l p –0.4 s i D –0.6
0 0.2 0.4 0.6 0.8 1.0 1.2 1.4
Time (s)
Solutions to Ap ply in g the Con ce pts
35
Section 7.2 10° 1. a) 0.17 rad 57.3°/rad 60° b) 1.0 rad 57.3°/rad 90° c) 1.6 rad 57.3°/rad 176° d) 3.07 rad 57.3°/rad 256° e) 4.47 rad 57.3°/rad 2. a) ( rad)(57.3°/rad) 180°
4
b) rad (57.3°/rad) 45°
(3.75 rad)(57.3°/rad) 675° (11.15 rad)(57.3°/rad) 639° (40 rad)(57.3°/rad) 2.3 103° Earth rotates 2 radians every 24 h. 2 rad 6.0 h 1.57 rad 24 h b) Earth moves 2 rad every 365 days. 2 rad 265 d 4.56 rad 365 d c) The second hand moves 2 rad every 60 s. 2 rad 2.62 rad 25 s 60 s d) A runner moves 2 rad for every lap. 2 rad 25.6 laps 161 rad 1 lap
c) d) e) 3. a)
2
v r v acr v (9.8 m/s2)(1200 m) v 108 m/s v 1.1 102 m/s v b) r
1.2 rev 2 rad 1 min 1 min 1 rev 60 s
0.12566 rad/s 0.13 rad/s b) r 1500 m ac r 2 ac (1500 m)(0.12566 rad/s)2 ac 24 m/s2 c) The angular acceleration is zero because
the angular velocity is constant. d) ac-space-station 24 m/s2 ac-Earth 9.8 m/s2 24 m/s2 2.4 9.8 m/s2
Section 7.4 1. a) (3.35 rev/s)(2 rad/rev) 21.0 rad/s 60 s t 2 min (50 sec) 1 min t 170 s
t t (21.0 rad/s)(170 s) 3.58 103 rad b) t (22.0 rad/s 0)
0.5 s
Section 7.3 2. a) ac
3. a)
108 m/s 1200 m
44 rad/s2 2. a) t (0 1.75 rad/s) t 0.21 rad/s2 t 8.3 s (1 2) t b)
2
(1.75 rad/s 0) (8.3 s) 2
7.3 rad c) There are 2 radians in one cycle.
0.090 rad/s
The angular acceleration is zero because the angular velocity is constant.
number of cycles
7.3 rad 2 rad/cycle
number of cycles 1.16 number of cycles 1.2
36
Solutions to Applying the Concepts
d)
1.16 cycles 0.58 cycles 2 (0.58 cycles)(2 rad/cycle) 3.6 rad 1 2t t 2 2 1 3.6 rad 0 (0.21 rad/s2)t 2 2 t 5.9 s
( )
3. a) t
2 1
t 2
2
92.2 rad (16.1 rad/s 14.5 rad/s)
t 6.026 s t 6.03 s b) t
14.5 rad/s 16.1 rad/s 6.026 s
0.266 rad/s2
2. a) I (0.045 kg·m2)(1.90 rad/s2) 0.086 N·m b) For 78 rpm: 1 0 78 rev 2 rad 1 min 2
1 min 1 rev 60 s
2 8.2 rad/s 22 12 2 (22 12) 2
(4.7 rad/s)2 0 2(1.90 rad/s2) 5.813 rad 5.8 rad
number of turns
5.813 rad 2 rad/turn
number of turns 0.93 1 For 33 rpm: 3 1 0 100 rev 3 2 rad 1 min 2 1 min 1 rev 60 s 2 3.5 rad/s 22 12 2 (22 12) 2 (3.5 rad/s)2 0 2(1.90 rad/s2) 3.223 rad 3.2 rad 3.223 rad number of turns 2 rad/turn number of turns 0.51
3. I
I
8.45 N·m 12.2 rad/s 2
I 0.693 kg·m2
1 2 1 I (5.55 kg)(1.22 m)2 2 I 4.13 kg·m2
4. a) I mr 2 (moment of inertia for a disk)
2
(8.2 rad/s) 0 2(1.90 rad/s ) 2
17.69 rad 18 rad
number of turns
Section 7.5
22 12 2 (22 12) 2
17.69 rad 2 rad/turn
number of turns 2.8 For 45 rpm: 1 0 45 rev 2 rad 1 min 2 1 min 1 rev 60 s 2 4.7 rad/s
b) rF (1.22 m)(15.1 N) 18.4 N·m c) I
18.4 N·m 4.13 kg·m 2
4.46 rad/s2
Solutions to Ap ply in g the Con ce pts
37
Section 7.6
b) v r v (0.320 m)(33.3 rad/s) v 10.7 m/s
1. a) rF (0.20 m)(23.1 N) 4.62 N·m 4.6 N·m W R W R (4.62 N·m)(2 rad) W R 29 J b) W R W R (4.62 N·m)(1.5 rad) W R 6.9 J c) 95° 1.66 rad W R W R (4.62 N·m)(1.66 rad) W R 7.7 J 2. a) 45° rad
1 2 1 Ek (1000 kg)(10.7 m/s)2 2 Ek 5.7 104 J Ek mv2
Section 7.8 v 1. a) r
78 rad/s
1 2 2 Erot 2(0.900 kg·m2)(78 rad/s)2 Erot 1.1 104 J 1 b) Ek mv2 2 1 Ek (1300 kg)(25 m/s)2 2 Ek 4.1 105 J Erot 4
4
W R W R rF W R (0.556 m)(12.2 N)
4 rad
W R 5.3 J b) The work done does not change.
Section 7.7 2 1. I mr 2 5 2 I (0.0350 kg)(0.035 m)2 5 I 1.7 105 kg·m2 1 Erot I 2 2 1 Erot (1.7 105 kg·m2)(165 rad/s)2 2 Erot 0.23 J 2. a) (5.3 rev/s)(2 rad/rev) 33.3 rad/s 1 Erot 4 I 2 2 1 Erot 4 (0.900 kg·m2)(33.3 rad/s)2 2 Erot 2.0 103 J
38
25 m/s 0.320 m
I
c) ET Ek Erot ET (4.1 105 J) (1.1 104 J) ET 4.2 105 J 2. v1 0 1 0 h1 12.0 m m 2.2 kg r 0.056 m I mr 2 (moment of inertia for a hollow
cylinder) 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 energy
halfway down: Eg mgh2 h Eg mg 1
2
Eg (2.2 kg)(9.8 m/s2) Eg 1.29 102 J
Solutions to Applying the Concepts
12.0 m 2
To find the velocity halfway down:
L I
2 5 2 L (85 kg)(0.9 m)2(25.7 rad/s) 5 L 7.1 102 kg·m2/s 3. At perihelion, v 5472.3 m/s r 4.4630 1012 m m 1.027 1026 kg
ET1 ET2
1 1 2 2 v2 1 1 mgh1 mv22 mr 2 r 2 2
L mr 2
mgh1 mv22 I 2 mgh2
2
mgh2
mgh1 mv22 mgh2 mv22 mgh1 mgh2 h v22 gh1 g 1
2
2v22 2 gh1 gh1 2v22 gh1 v2 v2
v r
2 (9.8 m/s )(12.0 m) 2 gh1
1.9 102 rad/s
Section 7.9 1 rev 1d 1h 2 rad 1. 365 d 24 h 3600 s 1 rev 7 1.99 10 rad/s 2 I mr 2 (moment of inertia for a sphere) 5
L I
2 L mr 2 5 2 L (5.98 1024 kg)(6.38 106 m)2 5 (1.99 107 rad/s) L 1.94 1031 kg·m2/s 4.5 cycles 2 rad 2. 1.1 s 1 cycle 25.7 rad/s 1.8 m r 2 r 0.9 m 2 I mr 2 (moment of inertia for a sphere) 5
2 5 2 L (1.027 1026 kg)(4.4630 1012 m)2 5 (1.2261 109 rad/s) L 1.003 1042 kg·m2/s At aphelion: v 5383.3 m/s r 4.5368 1012 m m 1.027 1026 kg L mr 2
10.8 m/s 0.056 m
12
1.2261 109 rad/s L I
2
v2 7.67 m/s v c) r
5472.3 m/s 4.4630 10 m
v r
5383.3 m/s 4.5368 10 m 12
1.1866 109 rad/s L I
2 5 2 L (1.027 1026 kg)(4.5368 1012 m)2 5 (1.1866 109 rad/s) L 1.003 1042 kg·m2/s L mr 2
Section 7.10 2. 1 1
t
2 rad 2.14 10 s 6
1 2.94 106 rad/s
Solutions to Ap ply in g the Con ce pts
39
I 11 I 22
2 5
2 5
mr 121 mr 222
r 121 r 222 r 2 2 1 2 1 r 2
2
8
2
6
(6.95 10 m) (2.94 10 rad/s) (5500 m)
2
2 4.69 104 rad/s 2 rad T 2 2
T 2
2 rad 4.69 10 rad/s 4
T 2 1.34 104 s 3. r a 1.52 1011 m r p 1.47 1011 m vp 30 272 m/s I aa I pp v vp mr a2 a mr p2 r a r p r ava r pvp r pvp va r a (1.47 1011 m)(30 272 m/s) va 1.52 1011 m va 2.93 104 m/s va 29.3 km/s
Section 7.11 3. R 0.040 m r 0.0070 m a
a
g I
1 mr 2 g
1 2
2
mR
mr 2
a
1
9.8 m/s2
1 m)2 2 1 (0.0070 m)2 a 0.64 m/s2 (0.040
40
Solutions to Applying the Concepts
From the force vector diagram we see that,
Section 8.4 1. q1 3.7 106 C, q2 3.7 106 C, d 5.0 102 m, k 9.0 109 N·m2/C2 F F
tan
mg F e mg tan kq1q2 mg tan r 2 r 2mg tan q1 kq2
kq q d 1 2 2
9
2
2
6
6
(9.0 10 N·m /C )(3.7 10 C)( 3.7 10 C) (5.0 10 m)
2
2
q1
r
r
(9.0 10 N·m /C )(3.7 10 C)( 3.7 10 C) 98 N 2
2
6
r 3.5 10 3. a) T
10
9
2
2
6
The dust balls are 0.20 m apart, and the charge on the tethered dust ball is 1.1 1015 C.
6
2
2
(0.20 m) (2.0 10 kg)(9.8 N/kg)(tan 21°) (9.0 10 N·m /C )(3.0 10 C)
q1 1.1 1015 C
kq1q2 F
9
F 49 N F 49 N (attraction) 2. F 2(49 N) F 98 N
F e
m
Section 8.5 1. a)
F e
mg
b)
F e
T
b) mg
c) How close do the dust balls get and what is
the charge on the tethered dust ball? m 2.0 1010 kg , l 0.42 m, dwall-1 0.35 m, q 3.0 106 C, 21° dwall-2 0.35 m 0.42 m(sin 21°) dwall-2 0.35 m 0.15 m dwall-2 0.20 m
c)
+
Solutions to Ap ply in g the Con ce pts
41
At 4 cm away:
Section 8.6 1. a) q 1.0 106 C, 1.7 106 N/C [right] Let right be the positive direction.
kq r 2
e F F e F e e F
q
(1.0 106 C)(1.7 106 N/C) 1.7 N 1.7 N [left]
b) q 1.0 106 C, 2(1.7 106 N/C) [right]
2. T
–
2
2
6
2
kq r 2 9
2
2
6
(9.0 10 N·m /C )(3.0 10 C) (6.0 10 m)
2
2
7.5 106 N/C
c) Doubling the distance, kq 1 (2r )2
1 4 Tripling the distance, 1
2 F e
2
At 6 cm away:
e q F F e (1.0 106 C)[2(1.7 106 N/C)] F e 3.4 N e 3.4 N [right] F
1.7 107 N/C
If right is still the positive direction,
9
(9.0 10 N·m /C )(3.0 10 C) (4.0 10 m)
–
kq (3r ) 2
1 9
2 mg
Stationary charge creating a field
1 1 4 9 the original strength. 1 d) 2 . The field strength varies as the
1 decreases to and 2 decreases to of
q 1.0 106 C, 1.7 106 N/C [right] F e mg tan e 1.7 N [left] F 3. a)
r
inverse square of the distance away from the charge. e) q1 1.0 106 C, q2 3.0 106 C, r 8.0 102 m
+
kq1 2 r
(9.0 109 N·m2/C2)(3.0 106 C) (8.0 102 m)2
The field lines radiate outward, away from the charge. b) k 9.0 109 N·m2/C2, q 3.0 106 C At 2 cm away from the charge: kq r 2
9
2
2
6
(9.0 10 N·m /C )(3.0 10 C) (2.0 10 m)
2
2
4.22 106 N/C e q F F e (1.0 106 C)(4.22 106 N/C) e 4.22 N [right] F 4. a) q1 q2 1.0 106 C, r 0.20 m
Let the positive direction be left. At point A: r 1 0.05 m, r 2 0.25 m
6.8 107 N/C
42
Solutions to Applying the Concepts
1 2 kq1 kq2 TA 2 r 1 r 22 TA (9.0 109 N·m2/C2)(1.0 106 C)
TA
1 (0.05 m) 2
TA
1 (0.25 m) 2
At point B: r 1 0.10 m, r 2 0.10 m The addition of these two distances as was done in the previous question will yield a zero quantity. TB 0 N/C At point C: r 1 0.15 m, r 2 0.05 m TC 1 2 kq2 kq1 TC 2 r 2 r 21 TC (9.0 109 N·m2/C2)(1.0 106 C)
1 1 (0.05 m) (0.15 m) 2
2
3.2 106 N/C [left] b) At the centre point, 1 is equal in TC
magnitude but opposite in direction to 2, therefore there is no net field strength as the fields cancel out. c) For all field strengths to cancel out, the q r
magnitudes of the ratio of 2 must be equal and pointing in opposite directions.
Section 8.7
Ee
kq q r 1 2
9
2
2
6
6
(9.0 10 N·m /C )( 5.0 10 C)(1.5 10 C) 10 10 m
2
On particle 2: W 2 qV W 2 (1.0 1010 C)(50 V) W 2 5.0 109 J b) W Ek
1 2
W mv2 v
2 2 The similar masses cancel. 2 2.0 10 J 5.0 10 J W m
v1 v2
W 1 m1 W 2 m2
v1 v2
W 1 W 2
v1 v2
8 9
v1 2 v2
3. a) Extensive: electric force, potential energy
Intensive: field strength, electric potential b) Electric force — Charge and the field strength Potential energy — Charge and the electric potential c) Extensive properties
Ee 6.8 101 J E b) V e q 6.8 101 J V 1.5 106 C
V 4.5 105 V kq c) V r (9.0 109 N·m2/C2)(5.0 106 C) V 5.0 102 m
V 9.0 105 V
W 1 qV W 1 (4.0 1010 C)(50 V) W 1 2.0 108 J
1. a) Ee
V 2 V 1 5 5 ∆V 9.0 10 V (4.5 10 V) 5 ∆V 4.5 10 V 2. a) m1 m2 5.0 109 g 5.0 1012 kg, q1 4.0 1010 C, q2 1.0 1010 C
On particle 1:
6
3.7 10 N/C [left]
∆V
Product cost (per package) Mass Volume Length Force of gravity Etc. Intensive properties
Unit product cost (per unit weight or measure) Density Heat capacity
Solutions to Applying the Concepts
43
Indices of refraction Gravitational field strength Etc.
5. a) V 20 kV 2.0 104 V, q 1.602 1019 C, m 9.11 1031 kg Ek Ee Ek Vq Ek (2.0 104 V)(1.602 1019 C) Ek 3.2 1015 J
Section 8.8 1. qA 2e, qB 79e, Ek 7.7 MeV (7.7 106 eV)(1.602 1019 J)
1 2
b) Ek mv2
Ee Ek kqAqB Ee r kqAqB r Ee
v
r
v
(9.0 109 N·m2/C2)(1.602 1019 C)2(2)(79) (7.7 106 eV)(1.602 1019 J)
r 2.96 1014 m r 3.0 1014 m 3. q 1.5 105 C
1 2
2
mv
v
2q(V V ) m
v
v
1
2(1.5 105 C)(12 V) (1.0 105 kg)
6.0 m/s [left] 4. a) V 1.5 103 V, m 6.68 1027 kg, q 2e 3.204 1019 C Ek Ee
1 2
2
mv
Vq
v v
2 Vq m
2(1.5 103 V)(3.204 1019 C) 6.68 1027 kg
1 2
2
mv
v
1 2
W q
V
V
2.4 10 J 6.5 10 C 7
V 3.7 102 V 2. d 7.5 103 m, V 350 V, V d
350 V 7.5 10 m 3
4.7 104 N/C 3. m 2.166 1015 kg, V 530 V, d 1.2 102 m F e F g qV mg d mgd q V
(1.5 10 V)(3.204 10 C) 6.68 10 kg Vq m
3
19
27
v 2.7 105 m/s
44
31
Section 8.9 1. W 2.4 104 J, q 6.5 107 C
15
Solutions to Applying the Concepts
2
(2.166 10 kg)(9.8 N/kg)(1.2 10 m) 530 V
q 4.8 1019 C
15
v 8.4 107 m/s
q
Vq
v
v 3.8 105 m/s b)
Ek m
4
q(V 2 V 1) 2
2 2(3.2 10 J) 9.11 10 kg
5. a) I 100 A L 50 m B 3.0 105 T 45° I r 2 B (4 107 T·m/A)(100 A) r 2(3.0 105 T)
Section 9.5 1. L 0.30 m I 12 A B 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 m F 9.2 102 N B 3.5 102 T 90° I
F BL
I
(9.2 10 N) (3.5 10 T)(0.15 m) sin 90°
r 0.67 m b) Referring to the diagram in question 3,
Earth’s field lies in a line that is crossing the wire at 45° below the horizontal. The magnetic field would form a circular ring in the clockwise direction (rising on the south side of the wire, descending on the north with a radius of 0.67 m). Therefore, the field will cancel that of Earth on the south side below the wire, as shown in the diagram.
sin 2
2
I 18 A 3. a) L 50 m I 100 A F 0.25 N 45° F B IL sin
0.67 m
(0.25 N) B (100 A)(50 m) sin 45°
45° x x
B 7.1 105 T b)
N
Wire (cross-section) 45°
45°
2
2 x (0.67 m) x 0.47 m The fields will cancel 4.7 101 m south and 4.7 101 m below the wire. 6. a) r 2.4 103 m I 13.0 A L 1 m
Earth's Magnetic Field
Direction of Force
Tower
S
5
4. B 3.0 10 T L 0.20 m N 200 4 107 T·m/A BL I N (3.0 105 T)(0.20 m) I (4 107 T·m/A)(200)
I 2.4 102 A
2
N
I 2 L F 2r (4 107 T·m/A)(13.0 A)2(1 m) F 2(2.4 103 m)
F 1.4 102 N/m 7. q 20 C B 4.5 105 T v 400 m/s 90° F qvB sin F (20 C)(400 m/s)(4.5 105 T) sin 90° F 0.36 N
Solutions to Applying the Concepts
45
8. q 1.602 1019 C v 4.3 104 m/s B 1.5 T 90° F qvB sin F (1.602 1019 C)(4.3 104 m/s)(1.5 T) sin 90° 1.0 1014 N [south] F
46
Solutions to Applying the Concepts
Section 10.2 1. a) T T
t cycles
6.7 s 10
T 0.67 s c) T
60 s 33.3
T 1.80 s d) T
57 s 68
T 0.838 s 2. a) f f
cycles t 120 2.0 s
f 60 Hz b) f
45 60 s
f 0.75 Hz c) f
40 1.2 60 60 s
f 0.009 26 Hz d) f
65 48 s
f 1.35 Hz 3. a) i)
1
f T f
1 75 60 s
f 2.22 104 Hz ii) f
1 0.67 s
f 1.49 Hz iii) f
1 1.80 s
f 0.556 Hz iv) f
1 0.838 s
f 1.19 Hz
1
T f T
375 min 5
T 75 min b) T
b) i)
1 60 Hz
T 0.0167 s ii) T
1 0.75 Hz
T 1.33 s iii) T
1 0.009 26 Hz
T 108 s iv) T
1 1.35 Hz
T 0.74 s 5. 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 v f 8
f
3.0 10 m/s 640 10 m 9
f 4.7 1014 Hz 3.0 108 m/s b) f
1.2 m
f 2.5 108 Hz 3.0 108 m/s c) f 2 109 m
f 1.5 1017 Hz 5. a) v f v f 3.0 108 m/s 1.5 1013 Hz 2.0 105 m
Solutions to Applying the Concepts
47
Section 10.5
8
b)
3.0 10 m/s 2.0 10 Hz 9
c n v
5. a)
0.15 m 3.0 108 m/s c) 3.0 1022 Hz 1.0 1014 m
vo ray
o ray
8
vo ray
Section 10.4
1.486
3.0 108 m/s v 1.33 v 2.26 108 m/s 3.0 108 m/s b) v 2.42 v 1.24 108 m/s 3.0 108 m/s c) v 1.51 v 1.99 108 m/s
ve ray 2.02 108 m/s ve ray 2.02 108 m/s b) 100% vo ray 1.81 108 m/s ve ray 111.6% vo ray Therefore, the speed of the e ray is 11.6% greater than the speed of the o ray.
c 5. a) n v 8
3.0 10 m/s 2.1 10 m/s 8
n 1.43 3.0 108 m/s b) n 1.5 108 m/s n 2.0 3.0 108 m/s c) n 0.79(3.0 108 m/s) n 1.27 6. a) n1 sin 1 n2 sin 2 n1 sin 1 2 sin1 n2
2 sin1
sin 25° 1.33
2 18.5°
b) 2 sin1
sin 25° 2.42
2 10.1°
c) 2 sin1
sin 25° 1.51
2 16.3°
48
3.0 10 m/s 1.658
vo ray 1.81 108 m/s c ve ray ne ray 3.0 108 m/s ve ray
c 4. a) n v c v n
n
c n
Solutions to Applying the Concepts
Section 11.4 2. d 5.6 m x2 28 cm L 1.1 m m2
Section 11.6 2. m 22 625 nm t
dxm mL
(5.6 10 m)(0.28 m) (2)(1.1 m)
2 (22)(6.25 107 m) t 2 6 t 6.87 10 m t 6.9 m 3. t 1.75 105 m 625 nm 1 2t m 2 2t 1 m 2 2(1.75 105 m) 1 m 7 (6.25 10 m) 2
6
7.13 107 m 713 nm 3. 510 nm d 5.6 m L 1.1 m L x d (5.10 107 m)(1.1 m) x 5.6 106 m
x 0.10 m x 10 cm 4. m 3 d 5.6 m L 1.1 m 713 nm m L xm d
(3)(7.13 107 m)(1.1 m) x3 (5.6 106 m)
x3 0.42 m x3 42 cm
Section 11.5 2. ∆ PD 3 ng 1.52 624 nm t
m
PD 2(n 1)
m 55.5 m 55
Section 11.8 1. w 5.5 106 m 550 nm L 1.10 m m2 a) sin m
m w
7
sin 2
(2)(5.50 10 m) 5.5 10 m 6
sin 2 0.2 2 11.5° b) xm L sin m xm (1.10 m)(0.2) xm 0.22 m xm 22 cm 2 L
2. a) x
g
(6.24 107 m)(3) t 2(0.52) t 1.8 106 m t 1.8 m
x
w 7
2(5.50 10 m)(1.10 m) (5.5 10 m)
6
x 0.22 m x 22 cm
Solutions to Ap ply in g the Con ce pts
49
1 sin 0.1 2 1 w 2 1 5.50 107 m sin 2 5.5 106 m
b) sin
11.5° L 3. x w
sin m
(5.50 10 m)(1.10 m) (5.5 10 m) 6
x 0.11 m x 11 cm 6. R 1 107 rad d 2.4 m d a) R
sin 2 0.410 2 24° sin m
m
sin 3
3(5.30 10 m) 2.59 10 m
7
6
sin 3 0.614 3 38°
m
m4 d b) m
2.59 106 m m 5.50 107 m
x b) sin = L
1 103 m) 2 L sin(1 107 rad) (1.0
m 4.7 m4 d c) m 6
m
2.59 10 m 4.50 10 m 7
m 5.7 m5 3. m 2 2 8.41o 614 nm m a) d sin m
L 5000 m L 5 km
Section 11.9 1. N 8500 w 2.2 cm 530 nm
(2)(6.14 10 m) d sin 8.41°
w d N
7
2.2 102 m d 8500 d 2.59 106 m
sin m
m
sin 1
5.30 10 m 2.59 10 m
d
7
sin 1 0.205 1 12°
50
2.59 106 m m 6.50 107 m
(1 107 rad)(2.4 m) 1.22 1.97 10 197 nm
d
d 2. a) m
1.22
7
d
2(5.30 107 m) sin 2 2.59 106 m
7
x
m
6
d 8.396 106 m d 8.40 m b) w 1.96 cm w N d 1.96 102 m N 8.396 106 m
N 2334 slits
Solutions to Applying the Concepts
Section 11.10 3000 lines 100 cm 1. 300 000 lines/m 1 cm 1m 20 000 lines 100 cm 100 000 lines/m 20 cm 1m Therefore, 3000 lines/cm produces the best resolution.
3. sin Red
m
6.
sin
2d sin m m
2d (5.2 1011 m)(2) sin 2(2.5 1010 m)
sin 0.208 168°, 192°
d
7
sin Red
(1)(7.30 10 m) 1.89 10 m 6
sin Red 0.386 Red 22.7° sin Violet
m
sin Violet
(1)(4.00 10 m) 1.89 10 m
d
7
6
sin Violet 0.211 Violet 12.2° sin Green
m
sin Green
(1)(5.10 10 m) 1.89 10 m
d
7
6
sin Green 0.269 Green 15.6° This can be similarly proven for the next 3 orders using the appropriate m. The sequence is violet, green, red. At the fourth order, green and red maxima are no longer visible. 5. d 2.5 1010 m 12o m2 2d sin
m 2(2.5 10 m) sin 12° 2 10
5.198 1011 m 52 pm
Solutions to Ap ply in g the Con ce pts
51
Section 12.2 1. T 12 000 K a) The maximum wavelength can be found using Wien’s law: 2.898 103 max
T 2.898 10 12 000 K
The peak wavelength of Rigel is 2.4 107 m. It is in the ultraviolet spectrum. 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 found using Wien’s law: 2.898 103 max
T 2.898 10 900 K
2
p o t s
1
8
7
9
10
11
12
13
f 0 (×1014 Hz)
2. a) Increasing the work function by 1.5 would
cause a vertical shift of the line. Hence, potential would have to be greater, but the frequencies would not change. h b) The term is constant and hence the e
slope would not change. 3. 230 nm 2.3 107 m The energy can be found as follows: hc E W 0 (8 1034 J·s)(3.0 108 m/s) E 2.3 107 m
3
max
) V (
0
max 2.4 107 m
vs. f 0
3
V
3
max
V stop
max 3.2 106 m
The peak wavelength of the light is 3.2 106 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 the visual spectrum.
Section 12.3
4.64 1019 J E 5.79 1019 J
Section 12.4 2. E 85 eV, 214 nm 2.14 107 m a) Momentum of the original electron can be found using: E p c
h W 1. V f 0 0 e e eV hf 0 W 0
19
p
Choosing two pairs of values from the table and subtracting, (1.6 1019 C)(0.95 V) h(7.7 1014 Hz) W 0 (1.6 1019 C)(0.7 V) h(7.2 1014 Hz) W 0 (1.6 1019 C)(0.25 V) h(0.5 1014 Hz) h 8 1034 J·s W 0 4.64 1019 J W 0 2.9 eV
(85 eV)(1.6 10 C) 3.0 10 m/s 8
p 4.53 1026 N·s b) Momentum of the resultant electron can be
found using: h p
6.626 1034 J·s p 2.14 107 m
p 3.1 1027 N·s c) The energy imparted can be found by: hc E E
52
Solutions to Applying the Concepts
E (85 eV)(1.6 1019 C)
The wavelength of the spectral lines is:
(6.626 1034 J·s)(3.0 108 m/s) 2.14 107 m
81
81
81
E 1.27 1017 J
The energy imparted to the electron was 1.27 1017 J. d) The energy imparted increased the speed of the electron. Hence, it can be found using: 2 E v v
2(1.27 10 J) 9.11 10 kg m
17
8
18
81 9.25 108 m
Similarly, the energy change when the electron transfers from 7 to 2 is: E72 E7 E2 2.18 1018 J 2.18 1018 J E72 2 2
7 2
E72 5 1019 J hc 72 E72
The speed increase of the electron is 5.27 106 m/s.
72 3.98 107 m
Hence the wavelength separation is
Section 12.5 1. v 1 km/s 1000 m/s The wavelength can be found using de Broglie’s equation:
∆
72 81 7 8 ∆ 3.98 10 m 9.25 10 m 7 ∆ 3.05 10 m
3. The change in energy can be computed using: E Ef Ei 13.6 eV 13.6 eV E 2 nf ni2
h mv
34
6.626 10 J·s (9.11 10 kg)(1000 m/s) 31
7.27 107 m
Hence, the wavelength of the electron is 7.27 107 m.
For the Lyman series, the lower boundary is when the electron jumps from the second to the first orbital: 13.6 eV 13.6 eV Emin 2 2 12 Emin 10.2 eV The higher boundary for the Lyman series is when the electron jumps from infinity to the first orbital: 13.6 eV 13.6 eV Emax 2 12 Emax 13.6 eV For the Balmer series, the lower boundary is when the electron jumps from the third to the second orbital: 13.6 eV 13.6 eV Emin 2 3 22 Emin 1.89 eV
Section 12.6 2. We shall first compute the change in energies and the wavelength of spectral lines emitted in each case. From that, the wavelength separation can be computed. The energy change when the electron transfers from 8 to 1 is: E81 E8 E1 2.18 1018 J 2.18 1018 J E81 2 2
8 1
E81 2.15 1018 J
(6.26 1034 J·s)(3.0 108 m/s) 72 5 1019 J
v 5.27 10 m/s
34
(6.626 10 J·s)(3.0 10 m/s) 2.15 10 J
31
6
hc E
Solutions to Ap ply in g the Con ce pts
53
The higher boundary for the Balmer series is when the electron jumps from infinity to the second orbital: 13.6 eV 13.6 eV Emax 2 22 Emax 3.4 eV For the Paschen series, the lower boundary is when the electron jumps from the fourth to the third orbital: 13.6 eV 13.6 eV Emin 2 4 32 Emin 0.66 eV The higher boundary for the Paschen series is when the electron jumps from infinity to the third orbital: 13.6 eV 13.6 eV Emax 2 32 Emax 1.51 eV For the Brackett series, the lower boundary is when the electron jumps from the fifth to the fourth orbital: 13.6 eV 13.6 eV Emin 2 5 42 Emin 0.31 eV The higher boundary for the Brackett series is when the electron jumps from infinity to the fourth orbital: 13.6 eV 13.6 eV Emax 2 42 Emax 0.85 eV Thus, the boundaries for the four series are: Lyman: 10.2 eV to 13.6 eV Balmer: 1.89 eV to 3.4 eV Paschen: 0.66 eV to 1.51 eV Brackett: 0.31 eV to 0.85 eV
Hence, the uncertainty in position is 6.3 102 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 1016 J, mp 1.673 1027 kg First we shall find the velocity using: 2 Ek v v
2(1.92 10 J) 1.673 10 kg
16
27
v 4.8 105 m/s
The uncertainty in position can be found using: y
h mv –
1.0546 1034 J·s y (1.673 1027 kg)(4.8 105 m/s)
y 1.32 1013 m
The uncertainty in the position is 1.32 1013 m. 7. The uncertainty does not affect the object at a macroscopic level.
Section 12.8 1. ∆v 1 m/s 1 106 m/s, mp 1.673 1027 kg The uncertainty in position can be found using: h mv –
1.0546 1034 J·s y (1.673 1027 kg)(1 106 m/s)
y 6.3 102 m
54
2
mp
y
h ,
Solutions to Applying the Concepts
2. For Phillip, at rest relative to the experiment: d t v 2h t 0 c
Section 13.1 1. For Nadia: mRLv0 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
t 0
For Jerry: mRLv0 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/s 2. The classical addition of velocities gives: kvp kvu uvp kvp 0.5c [R] 0.6c [R] kvp 1.1c
For Barb, the stationary observer watching the experiment travel by at v 0.6c : t
t 0 v2
1 c 2
t
8
2.0 10 s (0.6c ) 1 c 2
2
60 s/min 1.1538 s 52 beats/min The dilated time for the earthly observers is: t
1 2 mv Vq 2 v 2Vq
2(1.00 10 V)(1.6 10 C) (9.11 10 kg) m
6
19
31
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 time dilation from 2.2 s to 3.1 s that occurs at its speed of v 0.7c . The extra path length is: ∆d
t 0 2.0 108 s
3. For Marc, the time for one beat is:
3. Ek-gained Ee-lost
8
t 2.5 108 s
This answer violates the second postulate of special relativity.
v
2(3.0 m) 3.0 10 m/s
d2 d1 ∆ d vt 2 vt 1 ∆ d v(t 2 t 1) ∆ d (0.7c )(3.1 s 2.2 s) ∆ d 189 m
t 0 v2
1 c 2
t
1.1538 s (0.28c ) 2
1 2 c
t 1.2019 s
The new rate is: 60 s/min 49.9 bpm 1.20 s/beat 4. The contracted distance L, measured by Katrina, is given by: L 0.5 L0
1
v2 L L0 1 c 2
0.5 L0 L0
v2 c 2
v2 c
0.25 1 2 v 0.75 c v 0.866c v 2.60 108 m/s
Solutions to Applying the Concepts
55
6. L0 5.75 1012 m
3. L0 200 ca v 0.9986c
The time you take: distance measured t 0 velocity v2 c 2 v
v2 c
(vt 0)2 L02 1 2
c t 0 L0
1
L0 t 0
v2 L0 1 c 2 t 0 v 200 ca1 (0.9986) 2 t 0 0.9986c t 0 10.59 a
6. For Rashad: (∆s)2 c 2(∆t )2 (∆ x)2 (∆s)2 (3 108 m/s)2(1.5 s)2 02 (∆s)2 2.05 1017 m2
2
v2 c 2 v2
2
v
c c t 1
For Kareem: (s)2 c 2(t )2 ( x)2 )2 x c 2(t )2 (s 2 2 (3 1 08 m/s )2(2 s) ) x (2.025 1017 m 8 x 3.97 10 m
0
L0
v
(3.0 108 m/s)2 8 s) 1 (3.0 10 m/s)(3600 (5.75 1012 m)
Section 13.5 1. m0 5.98 1024 kg v 2.96 104 m/s
v 2.95 108 m/s
Section 13.4 1. L0 7 ca
m
L t 0 v v
7 ca 10 a v 0.7c 2. The age or time difference for the twins is: 5 a t S t T v
d v
d v
2 L0 v
5.98 1024 kg m (2.96 104 m/s)2 1 (3.0 10 8 m/s)2 m 5.980 000 03 1024 kg
2. m
m0 v2
1 c 2
m
v2 1 2 c v
2 L0
v2 c
10 ca v v 10 ca 1 2
v2 v 2c 2c 1 c 2
v2 v 2c 2c 1 c 2 v2 4vc 4c 2 4c 2 4v2 v(5v 4c ) 0 v 0.8c since v ≠ 0 56
1
At 0.9c :
T 5 a S
5 a
m0 v2 c 2
7 ca 7 a 3 a
5 a
m 1 (0 .9) 0
2
m 2.294m0 At 0.99c : m0 m 1 (0.99)2 m 7.089m0 At 0.999c : m0 m 1 (0.999)2 m 22.366m0
Therefore, there is a much greater increase in mass when accelerating from 0.99c to 0.999c .
Solutions to Applying the Concepts
3. Using the low-speed mass dilation approximation, 1 m0v2 m 2 2 c
1 (60 kg)(10 m/s)2 2 (3.0 108 m/s)2 m 3.3 1014 kg 4. Since cost, C , is proportional to energy3, E3, m
C C 1
3
E E1
2 2
C 2 ($100 million)
5000 MeV 500 MeV
3
3. The speed of the bullet relative to Earth is: bvc cvE bvE v cvE 1 bc c 2 c c 3 2 bvE c c 3 2 1 c 2 5c
bvE
5. The radius for charges moving at right angles mv to a magnetic field is r . The ratio of the Bq r m vf radius of a fast to slow electron is f f . r s msvs Assuming ms m0 (its rest mass), and m0 vf mf , the ratio becomes . 2 v v2 vs 1 1 2 c c 2 6. As in question 5, the ratio of radii is: r mv p p p and since vp ve: r e meve r p
6
5c
7 bvE 0.714c Therefore, the bullet will never reach the bandits because its speed is less than 0.75c . 4. Putting the limiting velocity v c into Hubble’s law: v Hr
gives the limiting case of:
r 1.76 1010 ca
Section 13.7 1. For momentum dilation, p
27
19
c
5
2
r 6.98 105 m
Section 13.6 2. Using the relativistic formula for velocity addition: v v v v N
N L
NN L
1 2 c
c 0.999c (c )(0.999c )
1
v c
c 2
m0v v2
1
c 2 At v 0.2c : m0(0.2c ) p (0.2c )2 1 c 2 p 0.204m0c At v 0.5c : m0(0.5c ) p (0.5c )2 1 c 2 p 0.577m0c At v 0.8c : m0(0.8c ) p (0.8c )2 1 c 2 p 1.33m0c
(1.67 10 kg)(0.996 ) r (1.6 10 C)(5.0 10 T) 1 (0 .996)
L
bvE
6 1 1
3.0 108 m/s r 1.7 102 m/s/ca
p e
r p 1833r e 7. m0 1.67 1027 kg v 0.996c B 5.0 105 T m0v r v2 qB 1 c 2
v
m r m
L
c r H
(1.67 1027 kg)r e r p 9.11 1031 kg
v
e
L
C 2 $100 billion
Solutions to Applying the Concepts
57
2. E m0c 2 Ek
3.
Case A: 125 J m0c 2 87 J m0 38c 2 J Case B: 54 J m0c 2 15 J m0 39c 2 J Therefore, B has the greater rest mass. 3. The energy used by the bulb is:
E mc 2 E Pt Pt m c 2 m
(80 W)(365 24 60 60 s) (3.0 10 m/s) 8
2
v
Section 13.8 1. E mc 2 E (106 MeV/c 2)c 2 E 106 MeV 1 106 eV 1.6 1019 J E 106 MeV
1 MeV
1 eV
E 1.696 1011 J
The equivalent mass is:
m
m0 v2
1 c 2
4 106
1 v2 1 c 2
Since v2 ≈ c 2, we can use the high-speed approximation:
v 2 1
1.696 1011 J m (3.0 108 m/s)2 m 1.88 1028 kg 2. A mass, m, is equivalent to an energy:
E mc 2 E (1.67 1027 kg)(3 108 m/s)2 E 1.503 1010 J 1 eV 1.6 1019 J 1.503 1010 J m 1.6 1019 J/eV m 939.37 106 eV/c 2 m 939.4 MeV/c 2
c 2
v2 1 c 2
E m 2 c
35 36 v 0.986c v 2.96 108 m/s 4. Given the dilated mass of the proton, m 4 106m0
m 2.80 108 kg m 2.80 105 g 4. E mc 2 E (65 kg)(3 108 m/s)2 E 5.85 1018 J
58
E2 (pc )2 (m0c )2 (mvc )2 (mc 2)2 (m0c )2 mc 2 (m0c 2 Ek) mc 2 m0c 2 5m0c 2 mc 2 6m0c 2 (mvc )2 (6m0c 2)2 (m0c 2)2 (mvc )2 35m02c 4 m2v2 35m02c 2 m0 Since m , v2 1 2 c 2 v 2 v2 35 1 c c 2 v2 35c 2 35v2 36v2 35c 2
4 106
c
1 v 1
2
c
1 v c (4 106) 2 v 1 3.13 1014 c c v 3.13 1014c c v 9.38 106 m/s 1
The protons are travelling 9.38 106 m/s slower than c .
Solutions to Applying the Concepts
Section 14.1 3. a) Binding energy is: B [ Zm(1H) Nmn m(2H)]c 2 B 938.78 MeV 939.57 MeV 1876.12 MeV B 2.23 MeV B 2.23 MeV b) 1.12 MeV/nucleon A 2 nucleons 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 A4 2. Since A Z X → Z 2 Y : 244 219 240 60 a) 234 90Th b) 94Pu c) 84Po d) 92U e) 27 Co A 3. Since A Z X → Z 1Y e : 23 35 45 64 a) 32 16S b) 11Na c) 17Cl d) 21Sc e) 30Zn A 4. Since A Z X → Z 1Y e : 46 239 64 a) 199F b) 22 10Ne c) 23V d) 92U e) 28 Ni Section 14.3 1. The amount eaten is: 1 1 1 1 1 1 1 2 4 8 16 32 128 256 255 255 1 . The amount left is 1 256 256 256 1 8 or . 2
2. T 12 1.28 109 a, N 0 5 mg, N 1 mg
1 N N 0 2
1 2
T 1
2
log 1 log 2
235
235
N 238 N
0.0044 log
0.0044 0.030
238
0
0
t T 235
t T 238
t
1 1 7.04 10 a 4.45 10 a ) 8
9
12
t (1.196 109 a1) log
t
0.8337 (0.3010)(1.196 10 a ) 9
1
t 2.3 109 a
Section 14.4 1. Bismuth or 209 83 Bi unobservable annual dosage 360 mSv 2. dosage per dental x-ray 0.20 mSv 1800 doses 4. annual dose dose equivalent activity time 6 D (1.3 10 eV)(1.602 1019 J/eV)(1) (29 000 Bq/kg) (365 24 60 60 s) D 0.1905 J/kg D 191 mSv
Section 14.5 2. In a head-on elastic collision with the target, 3 H at rest, the recoil velocity is: v v
mn mx mn mx
v v
v 0.5v
N N 0
t T 12
1 ( N) 2 1 ( N) 2 1 ( (0.030) 2
1u 3u 1 u 3 u
t
N t 1 log log N 0 2 T
3. T 235 7.04 108 a, T 238 4.45 109 a, 235 235 N N 0.030 0.0044, 238 238 N N 0
1 mg log 5 mg t (1.28 109 a) 1 log 2 9 t 2.97 10 a
Tritium is 50% effective in slowing down the fast neutrons. 4. power amount of energy/mole number of moles used/12 h (12 3600)1 h/s 400 g 600 g P (1699 GJ/mol) or 2 g/mol 3 g/mol 1 43200 s P 7.87 GW
Solutions to Applying the Concepts
59
5. Since 1 neutrino is created along with
1 deuterium atom, and 2 deuterium atoms are needed to create an 4He ion, 2 neutrinos are created.
Section 14.6 1. Using Einstein’s energy triangle: (mvc )2 (m0c 2 Ek)2 (m0c 2)2 mvc (0.511 MeV 310 0 MeV )2 (0 .511 M eV)2 mvc (3100.5 MeV)(1.602 1013 J/MeV) 4.9670 1010 J mv 3.0 108 m/s mv 1.6557 1018 N·s
The de Broglie wavelength is:
h
6.63 10 J·s 1.6557 10 N·s
mv
34
18
4.0 1016 m v 2. f 2r 3.0 108 m/s f 2(4300 m) f 11 kHz 3. a) In Einstein’s energy triangle, (mc 2)2 (m0c 2)2 (mvc )2 [see Chapter 13] m0c 2 938.27 MeV mc 2 m0c 2 Ek mc 2 938.27 MeV 10 MeV mc 2 948.27 MeV
In the triangle, m0c 2 cos mc 2
cos
938.27 MeV 948.27 MeV
b) r
2 f 4.35 107 m/s r 2(32 106 Hz) r 0.216 m
Section 14.7 2 2 1 2. a) uud 1 3 3 3 2 2 1 b) u u d 1 3 3 3 2 1 1 c) ud 3 3 2 1 1 d) udd 0 3 3 3 1 2 e) su 1 3 3 3. a) proton (baryon) b) antiproton (baryon) c) pion (meson) d) neutron (baryon) e) kaon (meson) 2 1 1 4. udd 0 3 3 3 5. The mass “defect” of a 0 meson is: md m b m (8 4700 5279) MeV/c 2 571 MeV/c 2 Section 14.8 1. i) An electron and a positron annihilate each other, releasing two gamma rays. ii) A neutron undergoes decay to an antineutrino, a positron, and an electron. iii)A planet orbits the Sun via the exchange of a graviton.
8.328° mvc sin mc 2 v sin 8.328° c v 0.1448c v 4.35 107 m/s
60
v
Solutions to Applying the Concepts
PART 2 Answers to End-of-chapter Conceptual Questions 6. Assume for all cases that north is positive and
Chapter 1 1. It is possible for an object to be accelerating
and at rest at the same time. For example, consider an object that is thrown straight up in the air. During its entire trajectory it is accelerating downward. At its maximum height it has a speed of zero. Therefore, at that point it is both accelerating and at rest. 2. A speedometer measures a car’s speed, not its velocity, since the speedometer gives no indication as to the direction of the car’s motion. 3.
Position-Time
m
s e r 5 t e M 4 n i 3 n o i t 2 i s 1 o P
1 2 3 4 5 6 7
t
Time in Seconds Time in Seconds 1
d n o –1 c e – 2 S r – 3 e p s – 4 e r – 5 t e M – 6 n – 7 i y t i – 8 c o l e – 9 V
2 3 4 5 6 7
t
– 10
m
4. Displacement, velocity, and acceleration are
all vector quantities. Therefore, a negative displacement, velocity, or acceleration is a negative vector quantity, which indicates that the vector’s direction is opposite to the direction designated as positive. 5. The seconds are squared in the standard SI unit for acceleration, m/s2, because acceleration is the change in velocity per unit of time. Therefore, the standard SI unit for acceleration is (m/s)/s, which is more conveniently written as m/s2.
south is negative. a) Positiontime graph: The object sits motionless south of the designated zero point. The object then moves northward with a constant velocity, crossing the zero point and ending up in a position north of the zero point. Velocitytime graph: The object moves southward with a constant velocity. The object then slows down while still moving southward, stops, changes direction, and speeds up northward with a constant acceleration. b) Positiontime graph: The object starts at the zero point and speeds up while moving northward, then continues to move northward with a constant velocity. Velocitytime graph: The object starts at rest and speeds up with an increasing acceleration while moving northward. The object then continues to speed up with a constant acceleration northward. c) Positiontime graph: The object starts north of the zero point and moves southward past the zero point with a constant velocity. The object then abruptly slows down and continues to move southward with a new constant velocity. Velocity time graph: The object slows down while moving northward, stops, changes direction, and speeds up southward with a constant acceleration. The object then abruptly reduces the magnitude of its acceleration and continues to speed up southward with a new constant acceleration. d) Positiontime graph: The object starts at the zero point and moves northward and slows down to a stop, where it sits motion less for a period of time. The object then quickly speeds up southward and moves
A n s w e rs t o E n d - of - c h a p te r C o n c e p t u a l Q u es t i o n s
61
southward with a constant velocity, going past the zero point. Velocitytime graph: The object starts at rest and speeds up while moving northward. The acceleration in this time period is decreasing. The object then continues to move northward with a constant velocity. The object then slows down while moving northward, stops, changes direction, and speeds up southward with a constant acceleration. 7. a) vavg
forces acting on an object is half of an actionreaction pair. If both the action forces and the reaction forces were included in a free-body diagram, then all the forces would cancel. For example, a free-body diagram for a ball being kicked must not include the reaction force provided by the ball on the foot, or else the forces would cancel and the ball would not accelerate. 10.
F n
dtot t tot
1000 m (5)(60 s)
F m
3.3 m s dtot b) vavg t tot
1000 m (4)(60 s)
11. Dear Cousin,
2000 m (9)(60 s)
3.7 m s
d) The answer for c) is the average speed of
the bus over the whole trip, whereas half the sum of its speed up the hill and its speed down the hill is an average of the average speeds up and down the hill. 8. In flying from planet A to planet B, you would need to burn your spacecraft’s engines while leaving planet A in order to escape its gravitational pull and then to make any necessary course corrections, and while arriving at planet B in order to slow down and stop. Assuming there were no forces acting on the craft in between, it would travel with constant velocity once the engines were turned off. 9. A free-body diagram shows the forces acting on an object, as these are the only forces that can cause the body to accelerate. Since, by Newton’s third law, for every action force there is a reaction force, equal in magnitude and opposite in direction, then each of the 62
F f
F g
4.2 m s dtot c) vavg t tot
Motorcycle
You asked me to explain Newton’s first law of motion to you. Newton’s first law of motion states that an object will keep moving at a constant speed in the same direction unless a force makes it slow down, speed up, or change direction. Here’s an example. Suppose you’re pushing a hockey puck across the carpet. When you let go, the puck quickly stops moving. This is because the carpet is not very slippery; we say that it has a lot of friction. The force of friction is making the puck slow down. What if you slide the puck across a surface with less friction, like ice? The puck will take longer to stop moving, because the force of friction is much less than on the carpet. Now suppose you slide the puck across an air hockey table. The force of friction is so small that the puck will slide for a much, much longer time. So, you can imagine sliding a puck on a surface with no friction at all. The puck never stops, because there is no force to slow it down! Perhaps you’re wondering about a motionless object that isn’t experiencing a force — why isn’t it moving at a constant speed in the same direction? But it is! Zero is a constant speed.
A n s w e rs t o E n d - of - c h a p te r C o n c e p t u a l Q u es t i o n s
F n
F n
Puck on carpet
F f
F g
Puck on ice
F f
F g
F n
F n
flight, respectively, and v2i and v2f are the initial and final velocities downward, respectively. Since d1 d2, we can write the following equation: t 1(v1i v1f ) t (v v2f ) 2 2i 2 2 On the left side, the final velocity upward, v1f , is equal to zero. On the right side, the initial velocity downward, v2i, is equal to zero. The equation simplifies: t 1(v1i) t (v ) 2 2f 2 2 But v1i is equal to v2f and is not zero, and therefore t 1 t 2. 15. The ball is undergoing uniform circular motion, as it is travelling in a circle at a constant speed. Because its trajectory is curved, it cannot be undergoing uniform motion, which requires an object to be travelling at a constant speed in a straight line.
Puck on air table
F f
F g
Puck on frictionless surface
F g
12. The gravitational force applied by the Moon
on Earth does not cancel with the gravitational force applied by Earth on the Moon because these forces act on different bodies. Only forces applied on the same body can possibly cancel one another. 13. When you fire a rifle, the forces applied to the bullet and the rifle make up an actionreaction pair. By Newton’s third law, the force applied to the bullet is equal and opposite to its reaction force, the force applied to the rifle. This reaction force causes the rifle and you to recoil in the opposite direction. 14. While in the air, the ball’s vertical acceleration is constant and equal to g 9.8 m/s2. The ball travels the same distance upward as downward, and therefore the ball’s speed is the same when it reaches the ground as when it leaves the ground, since its acceleration is constant. Suppose the lengths of time it takes the ball to travel upward and downward are t 1 and t 2, respectively. We can use the equations t (v v1f ) t (v v2f ) d1 1 1i and d2 2 2i 2 2 for the distances travelled upward and downward, respectively, where v1i and v1f are the initial and final velocities during the upward
Chapter 2 1. Frictional forces are forces that oppose motion.
A frictional force will only try to prevent an object from moving, it will not actually cause an object to move. 2. It is not possible to swing a mass in a horizontal circle above your head. Since gravity is always pulling down on the mass, an upward component of the tension force is required to balance gravity. As the speed of rotation increases, the angle relative to the horizontal may approach 0° but will never reach 0°. 3. If the gravitational force downward and the normal force upward are the only two vertical forces acting on an object, we can be certain that they are balanced if the object is not accelerating. If one of these forces were greater than the other, the object would accelerate in the direction of the greater force. 4. The most common way to describe directions in three dimensions is by the use of three unit vectors (and their opposites). Traditionally, the three unit vectors used are labelled as i, j , and k. One of these unit vectors will represent right, one will represent up, and one will
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represent coming out of the plane of the page toward you. 5. The bullets reach the ground in the same amount of time. Recall that the horizontal and vertical motions of each bullet are independent of each other. Since both identical objects are accelerating downward at the acceleration due to gravity and they are both dropped from the same height, it takes the same time for them to reach the ground. 6. Dear Wolfgang, You asked whether the time it takes to paddle a canoe across a river depends on the strength of the current. When you are paddling a canoe across a river, the variables that determine how long it takes are the width of the river and the forward velocity of the canoe due to your paddling. The canoe’s forward velocity and the current velocity are perpendicular to each other, so they don’t affect each other. As a result, the current does not affect the length of time required to cross the river. The only effect of the current on the motion of the canoe is to cause it to move downstream from where it would otherwise have landed. 7. The student who wants to apply the force above the horizontal has the better idea. The horizontal component of the applied force in the direction of motion will be the same regardless of whether the force is applied above or below the horizontal. It is in the students’ best interest to minimize the amount of friction. Recall that the frictional force is directly proportional to the normal force. If they apply the force above the horizontal, this will reduce the magnitude of the normal force needed to be supplied by the floor on the sofa, which will therefore reduce the frictional force and make it easier to move the sofa. On the other hand, if they apply the force below the horizontal, this will increase the normal force required and thereby increase the frictional force, making it harder to move the sofa.
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8. a) The baseball’s velocity will be upward with a
magnitude less than its initial velocity. The acceleration will be downward at 9.8 m/s2. b) The baseball’s velocity will be zero. The acceleration will be downward at 9.8 m/s2. c) The baseball’s velocity will be downward with the same magnitude as in a). The acceleration will be downward at 9.8 m/s2. 9. You would still need a pitcher’s mound on the Moon because the ball would still accelerate downward due to gravity. Since the Moon has a smaller mass than Earth, the acceleration due to gravity on the Moon is less than that on Earth. As a result, the height of the mound would not have to be as great as that on Earth. 10. She could jump twice as far on a planet that has one-half the gravity of Earth. If we assume that her initial speed and the direction for launch are the same, and that her initial vertical displacement is zero, we can write the following. 1 dy v1 t ayt 2 2 1 0 v1 ayt 2 2v1 y
y
y
t ay
If the acceleration, ay, is halved, then the time in flight, t , will be doubled. Therefore, the horizontal distance travelled will also be dou bled, assuming that her horizontal speed is constant. 11. As your bicycle’s rear tire spins, it takes water with it due to adhesion. Inertia causes the water to try to move in a straight line. As a result, the water leaves the wheel with a velocity tangential to the tire and may spray your back if your bicycle does not have a protective rear fender. 12. Inertia causes the water in your clothing to try to move in a straight line. If the drum in the washing machine were solid, it would apply a centripetal force on the water, which would keep it moving in a circle. Since the drum has holes in it, however, the water is able to leave the drum as it spins.
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13. The aircraft can be flown in one of two ways,
or a combination of these, to provide “weight lessness.” If the aircraft accelerates downward at the acceleration due to gravity, the astronauts inside the aircraft will experience “weightlessness.” The other possibility is to travel in a vertical arc. If the aircraft flies in a vertical arc at such a speed that at the top of the arc the gravitational force provides all the centripetal force required to keep the aircraft and its occupants travelling in a circle, they will experience “weightlessness.”
Chapter 3 1. Hydro lines and telephone cables cannot be
run completely horizontally because the force of gravity acts downward on the entire wire and there is very little means of counterbalancing this force using supports. 2. a) The ladder is pushing directly into the wall on which it is resting, normal to the surface of the wall. With no friction, there is no force to prevent the ladder from sliding down the wall. b) The force exerted by the ladder on the ground is exactly equal to the force of gravity (weight) of the ladder because there is no vertical force due to friction. The only force that acts vertically, upward or downward, is the force of gravity. 3. Standing with your feet together or wide apart makes no difference to the condition of static equilibrium, since in both cases all forces are balanced. In terms of stability, the wider stance is more stable. A wider stance means a lower centre of mass and a wider “footprint.” This means there is a greater tipping angle for this wider stance. 4. High-heeled shoes force the centre of mass of the person wearing to move forward from its normal position. To maintain balance, the person must move the centre of mass back again, usually by leaning the shoulders backward. This effort can cause fatigue in the back muscles.
5. Line installers allow a droop in their lines
when installing them because the droop allows a moderate upward vertical application of force as the wire curves upward to the support standards. This allows an upward force to support the wire when loaded with freezing rain and ice buildup. This droop means that the tension to support the load can be much less because of the greater angle. 6. A wrench can be made to more easily open a rusty bolt by adapting the wrench so as to apply more torque. More torque can be applied by the same force by adding length to the wrench handle. 7. The higher up on a ladder a person is, the farther he is from the pivot point, which is the point where the ladder touches the ground. Therefore, the ladder will be more likely to slide down the wall if the person stands on a higher rung. 8. The torque varies as sin , where is the angle between the pedal arm and the applied force. The torque is at a minimum (zero) when the pedals are vertical (one on top of the other), because the force (weight) is applied at 0° to the pedal arm, and sin 0° 0. The maximum torque is applied when the pedals are horizontal, because the angle between the pedal arm and the applied force is 90°, and sin 90° 1. 9. There is no extra benefit for curls to be done to their highest position. As the forearm is raised, the angle of the force of gravity vector decreases at the same rate as the angle between the muscle of effort and the arm. As the forearm is raised, the effort required to lift the arm decreases, but so does the muscle’s ability to provide the effort. 10. Your textbook is sitting in stable equilibrium when flat on your desk. When the book is balanced on its corner, it is in unstable equilibrium. Motion in any direction will cause a lowering of the centre of mass and a release of gravitational potential energy, making the tipping motion continue and thus making the book fall.
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11. In terms of stability, a walking cane provides a
18. Lumber is used this way to support greater
wider base (footprint) over which the person is balancing. It is harder to force the person’s centre of mass outside this wider support base. 12. When standing up from a sitting position, we first must lean forward to move our centre of mass over our feet to maintain stability. Unless we first lean forward, our centre of mass is already outside our support base and it is impossible to stand up. 13. A five-legged chair base is more stable because of the wider support base (footprint). The extra leg effectively increases the tipping angle, making the chair more stable. 14. Tall fluted champagne glasses must have a wide base to improve the stability of the glass. Recall that the tipping angle is given by the (0.5)(width of base) expression tan1 . height of centre of mass Therefore, the taller the glass, the greater the height of the centre of mass, and the smaller the tipping angle. A wider base increases the tipping angle by compensating for the taller glass. 15. The extra mass helps to mimic the mass of the cargo and lowers the centre of mass of the ship. Without this extra mass, the ship would be top-heavy and more prone to capsizing, especially in rough weather. 16. This figure is so stable because the design of the toy places the effective centre of mass below the balance point. A gentle push actually raises the centre of mass like a pendulum, which increases the gravitational potential energy, which tends to return the toy to its stable equilibrium position. 17. The bone that has the smaller length will fracture first if the same twisting stress is applied to two bones of equal radius but different lengths. This is due to the fact that the strain
spans because of the greater dimensions of wood in the vertical direction. More wood provides a means of supporting a greater weight through a tension force throughout the wood. 19. Concrete would not be an ideal material for a cantilevered structure because of the difference in the way that this material deals with tension and compression forces. A cantilever would require a great tensile strength in the upper layer and a great compressive strength in the lower layer. Concrete has great compressive strength but poor tensile strength.
Chapter 4 1. Momentum is the product of mass and velocity; p mv . Since velocity is a vector quantity, so
2.
3. 4. 5.
6.
on the longer bone will be much L L
smaller than that on the shorter bone, because the length term appears in the denominator of the expression for strain.
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7.
is momentum. A system represents all the objects involved in a collision. In a closed system, the boundary is closed (that is, there are no interactions with the external environment) and therefore the net external force acting on the system’s objects as a group is zero. In an open isolated system, the boundary is not closed but the net external force acting on the system is zero. The net force is used in the calculation of impulse; J F ∆t . Impulse is the change in momentum; J ∆p. In an isolated system, the net external force, F , acting on the system is zero. Therefore, the impulse, J , is zero ( J F ∆t ), and the change in momentum, ∆p, is zero ( J ∆p). The law of conservation of (linear) momentum states that the total momentum of an isolated system before a collision is equal to the total momentum of the system after the collision. This can be expressed algebraically as ptotalinitial ptotalfinal. Equivalently, in an isolated system the change in momentum is zero; ∆p 0. Yes, a ball thrown upward loses momentum as it rises because there is a net external force downward (gravity) acting on the ball, slowing it down.
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8. Assuming that the net external force acting on
the grenade during the explosion is zero (ignoring gravity), the sum of the 45 momentum vectors after the explosion is equal to the momentum vector of the grenade before the explosion, since ptotalinitial ptotalfinal. 9. Assume that the astronaut’s initial momentum is zero as he floats in space. By throwing the monkey wrench in the opposite direction of the space station, he would be propelled toward the space station. This is an example of Newton’s third law: The total momentum of the astronaut–wrench system would still be zero after he threw the wrench. 10. A rocket can change its course in space by ejecting any object or matter such as a gas. Assuming that the total momentum of the rocket–gas system is conserved, the momentum of the rocket will change as the gas is ejected. This change in momentum will correspond to an impulse, which will change the course of the rocket. 11. Assume that the total momentum of the system 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 total momentum before and after the collision is pTo 0 m(v1f v2f ) pTf . 12. As rain falls into the open-top freight car, the car will slow down. Assuming that momentum is conserved as the rain falls into the car, the combined mass of the car and the water will 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 the objects, vf , is equal to the average of their initial velocities, vAo and vBo:
pTo pTf mvAo mvBo mvAf mvBf vAo vBo vf vf vAo vBo vf
2
Since the angle between vBo and vf is greater than the angle between vAo and vf , the magnitude of vAo is greater than the magnitude of vBo. 14. The component method would be preferred for solving momentum problems in which trigonometry could not be used readily — for instance, problems involving more than two objects colliding, or non-linear problems. 15. a) Grocery clerks lean back when carrying heavy boxes so that their centres of mass stay in line with their feet. b) The centre of mass of a system of masses is the point where the masses could be considered to be concentrated or “balanced” for analyzing their motion. This concept can simplify momentum problems since the momentum of the centre of mass is equal to the total momentum before, and after, a collision, and is conserved during the collision.
Chapter 5 1. When you are holding your physics book
steady in your outstretched arm, there is no work done because there is no displacement (W F ∆d). 2. The momentum, p, of an object with mass m is related to its kinetic energy, Ek, according to the equation p 2mEk. If a golf ball and a football have the same kinetic energy then the football has the greater momentum, since the mass of the football is greater than the mass of the golf ball. 3. A negative area under a force–displacement graph represents negative work, which means that the displacement is in the opposite direction of the force applied. For example, when friction is slowing down a car, there is a positive displacement but a negative force. 4. After work is done on an object, it has gained energy.
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5. When a spring diving board is compressed by
6.
7.
8.
9.
a diver jumping on it, the diving board possesses elastic potential energy. As the diving board straightens out, it transfers its elastic potential energy to the diver, who gains kinetic and gravitational potential energy. As the diver rises in the air, her kinetic energy is transformed into potential energy until she only has gravitational potential energy as she reaches her highest point. As she descends toward the pool, her potential energy is transformed into kinetic energy and she increases her speed as she falls. As she enters the pool and slows down in the water, her kinetic energy is transferred to the water as kinetic energy, potential energy, and heat energy. 1 Ek mv2 2 ( J) (kg)[(m s)2] ( J) (kg·m2 s2) ( J) (kg·m s2·m) ( J) (N·m) ( J) ( J) The equation ∆ Ee ∆ Ek means that a loss of elastic potential energy becomes a gain in kinetic energy. Yes, since gravitational potential energy is measured relative to a point which could change. That point could be the ground level, the basement level, or any other arbitrary point. In an elastic collision the total kinetic energy is conserved, whereas in an inelastic collision the total kinetic energy is not conserved. An example of an (almost) elastic collision is a collision between two billiard balls. An example of an inelastic collision is a collision between two vehicles in which their kinetic energy is transferred to heat energy, sound energy, and energy used to permanently deform the vehicles.
2
10. No, the equation Ek
shows that if an 2m object has momentum then it must have kinetic energy. The converse is also true, as the equation also shows.
Chapter 6 1. We do not require the more general form of
2.
3.
4.
5.
6.
7.
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p
Newton’s law of universal gravitation because for situations on or near the surface of Earth, the values of G , M , and r can be assumed to be specified constants. After these simplifications are made, the general form becomes equivalent to the simpler form. Due to the direction in which Earth rotates, more energy would be required to reach the same orbit if a spacecraft was launched westward, since an eastward launch aids the spacecraft. The near side of the Moon is more massive than the far side, possibly due to impacted meteors. Over time this side was more attracted to Earth, so that eventually the more massive side came to face Earth all the time. This is also true for the moons of Jupiter and Saturn relative to their planets. The force of gravity is the derivative of gravitational potential energy, Ep. Equivalently, the force of gravity is the slope of the graph of Ep versus x. Assuming that the spacecraft is initially in orbit and that jettisoning a large piece of itself does not significantly alter its momentum, it will continue in the same orbit. The velocity of a spacecraft in orbit is constantly changing due to the centripetal force acting on it. Therefore, if one spacecraft points toward another and rockets in that direction, the two spacecraft will not meet because the “added velocity” vector of the first spacecraft does not change as is required for convergence. a) The escape speed required to leave Earth is approximately 11 km/s. The necessary upward acceleration, a, of a spacecraft during firing from an 80-m cannon is given
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by a
(11 (1 1 00 000 0 m/ m/s) s) 756 250 m/s . 2(80 m) 2
This is more than 77 000 times the magnitude of the acceleration due to gravity, and would be experienced for about 11 00 000 0 m/s m/s ∆t 0.015 s. The 756 75 6 25 250 0 m/s m/s2 mission would not be survivable. b) The downward force of the gun’s recoil would be roughly equal to the upward force on the spacecraft. If the spacecraft had a mass of 5000 kg, the force of the recoil would be approximatel approximately y 2 (5000 (500 0 kg)(7 kg)(756 56 250 250 m/s ) 3.781 109 N. 8. Given: hmax 2 m k 500 N/m xmax 0.45 m m 80 kg First, calculate the maximum energy that our knees can absorb without damage.
V g mgh V g (80 kg)(9.8 m/s2)(2 m) V g 1568 J
Next, calculate the maximum energy that the springs can absorb. 1 V e kx2 2 1 V e (500 N/m)(0.45 m2) 2 V e 51 J Finally,, calculate the maximum height from Finally which we could survive a fall without damage. V g mgh V g h mg h
1568 J 51 J (80 kg)(9.8 m/s )
2
h 2.07 m
With the springs attached, we could survive a fall 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 be only (500 N/m)(0.45 m) 225 N upward. The net force acting on us would act downward, so we would not bounce off the ground.
10. Three everyday examples of SHM are: an
idling engine, engine, as periodic power from combustion keeps piston movement in a state of SHM; someone rocking in a rocking chair, where periodic “foot pushes” or shifts in the centre of mass counteract dampening; the motion of a toy bird that “drinks” water, provided that there is a constant supply of water. 11. Three examples of damping in oscillatory systems are: engine braking (desired) — as the fuel supply to the cylinders is lessened, so is the power, which dampens piston movement; swinging on a swing (undesired) — the height of successive swings becomes smaller and smaller due to friction and air resistance; air bags (desired) — when deployed, they gradually dampen the effects of a collision on a person’s body, as opposed to a steering wheel or dashboard, which do so almost instantaneously.
Chapter 7 1. All objects on Earth that are stationary rela-
tive to Earth’s surface have the same angular velocity,, since they all complete one rotation velocity about Earth’s axis in the same amount of time. However, they do not all have the same tangential velocity, velocity, since they are not all the same distance from Earth’s axis of rotation. If is the angular velocity of an object on Earth’s surface and r is the object’s distance from Earth’s axis of rotation then the object’s tangential velocity, v, is given by v r . 2. A differential mechanism is necessary to allow a car to turn smoothly. The wheels on the inside of a turn move through a smaller radius than the wheels on the outside, thus travelling a smaller arc distance in the same amount of time. Therefore, the inside wheels rotate at a smaller angular speed. In the absence of a differential, however, the inside and outside drive wheels (connected to the motor) must rotate at the same angular speed. To turn, you would have to lock up the inside drive wheel, causing an uncontrolled turn.
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3.
4.
5.
6.
7.
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The differential allows the drive wheels to turn at different angular speeds. The top of the CN Tower, with the tower located at the highest altitude on the equator, would have the greatest tangential speed, because in that case the top of the tower would be the greatest distance from the axis of rotation of Earth. However However,, since the height of the CN To Tower wer is negligible compared to the radius of Earth, the variation in tangential speed among different parts of the tower is negligible. A larger car tire has a greater moment of inertia (greater radius and mass), thus in principle more energy would be needed to start turning the tire. Once the tire was moving, the law of inertia would apply and a greater force would be needed to slow and stop the tire, thus less energy would be needed to keep the tire moving. a) Yes, changing the tire size affects the odometer reading. For example, a tire with a larger radius than the calibrating tire covers a greater distance in the same number of turns. In that case, the car will travel a greater distance than what the odometer indicates. b) Yes, the speedometer reading is affected, for the same reason. For example, a car with larger tires will travel at a greater speed than what the speedometer indicates. The angular equivalents to force and displacement are torque and angular displacement. No linear work is done on an object if an applied force does not change the displacement of the object in the direction that the force is applied. No rotational work is done if an applied torque does not result in a change in angular displacement. No, angular momentum is conserved because the diver is in fact still rotating as she enters the water. There is no external torque applied to the diver after she leaves the diving board. Because the diver increases her moment of inertia by extending out straight from a tuck, her angular spin decreases. decreases. This is
not visually apparent apparent as the diver diver then enters enters the water out of sight of the spectators and judges. 8. No, the centripetal force acting on a rider varies depending on the radius of turn: the larger the radius, the larger the centripetal force. The riders on the outer part of the ride swing out farther than the inner riders because of the larger centripetal force. force. 9. Accor According ding to the law of conservation of angu lar momentum, the total angular momentum before the tape recorder was turned on high speed was equal to the total angular momentum after. When the tape recorder was turned on high speed, the angular momentum of the system had an added component in the angu lar direction of the turning tape. tape. Voyager 2 rotated in the opposite direction to compensate, although not as fast, since its moment of inertia was much larger than the tape recorder’s. 10. a) The hollow cylinder has a greater moment of inertia than the solid one because the hollow cylinder’s mass is concentrated farther from the axis of rotation. However, since there is no friction, there is no force available to create the torque necessary to turn the cylinders. Translational motion does not depend on the distribution of mass, so both objects accelerate at the same rate and reach the base of the incline at the same time. b) As in part a), in the absence of friction the cylinder does not roll. Therefore, both objects slide down the ramp, accelerating at the same rate (ignoring the effect of wind resistance on the different shapes). Translational motion does not depend on the distribution of mass, so both objects reach the ground at the same time. 11. A spinning projectile behaves like a gyroscope. The spin means that the object possesses angular momentum about its axis of rotation. This allows the object to resist forces acting on it as it travels, which in turn allows the object to maintain its projectile motion.
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12.
13.
14.
15.
16.
Without the spin, uneven airflow over the surface of the object would make it tumble, experience greater air resistance, and travel a shorter distance distance.. If the wheel does not slip as it rolls then the translational distance, d, that the axle moves is equal to the arc length, s, along the outside of the wheel. This is not true in the case of “squealing your tires.” Rotation axes can be anywhere, but for simplicity’s sake consider only some symmetric ones. Ranked from least to greatest moment of inertia, the rotation axis can pass through the centre of the top and bottom (shown), through the centre of the spine, through the centre of the front and back cover, or run diagonally from one corner to another another.. The angular momentum of a Sun–planet system is conserved. The force acting on the planet is that of gravity due to the Sun. At any instant in time, this force acts through the axis about which the planet instantaneously rotates. This means that the moment arm is zero and no torque acts on the planet. Therefore, the angular momentum of the planet remains constant and the total momentum of the system does not change. It is easier to balance on a moving bike than on a stationary one because of a combination of the aspect called “trail” and gyroscopic action. The law of conservation of angular momentum applies when a motorcycle is in mid-air. In the absence of an external torque, the increased angular momentum of the fasterspinning rear wheel causes the entire motorcycle to rotate in the other direction in order to keep the total angular momentum the same as it was when the motorcycle left the ground.
Chapter 8 1. A neutral object is attracted to a charged
object because the charged object induces a charge separation in the neutral object. The electrons in the neutral object are forced away from or toward the charged object, depending
on whether the charged object has a negative or positive charge, inducing an opposite charge which acts to attract the two objects by way of the law of electric forces. 2. The function of an electroscope is to detect an electric field. An electric field will cause the movement of electrons within an electroscope, inducing similar charges to cluster at each of the two pieces of dangling foil. The two pieces of foil will repel each other, indicating the presence of the electric field. 3. Rubbing the balloon against your dry hair charges the balloon electrostatically electrostatically.. When the balloon approaches the wall, the negative charge forces the electrons in the ceiling away, leaving the positive charges close to the surface. The result is that the negatively charged balloon attracts the now positively charged ceiling surface. –
–
–
+
+
+
Ceiling –
+
– –
+
+
–
–
–
+
+
+
Force of Attraction –
–
–
Balloon + + – – + –
–
–
4. The electrostatic series identifies silk as hav-
ing a greater affinity for electrons than acetate does. When acetate and silk are rubbed together,, electrons move from the acetate to together the silk because of the different affinity the materials have for electrons. 5. Choose two materials listed at either end of the electrostatic series, such as acetate and silk, and rub them together to place the predictable negative charge on the silk. Neutralize the acetate and then rub it with the mystery substance. Place the mystery substance next to the silk and judge whether the mystery substance has a negative charge (repulsion) or a positive charge (attraction). A negative charge charge would place the mystery mystery substance below acetate in the electrostatic series. Similarly, Similarly, rubbing the mystery substance with silk would help to place the
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mystery substance in the series compared to silk. By selectively choosing different substances, you could narrow down the appropriate spot for the mystery substance in the electrostatic series. 6. Computer technicians touch the metallic part of a computer before repair, assuming it is still plugged into the wall outlet, so that they ground themselves from any excess charge. Otherwise,, a static electric discharge could Otherwise damage the computer’s micro-circuitry. micro-circuitry. 7.
Criterion
Newton’s law of un univ iver ersa sall gr grav avit itat atio ion n
Coulomb’s law of el elec ectr tros osta tati ticc fo forc rces es
Equation
F
Constant of proportionality Type of force(s) Conditions for use
G 6.67 1011 N·m2 /kg2
k 9.0 109 N·m2 /C2
Attraction only
Attraction and repulsion
Acts between any two masses
Acts between any two electrostatic charges
Gm 1m 2 2
kq q 2 F 1 r 2
r
field lines. Otherwise, the rod will tend to rotate 180° and point in the opposite direcdirection (still parallel to the field lines). 11. Each point charge experiences an identical force of repulsion from all of the other point charges, so that they are all repelled symmetrically outward from the centre of the orientation. A test charge placed outside of the circle would experience a net force directed along radial lines inward to the centre of the circle, as shown in the diagram. A test charge placed inside of the circle would experience no net force, and therefore there would be no electric field inside the circle at all.
–
8. Field lines show the direction of the net force
on a test charge in an electric field. Two crossed field lines would mean that there would be two net forces acting on a test charge in two different directions at the same time. This is impossible, since there is only one net force at any point, which is only one force in one direction by definition. 9. In an electric field, charges always move along the direction described by the field lines. The direction in which a charge moves along a field line depends on the sign of the charge. A positive charge charge will move move in the direction described by the arrows in a field diagram, whereas a negative charge will move in the opposite direction. 10. a) When a polar charged rod is placed perpendicular to electric field lines, the rod will tend to rotate such that it will become parallel to the field lines. The positive end of the rod will point in the same direction in which the field lines are oriented. b) When a polar charged rod is placed parallel to electric field lines, the rod will tend to stay in the same orientation if its positive end is pointing in the same direction as the
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–
– –
– –
–
–
This charge distribution models the electric field inside a coaxial cable because the outer braided conductor in a coaxial cable acts as the site modelled by the ring of charge described above. above. This ring acts to eliminate the field within the entire cable. cable. 12. By definition, the electric potential is the same at any point along an equipotential line. Therefore, no force is required, and no work is done, to move a test charge along this line. In a situation like this, a constant force causes the constant acceleration of the test charge. 13. We use the term “point charge” to imply that the charge has no larger physical dimensions. Larger dimensions would mean that the charge would exist within a region of space instead of at a specific location. This implication reduces the number of variables and simplifies questions that deal with the distri bution of charges within a three-dimensional space. Any other approach would require some way of accounting for the variability of distances between charges.
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14. Statement: In each case the field gets
stronger as you proceed from left to right. False Reasoning: The field lines remain the same
distance apart as you move from left to right in the field in (b), so the field does not change in strength. Statement: The field strength in (a) increases from left to right but in (b) it remains the same everywhere. True Reasoning: The field lines become closer together as you move from left to right in the field in (a), so the field does increase in strength, whereas the field lines in (b) are parallel, so the field strength does not change. Statement: Both fields could be created by a series of positive charges on the left and negative ones on the right. False Reasoning: Although true for (b), (a) must be created by a single positive point charge at the base of the four arrows. Statement: Both fields could be created by a single positive point charge placed on the right. False Reasoning: As described above, a point charge could be responsible for (a), but (b) would require rows of parallel opposites such as those in oppositely charged parallel plates. 15. Electric fields are more complicated to work with because the forces that charges exert on each other are all significant. In contrast, the gravitational force between small masses is negligible compared with the gravitational force exerted on them by large masses like Earth. 16. The field shape around a single negative point charge is exactly like that around a single positive point charge with the exception that for a negative point charge, the arrows are all pointing inward instead of outward, as shown in the following diagram.
–
17. Doubling the value of the test charge will do
nothing to the measurement of the strength of the electric field. The force on the test charge will double because of the change to the test charge, but the field strength is measured as F qt
the force experienced per unit charge, .
Therefore, the doubling of the test charge and the doubling of the force will cancel, leaving the measurement of the field strength unchanged. 18. The stronger an electric field is, the closer together the field lines are. Therefore, a weak electric field has field lines that are farther apart than the field lines of a strong electric field. 19. Both gravitational fields and electric fields are made up of lines of force that are directed in a way that a test “item” would be forced. Gravitational fields are created by and influence masses, whereas electric fields involve charges. Gravitational fields are always attractive. Electric fields can be attractive or repulsive, since they can exert forces in opposite directions depending on the charge of the object that is experiencing the field. 20. The direction of an electric field between a positive charge and a negative charge is from the positive charge toward the negative charge, since electric fields are always directed the way that a positive test charge would be forced. 21. The electric potential energy is greater between two like charges than between two unlike charges the same distance apart because of the differing sign of the electric potential energy. The calculation of the
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electric potential energy involves multiplying the two charges. The product of two like charges is positive and therefore greater than the product of two unlike charges, which is negative. 22. A high-voltage wire falling onto a car produces a situation in which there is a highpotential source (the wire) very close to a low-potential region (the ground). The people in the car will be safe from electrocution as long as they do not complete a circuit between this high and low electric potential. They should not open the car door, for example, and step to the ground while maintaining contact with the car. 23. Although opposite electric charges occur at the two plates of a parallel-plate apparatus when it is connected to a power supply, the overall charge on the apparatus remains zero. For every charge at one plate, there is an opposite charge at the other plate, which balances the overall charge to zero. 24. a) If the distance between the plates is dou bled then the field strength between the plates will be halved. b) If the charge on each plate is doubled then the field strength will double. c) If the plates are totally discharged and neutral then the field strength will drop to zero. 25. Two point charges of like charge and equal magnitude should be placed side by side so that both the electric field strength and the electric potential will be zero at the midpoint between the charges. If one of the two like charges were doubled, the field strength and the potential would both be zero at a point two-thirds the separation distance away from the doubled charge. 26. In the presence of electric fields, a field strength and a potential of zero would exist at a point where the sum of all electric forces was zero. In question 25, the sum of the repulsive forces from each of the two like charges is zero at some point between the two charges.
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27. If a proton and an electron were released at a
distance and accelerated toward one another, the electron would reach the greater speed just before impact. The reason is that both particles would be acted upon by the same force of attraction, but the electron has less mass. The acceleration of each particle is F m
described by the formula a , which
shows that for the same force, the smaller mass would have the greater acceleration over the same time period and therefore the greater final speed. 28. q
This type of motion is like upside-down pro jectile motion, since the charge moves in a parabolic path. This is the type of motion that an object would take if it were thrown horizontally in Earth’s gravitational field. The only difference here is that this charge appears to be “falling upward” instead of downward. 29. No, a parallel-plate capacitor does not have uniform electric potential. It does have uniform field strength between the two plates, but the potential varies in a linear fashion from one plate to the other. By definition, the electric potential is uniform along any equipotential line, which in this case is any line parallel to the two plates. 30.
Charge Distribution (a) (b) (c)
Equipotential Lines (iii) (i) (ii)
31. a) The electrostatic interaction responsible
for the large potential energy increase at very close distances is the repulsion between the two positive nuclei. b) This repulsion of the nuclei, and the associated increase in electric potential energy, is one of the main stumbling blocks for generating energy through nuclear fusion. This repulsion between nuclei means that
A n s w e rs t o E n d - of - c h a p te r C o n c e p t u a l Q u es t i o n s
a very large amount of energy is required to begin the reaction process. c) The smaller increase in electric potential energy upon separation of the two atoms is caused by the attraction between the positively charged nucleus in each atom and the negatively charged electron in the other atom. d) A stable bond is formed when two hydrogen atoms are about 75 pm apart because this is the distance at which the electric potential energy is minimized — any closer and the repulsion between nuclei pushes the atoms apart, any farther away and the nucleus-electron attraction draws the atoms closer together. 32. A positive test charge moving along a line between two identical negative point charges would experience a topography similar to a vehicle moving up a hill (away from one charge), increasing the vehicle’s gravitational potential energy, and then rolling down the other side of the hill (toward the other charge). a) If the two identical point charges were both positive, the hill would change to a valley with the lowest part in the middle. b) If a negative test charge was placed between the two identical positive charges, the topography would still resemble a val ley but now there would be a very deep crater at the lowest part of the valley.
3. A material that is attracted to a magnet or that
can be magnetized is called ferromagnetic. Examples of ferromagnetic materials include materials made from iron, nickel, or cobalt. These materials are ferromagnetic because they have internal domains that can be readily aligned, due to the fact that these materials have unpaired electrons in their outermost electron energy level. 4. Magnets can lose their strength over time because their domains, which initially are aligned (pointing in the same direction), can become randomized and point in other directions. This randomizing of the domains reduces the overall strength of the entire magnet. 5. When a magnet is dropped or heated up, the domains of the magnet, which initially are aligned (pointing in the same direction), can be disrupted and forced to point in other, random directions. This randomizing of the domains reduces the overall strength of the entire magnet. 6. a) F
F T
Currents in the same direction wires forced together
b) F
F T
Chapter 9 1. The law of magnetic forces states that like
(similar) magnetic poles repel one another and different (dissimilar) poles attract one another, even at a distance. 2. A magnet can attract non-magnetic materials as long as they are ferromagnetic in nature. The magnet causes the internal domains (small magnets) of a ferromagnetic substance to line up in such a way that a new magnet is induced in the substance such that there are opposite magnetic poles which attract one another.
x
Currents in opposite directions — wires forced apart
7. The electrons in the beam that is illuminating
your computer monitor’s screen are directed from the back of the monitor forward to the front of the screen, toward your face. Therefore, conventional (positive) current points in the opposite direction, away from your face and back into the computer monitor. This is the direction of the thumb of the right
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hand when applying right-hand rule #1 for current flow. From your perspective, the magnetic field forms circular formations in the clockwise direction in and around the computer monitor. Relative to the direction of the electron beam, the magnetic field is directed in the counterclockwise direction around the beam. 8. A wire possessing an eastbound conventional (positive) current has an associated circular magnetic field that points upward on the north side of the wire and downward on the south side. 9. The magnetic field strength of a coil (an insu lated spring) varies inversely with the length of the coil. Therefore, a reduction in the coil length to half its original length will cause a doubling of the magnetic field strength. This all depends on the assumption that the length of the coil is considerably larger than its diameter. 10. a) For the force applied to a current-carrying conductor to be at a maximum, the magnetic field must cross the conductor at an angle of 90°. b) For the force applied to a current-carrying conductor to be at a minimum, the magnetic field must cross the conductor at an angle of 0°. 11. According to right-hand rule #3 for the motor principle, the direction of the force on the conductor will be to the north. 12. An electron moving vertically downward that enters a northbound magnetic field will be forced toward the west. 13. A current-carrying solenoid produces a magnetic field coming directly out of one end of the coil and into the other end. An electron passing by either end of this coil experiences a force that is at right angles to its motion. As this force changes the direction of motion (a centripetal force), the electron takes on a curved path (circular motion). Application of the appropriate right-hand rules predicts that the electron’s motion will curve in the same
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direction as the direction of conventional current flow through the coil. 14. The cathode rays will be deflected away from the current-carrying wire, moving in a plane that contains the wire. 15. Current passing through a helical spring will produce a situation very similar to having two parallel conductors with a current flowing in the same direction. Application of the appropriate right-hand rules predicts that the magnetic field interaction between each pair of the helical loops will force the spring to compress, reducing its length. 16. Current passing through a highly flexible wire loop will tend to result in magnetic field interactions that will force apart nearby sections of the wire, so that the wire loop will most likely (if the proper conditions exist) straighten out. 17. Faraday’s principle states that a magnetic field that is moving or changing in intensity in the region around a conductor causes or induces electrons to flow in the conductor. To improve the electromotive force induced in a conductor, we can increase the magnetic field strength, the length of the conductor, and the strength of the current flowing through the conductor. 18. Current can be induced to flow in a conductor if the conductor is moving with respect to a magnetic field. The maximum induced current occurs if the conductor and the field cross each other at right angles. 19. Lenz’s law states that the direction of the induced current creates an induced magnetic field that opposes the motion of the inducing magnetic field. Lenz derived this law by reasoning that a decrease in kinetic energy in the inducing magnetic field must compensate for the increase in the electric potential energy of the charges in the induced current, according to the law of conservation of energy. This decrease in kinetic energy is felt as an opposition to the inducing magnetic field by an induced magnetic field.
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20. The induced current can only create a mag-
netic field that opposes the action of movement (conductor or field) in order to follow the law of conservation of energy and Lenz’s law. If the motion were not opposed and the induced magnetic field instead “boosted” the motion, this would increase the kinetic energy of the moving conductor or magnet, which would violate the law of conservation of energy. 21. a) Electromagnetic brakes might work by using the undesirable motion of the vehicle to provide the energy to induce current flow in a conductor. The resulting creation of electrical energy would be at the expense of the kinetic energy of the vehicle, which would slow down. This would be a case of energy being transformed from one form to another, following the law of conservation of energy. b) Electromagnetic induction brakes would be capable of recovering some of the kinetic energy of a vehicle that is normally lost as heat in conventional brakes, thereby saving money. The electrical energy generated could be used to recharge the battery for an electric vehicle/hybrid.
4.
θi
Normal
5. When metallic objects are placed in a
6.
7.
Chapter 10 1. The motion of a vibrating spring can be mod-
elled mathematically by a sine wave, which resembles (visually) an electromagnetic wave. As well, both waves are periodic. 2. The magnetic field is induced by the electric field and thus they would both decrease. If one component vanishes then the electromagnetic radiation ceases to exist. 3. “Visible light” is relative to the human being perceiving it. Also, some other animals see in other regions such as the infrared and ultraviolet.
θr
8.
9.
microwave oven, they can absorb electromagnetic microwaves, which dislocate loose electrons in the metal and allow charges to build up on the surfaces, until the cumulative charge is large enough to “jump” across an air gap to another conductive material in the oven, causing a spark. Simple harmonic motion refers to a physical “state” where the restoring force, acting on an object when it is pulled away from some equi librium position, is proportional to the displacement of the object from the equilibrium position. Since there is a net force acting on the object, it experiences an acceleration, and thus the speed cannot be constant. If a circle is viewed edge-on, with a dot painted on the edge, and the circle is spun, the dot will seem to exhibit simple harmonic motion as it moves around the circle. From the edge it will seem as if the dot is moving back and forth, constantly passing the equilibrium position. Electron oscillators absorb energy from the incoming wave, causing it to be retarded. When this secondary wave interferes with the incident wave, a phase lag is created retarding the wavefront, slowing it down. Newton’s theory of refraction predicts that light speeds up as it changes direction. This is incorrect since light decreases its speed when bending toward the normal. You can show his theory by rolling a marble across a boundary between a flat area and an incline. As the marble crosses the boundary, it bends toward a line drawn perpendicular to the edge but it speeds up.
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10. One example of an invisible medium is a vac-
17. Yes, the effectiveness of Polaroid sunglasses
uum. The refractive index of a vacuum is 1.00. When the refractive index is 1.00, there is no component of an incoming light ray that is reflected. Since no light is reflected, the medium is invisible. Another possibility is that the medium is of the same refractive index as the environment. 11. Using a laser, which is a powerful coherent source of visible light, you can measure the refraction of the ray as it enters a medium, or the extent of polarization upon reflection and/or transmission, all of which can be com bined to calculate the optical density of the medium. 12. Because the refractive index is wavelength dependent, when white light refracts through a material, each component of light bends slightly differently. This separates the light. If the separation is great enough, dispersion occurs. 13. As light passes through a prism, both refractions cause the light to refract in the same spatial direction. This accentuates the spreading of the colours. 14. No, sound waves cannot be polarized. Sound waves are mechanical waves and refer to compressions and rarefactions within a medium. Sound waves have only one component, not two like electromagnetic waves, and thus polarization is impossible. 15. A polarizer and an analyzer are both thin pieces of film. They are given different names based on the order in which a wave enters them. If two pieces of thin film are positioned side by side, the first one struck by the wave is known as the polarizer and the second one the analyzer. If the two are flipped, the ana lyzer will become the polarizer and the polarizer will become the analyzer. 16. The lenses in polarized sunglasses are normally oriented in such a way as to restrict the passage of plane-polarized light reflecting off the surface of the ground and water (glare). If the lenses are rotated, they will no longer block the glare.
varies as the relative positions of the Sun and the horizon vary, since the distribution of scattered angles varies as well. The amount of polarization is angle dependent, hence the effectiveness of the glasses varies. 18. No, Polaroid sunglasses are not effective on circularly polarized light, which is composed of the two polarization directions combined in a specific phase relationship causing the direction of the electric field vector to rotate around. The linear polarizer cannot block out both components, hence light is transmitted. 19. With a powerful light source, you can easily notice that light reflects off dust particles in the air. Sometimes, depending on the size of the particles, certain frequencies of the electromagnetic spectrum are deflected/reflected more than others. This effect causes a certain colour to appear in the medium (for example, the blue colour of the sky). By noting the colour, you can determine the frequency and thus the wavelength of light associated with that colour. Once you know the wavelength, you can calculate the approximate size of the particles that would deflect waves of those frequencies. Also, you can use the intensity of the colour to estimate the density of the particles in the air. 20. Polarization: Electric fields of electromagnetic radiation behave sinusoidally. The direction of these fields is randomly oriented in any direction for unpolarized light. Two components are obtained by using plane polarizers. The two components can be combined using the wave equation ( sin t ) to form circularly or elliptically polarized light. Scattering: The wavelength of light, , comparable to the size of particles in the air creates the maximum scattering. The extent of scattering of light by air molecules is proportional to 4. Refraction: Using wavefronts, Snell’s law of refraction is derived. Based on phase relationships between the incident wave and the
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transmitted wave, light is bent and slowed down in different mediums.
Chapter 11 1. Refraction, polarization, interference, and
diffraction 2. Refraction, diffraction, and interference can be demonstrated using water waves in ripple tanks. Polarization cannot. 3. The film on a soap bubble is thicker at the bottom than at the top, forming a wedge shape, since gravity pulls the soap down. As the film’s thickness changes, the interference changes (destroys some wavelengths) and the colours change. 4. As the gasoline evaporates, it becomes thinner, changing the interference pattern and the colours. 5. A camera lens has a thickness and material designed to block out certain colours, whereas a car windshield does not. These properties of a lens produce interference patterns and a colour change. Camera lenses are designed to correct chromatic aberration caused by different wavelengths bending at different angles while being refracted. 6. a) Newton believed that light was a particle. b) Changing people’s environments through innovation can leave people feeling not in control, especially in cases where a new technology has the possibility of replacing people in jobs. c) Accepting theories prematurely hinders progress, since it discourages research. 8. No, there are no interference patterns because the two car headlights are not coherent light sources and do not form a double slit. 9. Any imperfections are in the order of magnitude of the wavelengths of light used for the experiment. This washes out the effect with its own random interference patterns. 10. Sound waves are comparable in wavelength size to the openings, increasing the diffractive effect. Light waves have much smaller wave lengths and hence do not show these effects.
11. The resolving power of your eyes restricts
your ability to distinguish between objects at great distances. This is because your pupils are circular, allowing diffraction to occur. 12. No, diffraction patterns place a limit on resolving power as well as the magnitude of the wavelength of light used. 13. Both spectroscopes separate white light into its colour components, but the prism spectroscope uses refraction and dispersion while the grating spectroscope uses diffraction. 14. Continuous spectra involve an extensive range of frequencies (example: sunlight spectrum). With line spectra, on the other hand, discrete frequencies are observed (example: molecular gas spectrum). 15. Each piece of a hologram contains the complete interference pattern of the object from which the hologram was created, whereas a piece of a normal photograph contains only local information and nothing about the complete photograph. 16. Diffraction gratings and interference gratings are really the same thing. Diffraction gratings actually use the interference superposition formula. Gratings show both effects—those due to the width of a single opening and the combination of all the openings. 17. Close spacing in a grating provides strong mutual coupling, increasing the effect of interference. The separation of the maxima increases. 18. Gratings with many slits have high resolving power. This means that the individual maxima become sharper. 19. Yes, because increasing the number of slits decreases the slit separation. If the slit separation is reduced beyond what is comparable to the wavelength of light, no light will get through. 20. A single slit has a double central maximum, with the intensities of the maxima dropping off dramatically with order number. A diffraction grating has a single central maximum and the intensities do not drop off as dramatically.
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21. Diffraction occurs as light enters the pupil.
8. Since the photons have detectable linear
This places a limit on the eye’s resolving power. As you move away from the picture, sooner or later you cannot distinguish between the dots and they blend together to form a continuous picture. 22. Electrons have a smaller wavelength to that of visible light, and therefore have a higher resolution. This also minimizes diffraction. In fact, the beams of electrons have an effective wavelength that is 105 times that of visible light. This is a 100 000-fold increase in resolution.
momentum, their mass equivalence can be computed. Momentum is an intrinsic property of matter, therefore we can assume that mass equivalency is correct. 9. An empirical relationship is a relationship that is determined experimentally. It is not backed up by theory. 10. Determinacy is a condition of a measurement being characterized definitely. An example of an everyday event could be a repetitive measurement of the length of a table. Each time the measurement is made, errors are encountered. If determinacy existed at the macroscopic level, we would get the same length every time. 11. The computation of uncertainties using Heisenberg’s uncertainty principle yields minute values for speed and position. The limitations of human perception prevent us from experiencing such minute variances at the macroscopic level. 12. Another device besides the STM that operates using the principle of quantum tunnelling is the electron tunnelling transistor, which is an on-off switch that uses the ability of an electron to pass through impenetrable energy obstacles. 13. The energy of an orbital varies as the inverse square of the radius. Hence, the spectral lines are closer together farther away from the nucleus. 14. a) The peak wavelength emitted by a mercury lamp lies in the visual spectrum. However, this implies that there is a tail in the ultraviolet spectrum. The ultraviolet photons are energetic enough to damage skin cells. b) An appropriate shielding that blocks ultraviolet light but allows photons in the visual spectrum to pass through could be used. 15. Consider two particles that have the same de Broglie wavelength and masses m1 and m2 such that m1 m2. According to de Broglie’s
Chapter 12 1. A photon is a unit particle (as opposed to
2.
3.
4.
5.
6. 7.
wave) of electromagnetic radiation that moves at the speed of light. Its energy is proportional to the frequency of the radiation. Ultraviolet radiation from the Sun is very energetic due to its high frequency. The photons that possess this energy are the cause of sunburn. These photons are energetic enough to remove electrons from our body cells, causing a change in our skin biology and in severe cases causing cancer. Visual light is mostly in the infrared-visual spectrum. The energy of these photons is not sufficient to damage skin cells. If h 0, quantization would not exist. There would be no energy levels in atoms. Electrons in atoms would therefore not attain any real value for energy, resulting in the absence of orbitals in atoms. The electron volt (eV) corresponds to the energy of an electron at a potential of one volt. Hence, one electron volt is the energy equalling the charge of an electron multiplied by the potential of one volt: 1 eV qe 1 V. Wien’s law relates the wavelength of photons to the temperature of the black body. W 0, the work function, is the amount of energy required to produce the photoelectric effect in a given metal. It is the minimum energy required to liberate electrons from a metal.
equation,
h h and , where v mv mv
1
1 1
2 2
and v2 are the velocities of the two particles. 80
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To swim straight across the stream (perpendicular to the current) and back, it would take a total time of: d 2dv2 w 2 . 2 v2 w2 v2 w 2 But v v2 w 2, so 2dv 2dv2 w 2 . 2 2 2 2
Since is the same for both particles, the fol lowing equation can be written: h h mv mv 1 1
2 2
This equation can be simplified:
5.
6.
7.
8.
Chapter 13 1. Your car is in an inertial frame when it is
stopped, or when it is moving at a steady speed in a straight line. Your car is in a non-inertial frame when it is accelerating, such as when you are braking, or when you are making a turn. 2. Donovan’s reference frame is inertial because the 100-m dash is in a straight line. Leah’s frame is non-inertial because the 400-m oval requires her to constantly change direction. 3. No, without reference to the outside world, it would be difficult to determine whether the cruise ship was at rest or moving with a constant velocity. 4. Suppose v swim speed and w water speed. To swim upstream and back down, it would take a total time of: d d 2dv 2 2.
v w v w
m1v1 m2v2 Since m1 m2, it follows that v2 v1. If the
mass of the first particle is much greater than that of the second particle, the velocity of the second particle must be much greater than that of the first particle. 16. According to Planck, the energy is quantized. The angular momentum is certainly related to the energy. Hence, the angular momentum needs to be quantized as well. To quantize L, Bohr had to quantize both the velocity, v, and the radius, r . 17. Although the initial and the final speed and the scatter angles are known, the manner in which the actual collision occurs cannot be precisely predicted, and the exact position of the particles during the collision is not known. Hence, the uncertainty principle is not violated.
9.
Therefore it would take longer to swim upstream and back down than to swim across the stream and back. The Michelson-Morley null result led to the development of special relativity, a tool needed in the understanding of high-energy physics. Analogous to the Doppler shift of sound, the constant speed of light in a vacuum, c , requires the wavelength of the approaching amber light to shorten or become more yellowish. In terms of Einstein’s first postulate involving relative motion, the two situations are equiva lent. The same physics occurs whether a magnet is moved into a stationary coil or a coil is moved around a stationary magnet. Proper time is the time measured by one watch between the beginning and the end of the experiment. This is the time measured by a watch moving with the muon. The scientists of Earth would require at least two watches, one at the birth of the muon and the other at its disintegration. The relativity equation for length is
2 v2 v . If v > c , then 1 2 is L L0 1 c c 2 negative and L becomes imaginary, which is
not physically reasonable. 10. Since the electrons would have a greater relative velocity than the protons, the space between the electrons would be more contracted. As a result, the concentration of electrons would exceed that of the protons, and the wire would seem negatively charged. For this reason magnetism is a result of special relativity.
vw vw v w
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11. No, you have not travelled faster than the
18. The starlight will pass you at a speed of c
speed of light. Instead, you have measured the Earthstar distance to be contracted and thus it took less time to travel at your speed of v 0.98c . 12. No, although time is dilated (events seem longer at relativistic speeds), it cannot slow to a complete standstill unless v c . 13. In both cases of the Doppler shift for sound, there is a shift to higher frequencies. However the physics of sound waves generated by a moving vibrating source colliding with air molecules and perceived by a stationary receiver is different from that of sound created by a stationary source and perceived by a moving source. For light, the frequency shift depends only on the relative speed of the source and receiver because the speed of light is always c , according to the second postulate. 14. Only Barb is correct in saying that Phillip’s clock ran slow, because his time was the proper time that was at the beginning and finish of his experiment. If Phillip observed stationary Barb doing a similar experiment beside the train, he would be correct in saying that her clock ran slow. 15. If the charge of an electron depended on its speed then the neutrality of atoms would be upset by the motion of electrons within the atoms. Experiments have shown that the charge on an electron is the same at all speeds. 16. The radius of the orbit becomes smaller as the magnetic field is increased because the radius
according to the second postulate of special relativity. 19. The occupants of the spacecraft would say that they observed the same things about us, due to relative motion. 20. No, according to the second postulate of special relativity, the light leaving the receding mirror travels with speed c relative to you. 21. Tachyons or particles that travel with a speed greater than c would seem to require infinite energy. Experiments do not support their existence. 22. Particle A would have the greater speed because its total energy due to mass dilation (mc 2) is three times its rest energy, whereas particle B has a total energy dilated by a factor of only two. 23. Since the ice and the water have the same mass, they have the same total energy (m0c 2 Ek). However, the kinetic energy, Ek, of the water is higher than that of the ice and for that reason the rest energy of the water is less than that of the ice. 24. If you consider that energy is equivalent to mass ( E mc 2), then electromagnetic energy in the form of light could be considered to have an equivalent mass. 25. A 100-eV electron has a dilated mass according to:
mv qB
is equal to , where B is the magnetic field strength. 17. No, because mass dilates as
v2 , but 1 2 c
mass density , and volume contracts as vo lume
1 .
v2 . Therefore, density dilates as 1 2 c
82
v2 c 2
mc 2 m0c 2 qV mc 2 (0.511 0.000 100) MeV mc 2 0.5111 MeV
This means that its mass is less than 0.02% greater than its rest mass. A 100-MeV electron has a mass equivalent to (0.511 100) MeV of energy, which means that its mass is about 197 times its rest mass. 26. When we say that the rest mass of a muon is 106 MeV/c 2, we mean that its rest energy is equivalent to the kinetic energy of an electron accelerated from rest through 106 million volts. 27. When a particle is travelling at an extremely high speed, say 90% of the speed of light, a lot of energy is needed to increase the particle’s velocity by a few percent. As a result, the
A n s w e rs t o E n d - of - c h a p te r C o n c e p t u a l Q u es t i o n s
mass of the particle increases by a large amount. Therefore, it may be more accurate to say that a particle accelerator increases the mass of electrons rather than their speed.
Chapter 14
3.
4.
5.
6.
7.
146 is or about 1.59, and the ratio of the 92 N 2 144 daughter nucleus, , is or about Z 2 90 1.60. This leads to greater nuclear stability by reducing the electrical repulsion of the protons relative to the nuclear attraction of
1. Every atom of the same element has the same
2.
8. During alpha decay of a uranium-238 nucleus, N for example, the ratio of the parent nucleus Z
number of protons, and the number of protons in the nucleus, Z , determines the chemical properties of the atom. However, atoms of different isotopes of the same element have different numbers of neutrons (and thus different A values), which results in different physical properties such as nuclear stability or decay. Many elements are composed of several naturally occurring isotopes, each with a different atomic mass number, A. The weighted average of the isotopes’ mass numbers often results in a non-integral value for the atomic mass of that element. Each nuclear isotope has a unique total binding energy determined by its nuclear structure. This binding energy is equivalent to the mass difference between the nucleus and its constituent nucleons (protons and neutrons) according to E mc 2. The missing mass was converted to energy of various forms such as gamma radiation emitted during the formation of the deuterium atom. Your body, composed of many elements, likely has more neutrons than protons, since stable atoms with A > 20 have more neutrons than protons. During a nuclear reaction, nucleons may be converted from one type to another, such as neutrons to protons in beta decay. However, the total nucleon number is conserved or remains constant. On the other hand, various forms of energy may be absorbed or emitted, resulting in an equivalent change in mass. The average binding energy per nucleon is greater in the more stable isotopes because it is the “glue” holding the nucleons together, or the average amount of energy needed to break them apart.
N
nucleons. During beta decay, the ratio of Z 146 the parent nucleus, or about 1.59, is 92 greater than the ratio of the daughter nucleus, N 1 145 , which is or about 1.56. Z 1 93 Although the greater ratio of protons to neutrons in the daughter tends to increase the electrical repulsive forces, the beta-decay process can lead to greater nuclear stability through the pairing of previously unpaired neutrons or protons in the nuclear shells. 9. During alpha decay, the daughter nucleus has a mass, M , that is much larger than the mass of the alpha particle, m. Since momentum is conserved, the velocity of the daughter nucleus, v, is much smaller than the velocity of the alpha particle, V ( Mv mV ). Therefore, the kinetic energy of the alpha particle, 0.5mV 2, is much greater than that of the daughter nucleus, 0.5 Mv2. 10. If an alpha particle had enough initial kinetic energy to contact a gold nucleus then a nuclear process such as fusion or fission could occur, because at that closeness the shortrange nuclear force would overpower the electrical force of proton repulsion that is responsible 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 force is very short-range, acting over distances of only a few femtometres (1015 m); (ii) the strong nuclear force is much stronger than the electrical force over nuclear distances of 1 or 2 fm; (iii) the strong nuclear force does
A n s w e rs t o E n d - of - c h a p te r C o n c e p t u a l Q u es t i o n s
83
1 not vary with distance r as 2 as does the r
electrical force; and (iv) the strong nuclear force is attractive only, acting between all nucleons (protonproton, protonneutron, and neutronneutron). 13. The rate of decay of radioactive isotopes was not affected by combining them in different molecules or by changing the temperature. These changes usually affect the rate of chemical reactions, thus radioactivity must be found deeper within the atom (in the nucleus). 14. The nuclear force only binds nucleons that are neighbours. This short-range energy is proportional to the number of nucleons, A, in the nucleus. On the other hand, the electrical repulsion of protons is long-range and acts between all proton pairs in the nucleus. The electrical energy is therefore proportional to Z 2. Repulsion would overcome attraction in a larger nucleus if there were not more neutrons than protons to keep the forces balanced and the nucleus stable. 15. Alpha particles are ions, since they are helium atoms stripped of their electrons. 16. If human life expectancy were a random process like radioactive decay then you would expect 25% of the population to live to 152 years. However, this is not the case. As humans age, their expected number of years left to live decreases. 17. Carbon-14 undergoing beta decay results in the daughter isotope nitrogen-14. 18. Industrialization and automobile emissions have effected changes in our atmosphere such as global warming and ozone-layer depletion. Such changes in the past 100 years may be altering the 14C:12C ratio in the air. 19. Potassium salts are rapidly absorbed by brain tumours, making them detectable. The short half-life of potassium-42 means that the dosage decays to a safe, insignificant level quickly. The transmutation to a stable calcium salt by beta decay is not harmful to the body.
84
20. Aquatic creatures do not respire or breathe
atmospheric gases directly. The 14C:12C ratio in the ocean is different than in the air. 21. Relics that are more than 60 000 years old have lasted more than 10.5 half-lives of car bon-14. The 14C:12C ratio in these relics is about 1500 times smaller now and is difficult to determine. 22. The more massive lead atoms scatter the radiation particles more effectively than do the less massive water molecules, and may also present a larger “target” for a high-speed electron or alpha particle. 23. Transmutation involves a change in the proton number, Z . This occurs during alpha and beta decay but does not occur during gamma decay, in which a nucleus merely becomes less energized. 24. Alpha particles are more massive than beta or gamma particles and transfer more energy to a molecule of the body during a collision. This has a much more disastrous effect upon the cells of the body. 25. Yes, 4.2 MeV of kinetic energy is sufficient for an alpha particle to overcome the electrical repulsion of the positively charged nuclei (see problem 72) and contact the nitrogen-14 nucleus, thus a nuclear interaction or process is possible. 26. The matches are as follows: gaseswind; liquidswater; plasmasfire; and solidsearth. 27. For fission to occur in naturally occurring deposits of uranium, a minimum concentration of uranium would be needed in order to sustain a source of slow neutrons necessary to maintain the fission process. This concentration is not present in uranium deposits. 28. The huge inward pull of the Sun’s gravitational field confines the solar plasma. Lacking this huge confining force on our less massive Earth, scientists instead use strong electromagnetic fields to confine plasmas. 29. In a fusion reactor, the major problem is to create the exact and difficult conditions of high temperatures and plasma concentrations
A n s w e rs t o E n d - of - c h a p te r C o n c e p t u a l Q u es t i o n s
needed to initiate a fusion process. The moment these conditions are not met, the process stops. 30. The high temperature in fusion means that the ions have a very high speed, which allows them to approach one another very closely during collisions. If the ions’ kinetic energy is sufficient to overcome the electrical repulsion of the nuclei, and the nuclei touch, then fusion is possible. 31. Critical mass in fission involves the existence of enough fuel so that the fast neutrons emitted during fission are slowed and absorbed within the fuel itself before they escape. In this way the reaction is sustained by a continual source of slow neutrons. 32. Natural uranium is not concentrated enough (“it’s too wet”) to provide the critical mass needed to slow down any fast neutrons (“the spark needed”) and capture them to create a sustainable reaction. 33. A bubble chamber is superheated almost to the point of instability. When a charged particle passes through, it triggers the formation of a fine stream of bubbles in its wake. Neutral particles such as neutrons carry no electric field and leave no visible tracks in the chamber. 34. High-energy accelerators provide ions with enough kinetic energy that each ion’s total energy, E mc 2, becomes many times greater than its rest mass. In a collision there is a probability that this energy could be converted to a massive elementary particle. 35. In the high-energy accelerator at UBC, the strong nuclear force, acting over a very brief period of time (1023 s) during collisions, produces pi-mesons or pions. 36. The weak nuclear force is usually masked by the stronger (by a factor of 103) electromagnetic force or (by a factor of 105) strong nuclear force, unless these forces are forbidden. Any process involving the neutrino, such as beta decay, involves the weak force. The neutrino reacts rarely, or weakly, with other elementary particles over a longer time span (108 s) compared with the shorter interaction
times (1023 s) of the strong nuclear force. 37. Gravitational interactions are the weakest of the four forces. At elementary particle distances the gravitational force is 1040 times as great as the strong nuclear force. For this reason the graviton or messenger of the gravitational force is extremely difficult to detect. 38. The weak force is 103 times as great as the electromagnetic force at elementary particle distances. The weak force is mainly involved in neutrino interactions or processes where the electric and strong forces are forbidden. The exchange bosons of the weak force are W and Z bosons of mass 80 GeV/c 2 and 91 GeV/c 2 respectively, as compared to the photons of the electric force. The range of the weak force is about 1018 m, compared to infinity for electromagnetism. The weak force acts on both leptons (particles not affected by the strong force, such as electrons) and hadrons (particles affected by the strong force), whereas electromagnetism acts only on charged particles. 39. Strong nuclear processes are the fastest (or shortest), with a lifetime of about 1023 s. 40. A high-energy particle travels close to the speed of light, c 3 108 m/s. Thus, in a strong nuclear interaction of 1023 s, the cloud-chamber track would be 3 1015 m, too small to measure. 41. No, a heavier, unstable version of the electron, the tauon or tau, , has a mass of 1777 MeV/c 2, greater than the mass of the proton or neutron (931.5 MeV/c 2). 42. Since gluons, the quanta of the quark force field, 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 differently because of what are known as symmetry laws. For this reason only two possible neutral couplings exist, not three, making a total of eight colour gluons to act as the source of quarkquark interactions.
A n s w e rs t o E n d - of - c h a p te r C o n c e p t u a l Q u es t i o n s
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86
A n s w e rs t o E n d - of - c h a p te r C o n c e p t u a l Q u es t i o n s
PART 3 Solutions to End-of-chapter Problems Chapter 1 16. 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, your displacement is 0 m. 17. a) Since distance is a scalar, his total distance travelled 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] 9.8 m 100 cm 0.394 in 0.083 ft 18. g 2 1s 1m 1 cm 1 in 2 g 32 ft/s 10 nm 6080 ft 12 in 19. a) 10 knots 1h 1 nm 1 ft 2.54 cm 1.0 105 km 1 in 1 cm 10 knots 18.5 km/h
b) from a), 18.5 km 1000 m 10 knots 1h 1 km 2.78 104 h 1s 10 knots 5.14 m/s 20. To find the number of centimetres in one light year, simply express the speed of light in centimetres per year: 3.0 108 m 100 cm 60 s 1s 1m 1 min 60 min 24 h 365 d 1h 1d 1y 9.5 1017 cm/y Therefore, there are 9.5 1017 cm in one light year.
21. Catwoman: d vavg t 100 m vavg 15.4 s vavg 6.5 m/s Robin: d vavg t 200 m vavg 28.0 s vavg 7.1 m/s 22. a) The speed of the sweep second hand at the 6 o’clock position is the same as anywhere else on the clock: d vavg t 2r vavg t 2(0.02 m) vavg 60 s vavg 2.1 103 m/s b) The velocity of the second hand at the 6 o’clock position is 2.1 103 m/s [left] because the velocity is always tangent to the face and perpendicular to the hand. 23. a) The time it would take the shopper to walk up the moving escalator is: d t vt where vt is the sum of the velocity of the escalator and the woman: d d vt 15 s 8.0 s 23d vt 120 s d Therefore, t 23d 120 s t 5.2 s
Solutions to End-of-chapter Problems
87
b) In this case, vt is equal to her walking speed minus the speed of the escalator: d d vt 8.0 s 15 s 7d vt 120 s Since this velocity is positive, she could walk down the escalator. Also, intuitively, if the escalator takes 15 s to go the same distance that the woman can in 8 s, then she is faster and will therefore make it down the escalator. To find how long it will take her, solve for time: d t 7d 120 s t 17 s 24. Because the rabbit accelerates at a constant rate, v2 v1 at v2 (0.5 m/s) (1.5 m/s2)(3.0 s) v2 5.0 m/s 25. Mach 1 332 m/s Mach 2 2(332 m/s) Mach 2 664 m/s Because the jet accelerates at a constant rate: v2 v1 at (v v1) t 2 a (664 m/s 332 m/s) t 50 m/s2 t 6.6 s ( v v1) 26. a 2 t a (25 m/s [E] 15 m/s [W]) 0.10 s a (25 m/s [E] 15 m/s [E]) 0.10 s 2 a 400 m/s [E] 27. Let t be the time when the two friends meet. Let x be the distance travelled by the second friend to reach the first friend. For the first friend: d v t d vt
88
and d 50 x
Therefore: x 50 vt For the second friend: 2
x v1t x
at 2
at 2
2
Now we set these two expressions for x equal to each other and solve for time: 2
50 vt
at 2
t 2 t 100 0 1 1 4( 100 ) t
2
t 9.5 s 28. a) v2 v1 at and v1 equals zero, so v2 at v t 2 a t
60 km/h 10 km/h/s
t 6.0 s
b) To find Batman’s distance travelled, we must first convert his acceleration into standard SI units: 10 km 1h 1000 m 1 hs 3600 s 1 km 2 2.78 m/s Now: at 2 d 2 (2.78 m/s)(6.0 s) 2 d 2 d 50 m c) Robin’s speed in SI units is: 60 km 1h 1000 m 1h 3600 s 1 km 16.7 m/s When Batman catches up with Robin, Robin will have travelled: d v1t d (16.7 m/s)t relative to Batman’s initial position.
Solutions to End-of-chapter Problems
Similarly, Batman will have travelled: 2
d
at 2
d (1.39 m/s2)t 2
Setting these two expressions equal and solving for t gives: ( 16.7 m/s)t (1.39 m/s2)t 2 0 t 12.0 s 29. If the child catches the truck, she will have travelled 20 m d, and the truck will have travelled d in the same amount of time, t . For the truck: 2
d
at 2
d (0.5 m/s2)t 2
For the child: (d 20 m) vavg
above the window from which the pot must have been dropped. Since the pot accelerates at a constant rate of 9.8 m/s2, we can write: 2
d (4.0 m/s)t 20 m
Setting these two equations equal and solving for t gives: (4.0 m/s)t 20 m (0.5 m/s2)t 2 t 2 8t 40 0 This expression has no real roots, therefore the 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:
d
v1
19 m (9.8 m/s )(0.20 s) 0.20 s 2
t
at 2 2
v1 8.5 m/s
Now we can find the distance above the window: v12 vo2 2ad (v12 vo2) d 2a ((8.5 m/s)2 (0 m/s)2) d 2(9.8 m/s2) d 3.7 m 32. a) The only force acting on the ball while it is falling is that of gravity, so its acceleration is 9.8 m/s2 downward. b) Since the ball is being constantly accelerated downward, it cannot slow down. c) d v1t
at 2
2
d (8.0 m/s)(0.25 s) 2
2
(9.8 m/s )(0.25 s) 2
d vavg t
d 2.3 m
3200 m 600 s
33.
vavg 5.33 m/s
If she then accelerates at 0.40 m/s 2 for the final 800 m, it will take: 2
d v1t
v1
t
vavg
at 2
d v1t
at 2
800 m (5.33 m/s)t (0.20 m/s2)t 2 5.33 (5.33)2 4(0.20)(800) t 2(0.20) t 51 s b) 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 window, and then use the speed to find the height
d v1t
at 2
2
(4.0 m) (4.0 m/s)t (9.8 m/s2)t 2 2 2 2 (4.9 m/s )t (4.0 m/s)t 4.0 m 0 4 16 4(4.9)( 4) t 9.8 t 1.4 s 34. For the first stone, the distance it falls before reaching the second stone is:
2
h d
at 2
d (4.9 m/s2)t 2 h
Solutions to End-of-chapter Problems
89
For the second stone, its distance travelled is given by: 2
d vit
at 2
d vit 4.9t 2
Setting these expressions equal to each other and solving for t gives: 4.9t 2 h vit 4.9t 2 h t vi
35. a) Because the jackrabbit’s distance vs. time is changing at a constant rate during segments B, C, and D, he is undergoing uniform motion at these times. b) Because the jackrabbit’s distance vs. time is not changing at a constant rate during segment A but is increasing exponentially, his velocity vs. time must be increasing at a constant rate, and he is undergoing uniform acceleration during this segment. c) The average velocity during segment B is: d vavg t 150 m 100 m vavg 20 s 10 s vavg 5 m/s During segment C: d vavg t 150 m 150 m vavg 30 s 20 s vavg 0 m/s During segment D: d vavg t 50 m 150 m vavg 50 s 30 s vavg 10 m/s d d) vavg t 150 m 0 m vavg 17.5 s 1.0 s vavg 9.1 m/s
e) Since the jackrabbit’s displacement is not changing between 20 s and 30 s, his velocity over this interval, and at 25 s, is 0 m/s. f) The jackrabbit is running in the opposite direction. 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 segment can be found by taking the slope of the graph during that segment: During segment A: v a1 t (5 m/s 0 m/s) a1 (5 s 0 s) a1 1 m/s2 During segment B: v a2 t (13 m/s 5 m/s) a2 (9 s 5 s) a2 2 m/s2 During segment C: v a3 t (1 m/s 13 m/s) a3 (15 s 9 s) 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 of rectangles and triangles: vt A: d1 2 (5 m/s 0 m/s)(5 s 0 s) d1 2 d1 12.5 m vt B: d2 v1t 2 (12.5 m/s 5 m/s)(9 s 5 s) d2 2 (5 m/s)(9 s 5 s) d2 35 m
90
Solutions to End-of-chapter Problems
C: d3 d3
vt v t 2
a
v
(1 m/s 12.5 m/s)(15 s 9 s) 2
a
(20 m/s 0 m/s) (4.0 s 0 s)
2
dtotal d1 d2 d3 dtotal 12.5 m 35 m 40.5 m dtotal 88 m
NOTE: The solutions to problem 37 are based on the 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 portion of the graph must describe his upward motion. b) Since the skateboarder has a negative velocity from 5 to 10 seconds on the graph, he must be descending during this portion of the graph. c) The skateboarder is undergoing uniform acceleration. d) The skateboarder is at rest when his velocity equals zero, at t 5 s. When his velocity equals zero, he is at the top of the side of the swimming pool, or ground level. e) The skateboarder’s acceleration can be found from the slope of the graph. It should be equal to g : v
t (50 m/s 50 m/s) a (10 s 0 s) a 10 m/s2 38. a) At t 4.0 s, each Stooge’s acceleration is:
Curly: a
v t
a 5.0 m/s2
(1 m/s)(15 s 9 s) d3 40.5 m
a
t
b) To find their distance travelled, we take the area under the graph for each Stooge: Curly: d vt d (25 m/s)(4.0 s) d 100 m Larry: vt d 2 (10 m/s)(4.0 s) d 2 d 20 m Moe: vt d 2 (20 m/s)(4.0 s) d 2 d 40 m c) Since Curly is travelling at a constant velocity: d v t d t v (600 m) t (25 m/s) t 24 s Larry accelerates for the first 18 s of the race. His distance travelled at this point is: vt d 2 (45 m/s)(18 s) d 2 d 405 m
a 0 m/s2
Larry: v t
a
a
(10 m/s 0 m/s) (4.0 s 0 s)
a 2.5 m/s2
Moe: Solutions to End-of-chapter Problems
91
Then, travelling at a constant velocity of 45 m/s, he will traverse the last 195 m in: t
d
t
(195 m) (45 m/s)
F n
v
Box #2
F f
F 2,1
t 4.3 s
His total time is: 18 s 4.3 s 22.3 s Moe accelerates for the first 8 s of the race. His distance travelled in this time is: d
F g
41.
F support
vt
2 (40 m/s)(8.0 s) d 2 d 160 m Then, travelling at a constant velocity of 40 m/s, he will traverse the last 440 m in:
Baby
t
d
t
(440 m) (40 m/s)
F g
42.
F n
v
t 11 s
His total time is: 8 s 11 s 19.0 s Therefore, Moe wins with the fastest time of 19.0 s. 39.
F g F tension
The gravitational force downward is equal in magnitude to the tension in the elevator cable.
Elevator
F g
40.
F f
43. a)
F bat
Ball
Textbook
F n
F g
F f F 1,2
Box #1
F g
F applied
b)
F tension
Elevator
F g
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Solutions to End-of-chapter Problems
The gravitational force downward is equal in magnitude to the tension in the elevator cable.
c) Elevator
The acceleration downward is 9.8 m/s2.
45. F net ma, and (v 2 v12) , so a 2 2d m(v22 v12) F net 2d (10 000 kg)[(150 m/s) 2 (100 m s)2] F net 2(1000 m) 4 F net 6.2 10 N 46. F net ma F net a m 400 N a 200 kg a 2.0 m/s2 We also know that: (v v1) t 2 a (0 m/s 0.5 m/s) t 2.0 m/s2 t 0.25 s 47. Since F ma(6.0 m/s2) and F m b(8.0 m/s2), it follows that: ma(6.0 m/s2) m b(8.0 m/s2) m b 0.75ma If the same force were used to accelerate both masses together, we would have: F (ma m b)a F (ma 0.75ma)a F 1.75ama But we already know that F 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: F net ma But we also know that: at 2 d 2 2d a t 2
F g
d)
F n
The gravitational force downward is equal in magnitude to the normal force upward.
Car
F g
e)
F n
F-14
The gravitational force is equal in magnitude to the normal force, and F catapult the force due to the catapult must be large in order to accelerate the jet.
F g
44. The driver’s initial velocity is the same as that of the car: 50 km 1h 1 min 1000 m v1 1h 60 min 60 s km v1 13.9 m/s His final velocity is zero, and the distance he travels is 0.6 m: v22 v12 2ad (v 2 v12) a 2 2d ((0 m/s)2 (13.9 m/s)2) a 2(0.6) a 161 m/s2 a 16.4 g
Solutions to End-of-chapter Problems
93
Therefore: 2md
F net F net
t 2(1.8 kg)(0.013 m) (0.10 s) 2
2
F net 4.7 N
Therefore, the force applied to the nail by the hammer is 4.7 N and the force applied to the hammer by the nail is 4.7 N. 49. The force due to the cows on the plate is: F net (m1 m2 m3 m4 m5) g F net 5(200 kg)(9.8 m/s2) F net 9800 N From Newton’s third law, the steel plate exerts a force of 9800 N upward. 50. a) The acceleration of the water skiers can be found using: F net ma F net a (m1 m2 m3)
10 000 N a (75 kg 80 kg 100 kg) a 39.2 m/s2
51. The forces in the vertical direction are balanced: F g mg F g F n, therefore: F f F n F f mg F f (0.16)(2.0 kg)( 9.8 m/s2) F f 3.1 N 52. F net F f F net ma F net F n (v22 v12) We also know that a , therefore: 2d m(v22 v12) F n 2d m(v22 v12) mg 2d (v 2 v12) d 2 2 g [(0 m/s)2 (2.0 m/s)2] d 2(3.0)(9.8 m/s2) d 6.8 102 m engine F friction 53. F net F F engine F net F friction F net ma F friction F n F engine ma F n F engine ma mg (v v1) , therefore: a 2 t m(v2 v1) F engine mg t (800 kg)(27.8 m/s 13.9 m/s) F engine 6.0 s (0.3)(800 kg)(9.8 m/s2) F engine 4.2 103 N Gm1m2 54. F g r 2 (6.67 1011 N m2/kg2)(300 000 kg) 2 F g (1000 m)2 F g 6.0 106 N Their acceleration would be: F a g m (6.0 106 N) a (300 000 kg) a 2.0 1011 m/s2
b) The force applied by the first skier on the second two skiers is equal to the sum of their masses times their acceleration: F net mt a F net (75 kg 80 kg)(39.2 m/s2) F net 6.1 103 N The force applied by the third skier on the first two skiers is equal to his mass times his acceleration: F net mt a F net (75 kg)(39.2 m/s 2) F net 2.9 103 N From Newton’s third law, the forces applied by the second skier on the first and third skiers are 6.1 103 N and 2.9 103 N, respectively.
94
Solutions to End-of-chapter Problems
55. If the mass of Earth where doubled, the acceleration due to gravity would be: Gm12mE F g r 2 Gm12mE m1 g r 2 G 2mE g r 2
g
11
2
2
24
2(6.67 10 N m /kg )(5.97 10 kg) (6.38 10 m) 6
2
g 19.6 m/s2
net F x F y F Gmzmx Gmzmy F net 2 r zx r zy2 F net (6.67 1011 N m2/kg2)(5.0 1024 kg) 3.0 1024 kg (6.0 1010 m 5.0 1010 m)2
4.0 10 kg (5.0 10 m)
24 10 F net 6.16 1017 N
2
57. The astronaut’s weight, or his mass times the acceleration due to gravity, is: GmAmE W r AE2
W
11
2
2
24
(6.67 10 N m /kg )(100 kg)(5.98 10 kg) (6.38 10 m 3.0 10 m)
W 894 N
14. a) Horizontal: dx (25 km) cos 20° dx 23 km [E] Vertical: dy (25 km) sin 20° dy 8.6 km [N] b) Horizontal: F x (10 N) sin 30° x 5.0 N [E] F Vertical: F y (10 N) cos 30° y 8.7 N [S] F c) Horizontal: ax (30 m/s) cos 45° ax 21 m/s2 [W] Vertical: ay (30 m/s) sin 45° ay 21 m/s2 [S] d) Horizontal: px (42 kg·m/s) sin 3° px 2.2 kg·m/s [W] Vertical: py (42 kg·m/s) cos 3° py 42 kg·m/s [N] 15. a) l (10 m) cos 40° l 7.7 m b) h (10 m) sin 40° h 6.4 m 16. 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:
56. The net gravitational force on planet Z would be equal to the sum of the gravitational forces caused by each planet:
Chapter 2
6
5
2
d (2.0 km) (3.0 km) cos 20° d 4.8 km [W] d (3.0 km) sin 20° d 1.0 km [N] (4.8 k m) (1.0 km ) d d 4.9 km x
x y
y
2
tan 1
2
1.0 km 4.8 km
12°
Therefore, d 4.9 km [W12°N]. /s)2 18.|vi| (10 m/s)2 (20 m |vi| 22 m/s 20 m/s tan 1 10 m/s 63° Therefore, vi 22 m/s inclined 63° to the horizontal.
Solutions to End-of-chapter Problems
95
19. Adding by components:
d d d
x
(50 cm) sin 35° (100 cm) cos 15°
x
67.9 cm
y
(20 cm) (50 cm) cos 35°
(100 cm) sin 15°
d 46.8 cm (67.9 cm) (46.8 cm) d d 82 cm y
2
2
67.9 cm 46.8 cm
tan 1 55°
v c
d 82 cm [S55°W].
20. v v f vi 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 vf vi 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] v a t a 3.3 m/s [N2°W] 0.10 s a 33 m/s2 [N2°W] 22. a) The current velocity has no effect on the vertical component of the swimmer’s velocity, which is needed for crossing the river. Therefore: d t vs 0.80 km t 1.8 km/h t 0.44 h b) The current velocity determines how far the swimmer travels downstream, therefore: dd (vc)(t ) dd (0.50 km/h)(0.44 h) dd 0.22 km
Therefore,
2 c) vg vs2 v c vg (1.8 km/h)2 (0.5 km/ h) 2 vg 1.9 km/h v tan c vs 0.5 km/h tan1 1.8 km/h 16° The ground velocity is vg 1.9 km/h [N16°E]. 23. a) In order to go north, his ground velocity must be north.
v s
v g
θ
Since vs and vc are known, v vs
sin c
sin1
96
0.5 km/h 1.8 km/h
16°
The swimmer must swim [N16°W] in order to go straight north. 2 b) vg vs2 v c vg (1.8 km/h)2 (0.5 km/ h) 2 vg 1.7 km/h His ground velocity is vg 1.7 km/h [N].
d c) t vg 0.8 km t 1.7 km/h t 0.46 h
Solutions to End-of-chapter Problems
24. The time it takes the sandwich to reach the road is: d t p vs 10 m t 2.0 m/s t 5.0 s The distance of the pick-up truck when the sandwich is released is: dt (vt)(t ) dt (60 km/h)(5.0 s) dt (17 m/s)(5.0 s) dt 83 m 25. v
g
v w
27. a) Since the velocities are given relative to the deck, they are the velocities relative to the ship. Walking towards stern: v 0.5 m/s [S]; walking towards port, v 0.5 m/s [W]. b) velocity of velocity of velocity of
pass. relative to water v
pw
v
ship relative to water
ps
v
sw
Walking towards stern: v
vps vsw v 0.5 m/s [S] 10 km/h [N] pw v 0.5 m/s [N] 2.78 m/s [N] pw v 2.3 m/s [N] pw
pw
Walking towards port: vps vsw v 0.5 m/s [W] 2.78 m/s [N] pw vpw vps2 vsw2 vpw (0.5 m/s)2 (2.78 m/s)2 vpw 2.8 m/s vps tan vsw
pw
v h
vw cos v h cos1
v θ
pass. relative to ship
20 km/h 150 km/h
82°
The pilot must fly [N82°E] or [E7.7°N]. 26.
tan1
0.5 m/s 2.78 m/s
10°
γ v c
Walking towards stern, v 2.3 m/s [N]; walking towards port, v 2.8 m/s [N10°W].
β
v s
v g
θ
d 28. a) vf g t d t g vf 6.0 m t 5.0 m/s t 1.2 s To reach the pail, the quarterback must be 1.2 s away from reaching the garbage pail, therefore: dqg (vq )(t ) dqg (4.0 m/s)(1.2 s) dqg 4.8 m The quarterback must release the ball 4.8 m in advance. b) 1.2 s as calculated in part a. 2 c) vg vf 2 v q /s)2 vg (5.0 m/s)2 (4.0 m vg 6.4 m/s
45°
Use the sine law to find : sin sin v v c
s
sin1
sin vc
vs
6.8°
Use the sum of the interior angles of a triangle to find : 180° 180° 180° 135° 6.8° 38° The ship’s required heading is [N38°E].
Solutions to End-of-chapter Problems
97
v vg
tan f tan1
5.0 m/s 4.0 m/s
x
51°
The ground velocity is vg 6.4 m/s [E51°N]. 29. a) In order for the football to reach the garbage pail, the football’s ground velocity must be pointing north at the time of release.
v vf
cos q cos1
4.0 m/s 5.0 m/s
37°
y
x
The ball must be thrown [W37°N]. b) Calculate the magnitude of vg:
vf 2 vg2 vq 2 vg2 vf 2 vq 2 2 vg vf 2 v q /s)2 vg (5.0 m/s)2 (4.0 m vg 3.0 m/s d vg t d t vg
t
31. a) Find the time it takes the rock to reach the ground: 1 dx vi t axt 2 2 1 20.0 m (10.0 m/s)t (0 m/s2)t 2 2 t 2.00 s Find the height of the water tower: 1 h vi t ayt 2 2 1 h (0 m/s)(2.00 s) (9.8 m/s2)(2.00 s)2 2 h 19.6 m b) In the horizontal direction, vf vi axt vf 10.0 m/s (0 m/s2)(2.00 s) vf 10.0 m/s In the vertical direction, vf vi ayt vf 0 m/s (9.8 m/s2)(2.00 s) vf 19.6 m/s vf vf 2 vf 2 )2 vf (10.0 m/s)2 (19.6 m/s vf 22.0 m/s v tan f vf x
x
x
y
y
y
y
x
y
x
10 m 3.0 m/s
tan1
c) The ball is thrown such that its direction is north. The ground velocity is vg 3.0 m/s [N]. 30. The time it takes the ball to reach the ground is: 1 h vi t ayt 2 2 1 10 m (0 m/s)t (9.8 m/s2)t 2 2 t 1.4 s The horizontal distance travelled in 1.4 s is: 1 dx vi t axt 2 2 1 dx (3.0 m/s)(1.4 s) (0 m/s2)(1.4 s)2 2 dx 4.2 m The friend must be 4.2 m away to catch the ball at ground level.
x
98
19.6 m/s 10.0 m/s
63.0°
t 3.3 s
y
y
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: 1 dx vi t axt 2 2 1 100 m [(20 m/s) cos 15°]t (0 m/s2)t 2 2 t 5.2 s Find the drop in height during the 5.2 s: 1 h vi t ayt 2 2 h [(20.0 m/s) sin 15°](5.2 s) 1 2 2 (9.8 m/s )(5.2 s) 2 h 26.9 m 132.5 m h 105 m
Solutions to End-of-chapter Problems
x
y
Find the height of the second building: h2nd building h1st building h h2nd building 200 m 105 m h2nd building 95 m
The second building is 95 m high. 1 h vi t ayt 2 33. a) 2 1 1.3 m (0 m/s)t (9.8 m/s2)t 2 2 t 0.52 s b) The cup lands at the tourist’s feet, since both the cup of coffee and tourist are not moving 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 m The train is 26 m closer to Montreal. 34. Find the time it takes the Humvee to drop down to the other ramp: 1 h vi t ayt 2 2 Since both ramps are the same height, h 0 m. 0 m [(30 m/s) sin 20°] t 1 2 2 (9.8 m/s )t 2 1 [(30 m/s) sin 20°]t (9.8 m/s2)t 2 2 t 2.1 s Find the maximum horizontal distance the Humvee can travel in 2.1 s: 1 dx vi t axt 2 2 dx [(30 m/s) cos 20°](2.1 s) 1 2 2 (0 m/s )(2.1 s) 2 dx 59 m The maximum width of the pool is 59 m. 35. Find the time required to reach maximum height: 1 h vi t ayt 2 2 1 (eq. 1) h (vi sin )t ayt 2 2 y
y
x
y
Since the ball’s motion is symmetrical, it will take twice the time for the soccer ball to reach the ground: 1 0 m (vi sin )2t ay(2t )2 2 2 ay2t (vi sin )2t ayt vi sin vi sin (eq. 2) ay
t
For the range, R: 1 R vi 2t ax(2t )2 2 1 R vi 2t (0 m/s2)(2t )2 2 R vi 2t R (vi cos )2t x
x
x
t
R
(eq. 3)
2 cos vi
Substitute equation 1 into equation 2: 1 vi sin 2 h (vi sin )t t 2 t 1 h (vi sin )t (vi sin )t 2 1 (eq. 4) h (vi sin )t 2 Substitute equation 3 into equation 4, 1 R h (vi sin ) 2 2vi cos 1 sin h R 4 cos 1 h (tan ) R 4 h 0.25 R 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: 1 h vi t ayt 2 2 1 1.7 m vi (sin 45°)t (9.8 m/s2)t 2 (eq. 1) 2 Find the time it takes the ball to clear the wall: 1 dx vi t axt 2 2 130 m vi (cos 45°)t 130 m (eq. 2) t vi cos 45°
y
x
Solutions to End-of-chapter Problems
99
Substitute equation 2 into equation 1: 130 m 1.7 m (vi sin 45°) vi cos 45° 1 130 m 2 2 (9.8 m/s ) 2 vi cos 45° 1.7 m (tan 45°)(130 m) 33 800 m (4.9 m/s2) 2
F a F a 2 F a 2 F a (59 N)2 (394 N) 2 F a 399 N F a tan F a
x
vi
vi 36 m/s
x
y
tan1
38.6 N [W] F
8.5° a 399 N [S8.5°E] F F f k F n F f kmg F f (0.10)(300 kg)(9.8 m/s 2) F f 294 N 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] F net 105 N [S8.5°E] The net force is F b) F net ma net a F m
x
Vertical components: F y (60 N) cos 40° 80 N
34.0 N [S] F
a
105 N [S8.5°E] 300 kg
a
0.35 m/s2 [S8.5°E]
y
net 51 N [S49°W] F
c) Horizontal components:
F (50 N) cos 60° 10 N 15 N [E] F x
x
Vertical components: F y (50 N) sin 60° 60 N
16.7 N [S] F
y
net 22 N [S42°E] F a F f 38. a) F net F The sum of the x components is: a F 1 F 2 F 3 F a (100 N) cos 20° [W] F
x
x
x
x
x
(200 N) cos 40° [E] a 59 N [E] F The sum of the y components is:
net F applied force in the x direction F kinetic friction 39. F net F a F k F Find the kinetic frictional force, F k: F k k F n F k (0.30)( F g F a sin 50°) F k (0.30)[(20 kg)(9.8 m/s 2) (100 N) sin 50°] F k 36 N F net (100 N) cos 50° 36 N F net 28 N F net ma 28 N ma 28 N a 20 kg a 1.4 m/s2
x
a F 1 F 2 F 3 F a (100 N) sin 20° [N] F (200 N) sin 40° [S] 300 N [S] a 394 N [S] F
y
y
y
y
y
y
100
394 N 59 N
The player strikes the ball at 36 m/s, 45° above the horizontal. 37. a) F net (30 N)2 (10 N)2 F net 32 N 30 N tan1 10 N 72° net 32 N [N72°E] So F b) Horizontal components: F x (60 N) sin 40°
y
Solutions to End-of-chapter Problems
x
40. The hockey stick provides the only force on the puck, therefore it is the net force acting on the puck: F s F net net ma F net a F m a 300 N [N25°E] 0.25 kg a 1200 m/s2 [N25°E] Find vf : a vf vi t at vf vi v at v f i v (1200 m/s2 [N25°E])(0.20 s) 12 m/s [S] f v 240 m/s [N25°E] 12 m/s [S] f The vector sum of the x components is: v (240 m/s) sin 25° [E] fx v 101 m/s [E] fx The vector sum of the y components is: v (240 m/s) cos 25° [N] 12 m/s [N] fy v 206 m/s [N] fy F f 2 vf F f 2 vf (101 m/s)2 (206 m/s)2 vf 229 m/s vf tan vf 206 m/s tan1 101 m/s 26° The final velocity is vf 229 m/s [N26°E]. 41. a) F k k F n F k (0.50)(100 kg)(9.8 m/s 2) F k 4.9 102 N b) The frictional force is the only force acting on the baseball player, therefore it is also the net force. F net F k ma F k 490 N a 100 kg a 4.9 m/s2
x
y
x
y
Find vi: a
v v t f
i
at vi vi (4.9 m/s2)(1.3 s) vi 6.4 m/s 42. F net F k ma k F n ma k mg a k g a (0.3)(9.8 m/s2) a 2.94 m/s2 Find distance, d: vf 2 vi2 2ad vi2 2ad vi2 d 2a (2.0 m/s)2 d 2(2.94 m/s2) d 0.68 m
The key will slide 0.68 m across the dresser. 43. F net F a F k The horizontal acceleration of 1.0 m/s 2 is the net acceleration of the mop, therefore: F net max max F a k F n max (30 N) cos 45° [(0.1)( F g F a 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 0.1mg m(1.0 m/s2 0.1 g ) 19.09 N m(1.98 m/s2) 19.09 N m 9.6 kg 44. Let be the angle of the inclined plane when the box starts to slide. At this angle, F s s F n (eq. 1) F s (0.35)(mg cos ) (eq. 2) F x mg sin
Solutions to End-of-chapter Problems
x
101
Set equation 1 equal to equation 2: (0.35)(mg cos ) mg sin sin 0.35 cos tan 0.35 tan1 (0.35) 19° The minimum angle required is 19°. 45. a) The acceleration for child 1:
F net F x m1a1 m1 g sin a1 g sin a1 (9.8 m/s2) sin 30° a1 4.9 m/s2
The acceleration for child 2: F net F x m2a2 m2 g sin a2 g sin a2 4.9 m/s2
Both children accelerate downhill at 4.9 m/s2. b) They reach the bottom at the same time. 46. a) F net F x F k ma mg sin k F n 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/s 2. b) vf 2 vi2 2ad vf 2 2(0.14 m/s2)(200 m) vf 7.6 m/s The box reaches the bottom of the hill at 7.6 m/s2. c) a
v v t f
i
v t f a t
47. Find his final speed, vf , at the bottom of the ramp by first finding his acceleration: F net F x 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 vf 2 2(5.6 m/s2)(50 m) vf 23.6 m/s vf will be the initial speed, vi2, for the horizontal distance to the wall of snow. Find the deceleration caused by the snow: F net F k ma k F n ma (0.50)(9.8 m/s2)m a 4.9 m/s2 Find the distance Boom-Boom will go into the wall of snow: vf 2 vi2 2ad 0 vi2 2ad vi2 2ad vi2 d 2a (23.6 m/s)2 d 2(4.9 m/s2) d 57 m Boom-Boom will go 57 m into the wall of snow. 48. Find the net force on Spot, then solve for the net acceleration: F net F r F x F net 2000 N mg sin ma 2000 N (250 kg)(9.8 m/s 2)(sin 20°) ma 2000 N 838 N ma 1162 N 1162 N a 250 kg a 4.6 m/s2
7.6 m/s 0.14 m/s 2
t 53 s
It takes the box 53 s to reach the bottom of the hill.
102
Solutions to End-of-chapter Problems
Find time, t :
Find T 1: 1 2
(eq. 1) m1a T 1 m1 g 2 T 1 (10 kg)(3.9 m/s )
d vit at 2
1 2 1 250 m (4.6 m/s2)t 2 2 2 t 108 s2 t 10 s 49. a) (a) For m1:
(10 kg)(9.8 m/s 2) T 1 137 N
d at 2
Find T 2: (eq. 3) m3a m3 g T 2 2 T 2 (30 kg)(9.8 m/s )
F net1 T T m1a
(30 kg)(3.9 m/s2) T 2 176 N (eq. 1)
For m2: F net2 F g T m2a m2 g T
(eq. 2) Substitute equation 1 into equation 2: m2a (m1 m2)a (40 kg)a a
m2 g m1a m2 g (20 kg)(9.8 m/s2) 4.9 m/s2 [left] For tension T , substitute acceleration
into equation 1: T m1a T (20 kg)(4.9 m/s2) T 98 N
(c) For m1: F net1 T F x (eq. 1) m1a T mg sin For m2: F net2 F 2g T (eq. 2) m2a m2 g T Add equations 1 and 2: (eq. 1) m1a T mg sin (eq. 2) m2a m2 g T (m1 m2)a m2 g m1 g sin (25 kg)a (15 kg)(9.8 m/s2) (10 kg)(9.8 m/s 2) sin 25° a 4.2 m/s2 [right] For tension T , substitute acceleration into equation 2: m2a m2 g T T (15 kg)(9.8 m/s2) (15 kg)(4.2 m/s 2) T 84 N b) (a) For m1: F net1 T F k (eq. 1) m1a T km1 g For m2: F net2 F g T (eq. 2) m2a m2 g T Add equations 1 and 2: m1a T km1 g (eq. 1) (eq. 2) m2a m2 g T m1a m2a m2 g km1 g a(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]
(b) Assume the system moves towards m3: For m1: F net1 T 1 F 1g (eq. 1) m1a T 1 m1 g For m2: F net2 T 2 T 1 (eq. 2) m2a T 2 T 1 For m3: F net3 F 3g T 2 (eq. 3) m3a m3 g T 2 Add equations 1, 2, and 3: m1a T 1 m1 g (eq. 1) (eq. 2) m2a T 2 T 1 m3a m3 g T 2 (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/s 2) (50 kg)a 196 N a 3.9 m/s2 [right]
Solutions to End-of-chapter Problems
103
For tension T , substitute acceleration into equation 2:
Add equations 1 and 2: m1a T m1 g sin (eq. 1) km1 g cos (eq. 2) m2a m2 g T (m1 m2)a m2 g m1 g sin km1 g cos (25 kg)a (9.8 m/s2)[15 kg (10 kg) sin 25°
m2a m2 g T T (20 kg)(9.8 m/s2)
(20 kg)(3.9 m/s 2) T 118 N (b) Assume the system moves towards m3: For m1: F net1 T 1 F 1g m1a T 1 m1 g
0.2(10 kg) cos 25°] a 3.5 m/s2 [right] For tension T , substitute acceleration into equation 2:
(eq. 1)
For m2: F net2 T 2 T 1 F k m2a T 2 T 1 km2 g (eq. 2)
m2a m2 g T T (15 kg)(9.8 m/s2)
For m3: F net3 F 3g T 2 m3a m3 g T 2
(eq. 3)
Add equations 1, 2, and 3:
50. For m1:
m1a T 1 m1 g (eq. 1) m2a T 2 T 1 (eq. 2) km2 g m3a m3 g T 2 (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) a
(15 kg)(3.5 m/s2) T 94 N
10 kg] 3.5 m/s2 [right]
Find T 1: (eq. 1) m1a T 1 m1 g T 1 (10 kg)(3.5 m/s 2) (10 kg) (9.8 m/s2) T 1 133 N Find T 2: m3a m3 g T 2 T 2 (30 kg)(9.8 m/s2)
(eq. 3)
F net2 F 2g T 1 m2a m2 g T 1
(eq. 2)
Add equations 1 and 2: (eq. 1) m1a T km1 g (eq. 2) m2a m2 g T (m1 m2)a m2 g km1 g (eq. 3) 2 (9.0 kg)a (4.0 kg)(9.8 m/s ) (0.10)(5.0 kg)(9.8 m/s 2) a 3.8 m/s2 [right] 51. For the system to be NOT moving, the acceleration 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 0.80
(30 kg)(3.5 m/s ) T 2 188 N (c) For m1: F net1 T F x F k m1a T m1 g sin (eq. 1) km1 g cos
For m2:
104
(eq. 1)
For m2:
2
F net2 F 2g T m2a m2 g T
F net1 T F f1 m1a T k F n m1a T km1 g
(eq. 2)
52. First find the system’s acceleration: For Tarzana: F netTA T mTAa T For Tarzan: F netTZ F TZg T mTZa mTZ g T
Solutions to End-of-chapter Problems
(eq. 1)
(eq. 2)
Add equations 1 and 2:
56.
(eq. 1) mTAa T (eq. 2) mTZa mTZ g T (mTA mTZ)a mTZ g (65 kg 80 kg)a (80 kg)(9.8 m/s2) 2
a
(80 kg)(9.8 m/s ) (65 kg 80 kg)
a 5.4 m/s
2
F c F f mac F n mv 2 mg r v gr v 21 m/s
It is not necessary to know the mass. 57. Vertically: F n cos mac
To find time t :
F n
1 2
d vit at 2
Horizontally:
1 15 m at 2 2 30 m 2
c
v2 g tan r 25° v rg tan v 19 m/s
30 m 5.4 m/s
2
t 2.4 s 42r Assuming ac is a constant, 53. ac T 2
T
F c F n sin mac F n sin mg cos sin ma
t a t
mg cos
42r ac
a) If the radius is doubled, the period 2. increases by a factor of b) If the radius is halved, the period decreases by a factor of 2. 2 4 r 54. a) ac T 2 42(0.35 m) ac (0.42 s)2 ac 78 m/s2 b) The clothes do not fly towards the centre because the wall of the drum applies the normal force that provides the centripetal force. When the clothes are not in contact with the wall, there is no force acting on them. The clothes have inertia and would continue moving at a constant velocity tangential to the drum. The centripetal force acts to constantly change the direction of this velocity. 42r 55. ac T 2 T 365 days 3.15 104 s 42(1.5 1011 m) ac (3.15 107 s) ac 6.0 103 m/s2
58.
F c F g mac mg v2 g r v gr v 9.9 m/s 59. a) T mg T (0.5 kg) g T 4.9 N mv2 b) T mg r mv2 T mg r
(0.5 kg)(2.4 m/s) 2 T (0.6 m) (0.5 kg)(9.8 m/s2) T 9.7 N 60. Maximum tension occurs when the mass is at its lowest position. Tension acts upward, and gravity acts downward. The difference between these forces is the centripetal force:
2
T max mg
mv r
2
T max
mv mg r
(2.0 kg)(6.6 m/s) 2 T max (2.0 kg)(9.8 m/s2) 3.0 m T max 49 N
Solutions to End-of-chapter Problems
105
The tension is minimized when the mass is at the top of its arc. Tension and gravity both act downward, and their sum is the centripetal force: 2
T min min mg
mv r 2
T min min
mv mg r
T min min
(2.0 kg)(6.6 m/s) 3.0 m
2
63. On mass 2: 42r F c m2 T 2 42r T 2 m2 T 2 42( L1 L2) T 2 m2 T 2 On mass 1: 42r F c m1 T 2 42r T 1 T 2 m1 T 2 42 L1 42( L1 L2) T 1 m1 m 2 T 2 T 2 42 (m1 L1 m2( L1 L2)) T 1 T 2
2
(2.0 kg)(9.8 m/s ) T min min 9.4 N 61. a)
F net net ma F n mg m(9 g ) F n 9mg mg F n 10mg F n 5.9 103 N v2 b) ac r v2 9 g r v2 r 9 g
(91.67 m/s)2 r 9(9.8 m/s2) r 95 m 62. a) G 6.67 1011 N·m2/kg2, T 365 days 3.15 107 s 4 2r GmEmS
m T r E
2
2
2 3
mS
4 r GT
mS
4 (1.5 10 m) (6.67 10 N·m /kg )(3.15 10 s)
2
2
11
11
2
3
2
7
2
mS 2.0 1030 kg m b) Density of the Sun V 30
2.0 10 kg 4 3
r
3
1.4 103 kg/m3
mEarth 5.98 1024 kg 24
Density of Earth
5.98 10 kg 4 r
3
3 5.5 103 kg/m3 1 The Sun is about as dense as Earth. 4 106
Solutions to End-of-chapter Problems
24.
Chapter 3
60°
60°
21.
F s = 2500 N
2500 N
T cable F strut
30°
F g
T
T
m
mg
30°
F g
F s
flower pot
F s = 2500 N
F T F g
30°
sin 30° g T
T
sin 30° mg T sin 30° T
60° mg
mg
sin 30°
F s F sin 30° m s g
(10 kg)(9.8 N/kg) sin 30°
T 196 N F 22. tan g F s F g F s tan
m
(2500 N) sin 30° 9.8 N/kg
m 128 kg
25.
F s
98 N tan 30°
T cable
F s 169.7 N F s 170 N
12°
T cable
m g 12°
23.
500 kg
30°
30° 30°
T 1
T rope T 1
T 2 60°
F g — 2
F g
30°
2 F g
T
cos 12°
mg T cable cable
T 2
T 1 T 2 T F g 2 cos 30° T
T rope
F g
(cos 30°)
(100 kg)(9.8 N/kg) 2
T cable cable
mg cos 12°
T cable cable
(500 kg)(9.8 N/kg) cos 12°
T cable cable 5009.5 N 3 T cable cable 5.01 10 N F rope rope tan 12° mg F rope rope mg tan 12° F rope rope (500 kg)(9.8 N/kg) tan 12° 3 F rope rope 1.04 10 N
T
(cos 30°)
T 566 N Solutions to End-of-chapter Problems
107
26. a)
27.
F app
F app
1.5 m
F n
F f
100 kg
1.5 m
25.0 m
mg
25.0 m ——— 2
θ
0.63 F app app F f f 0 F app app F f f F app app F n F app app mg 0.63(100 0 kg)(9.8 N/kg) F app app 0.63(10 F app app 617.4 N
T
F n
425 N
θ
θ
car
tan
b)
1.5 m 25.0 m
2
tan 0.12 6.8°
L = 10 m
F app app T F app T app sin
sin 250 kg
d
T
T 3.59 103 N
T θ
θ
10 m
mg
250 kg
T
425 N sin 6.8°
F app
The rope pulls with a force of 3.59 103 N. 28. T
d
F app = 617 N
Bird mg
Using similar triangles, find T first: 2 T 2 (mg )2 F app app N)2 T [(250 kg)(9.8 N/kg)]2 (617.4 T 2526.6 N T 2.53 103 N
d
9.0 m
θ
F app app L T F app L d app T
0.52 m
(617.4 N)(10 m) 2526.6 N
d 2.4 m
108
m Bg
18.0 m
d
T
tan
0.52 m 9.0 m
tan 0.058 3.3°
Solutions to End-of-chapter Problems
30.
T
T h
F f
T v
θ θ = 3.3°
m Bg
Pulley
T
T
2 2 sin 2(90 N) sin 3.3° 9.8 N/kg m g B
sin mB mB
T
T T g
T h
mg
T
mB 1.1 kg
29.
L – 2
T 1 x – 2 80°
m Lg
40° 40°
L – 2 T 2
T h F f 0 With left taken to be the positive direction, T h F f f 0 T h F f f T h F n mg T h 2 From Pythagoras’ theorem: mg 2 T 2 T h2 2
T 1
Leg T 2
T (5.0 kg)(9.8 N/kg) T 49 N 40° T 2
T 1
2 2 2 ( 1) ( 1) 2
T 2 2 40°
T 2
F app
T
app T1 T2 F F app
cos 40°
2
2
mg
mg
2
mg
2
2
mg
2
From similar triangles:
T F app app 2(T cos 40°) F app app 2mg cos 40° F app app 2(5.0 kg)(9.8 N/kg) cos 40° [left] ft] F app app 75 N [le
2 2 x
L
T h T
x T h L T T h L x T
Solutions to End-of-chapter Problems
109
Substituting for T h and T ,
b)
2
T
m gL
x m g
2 1
2
x
2
x
L 2
31. a)
m T = m 1 + m 2
L
1 1
m Tg
T ? If up is positive, T m T g T (4.0 kg)(9.8 N/kg) T 39.2 N [up] 32. T 0 The pivot is the left support. 1 0 2 Board Duck 0
+
P
1 3 kg
2 1 kg 2.0 m
centre of mass ? net 0 With clockwise as the positive rotation, 1 2 0
1 2 r 1m1 g sin r 2m2 g sin m2 g r 1 r 2 m1 g
r 1 r 2
m2 m1
r r 1 2
3
4 2.0 m r 1 4 r 1 0.5 m The centre of mass is 0.5 m from m1 and 1.5 m from m2.
2 B D 2 r B F gB r D F gD 2 r BmB g r DmD g 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 1313.2 Nm F 2 0.8 m F 2 1641.5 N 2 1.6 103 N [up] F For F 1: F T 0 With down as positive, 0 F 1 F 2 F B F D
But r 2 r 1 r T 3r 1 r T r 1 4r 1 r T r T r 1
F 1 F B F D F 2 F 1 (mB g ) (mD g ) F 2 F 1 (50 kg)(9.8 N/kg) (8.5 kg)(9.8 N/kg) 1.6 103 N F 1 1068.2 N 1 1.1 103 N [down] F
and 2 1.6 103 N [up] F
110
Solutions to End-of-chapter Problems
33.
x 5.0 m 3.75 m x 1.25 m 35. T 0 man L(left) L(right) rock 0
m 5 . 0 m 0 . 1
=
2
y
=
X 1 = 0.5 m X 2 = 2.5 m
xcm xcm
rock man L(left) L(right) r rockmrock g sin r manmman g sin r L(left)mL(left) g sin r L(right)mL(right) g sin r rockmrock [(1.90 m)(86 kg)]
x1 x2
2 0.5 m 2.5 m 2
xcm 1.5 m [right] y y2 ycm 1 ycm
With clockwise as the positive direction of rotation, 0 man L(left) L(right) rock
1
y
1.90 m 2 1.90 m (2.0 kg) 2.40 m 0.5 m 0.50 m (2.0 kg) 2 2.40 m
2 0.5 m 1.0 m 2
ycm 0.75 m [up] Centre of mass 1.5 m [right], 0.75 m [up]
34.
r rockmrock 163.4 kg·m 1.504 kg·m
0.104 kg·m r rockmrock 164.8 kg·m 164.8 kg·m mrock 0.50 m mrock 329.6 kg mrock 3.3 102 kg
x
5.0 m
F 23
F 1
2.5 m
36. a)
20 kg 3
17 kg 1
27 kg 2 +
P
x
3.8 kg F g
Let F 1 be the pivot. T 0 0 2 3 L With clockwise as positive, 223 L 0 2 r 23 m g r Lmg 3 3r L r 23 2 5.0 m 3 2 r 23 2 15.0 m r 23 4 r 23 3.75 m
T 0 1 2 3 TL TR 0
With clockwise as the positive rotation, 1 2 3 TL TR 0 3 2 1 r 3m3 g r 2m2 g r 1m1 g
r 3m3
3.8 m 2 (27 kg) 3.8 m 2 (17 kg)
r 3m3 51.3 kg·m 32.3 kg·m r 3m3 19 kg·m r 3
19 kg·m 20 kg
r 3 0.95 m
Solutions to End-of-chapter Problems
111
The third child of mass 20 kg must sit 0.95 m from the centre of the teeter-totter and on the same side as the 17.0-kg child. b) No, the mass of the teeter-totter does not matter. 37.
F 1
Back D r B F B r D F gD r B F B r DmD g r B F B (0.30 m)(30 kg)(9.8 N/kg) r B F B 88.2 N·m
F 2
5.0 kg
+
2.5 m
Let F 2 be pivot.
Let up be positive. 0 F F F D F B
net 0 0 1 B C
With clockwise as the positive rotation, 1 p c 0 1 p c r 1 F 1 r p F gp r c F gc r F r c F gc F 1 p gp r 1
2.5 m (2.0 kg)(9.8 N kg) [(2.5 m 1.5 m)(5.0 kg)(9.8 N/kg)] 2 F
1
F F F D F B F F mD g F B F F (30 kg)(9.8 N/kg) 88.2 N F F 205.8 N F F 2.1 102 N Front legs: 1.05 102 N each; back legs: 4.4 101 N each (each divided by 2).
39. a)
P
2.5 m
F 1 29.4 N But F net 0 F 1 F gB F gC F 2 0
C of m
2.4 m
With up as the positive direction, 0 F 1 F gB F gC F 2 F 2 F gB F gC F 1 F 2 mB g mC g F 1 F 2 (2.0 kg)(9.8 N/kg) (5.0 kg)(9.8 N/kg) 29.4 N F 2 39.2 N The man farthest from the cement bag ( F 1)
lifts with 29.4 N and the second man lifts with 39.2 N of force. 38. Take front two and back two legs as single supports. net 0 with front legs as pivot D Back 0
20 kg
0.8 m
net 0 F T F F D 0
Taking up to be positive, 0 F T F D F T mD g F T (20 kg)(9.8 N/kg) T 196 N [up] F
112
88.2 N·m 1.0 m
F B 88.2 N F B 8.8 101 N net 0 But F F F F D F B 0
1.5 m
F B
P
m p = 2.0 kg
Let clockwise be positive. D Back 0
Solutions to End-of-chapter Problems
b)
41. a)
P
1.0 m
r D
1.2 m θD
75 kg
1.6 m 0.4 m
C of m
C of m
P
net 0 F app-h F f 0 F
tan1
0.4 m 1.2 m
Taking the direction of force application to be positive,
18.4°
F app-h F f F app-h F n F app-h mg F app-h 0.42(75 kg)(9.8 N/kg) app-h 308.7 N [horizontally] F app-h 3.1 102 N [horizontally] F
Assume the upper hinge is the pivot. B door 0 B door 0
B door r B F B sin B r DmD g sin D r m g sin D F B D D r B sin B
b)
(1.26 m)(20 kg)(9.8 N/kg) sin 18.4° (2.4 m) sin (90° 18.4°)
F B
B 34.2 N [out horizontally] F
θbox θa
40.
P 1.2 m
+
Just to the tip the box, net 0 a box 0 Taking the direction of force application to be positive, a box 0
72 kg
7.0 m
θp
65°
a box
P
p 90° 65° p 25°
Choose bottom as pivot. net 0 wall p 0 Taking right (horizontally) as positive, wall p 0
Take bottom corner as pivot. 1.6 m 2 tan a 1.0 m 2 a 58° r a F a sin a r boxm box g sin box r m g sin box r a box box F a sin a
wall p r w F w sin w r pmp g sin p r m g sin p F w p p r w sin w
(0.8 m )2 (0 .5 m)2(75 kg)(9.8 N/kg) sin (90° 58°) r a
(308.7 N) sin 58°
F w
r a 1.40 m
[(7.0 m 1.2 m)(72 kg)(9.8 N/kg)(sin 25°)] (7.0 m) sin 65°
w 272.6 N [horizontal] F F w 2.7 102 N But F h(bottom) F h(top) so 2.7 102 N is
But: h r a sin 58° h 1.2 m
required to keep the ladder from sliding. Solutions to End-of-chapter Problems
113
Let the contact point of F 2 be the pivot P . 0 1 w With clockwise being the positive torque direction, 1 w 0
42.
+ 10 kg
P
1 w r 1 F 1 r wmg r mg F 1 w r 1
5.0 cm 16 cm
35 cm
0 muscle arm water 0
F 1 1911 N 1 1.9 103 N [up] F net 0 F 1 F F 2 F g 0
net
With clockwise as the direction of positive rotation, m a w 0
F m 780.1 N m 7.8 102 N [up] F
F 2 F 1 F g F 2 1911 N (65.0 kg)(9.8 N/kg) F 2 2548 N 2 2.5 103 N [down] F
(0.16 m)(3.0 kg)(9.8 N/kg) (0.35 m)(10 kg)(9.8 N/kg) 0.050 m
With up taken to be the positive direction,
m a w r m F m sin r ama g sin r wmw g sin r m g r wmw g F m a a r m
F m
(0.12 m)(65 kg)(9.8 N/kg) 0.04 m
F 1
45.
F = 0.5 N 0.01 m
43.
1.9 kg
1.2 kg
P
0.4 kg
+
0.15 m 0.40 m
0.02 m 0.60 m
The total of all three torques must be equiva lent to the total torque through the centre of mass.
ua fa hand r cmm T g r uamua g r famfa g r handm hand g r m r famfa r handm hand r cm ua ua mT
cm
r cm
(0.15 m)(1.9 kg) (0.40 m)(1.2 kg) (0.60 m)(0.4 kg) 3.5 kg
r cm 0.29 m from shoulder
44.
F 1
W
P 4.0 cm
12 cm
+
F 2
Use pivot P as the point of contact of F 1. net 0 0 F2 F With clockwise taken to be the positive torque direction, F2 F 0
F2 F r F2 F 2 r F F r F F F 2 r F2
F 2
(0.01 m)(0.5 N) 0.02 m
F 2 0.25 N
114
Solutions to End-of-chapter Problems
net 0 F F F F 2 0
48.
1
With right taken to be the positive direction, F 1 F 1 F 1 1 F
h
F F 2 0.5 N 0.25 N 0.75 N 2 0.25 N [right] 0.75 N [left], and F
h cm
46.
θ
1.00 m
θ L — 2
1.00 m
28 cm
2 L
11 cm
tan
7.25 kg
2.4 cm P
C of m
hcm
2 L
F T
hcm
+
tan 1.00 m 2 hcm tan 30° hcm 0.8660 m But: h 2hcm h 2(0.8660 m) h 1.73 m NOTE: The solution to problem 49 is based on the pivot point of the glass being at the corner of the base.
Set P at elbow joint. net 0 T arm sp 0 With clockwise taken to be the positive torque direction, T arm sp 0
T arm sp r T F T r arm F g(arm) r sp F g(sp) r m g r spmsp g F T arm arm r T
F T
(0.11 m)(2.7 kg)(9.8 N/kg) (0.280 m)(7.25 kg)(9.8 N/kg) 0.024 m
49.
d x
2
F T 9.5 10 N
47.
θ
0.6 m
θ
0.050 m
θ
0.3 m θ
0.020 m
0.3 m 0.6 m 26° The tipping angle is 26° from the horizontal. tan
0.14 m
tan
θ
0.020 m 0.050 m
21.8°
sin
x h 0.050 m
x (0.14 m 0.050 m) sin 21.8° x 0.033 m d x r d 0.033 m 0.020 m d 0.053 m Solutions to End-of-chapter Problems
115
Use hinge as pivot. net 0 0 s m With clockwise taken to be the positive torque direction, s m 0
50.
θ 2.5 m
m 2. 5
θ
s m r s F s sin s r m F m sin 90° r m g F s m m r s sin s
2 base
tan
hcm
F s
2.5 m 2 tan 2.5 m 26.6° 51.
(1.0 m)(10.0 kg)(9.8 N/kg) (0.75 m) sin 10°
F s 752.5 N
But: F s kx F k s x
F s
k
752.5 N 4.0 10 m 2
k 1.88 104 N/m
nails
53.
x 2
F g x 1
s F g 0 F
45° 0.50 m
With up taken to be the positive direction, F s mg
So:
kx mg mg k x
tan
(3.0 kg)(9.8 N/kg) 1.8 10 m
52. +
10°
P 0.75 m 1.0 m
1.50 m 0.50 m
71.56° ∆ x 1.58 m 0.7071 m ∆ x 0.874 m
2
k 1.6 103 N/m
116
0.50 m
m)2 x1 (0.50 m)2 (0.50 x1 0.7071 m m)2 x2 (0.50 m)2 (1.50 x2 1.58 m
F s kx
k
θ 0.50 m
But:
1.5 m
θ
T
F s
T
bar 10 kg
F gm
Solutions to End-of-chapter Problems
F g
θ
T F s T k∆ x T (1.5 102 N/m)(0.874 m) T 1.311 102 N
55. a) A 0.1 m2 F Stress A (100 kg)(9.8 N/kg) Stress 0.1 m2 Stress 9.8 103 N/m2 Strain Eiron 100 109 N/m2 Stress E Strain Stress Strain E 9.8 103 N/m2 Strain 100 109 N/m2 Strain 9.8 108 b) ∆ L ? L 2.0 m ∆ L L(Strain) ∆ L (2.0 m)(9.8 108) ∆ L 1.96 107 m ∆ L 2.0 107 m c) Maximum stress is 17 107 N/m2. F max Stress( A) F max (17 107 N/m2)(0.1 m2) F max 1.7 107 N mg F max 1.7 107 N m 9.8 N/kg m 1.7 106 kg 56. Maximum stress for femur is 13 107 N/m2. A 6.40 104 m2 F Stress A F max A(Stress) F max (6.40 104 m2)(13 107 N/m2) F max 8.32 104 N 57. F c 200 N A 1 105 m2 L 0.38 m E 15 109 N/m2 F A E L L FL L AE (200 N)(0.38 m) L (1 103 m2)(15 109 N/m2)
T
θ
mg — 2
T
2 T m g
sin m
2T sin g
m
2(1.311 10 N) sin 71.56° 9.8 N/kg
2
m 25.4 kg 54. L 20 m r 2.0 103 m Limit F L 6.0 107 N/m2 F a) Stress A F A(Stress) F r 2(Stress) F (2.0 103 m)2(6.0 107 N/m2) F 753.6 N F 7.5 102 N b) E for A l is: EAl 70 109 N/m2 E
E
Stress Strain
F A L L
L
L
F A E
(20 m)(6.0 107 N/m2) 70 109 N/m2 L 0.017 m L 1.7 102 m L
L 5.067 106 m Solutions to End-of-chapter Problems
117
k
F
k
200 N 5.067 10 m
x
6
k 3.95 107 N/m
58.
2.0 m
L
R θ
R
τ
9
r r
125.6 m
5.73°
R L
R R L R cos
20 m R L cos 5.73° R L 20.1005 m L 20.1005 m R L 20.1005 m 20 m L 0.1005 m G
F A L L
A
3
2
E
Stress Strain
Strain
Stress E
F A Strain E [m(a g )] Strain A E
2
Arod (0.01 m) G LA F L (80 109 N/m2)(0.1005 m)[ (0.01 m)2] F
2.0 m
F 1 262 920 N F 1.26 106 N rod rF sin rod 2.0 m(1 262 920 N) sin 90° rod 2.52 106 N·m The torque on rod is 2.5 106 N·m.
118
1.96 10 m
r 0.025 m r 2.5 102 m b) a 2.0 m/s2 F net F app mg F net ma mg F net m(a g ) Esteel 200 109 N/m2
(360°)(2 m) 125.6 m
cos
2
G steel 80 10 N/m rF C 2r C 2(20 m) C 125.6 m 2.0 m
360°
59. Stress is 10% of T max. Stress 0.10(50 107 N/m2) Stress 5.0 107 N/m2 a) A ? F Stress A F A Stress mg A Stress (1.00 104 kg)(9.8 N/kg) A 5.0 107 N/m2 A 1.96 103 m2
Strain
4
2
2
[(1.00 10 kg)(2.0 m/s 9.8 m/s )] 1.96 1 0 m 200 10 N/m 3
9
2
2
Strain 3.01 104 60. L ? Epine 10 109 N/m2 L 3.0 m A (10 102 m)(15 102 m) A 1.5 103 m2 F g 1000 N
Solutions to End-of-chapter Problems
a)
E
E
For column 2: Strain Stress
F A L L
E
Stress
Stress Strain
2
5
Stress
F Stress A
Stress
F A E
2.45 10 N (0.5024 m )(50 10 N/m ) 2
9
2
Stress 9.75 106
1000 N 1.5 10 m 3
2
Stress 6.67 105 N/m2 Stress Strain
E
6.67 105 N/m2 Strain 10 109 N/m2 Strain 6.67 105 b) ∆ L L(Strain) ∆ L (3.0 m)(6.67 105) ∆ L 2.0 104 m 61. m 2.5 104 kg F app (2.5 104 kg)(9.8 N/kg) F app 2.45 105 N
L2 9.75 10 L
6
2
L2 L2 (9.75 106) But: L2 L2 21.999 863 m L2 L2(9.75 106 m) 21.999 863 m
L2
21.999 863 m 1 9.75 10 m 6
L2 22.000 0775 m
The narrower column needs to be only 7.8 105 m longer than the wider column.
A1 r 2 A1
1.00 m 2
2
A1 0.785 m2 A2 r 22 A2
0.80 m 2
2
A2 0.5024 m2 Emarble 50 109 N/m2 ∆ L1 ? Stress L1 L1 E
L1 L 1
L1
F A1
E
FL1 A1 E
(2.45 105 N)(22.0 m) L1 (0.785 m2)(50 109 N/m2) L1 1.37 104 m Column 1 final loaded: Loaded 22.0 m L1 Loaded 22.0 m 1.37 104 m Loaded 21.999863 m
Solutions to End-of-chapter Problems
119
Chapter 4
16. p mv p (7500 kg)(120 m/s) p 9.0 105 kg·m/s 17. p mv p (0.025 kg)(3 m/s) p 0.075 kg·m/s 18. 90 km/h 25 m/s, m 25 g 0.025 kg p mv p (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 p m v 23 000 kg·m/s m 138.89 m/s m 165.6 kg p 20. v m 1.00 kg·m/s v 1.6726 1027 kg
v 6.00 1026 m/s, which is much greater
than the speed of light. 21. p mv p (0.050 g)(10 m/s [down]) p 0.5 kg·m/s [down]
p = 0.5 kg·m/s [down]
22. v (300 km/h)
1000 m 1h 1 km 3600 s
83.3 m/s p mv p (6000 kg)(83.3 m/s [NW]) p 5 105 kg·m/s [NW]
p = 5 x 105 kg·m/s [NW] 45°
120
250 N [forward], 23. m 50 kg, F ∆t 3.0 s, v1 0 t m v F (250 N [forward])(3.0 s) (50 kg)( v2 v1) 740 N [forward] v2 50 kg v 15 m/s [forward] 2 24. m 150 kg, v1 0, a 2.0 m/s2, ∆t 4.0 s a) v2 v1 a∆t v2 0 (2.0 m/s2)(4.0 s) v2 8.0 m/s p mv p (150 kg)(8.0 m/s) p 1200 kg·m/s b) J ∆p J 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 1 h v1t at 2 a) 2 1 17 m 0 (9.8 m/s2)t 2 2 t 1.86 s b) F ma F (1.5 kg)(9.8 N/kg) F 14.7 N c) J F ∆t J (14.7 N)(1.86 s) J 27.3 kg·m/s 26. F 700 N, ∆t 0.095 s a) 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
Solutions to End-of-chapter Problems
29. p p2 p1
p p 2
33°
c) F ma F (0.03 kg)(1.3 106 m/s2) F 3.9 104 N v v1 d) a 2 t v v1 t 2 a 0 360 m/s t 1.3 106 m/s2 t 2.8 104 s e) J p J m2v2 m1v1 J (0.03 kg)(0) (0.03 kg)(360 m/s) J 11 kg·m/s f) – 4
p 1
30. v1 0, v2 250 m/s, m 3.0 kg, F 2.0 104 N a) J ∆p J m2v2 m1v1 J (3.0 kg)(250 m/s) 0 J 750 kg·m/s b) J F t J t F 750 kg·m/s t 2 104 N t 0.038 s 31. m 7000 kg, v1 110 km/h 30.56 m/s, ∆t 0.40 s, v2 0 p a) F t m v m1v1 F 2 2 t 0 (7000 kg)(30.56 m/s) F 0.40 s F 5.3 105 N p b) F t m v m1v1 F 2 2 t 0 (7000 kg)(30.56 m/s) F 8.0 s 4 F 2.7 10 N 32. m 30 g 0.03 kg, v1 360 m/s, ∆d 5 cm 0.05 m a) p mv p (0.03 kg)(360 m/s) p 11 kg·m/s b) v22 v12 2a∆d 02 (360 m/s)2 2a(0.05 m) a 1.3 106 m/s2
t (× 10
) 0 N – 1
s)
2
3
4
0 1 – 2 × ( – 3 F
1
– 4
33. a)
8 7
) 6 N
6
0 5 1 × 4 ( F 3
2 1
0 5 10 15 t (s)
1 b) Area h(a b) 2 1 J (15 s)(5 106 N 8 106 N) 2 J 9.8 107 N·s 34. J area under the curve 1 J (90 N)(0.3 s) (120 N)(0.2 s) 2 1 (75 N)(0.4 s) 2 J (13.5 N·s) (24 N·s) (15 N·s) J 25.5 N·s 35. J area under the graph Counting roughly 56 squares, J 56(0.5 103 N)(0.05 s) J 1.4 103 N·s
Solutions to End-of-chapter Problems
121
J ∆p J mv2 mv1, where v1 0, 3 1.4 10 N·s (0.250 kg)(v2) v2 5.6 103 m/s pTo pTf 37. m1 v1o m2 v2o (m1 m2) vf , where v2o 0 (5000 kg)(5 m/s [S]) (10 000 kg)( vf ) v 2.5 m/s [S] f pTo pTf 38. m1 v1o m2 v2o (m1 m2) vf , where v2o 0 (45 kg)(5 m/s) (47 kg)( vf ) v 4.8 m/s [in the same f direction as v1o] pTo pTf 39. m1 v1o m2 v2o m1 v1f m2 v2f (65 kg)(15 m/s) (100 kg)(5 m/s)
36.
1 (65 kg) (15 m/s) (100 kg)(v2f ) 3 (975 500 325) kg·m/s (100 kg)(v2f ) v2f 1.5 m/s pTo pTf
40.
m1 v1o m2 v2o (m1 m2) vf , where v2o 0 (0.5 kg)(20 m/s) 0 (30.5 kg)(vf ) vf 0.33 m/s pTo pTf 41. m1 v1o m2 v2o m1 v1f m2 v2f (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/s m1(25 m/s 22.2 m/s) 133 333.3 kg·m/s m1
133 333.3 kg·m/s 2.8 m/s 4
m1 4.8 10 kg
122
43.
F 1 F 2 ma1 ma2 m
v v v v m t t 1f
1o
2f
2o
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 1027 kg, m2 4m1, v1 2.2 107 m/s pTo pTf m1v1o m2v2o (m1 m2)vf , where v2o 0 7 m1(2.2 10 m/s) (5m1)vf 2.2 107 m/s vf
5
vf 4.4 106 m/s 45. m1 3m, m2 4m, v1o v pTo pTf m1v1o m2v2o (m1 m2)vf , where v2o 0 (3m)v (7m)vf
3 7
vf v
46. m1 99.5 kg, m2 0.5 kg, v1f ?, v2f 20 m/s pTo pTf 0 (99.5 kg)(v1f ) (0.5 kg)(20 m/s) 10 kg·m/s v1f 99.5 kg v1f 0.1 m/s d t v 200 m t 0.1 m/s t 2 103 s 47. p1o 375 kg·m/s [E], p2o 450 kg·m/s [N45°E] a) pTo p1o p2o
p To
θ p 1o = 375 kg·m /s
Solutions to End-of-chapter Problems
p 2o = 450 kg·m /s 45°
b) pTf pTo Using the cosine and sine laws, pTo2 (375 kg·m/s)2 (450 kg·m/s)2 2(375 kg·m/s)(450 kg·m/s) cos 135° pTo 762.7 kg·m/s pTf 762.7 kg·m/s sin sin 135° 450 kg·m/s 762.7 kg·m/s 24.7° Therefore, pTf 763 kg·m/s [E24.7°N] 48. m1 3.2 kg, v1o 20 m/s [N], p1o 64 kg·m/s [N], m2 0.5 kg, v 5 m/s [W], p 2.5 kg·m/s [W] 2o 2o pTo pTf p1o p2o pTf m1 v1o m2 v2o (m1 m2) vf Using the diagram and Pythagoras’ theorem,
v2o
p 2o = 2.5 kg·m /s
103 923 kg·m/s 5000 kg
v
20.8 m/s [E] 50. mo 1.2 1024 kg, vo 0, po 0, m1 3.0 1025 kg, v1 2.0 107 m/s [E], p 6 1018 kg·m/s [E], 1 m2 2.3 1025 kg, v2 4.2 107 m/s [N], p2 9.66 1018 kg·m/s [N] m3 1.2 1024 kg 3.0 1025 kg 2.3 1025 kg m3 6.7 1025 kg pTo 0 pTo pTf 0 p1 p2 p3
2o
Drawing a momentum vector diagram and using Pythagoras’ theorem,
p Tf
θ
2o
p2o (60 000 kg·m/s)(tan 60°) p2o 103 923 kg·m/s m2v2o 103 923 kg·m/s
p 60 000 kg·m/s
tan 60°
p 1o = 64 kg·m /s
θ p 3 = m 3v 3
p 2o = 9.66 × 10 – 18 kg·m /s p 1o = 6.0 × 10 – 18 kg·m /s
pTf (2.5 kg ·m/s)2 (6 4 kg· m/s)2 pTf 64.05 kg·m/s
tan
2.5 kg·m/s 64 kg·m/s
m3
(m1 m2) vf 64.05 kg·m/s [N2.2°W] (3.7 kg) vf v 17 m/s [N2.2°W] f 49. m1 3000 kg, v1o 20 m/s [N], p1o 60 000 kg·m/s [N], m2 5000 kg, v ? [E], p ? [E], v ? [E30°N], 2o 2o f pf ? [E30°N] pTo pTf p1o p2o pTf
(6 1018 kg·m/s)2 p3 1.1372 1017 kg·m/s
p3 v3 pT
2.2°
p32 (9.66 1018 kg·m/s)2
Using the following momentum diagram,
1.1372 1017 kg·m/s 3 6.7 1025 kg v3 1.7 107 m/s 6 1018 kg·m/s tan 9.66 1018 kg·m/s 31.8° Therefore, v3 1.7 107 m/s [S32°W] 60 m 12.5 m/s 51. m1 m2 m, v1o 4.8 s [ R], v2o 0, v2f 1.5 m/s [R25°U] v
30°
m1 v1o m2 v2o m1 v1f m2 v2f Since m1 m2 and v2o 0, v v v 1o 1f 2f
p Tf
pTo pTf
p 2o = m 2 v 2o p 1o = 60 000 kg·m /s
Solutions to End-of-chapter Problems
123
Drawing a vector diagram and using trigonometry, v 1f
v 2f = 1.5 m/s 25°
Using the vector diagram and trigonometry, v 1f
θ
θ v 1o = 6.0 m / s
v 1o = 12.5 m/s
v1f 2 (1.5 m/s)2 (12.5 m/s)2
2(1.5 m/s)(12.5 m/s) cos 25° v1f 11.16 m/s sin sin 25° 1.5 m/s 11.6 m/s 3.3° Therefore, the first stone is deflected 3.3° or [R3.3°D]. 52. m1 10 000 kg, v 3000 km/h [E] 833.3 m/s [E], 1 p1 8.333 106 kg·m/s [E], m2 ?, v 5000 km/h [S] 1388.9 m/s [S], 2 p2 m2(1388.9 m/s) [S], m3 10 000 kg m2, v3 ? [E10°N], p3 (10 000 kg m2)(v3) [E10°N]
v 2f = 4.0 m / s 25°
v1f 2 (6.0 m/s)2 (4.0 m/s)2
2(6.0 m/s)(4.0 m/s) cos 65° v1f 5.63 m/s sin sin 65° 4.0 m/s 5.63 m/s 40° Therefore v1f 5.63 m/s [U40°R] 54. 2m1 m2, v1o 6.0 m/s [U], v2o 0, v 4 m/s [L25°U], v ? 2f 1f
pTo pTf
p1
p2 p3
m1 v1o m2 v2o m1v1f m2v2f , since 2m1 m2 and v2o 0 v v 2 v 1o 1f 2f
Using the vector diagram and trigonometry,
Drawing a momentum diagram and using trigonometry,
v 1f
θ V 1o = 6.0 m /s
p 1 = 8.33 × 106 kg·m /s v 2f = 8.0 m /s
p 2
p 3
10°
25°
v1f
p2
tan 10° p1
p2
6
(8.33 10 kg·m/s)(tan 10°) p2 1.47 106 kg·m/s p m (1388.9 m/s) [S] 2 2 1.47 106 kg·m/s m2
1388.8 m/s
The mass of the ejected object is 1.058 103 kg. 53. m1 m2 m, v1o 6.0 m/s [U], v2o 0, v 4 m/s [L25°U], v ? 2f 1f
pTo pTf
m1 v1o m2 v2o m1 v1f m2 v2f Since m1 m2 and v2o 0, v v v 1o 1f 2f
2
(6.0 m/s)2 (8.0 m/s)2
2(6.0 m/s)(8.0 m/s) cos 65° v1f 7.7 m/s sin sin 65° 8.0 m/s 7.7 m/s 70° Therefore, v1f 7.7 m/s [R20°U] 55. Counting ten dots for a one-second interval and measuring the distance with a ruler and the angle with a protractor gives: 33 mm a) v1o 1s v1o 0.033 m/s v2o 0 v1f 33 mm 1s v1f 0.033 m/s v2f 33 mm 1s v2f 0.033 m/s
m2 1057.6 kg
124
Solutions to End-of-chapter Problems
b) v1o 0.033 m/s [E] v 0 2o v 0.033 m/s [E45°S] 1f v 0.033 m/s [E45°N] 2f c) p1o (0.3 kg)(0.033 m/s [E]) p1o 9.9 103 kg·m/s [E] p2o (0.3 kg)(0) 0 pTo 9.9 103 kg·m/s [E] p1f (0.3 kg)(0.033 m/s [E45°S]) p1f 9.9 103 kg·m/s [E45°S] p (0.3 kg)(0.033 m/s [E45°N]) 2f p2f 9.9 103 kg·m/s [E45°N] The vector diagram for the final situation is shown below.
Drawing a vector diagram and using trigonometry, B p 3 = 7.5 kg·m /s
A
p 4
p 2 = 5.4 kg·m /s
θ
C p = 4.80 kg·m /s 1
45°
p 1f = 9.9 × 10 – 3 kg·m /s
Using triangle ABC, 2.7 kg·m/s tan 5.4 kg·m/s 26.6° p42 (2.7 kg·m/s)2 (5.4 kg·m/s)2 p4 6.037 kg·m/s v4 6.037 kg·m/s 0.25 kg v4 24.1 m/s Therefore, v4 24.1 m/s [S26.6°E] 57. a) Masstotal 5000 kg 10 000 kg Masstotal 15 000 kg 5000 kg 1 b) the distance to the larger 15 000 kg 3 1 truck, or (400 m) 133.3 m from the 3 larger truck. 58. a) m1 2000 kg, v1o 200 m/s [E], p1o 4 105 kg·m/s [E]
p 2f = 9.9 × 10 – 3 kg·m /s
45°
p Tf
Using Pythagoras’ theorem, pTf 2 (9.9 103 kg·m/s)2 (9.9 103 kg·m/s)2 pTf 1.4 102 kg·m/s [E] d) p1oh 9.9 103 kg·m/s p1ov 0 p2oh 0 p2ov 0 p1fh (9.9 103 kg·m/s)(cos 45°) p1fh 7.0 103 kg·m/s p1fv (9.9 103 kg·m/s)(sin 45°) p1fv 7.0 103 kg·m/s p2fh (9.9 103 kg·m/s)(cos 45°) p2fh 7.0 103 kg·m/s p2fv (9.9 103 kg·m/s)(sin 45°) p2fv 7.0 103 kg·m/s e) Momentum is not conserved in this collision. The total final momentum is about 1.4 times the initial momentum. 56. m1 0.2 kg, v1 24 m/s [E], p1 4.8 kg·m/s [E], m2 0.3 kg, v 18 m/s [N], p 5.4 kg·m/s [N], 2 2 m3 0.25 kg, v3 30 m/s [W], p3 7.5 kg·m/s [W], m4 0.25 kg, v ?, p ? 4 4 pTo 0 p1 p2 p3 p4 0
p 1o = 4 × 105 kg·m/s [E]
b) m2 1000 kg, v2o 200 m/s [S30°E], p2o 2 105 kg·m/s [S30°E]
p 2o = 2 × 105 kg·m/s [S30°E] 30°
c) pcmo p1o p2o
p 2o p cmo
d) pcmf pcmo
p cmf
p 1o
59. a) pcmo pTo pcmo 9.9 103 kg·m/s [E] (see 55c) b) pcmf pTf pcmf 1.4 102 kg·m/s [E] (see vector diagram for 55c)
Solutions to End-of-chapter Problems
125
The force that the plow applies in 1 s is:
Chapter 5 11. 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
1 2 1 W (9.1 1031 kg)(1.6 108 m/s)2 2 W 1.2 1014 J W mv2 0
12. 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.
F app F g F app mg F app (3556 kg)(9.8 N/kg) F app 34 848.8 N
This force is applied over a distance of 5 m: W F d W (34 848.8 N)(5.0 m) W 174 244 J
To find the number of seconds it takes to plow the road: d t v t
8000 m 10 m/s
t 800 s W T (800 s)(174 244 J/s) W T 1.4 108 J 15. The two campers must overcome 84 N of fric-
tion, or 42 N each in the horizontal direction since both are at the same 45° angle. 45°
F c
25.0 m h
θ 45°
W Eg F d mg h (350 N)(25.0 m) (50.0 kg)(9.8 N/kg)h h 18 m h sin
25.0 m 18 m sin 25.0 m 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/m 3) 3556 kg
126
F h
The horizontal component of F c, called F h, must be equal to 42 N. W F h d W (42 N)(50 m) W 2.1 103 N·m
Each camper must do 2.1 103 J of work to overcome friction. 16. W F d ,where d for each revolution is zero. Therefore, W 0 J. 17. 350 J indicates that the force and the displacement are in the opposite direction. An example would be a car slowing down because of friction. Negative work represents a flow or transfer of energy out of an object or system.
Solutions to End-of-chapter Problems
18. dramp 5 m m 35 kg d height 1.7 m a) F ma F (35 kg)(9.8 N/kg) F 343 N F 3.4 102 N b) W F d W (3.4 102 N)(1.7 m) W 583.1 J W 5.8 102 N c) W F dramp 583.1 J F (5 m) F 116.62 N F 1.2 102 N 19. W Area under the graph W
b) v v
1 2 1 850 J (120 kg)v2 2 v 3.8 m/s 1 21. a) Ek mv2 2 1 Ek (45 kg)(10 m/s)2 2 Ek 2.3 103 J Ek mv2
1 2 1 Ek (0.002 kg)(0.628 m/s)2 2 Ek 3.9 104 J 100 km 1h 1000 m c) v 1h 3600 s 1 km v 27.7778 m/s 1 Ek mv2 2 1 Ek (15 000 kg)(27.7778 m/s)2 2 Ek 5.8 106 J Ek mv2
(20 m)(200 N) (10 m)(200 N) 2 (20 m)(600 N) (20 m)(200 N) 2 (20 m)(400 N) (10 m)(800 N) 2
W Ek
2(0.1 m) 1s
v 0.628 m/s
(20 m)(800 N) (10 m)(1200 N) W 5.4 104 J 20. a) W area under the graph (1 m)(100 N) (1 m)(200 N) W 2 2 (1 m)(100 N) (2 m)(300 N) W 8.5 102 J b) The wagon now has kinetic energy (and may also have gained gravitational potential energy). c)
2r t
22. v v
d t
5.0 m 2.0 s
v 2.5 m/s
1 2 1 450 J m(2.5 m/s)2 2 m 1.4 102 kg 1 23. Ek mv2 2 1 5.5 108 J (1.2 kg)v2 2 Ek mv2
v
2(5.5 10 J) 1.2 kg 8
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 2 2 2 kg·m/s kg m /s kg·m/s kg·m/s
Solutions to End-of-chapter Problems
127
1000 eV 1.6 1019 J 26. 5 keV 1 keV 1 eV 8 1016 J 1 Ek mv2 2 1 8 1016 J (9.1 1031 kg)v2 2 v 4.2 107 m/s As a percentage of the speed of light: 4.2 107 m/s 100 14% 3 108 m/s (v 2 v12) 27. a) a 2 2d 0 (350 m/s)2 a 2(0.0033 m) a 1.86 107 m/s2
F ma F (0.015 kg)(1.86 107 m/s2) F 2.8 105 N b) F force of bullet F 2.8 105 N 28. For 1 m: W (50 N)(1 m) W 50 J W Ek
1 2 1 50 J (1.5 kg)v2 2 v 8 m/s For 2 m: Ek mv2
1 2
W 50 J (50 N)(1 m) (250 N)(1 m) W 225 J
1 2 1 225 J (1.5 kg)v2 2 v 17.3 m/s For 3 m: 1 W 225 J (50 N) 2 1 (300 N) m 6 W 425 J Ek mv2
128
1 6 1 5 (350 N) m 2 6 m
1 2 1 425 J (1.5 kg)v2 2 v 23.8 m/s 29. p 2mEk p 2(5 kg)(3.0 102 J) p 55 N·s 30. m1 0.2 kg m2 1 kg v1o 125 m/s v1f 100 m/s v2o 0 v2f ? d2 3 m Ek mv2
a)
pTo pTf m1v1o m2v2o m1v1f m2v2f (0.2 kg)(125 m/s) 0 (0.2 kg)(100 m/s) (1 kg)v2f v2f 5 m/s
1 2 1 Ek (1 kg)(5 m/s)2 2 Ek 12.5 J c) This collision is not elastic since some kinetic energy is not conserved. Some energy may be lost due to the deformation of the apple. d) v22 v12 2ad 0 (5 m/s)2 2a(3.0 m) a 4.1667 m/s2
b) Ek mv2
F ma F (1.0 kg)(4.1667 m/s2) F 4.2 N 31. a) Eg mg h Eg (2.0 kg)(9.8 N/kg)(1.3 m) Eg 25 J b) Eg mg h Eg (0.05 kg)(9.8 N/kg)(3.0 m) Eg 1.5 J c) Eg mg h Eg (200 kg)(9.8 N/kg)(469 m) Eg 9.2 105 J d) Eg mg h Eg (5000 kg)(9.8 N/kg)(0) Eg 0 J
Solutions to End-of-chapter Problems
36. a) The greatest potential energy is at point A
F 32. a) m a m
(highest point) and point F represents the lowest amount of potential energy (lowest point). b) Maximum speed occurs at F when most of the potential energy has been converted to kinetic energy.
4410 N 9.8 N/kg
m 4.5 102 kg b) W F d W (4410 N)(3.5 m) W 1.5 104 J
Eg-lost Ek-gained
1 2 1 (1000 kg)(9.8 N/kg)(75 m) (1000 kg)v2 2 v 38 m/s c) At point E, 18 m of Ep is converted to Ek. 1 mg h mv2 2 1 (1000 kg)(9.8 N/kg)(18 m) (1000 kg)v2 2 v 19 m/s d) Find the acceleration, then use F ma. v22 v12 2ad 0 (38 m/s)2 2a(5 m) a 144.4 m/s2
33. Using conservation of energy: ETo ETf
mg h mv2
1 2
1 2 1 1 (9.8 m/s2)(1.8 m) (4.7 m/s)2 v2 2 2 1 17.64 m2 s2 11.045 m2 s2 v2 2 v 7.6 m/s mgh mvo2 mvf 2
Ee Eg
34.
1 2
2
kx
1 2
(1200
mg h
N/m) x2 (3.0 kg)(9.8 N/kg)(0.80 m) x 0.2 m x 20 cm
35. m 0.005 kg h 2.0 m
Initial:
37.
E mg h E (0.005 kg)(9.8 N/kg)(2.0 m) E 0.098 J
F ma F (1000 kg)(144.4 m/s2) F 1.4 105 N Ee Ek
1 2 1 mv2 2 2 (890 N/m) x2 (10 005 kg)(5 m/s)2 x 16.8 m x 17 m 2 v1 sin 2 kx
At half the height: E (0.005 kg)(9.8 N/kg)(1.0 m) E 0.049 J After the first bounce: E (0.80)(0.098 J) E 0.0784 J After the second bounce: E (0.80)(0.0784 J) E 0.062 72 J After the third bounce: E (0.80)(0.062 72 J) E 0.050 176 J After the fourth bounce: E (0.80)(0.050 176 J) E 0.040 140 9 J Therefore, after the fourth bounce, the ball loses over half of its original height.
38.
d h
15 m
g 2 1
v sin 90° 9.8 m/s 2
v1 12.1 m/s Ek Ee
1 2 1 2 kx 2 2 (0.008 kg)(12.1 m/s) (350 N/m) x2 x 0.058 m x 5.8 cm mv
Solutions to End-of-chapter Problems
129
39. 85% of the original energy is left after the
first bounce, therefore, (0.85)mg htree mg h bounce (0.85)(2 m) h bounce h 1.7 m
1 2 1 2 kx 2 2 (3 kg)v2 (125 N/m)(0.12 m)2 v 0.77 m/s mv
Ee Ek
40.
b) F f F n F f (0.1)(3 kg)(9.8 N/kg) F f 2.94 N F ma F a m 2.94 N a
1 2 1 2 mv 2 2 2 (35 000 N/m)(4.5 m) (65 kg)v2 v 104.4 m/s 2 v1 sin 2 kx
d h
g
(104.4 m/s)2 sin 90° d h 9.8 m/s2 d h 1.1 103 m 41. k slope rise k run
3 kg a 0.98 m/s2 v22 v12 2ad 0 (0.77 m/s)2 2(0.98 m/s2)d d 0.3 m d 30 cm
F k x
Ek Ee
1 2 1 2 kx 2 2 2 (0.05 kg)v (400 N/m)(0.03 m)2 v 2.7 m/s mv
44.
Ek Ee
1 2 1 2 kx 2 2 (2.5 103 kg)(95 m/s)2 k(35 m)2 k 1.8 104 N/m mv
45.
Ee Ek
1 1 2 2 7 2 (5 10 N/m)(0.15 m) (1000 kg)v2 v 34 m/s kx2 mv2
130
Ek Ee
47.
120 N k 0.225 m k 5.3 102 N/m 42. W area under the graph 1 a) W (0.05 m)(2 103 N) 2 W 50 J b) W 50 J (0.02 m)(2 103 N) 1 (0.02 m)(4.5 103 N) 2 W 135 J W 1.4 102 J 43.
Ek Ee
46. a)
1 1 2 2 2 (3.0 kg)v (350 N/m)(0.1 m)2 v 1.1 m/s mv2 kx2
48. F kx F (4000 N/m)(0.15 m) F 600 N 49. F ma F (100 kg)(9.8 N/kg) F 980 N
Divided into 20 springs: 980 N F 20 F 49 N per spring
F kx 49 N k(0.035 m) k 1.4 103 N/m 50. F kx mg kx (10 kg)(9.8 N/kg) k(1.3 m) k 75.3846 N/m
1 2 1 2 106 J (75.3846 N/m) x2 2 x 2.3 102 m
Solutions to End-of-chapter Problems
Ee kx2
51. 3 d
24 h 60 min 60 s 259 200 s 1d 1h 1 min
60 min 60 s 8h 28 800 s 1h 1 min 60 s 900 s 15 min 1 min t 259 200 s 28 800 s 900 s t 288 900 s
In 1 s: 255.78 J P 1.0 s P 256 W 56. Using the conservation of momentum:
m1v1o m2v2o m1v1f m2v2f m1v1o m1v1f m2v2f m1(v1o v1f ) m2v2f
(eq. 1) Using the conservation of kinetic energy: m1(v1o2 v1f 2) m2v2f 2 (eq. 2) Dividing equation 2 by equation 1: m1(v1o2 v1f 2) mv 2 2 2f m1(v1o v1f ) m2v2f
E P t E P t E (60 W)(288 900 s) E 1.7 107 J
1.7 104 kJ
1 kWh 4.8 kWh 3600 kJ
v1o v1f v2f v1f v2f v1o
52. a) E E E E
Eg mgh (3500 kg)(9.8 m/s2)(13.4 m) 459 620 J E P t
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) 2m1v1o v2f
459 620 J P 23 s P 19 983 W 19 983 W P E 0.46 P E 4.3 104 W 1 hp b) 4.3 104 W 58 hp 746 W
57. a) v1f v1o
m1 m2
m1 m2 m1 m2
v1f (3 m/s)
15 kg 3 kg 15 kg 3 kg
v1f 2 m/s v2f v1o
54. a) P Fv P F gv P mgv P (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, the
cyclist 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 m The increase in potenial energy is: Ep mgh Ep (75 kg)(9.8 m/s2)(0.348 m) Ep 255.78 J
(eq. 3)
2 m1 m1 m2
v2f (3 m/s)
2(15 kg) 15 kg 3 kg
v2f 5 m/s
1 2 1 Ek (3 kg)(5 m/s)2 2 Ek 37.5 J Ek 38 J
b) Ek mv2
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
Solutions to End-of-chapter Problems
131
59. a) pTo m1v1om2v2o pTo (3.2 kg)(2.2 m/s) (3.2 kg)(0) pTo 7.0 kg·m/s
1 Ek-To mv2 2 1 Ek-To (3.2 kg)(2.2 m/s)2 2 Ek-To 7.7 J b) 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
1 1 2 2 1 Ek-Tf (3.2 kg)(1.1 m/s)2 2 1 2 (3.2 kg)(1.1 m/s) 2 Ek-Tf 3.8 J d) The collision is not elastic since there was a loss of kinetic energy from 7.7 J to 3.8 J. c) Ek-Tf mv1f 2 mv2f 2
pTo pTf m1v1o m2v2o m1v1f m2v2f (0.015 kg)(375 m/s) 0 (0.015 kg)(300 m/s) (2.5 kg)v2f v2f 0.45 m/s 61. m1 6m v1o 5 m/s m2 10m v2o 3 m/s Changing the frame of reference so that v2f 0: v1o 8 m/s m m2 v1f v1o 1 m1 m2 60.
v1f (8 m/s)
6m 10m 6m 10m
v1f 2 m/s 2m1 v2f v1o m1 m2
v2f (8 m/s)
2(6m) 6m 10m
v2f 6 m/s
Returning 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 132
62. a) v1f v1o
m1 m2 m1 m2
v1f (5 m/s)
3m m 3m m 2
2
2
2
v1f 2.5 m/s 2m1 b) v2f v1o m1 m2
v2f (5 m/s)
2(3m ) 3m m 2
2
v2f 7.5 m/s 63. mw 0.750 kg k 300 N/m m b 0.03 kg x 0.102 m a)
2
Ee-gained Ek-lost
1 1 2 2 (300 N/m)(0.102 m)2 (0.78 kg)v2 v 2 m/s Using the conservation of momentum: kx2 mv2
pTo pTf m bv bo mwvwo m(bw)vf (0.03 kg)v bo 0 (0.78 kg)(2.0 m/s) v bo 52 m/s b) The collision is inelastic since:
1 2 Eko 40.56 J and Ekf 0 The kinetic energy is not conserved. 1 64. a) mgh mv2 2 1 (2.05 kg)(9.8 m/s)(0.15 m) (2.05 kg)v2 2 v 1.7 m/s b) 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 : Eko (0.03 kg)(52 m/s)2
pTo pTf mv1o mv2o mv1f mv2f v1o 0 v1f v1f (eq. 1)
Solutions to End-of-chapter Problems
Using the conservation of kinetic energy: EkTo EkTf
1 2
1 2
1 2
1 2
mv1o2 mv2o2 mv1f 2 mv2f 2
v1o2 0 v1f 2 v2f 2 v1o2 v1f 2 v2f 2
(eq. 2) Equation 1 can be represented by the vector diagram: v 1o
v 1f
θ
v 2f
The angle is the angle between the final velocity of the eight ball and the cue ball after the collision. Using the cosine law and equation 2: v1o2 v1f 2 v2f 2 2(v1f )(v2f ) cos v1o2 v1o2 2(v1f )(v2f ) cos 0 2(v1f )(v2f ) cos 0 cos 90° Therefore, the angle between the two balls after collision is 90°.
Solutions to End-of-chapter Problems
133
1 2 1 E (920 kg)(1.12 104 m/s)2 2 E 5.75 1010 J c) The initial speed needed to keep going indefinitely should be greater than the escape speed, i.e., greater than 11.2 km/s. 15. ms 550 kg, mE 5.98 1024 kg, h 6000 km 6 106 m, r E 6.38 106 m E mv2
Chapter 6 13. vi 4 km/s 4 103 m/s, vf 80 m/s
1 1 2 2 1 E (100 000 kg)[(80 m/s)2 (4000 m/s)2] 2 E 7.9968 1011 J It has released 7.9968 1011 J of energy to the atmosphere. The shuttle’s initial height was 100 km, and it landed on Earth’s surface, therefore its change in height is 100 km. 14. mE 5.98 1024 kg, msat 920 kg, Ek 7.0 109 J a) At the start, the height is r E 6.38 106 m. Therefore, the total energy is E mvi2 mvf 2
1 1 r h r
Ep GMm
Ep (6.67 1011 N·m2/kg2) (5.98 1024 kg)(550 kg) 1 6 6.38 10 m 6 106 m
1 6.38 10 m
ET Eki Epi ET 7.0 109 J
GMm r
6 Ep 1.67 1010 J
9
ET 7.0 10 J 11
2
2
24
(6.67 10 N·m /kg )(5.98 10 kg)(920 kg) 6.38 10 m
b) At the maximum height of 6000 km, the
6
ET 5.051 672 1010 J
Since velocity is 0 at maximum height, total energy at maximum height Epf ET
GMm r h
11
2
2
24
ET
(6.67 10 N·m /kg )(5.98 10 kg)(920 kg) 6.38 10 m h
ET
3.67 10 N·m 6.38 10 m h
6
17
2
6
The total energy is constant, therefore, 3.67 1017 N·m2 5.051 672 1010 J 6.38 106 m h h 884.1 km b) Escape velocity from Earth’s surface is given by: v 2GM
esc
vesc
kinetic energy is 0 since the velocity is zero. Therefore, the change in Ep is the initial kinetic energy, i.e., Eki 1.67 1010 J. 16. mE 5.98 1024 kg, r E 6.38 106 m, mm 20 000 kg, vi 3.0 km/s 3.0 103 m/s, h 200 km 2.0 105 m Since the meteorite is headed from outer space, 1 Epi 0 and Eki mv2i 2 1 Therefore, ET mv2i 2 At 200 km, ET Ekf Epf
1 2
r
2(6.67 1011 N·m2/kg2)(5.98 1024 kg) (6.38 106 m)
vesc 1.12 104 m/s
GMm 1 h r 2 GM 1 2 1 2 vi vf h r 2 2 2
mvi
mvf 2
1 2
(3.0
Therefore, the initial kinetic energy required for the escape should be greater than:
GMm GMm r r h
a) Ep
1 2
103 m/s)2 vf 2 11
2
2
24
(6.67 10 N·m /kg )(5.98 10 kg) 6.58 10 m 6
vf 11 412.1 m/s
The meteorite’s speed 200 km above Earth’s surface is approximately 11.4 km/s. 134
Solutions to End-of-chapter Problems
17. vesc c 2.99 108 m/s, mE 5.98 1024 kg
2
GM r 2GM r (vesc)2 vesc
Orbital speed is given by:
11
2
2
24
2(6.67 10 N·m /kg )(5.98 10 kg) (2.99 10 m/s) 8
GM r h
v 7.67 km/s
Equating the forces of gravity between Earth and the Moon, using the distance from Earth as r , GM m GM m r (3.82 10 m r ) M M r (3.82 10 m r ) Moon 8
2
Moon 8
2
Earth 2
Earth 2
2
8
2
M Moonr M Earth(3.82 10 m r ) 0 M Earth(1.46 1017 m 7.64 108r r 2) M Moonr 2 0 8.73 1041 m 4.57 1033r 5.98 1024r 2 7.36 1022r 2 0 5.91 1024r 2 4.57 1033r 8.73 1041 m r 4.29 108 m, 3.45 108 m
The forces of gravity from Earth and the Moon 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, r E 6.38 106 m, r M 1.738 106 m Let m be the mass of the payload. GM m GM m r R Moon
2
The period of the orbit is the time required by the satellite to complete one rotation around Earth. Therefore, the distance travelled, d, is the 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 m Hence, speed is given by, d v T d T v T
42 599 996 m 7670 m/s
T 5552 s
The period of the orbit is 5552 s or 92.5 min. 21. mE 5.98 1024 kg, r E 6.37 106 m Since the orbit is geostationary, it has a period of 24 h 86 400 s. Using Kepler’s third law, r 3 GM 2 2 T 4 r
GMT 4
r
2
1 3
2
11
2
2
24
2
(6.67 10 N·m /kg )(5.98 10 kg)(86 400 s) 4(3.14)
2
1 3
r 4.22 107 m
Earth
11
(6.67 1011 N·m2/kg2)(5.98 1024 kg) 6.38 106 m 4.0 105 m
v
2
r 8.92 103 m 18. Given: dEM 3.82 108 m, mMoon 7.35 1022 kg, mEarth 5.98 1024 kg
E
v
r
20. mE 5.98 1024 kg, r E 6.38 106 m, h 400 km 4.0 105 m
2
E (6.67 10 N·m /kg ) 5.98 1024 kg 7.35 1022 kg 1.738 106 m 6.38 106 m E 5.97 107 J
The energy required to move a payload from Earth’s surface to the Moon’s surface is 5.97 107 J/kg.
Subtracting Earth’s radius, r 4.22 107 m 6.37 106 m r 3.59 107 m The satellite has an altitude of 3.59 104 km. 22. mE 5.98 1024 kg, r E 6.37 106 m, r 1 320 km 3.2 105 m, r 2 350 km 3.5 105 m
Solutions to End-of-chapter Problems
135
Energy added to the station’s orbit is given by: E
GMm GMm r r r r 2
E
E GMm
1
Equating two equations for kinetic energy, GMm 1 mv2 2 2r
E
1
1
r 2 r E r 1 r E E (6.67 1011 N·m2/kg2) (5.98 1024 kg)m
6.73 10 m 1
6 E 2.65 105m J
6.70 10 m 1
r 3 b) In Kepler’s third law equation 2 K , T 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 m
The Moon’s total energy in its orbit around Earth is given by: 1 ET Ep 2 1 GMm ET 2 r 1 (6.67 10 N·m /kg )(5.98 10 kg)(7.35 10 kg) ET 3.82 10 m 2 28 ET 3.84 10 J
24
v
(6.67 10 N·m /kg )(5.7 10 kg) 6 10 m
GM r
11
2
26
22
If an object is orbiting Saturn, it must have a minimum speed of 2.5 104 m/s. 26. mM 7.35 1022 kg, r r M 100 km r 1.738 106 m 1 105 m r 1.838 106 m vesc vesc
2 GmMoon r
2(6.67 1011 N·m2/kg2)(7.35 1022 kg) 1.838 106 m
vesc 2.31 103 m/s
The escape speed from the Moon at a height of 100 km is 2.31 km/s. 27. According to Kepler’s third law, r 3 GM 2 2 T 4 42r 3 T 2 GmMoon
T
4(3.14) (1.838 10 m) (6.67 10 N·m /kg )(7.35 10 kg) 2
11
2
2
6
3
22
T 7071 s
It would take the Apollo spacecraft 7071 s or 1 h 58 min to complete one orbit around the Moon. 28. dMS 2.28 1011 m, r M 3.43 106 m, mM 6.37 1023 kg, mS 2.0 1030 kg a) Orbital speed is given by: v
v
(6.67 10 N·m /kg )(2.0 10 kg) 2.28 10 m
8
GM r
11
2
v 24.2 km/s
136
2
7
v 2.5 10 m/s
increases as it becomes less negative. Thus, when potential energy increases, kinetic energy decreases to maintain the total 1 energy a constant. Since Ek mv2, if 2 kinetic energy decreases, v also decreases and when r increases, v decreases.
2
4
G Mm Ep . As r increases, the energy r
2
v
6
The shuttle has added 2.65 105m J of energy to the station’s orbit. 23. a) The total energy of a satellite in an orbit is the sum of its kinetic and potential energies. In all cases, total energy remains constant. Therefore, when r is increased, the gravitational potential energy increases as
11
25. mSaturn 5.7 1026 kg, r Saturn 6.0 107 m
Solutions to End-of-chapter Problems
2
11
30
b) h 80 km 8 104 m
32. m 2.0 kg, x 0.3 m, k 65 N/m
v
Gm r h
v
(6.67 10 N·m /kg )(6.37 10 kg) 3.43 10 m 8 10 m
11
2
6
2
23
4
v 3.48 km/s
The speed required to orbit Mars at an altitude of 80 km is 3.48 km/s. 29. mM 7.35 1022 kg, r M 1.738 106 m Escape speed is given by: vesc vesc
2 GM r
2(6.67 1011 N·m2/kg2)(7.35 1022 kg) 1.738 106 m
vesc 2.38 km/s 30. Three waves pass in every 12 s, with 2.4 m
between wave crests. number of waves f time 3 f 12 s f 0.25 Hz 31. k 12 N/m, m 230 g 0.23 kg, A 26 cm 0.26 m At the maximum distance, i.e., A, v 0, therefore the total energy is: 1 E kA2 2 Also, at the equilibrium point, the displacement is zero, therefore the total energy is the kinetic energy: 1 E mv2 2 Hence, 1 2 1 2 kA mv 2 2
v v
1 2 1 E (65 N/m)(0.3 m)2 2 E 2.925 J Initial potential energy of the spring is 2.925 J. b) Maximum speed is achieved when the total energy is equal to kinetic energy only. Therefore, 1 E mv2 2 1 2.925 J (2.0 kg)v2 2 v 2.925 v 1.71 m/s The mass reaches a maximum speed of 1.71 m/s. c) x 0.20 m Total energy of the mass at this location is given by: 1 1 E mv2 kx2 2 2 1 2.925 J (2.0 kg)v2 2 1 2 (65 N/m)(0.2 m) 2 v 1.625 v 1.275 m/s The speed of the mass when the displacement 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 since F kx and F ma Therefore, maximum displacement is x 0.30 m Hence, a) E kx2
(12 N/m)(0.26 m) 0.23 m kA2 m
2
v 1.88 m/s
The speed of the mass at the equilibrium point is 1.88 m/s.
ma kx kx a m a
(65 N/m)(0.30 m) (2.0 kg)
a 9.75 m/s2
The mass’ maximum acceleration is 9.75 m/s2. Solutions to End-of-chapter Problems
137
b) x 0.2 m kx a m a
(65 N/m)(0.2 m) 2.0 kg
a 6.5 m/s2
The mass’ acceleration when the displacement is 0.2 m is 6.5 m/s2. 34. dtide 15 m, mfloats m, spanfloats 10 km, T tide 12 h 32 min 45 120 s a) Finding the work done by the upward movement of the floats, W up F g d W up m(9.8 m/s2)(15 m) W up 147m J
Since there is a downward movement as well, W up, down 2W up W up, down 294m J Since the linkages are only 29% efficient, W actual 0.29(294m J) W actual 85.26m J To find power: W P t P
85.26m J 45 120 s
P 1.89 103m W P 1.89m mW
The power produced would be 1.89m mW. b) 1.89m mW from the hydroelectric linkages is not even comparable to 900 MW from a reactor at Darlington Nuclear Power Station. In order for the linkages to produce the same power, the total mass of the floats would have to be 4.76 1011 kg, or 476 million tonnes. 35. m 100 kg, d 12 m, x 0.64 cm 0.0064 m First, we must find the speed at which the mass first makes contact with the spring. Using kinematics, v2 vo2 2ad v vo2 2 ad v 0 2(9.8 m/s2)(12 m) v 15.34 m/s
Finding the maximum kinetic energy of the mass (instant before compression of spring), 1 Ekmax mv2 2 1 Ekmax (100 kg)(15.34 m/s)2 2 Ekmax 11 760 J Since kinetic energy is fully converted to elastic potential energy when the spring is fully compressed, 1 Epmax kx2 2 2 Epmax k
x 2(11 760 J) k (0.0064 m) 2
k 5.7 108 N/m
The spring constant is 5.7 108 N/m. 36. k 16 N/m, A 3.7 cm Since total energy is equal to maximum potential energy, maximum amplitude x at the point of maximum potential energy: Ep Etotal
1 2 1 Ep (16 N/m)(0.037 m)2 2 Ep 0.011 J The total energy of the system is 0.011 J. 37. m bullet 5 g 0.005 kg, mmass 10 kg, k 150 N/m, vo bullet 350 m/s To find the final velocity, use the law of conservation of linear momentum: Ep kx2
po pf (0.005 kg)(350 m/s) 0 (10.005 kg)v v 0.175 m/s
Therefore, the mass and bullet’s kinetic energy is: 1 Ek mv2 2 1 Ek (10.005 kg)(0.175 m/s)2 2 Ek 0.153 J
138
2
Solutions to End-of-chapter Problems
Since all of this energy is transferred to elastic potential energy, Ep Ek
1 2
2
kx
0.153 J
x
2(0.153 J) 150 N/m
x 0.045 m 38. b 0.080 kg/s, m 0.30 kg, xo 8.5 cm, bt
x xoe 2m a) t 0.1 s x (8.5 cm)e x 8.39 cm b) t 1.5 s x (8.5 cm)e x 6.96 cm c) t 15.5 s
(0.080 kg/s)(0.1 s) 2(0.30 kg)
(0.080 kg/s)(180 s)
x (8.5 cm)e 2(0.30 kg) x 3.21 1010 cm e) t 5.2 h 18 720 s
(0.080 kg/s)(18 720 s) 2(0.30 kg)
1 2 Hence,
39. x xo
(0.080 kg/s)(0.1 s)
0.30 kg
bt
x xoe 2m
(0.080 kg/s)(22.3 s)
bt
0.30 kg
xoe2m
(0.080 kg/s)t
0.5 e 2(0.30 kg)
ln 0.5
bt m
1 2
bt
bt m
(0.080 kg/s)(1.5 s) 2(0.30 kg)
(0.080 kg/s)(15.5 s)
xo
1 2 1 Eo kxo2e0 2 1 Eo kxo2 2 One-half of the initial energy is: 1 1 2 1 2 kxo kxo 2 2 4 Therefore, the time required for the energy to reach this value is: 1 Ef kxo2e 2 1 2 1 2 kxo kxo e 4 2 1 e 2 1 bt ln m 2 (0.080 kg/s)t 1 ln 2 0.3 kg t 2.6 s Therefore, it takes 2.6 s for the mechanical energy to drop to one-half of its initial value. b) i) t 0.1 s 1 E (100 N/m)(0.085 m)2e 2 E 0.352 J ii) t 22.3 s 1 E (100 N/m)(0.085 m)2e 2 E 9.45 104 J iii) t 2.5 min 150 s 1 E (100 N/m)(0.085 m)2e 2 E 1.53 1018 J iv) t 5.6 a 176 601 600 s 1 E (100 N/m)(0.085 m)2e 2 E 0 J Eo kxo2e m (t 0 s)
bt m
x (8.5 cm)e 2(0.30 kg) x 1.076 cm d) t 3.0 min 180 s
x (8.5 cm)e x 0 cm
a) Initial energy:
(0.080 kg/s)t 0.30 kg
(0.080 kg/s)(150 s)
0.30 kg
t 5.2 s
Therefore, the time required for the amplitude to reach one-half its initial value is 5.2 s. 40. k 100 N/m 1 E kxo2e 2 bt m
Solutions to End-of-chapter Problems
(0.080 kg/s)(176 601 600 s)
0.30 kg
139
Chapter 7 17. a)
1° 0.0175 rad 57.3°/rad 1 4
b) 90° 2 rad
c) d) e) 18. a) b) c) d) 19. a) b) c) d) 20. a) b) c) d)
90° rad 2 220° 3.84 rad 57.3°/rad 459° 8.01 rad 57.3°/rad 1200° 20.9 rad 57.3°/rad (15.3 rev)(2 rad/rev) 96.1 rad 3 2 rad 3 turn rad 4 turn 2 2 rad 2.3 rad 4.4 h 12 h 2 rad 28.5 h 7.46 rad 24 h 0 rad 0° 2 rad (57.3°/rad) 120° 3 (20 rad)(57.3°/rad) 3600° (466.6 rad)(57.3°/rad) 2.67 104° 3.5 rad 0.56 cycles 2 rad/cycle 1 rad cycle 2 1 cycle 50° 0.14 cycle 360° 1 cycle 1.25 cycles 450° 360°
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)
140
560° 9.77 rad 57.3°/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 t 30 rad 0
3.5 s
27 rad/s c) and become negative. 23. t 26 s (4 cycles)(2 rad/cycle) 8 rad t 8 rad 0
26 s
0.97 rad/s
24. a)
1700 rev 2 rad 1 min rev
1 min 60 s
178.0 rad/s b) t t (178.0 rad/s)(0.56 s) 1.0 102 rad 25. a) 1 0 2 2.55 rad/s t 115 s (2 1) t 2.55 rad/s 0
115 s
0.0222 rad/s2
124° 2.16 rad 57.3°/rad s r s (40 m)(2.16 rad) s 86 m
d)
b) f max
2.55 rad/s 2 rad/cycle
f max 0.406 Hz 45 rev 2 rad 60 s 26. 1
1 min 1 rev 1 min
1 4.7 rad/s 2 0 t 22.5 s
Solutions to End-of-chapter Problems
( ) t
0 4.7 rad/s 22.5 s
2
Therefore, the angular speed of the reel is approximately 4.3 rad/s. 30. r 0.50 m (3.5 rev/s)(2 rad/rev) 22 rad/s
1
0.21 rad/s2 27. 1 18.0 rad/s 2 0 t 22.0 s (2 1) a) t (0 18.0 rad/s)
ac r 2 ac (0.50 m)(22 rad/s)2 ac 2.4 102 m/s2 31. ac 7.98 m/s2 2.50 103 m r
2
22.0 s
r 1.25 103 m v2 a) ac r v acr v (7.98 m/s2)(1.25 103 m ) v 99.9 m/s b) 0 v c) r
0.818 rad/s2 b) 22 12 2 12 2 (18.0 rad/s)2 2(0.818 rad/s2) 198 rad
c) number of cycles
198 rad 2 rad/cycle
number of cycles 31.5 d) 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/s a) 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 3.0 103 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 m v 0.12 m/s v r v r
0.12 m/s 0.028 m
99.9 m/s 1.25 10 m
3
0.0799 rad/s
number of revolutions 0.0799 rad/s 3600 s 24 h 2 rad/rev 1h number of revolutions 1.10 103 0.0799 rad 60 s d) 45 min 1s 1 min 216 rad
s r s (1.25 103 m)(216 rad) s 2.70 105 m 32. c 3.0 108 m/s r 0.80 m d 2(10.0 km) d 20 000 m d a) t c t
20 000 m 3.0 10 m/s 8
t 6.7 105 s
1° 57.3°/rad
0.0174 rad
4.3 rad/s Solutions to End-of-chapter Problems
141
0.0174 rad 6.7 10 s 5
2.6 102 rad/s b) 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 angular
speed but in the opposite direction. Therefore, they will meet halfway, after each
person has travelled rad. 2 t t
2
1
42 rad/s2 (2 1) b) t
rad 2 t 1.3 rad/s t 1.2 s 34. A 1.3 rad/s B 1.6(1.3 rad/s) B 2.08 rad/s
(80 rad/s 190 rad/s) (6.4 s) 2
3.5 102 rad c) (3.5 102 rad)(57.3°/rad) 2.0 104° d) 2 0 42 rad/s2 1 t 2 0 190 rad/s t 42 rad/s2 t 4.5 s 37. 3.8 rad/s2 t 3.5 s 110 rad
A B B A t A t B A B A B
A A 1.3 rad/s 2.08 rad/s
(2.08 rad/s)A (1.3 rad/s) (1.3 rad/s)A A 1.208 rad t t A A t A
1 2
a) 1t t 2 1
1.208 rad 1.3 rad/s
1 2
2
t
t
1 110 rad (3.8 rad/s2)(3.5 s)2 2 1 3.5 s 1 24.77 rad/s 1 25 rad/s b) 22 12 2 2 (25 rad/s)2 2(3.8 rad/s2)(110 ra d) 2 38.22 rad/s 2 38 rad/s
t 0.93 s 35. 1 4.2 rad/s 1.80 rad/s2 t 2.8 s a) 2 1 t 2 4.2 rad/s (1.80 rad/s2)(2.8 s) 2 9.2 rad/s
142
t 80 rad/s 190 rad/s 6.4 s
a)
2
t
1 2 (4.2 rad/s)(2.8 s) 1 2 2 (1.8 rad/s )(2.8 s) 2 19 rad 36. 1 190 rad/s 2 80 rad/s t 6.4 s b) 1t t 2
t
Solutions to End-of-chapter Problems
c)
t 2
1
b)
38.22 rad/s 24.77 rad/s 3.5 s 2 3.8 rad/s (2 1) d) t 2 (38.22 rad/s 24.77 rad/s) (3.5 s) 2 110 rad 1 2t t 2 2 (38.22 rad/s)(3.5 s) 1 2 2 (3.8 rad/s )(3.5 s) 2 110 rad 400 rev 2 rad 1 min 38. 1 1 min rev 60 s 1 41.9 rad/s t 1.2 s a) (10 turns)(2 rad/turn) 20 rad (1 2) t b) 2 2
2
1 t
2
2(20 rad) 41.9 rad/s 1.2 s
2 62.8 rad/s 2 63 rad/s 1 c) 2 t 62.8 rad/s 41.9 rad/s
t
1
3.5 104 rad/s 0 t 1.5 104 rad/s2 t 2.3 s 41. 2 15 rad/s t 3.4 s 2.3 rad/s2 1 a) 2t t 2 2
1 2
(15 rad/s)(3.4 s) (2.3 rad/s2)(3.4 s)2 38 rad 1 b) 2 t 1 2 t 1 15 rad/s (2.3 rad/s2)(3.4 s) 1 7.2 rad/s 42. T M 5.94 107 s T E 3.16 107 s headstart 30° headstart rad
rad
t
6 M E
rad
t
6 2 rad 2 rad
5.94 10 s 3.16 10 s
2 2
4
2(2.0 10 rad) (3.5 10 rad/s) (2.5 10 rad/s)
t 1.4 s
2
6
2
3
rad Mt Et
17 rad/s 39. 2.0 104 rad 1 3.5 103 rad/s 2 2.5 104 rad/s (1 2) a) t
t
1
(2.5 104 rad/s) (3.5 103 rad/s) 1.4 s 4 2 1.5 10 rad/s 40. 1.5 104 rad/s2 (from 39b) 1 0 2 3.5 104 rad/s
2
1
2
6
1.2 s
t
t
4
7
7
t 5.63 106 s 43. A 0.380 rad/s B 0.400 rad/s A 0.080 rad/s2 B 0
Solutions to End-of-chapter Problems
143
headstart
25° 57.3°/rad
headstart 0.436 rad A 0.436 rad B 1 At At 2 0.436 rad Bt 2 1 0 At 2 At Bt 0.436 rad 2 1 0 (0.080 rad/s2)t 2 (0.380 rad/s)t 2 (0.400 rad/s)t 0.436 rad 0 (0.040 rad/s2)t 2 (0.020 rad/s)t 0.436 rad Use the quadratic formula: t
0.020 rad/s (0.0 20 rad /s) 4(0.0 40 rad /s )( 0.436 rad) 2(0.040 rad/s ) 2
2
2
t 3.56 s
1 2 I a 14 kg·m2 I b 4.8 kg·m2 I c 6.8 kg·m2 The order is a, c, b.
44. I mr 2
45. I a mr 2
1 2 1 I c mr 2 2 1 I d (3m)l2 12 1 I d (3m)(2r )2 12 I b mr 2
I d mr 2
2 5 2 1 2 I e (2m) r 5 2 1 I e mr 2 5 The order is a and d, b and c, e. 46. m 4200 kg r 0.3 m 1 I mr 2 2 1 I (4200 kg)(0.3 m)2 2 I 189 kg·m2 I e mr 2
144
47. m 3.5 kg a) I mr 2 I (3.5 kg)(0.21 m)2 I 0.15 kg·m2
1 2 1 I (3.5 kg)(0.21 m)2 2 I 0.077 kg·m2 2 c) I mr 2 5 2 I (3.5 kg)(0.25 m)2 5 I 0.088 kg·m2 1 d) I mr 2 2 1 I (3.5 kg)(0.50 m)2 2 I 0.44 kg·m2 48. m 1.4 kg r 0.12 m 1 a) I mr 2 2 1 I (1.4 kg)(0.12 m)2 2 I 0.010 kg·m2 b) (60 rev/s)(2 rad/rev) 120 rad/s 377 rad/s 49. m 10.0 kg 1 r i (0.54 m) 2 r i 0.27 m 1 r e (0.54 m)(1.4) 2 r e 0.378 m 1 I m(r i2 r e2) 2 1 I (10.0 kg)[(0.27 m)2 (0.378 m)2] 2 I 1.08 kg·m2 50. m 2.0 kg r 1.5 m 2 a) I mr 2 3 2 I (2.0 kg)(1.5 m)2 3 I 3.0 kg·m2 b) I mr 2
Solutions to End-of-chapter Problems
b)
200 rev 2 rad 1 min 1 min 1 rev 60 s
number of turns 1.4 53. m 5.98 1024 kg r 6.38 106 m t 3.16 107 s 2 rad 1 a) Erot I 2 2 1 2 Erot mr 2 2 2 5 1 2 Erot mr 2 5 t 1 Erot (5.98 1024 kg)(6.38 106 m)2 5 2 2 rad 3.16 107 s Erot 1.92 1024 J
20.9 rad/s
2 5 2 I (2.0 kg)(1.5 m)2 5 I 1.8 kg·m2 51. m 20 kg r 0.9 m 2 0 1 (12.3 rev/s)(2 rad/rev) 1 77.3 rad/s c) I mr 2
W R W R I W R (mr 2)
( ) 2 t
1 2
t
1 2
W R mr 2(0 1)(1 0)
b) v r
1 W R (20 kg)(0.9 m)2(77.3 rad/s)2 2 W R 4.8 104 J 52. m 1450 kg 1.35 m r 2 r 0.675 m 1.40 rad/s 1 a) I mr 2 2 1 I (1450 kg)(0.675 m)2 2 I 330 kg·m2 1 b) Erot I 2 2 1 Erot (330 kg·m2)(1.40 rad/s)2 2 Erot 3.24 102 J
c) v r v (0.675 m)(1.40 rad/s) v 0.945 m/s d) t t (1.40 rad/s)(6.5 s) 9.1 rad
9.1 rad 2 rad/turn
number of turns
2 rad 3.16 10 s
v (6.38 106 m)
7
v 1.27 m/s 54. m 8.30 1025 kg r 3.5 m a) I e mr 2 I e (8.30 1025 kg)(3.5 m)2 I e 1.0 1023 kg·m2 (1000 cycles)(2 rad/cycle) b)
1.0 s
6.3 103 rad/s
1 2 1 Ek m(r )2 2 1 Ek (8.30 1025 kg)(3.5 m)2 2 (6.3 103 rad/s)2 Ek 2.0 1016 J 55. me 9.11 1031 kg mn 1.67 1027 kg r 5.0 1011 m L 1.05 1034 kg·m2/s c) Ek mv2
a) I mer 2 I (9.11 1031 kg)(5.0 1011 m)2 I 2.3 1051 kg·m2
Solutions to End-of-chapter Problems
145
1 2
4.6 1016 rad/s
1 2 1 Erot (2.3 1051 kg·m2) 2 (4.6 1016 rad/s)2 Erot 2.4 1018 J 56. r 0.20 m h1 2.5 m h2 0 v1 0 1 0 1 1 a) mgh1 mv12 I 12 2 2 1 1 mgh2 mv22 I 22 2 2 1 1 1 mgh1 mv22 mr 2 22 2 2 2 v 2 1 1 gh1 v22 r 2 2 r 2 4 1 1 gh1 v22 v22 2 4 3 gh1 v22 4 4 v2 gh1 3 4 v2 (9.8 m/s2)(2.5 m) 3 v2 5.7 m/s c) Erot I 2
2
v2 r
5.7 m/s 0.20 m
2 29 rad/s 57. r 0.20 m h1 2.5 m h2 0 v1 0 1 0
146
1 2
1 2
mgh2 mv22 I 22
1.05 1034 kg·m2 s 2.3 1051 kg·m2
b) 2
1 2
a) mgh1 mv12 I 12
b) L I L I
1 1 2 2 v 1 1 gh1 v22 r 2 2 r 2 2 1 1 gh1 v22 v22 2 2
mgh1 mv22 (mr 2)22
2
gh1 v22 v2 gh1 v2 (9.8 m/s2)(2.5 m) v2 4.9 m/s v b) 2 2 r
2
4.9 m/s 0.20 m
2 25 rad/s
1 2
1 2 1 1 mgh2 mv22 I 22 2 2 v 2 1 1 2 mgh1 mv22 mr 2 2 r 2 2 5 1 1 gh1 v22 v22 2 5 10 v2 gh1 7 59. l 2.8 m r 2.8 m h1 2.8 m h2 0 v1 0 1 0 1 1 mgh1 mv12 I 12 2 2 1 1 mgh2 mv22 I 22 2 2 v 2 1 1 1 mgh1 mv22 ml2 2 r 2 2 3 1 1 gh1 v22 v22 2 6 2 gh1 v22 3 3 v2 gh1 2 3 v2 (9.8 m/s2)(2.8 m) 2 v2 6.4 m/s 58. mgh1 mv12 I 12
Solutions to End-of-chapter Problems
60. m 3.9 kg r 0.13 m 150 rad/s
2 5 2 I (3.9 kg)(0.13 m)2 5 I 0.0264 kg·m2
a) I mr 2
b) L I L (0.0264 kg·m2)(150 rad/s) L 4.0 kg·m2/s 61. m 2.4 kg r 0.30 m 1 0 2 250 rad/s t 3.5 s
1 2 1 I (2.4 kg)(0.30 m)2 2 I 0.108 kg·m2
a) I mr 2
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) t
250 rad/s 3.5 s
71.4 rad/s2 e) I (0.108 kg·m2)(71.4 rad/s2) 7.7 N·m 62. I 1.50 103 kg·m2 d 4.5 m (3.0 turns)(2 rad/turn) 6.0 rad a) v 17.0 m/s d t v t
4.5 m 17.0 m/s
b)
t
6.0 rad 0.2647 s
71 rad/s c) L I L (1.50 103 kg·m2)(71 rad/s) L 0.11 kg·m2/s 63. l 2.5 m m 3.2 kg t 13 s r 0.010 m (28 turns)(2 rad/turn) 56 rad
1 2 1 I (3.2 kg)(0.010 m)2 2 I 1.6 104 kg·m2
a) I mr 2
b) L I
56 rad 13 s
L (1.6 104 kg·m2)
L 2.2 103 kg·m2 s 64. l 2.5 m m 3.2 kg t 13 s (28 turns)(2 rad/turn) 56 rad
1 12 1 I (3.2 kg)(2.5 m)2 12 I 1.667 kg·m2 I 1.7 kg·m2
a) I ml2
b) L I
56 rad 13 s
L (1.667 kg·m2)
L 22 kg·m2 s 65. m 3.2 kg l1 2.5 m l2
2.5 m 0.5 m 2
l2 0.75 m
t 0.2647 s t 0.26 s Solutions to End-of-chapter Problems
147
I I cm ml22
1 12 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/s I 1 xI 2 where x is the factor by which the moment of inertia changes. I ml12 ml22
x 1 2
69. r p 4.3 m r t 4.3 m mp 600 kg p 6.4 rad/s mt 35 kg a) I pp ( I p I t )f I pp f ( I p I t)
f
f
1 I 11 I 1 2 2 1 1 2 2 21 2 Therefore, the angular speed will increase by a factor of 2. 68. I m 1.5 103 kg·m2 I s 8.5 kg·m2 s 10 rad/s a) I ss I mm I ss m
m 5.7 104 rad/s b) (10 rad)(57.3°/rad) 573° c) (5.7 104 rad)(57.3°/rad) 3.3 106° d) 45° rad
4
2 (600 kg)(4.3 m)
2
(35 kg)(4.3 m)2
f 5.7 rad/s b) t 3.1 rad/s I pp I t t ( I p I t )f I I t t f p p ( I p I t)
1 2
2
p mt r t2 t
mpr p
f
1
2
2
mpr p
1 2
+ mt r t2
kg)(4.3 m)2(6.4 rad/s) + (35 kg)(4.3 m)2(3.1 rad/s)
1
2 (600 kg)(4.3 m)
2
(35 kg)(4.3 m)2
f 6.0 rad/s c) t 6.4 rad/s I I t t f p p ( I p I t)
1 2
2
mpr p
f
p mt r t2 t
12 2
mpr p
1 2
f
mt r t2
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) (35 kg)(4.3 m)
4 10 rad/s
1
(600
rad
kg)(4.3 m)2(6.4 rad/s)
s
t 0.0785 s m mt m (5.7 104 rad/s)(0.0785 s) m 4.45 103 rad 4.45 103 rad number of rotations 2 rad/rotation number of rotations 7.1 102
f 5.0 rad/s 70. m1 30 kg r 1 1.5 m m2 20 kg r 2 1.0 m 1 12 rad/s
148
mt r t2
1 2
f
t
2
(600
Im (8.5 kg·m2)(10 rad/s) m (1.5 103 kg·m2)
2
p
1 (600
67. I 11 I 22
s
2
mpr p
6.85 rad/s x 4.40 rad/s x 1.56
t
1 2
mpr p
Solutions to End-of-chapter Problems
2
2
a) I 11 ( I 1 I 2)f I 11 f ( I 1 I 2) m1r 121 f (m1r 12 m2r 22)
2
f
(30 kg)(1.5 m) ( 12 rad/s) ((30 kg)(1.5 m) (20 kg)(1.0 m) ) 2
2
f 9.2 rad/s b) I 11-i I 22-i ( I 1 I 2)f I I 22-i f 1 1-i ( I 1 I 2) f
f
2 2 2 2 2 2
1-i 2 1 1
2-i
a 0.138 m/s2
2
2
(30 kg)(1.5 m) (12 rad/s) (20 kg)(1.0 m) (12 rad/s) ((30 kg)(1.5 m) (20 kg)(1.0 m) ) 2
2
f 12 rad/s I I 22-i c) f 1 1-i ( I 1 I 2)
f
2 1 1
2 2 2 2 2 2
1-i 2 1 1
t
2-i
(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)
f 6.5 rad/s d) I 11-i I 22-i ( I 1 I 2)f I 11-i 2-i I 2
2 1 1 1-i 2 2 2
2-i
m r m r
2-i
(30 kg)(1.5 m) (12 rad/s) (20 kg)(1.0 m) 2
2
v2 v1 t v2 v1 at v2 0 (0.138 m/s2)(3.99 s) v2 0.551 m/s d) v2 r 2 v 2 2 r c) a
2
0.551 m/s 0.0030 m
1 2 1 Ek(final) (0.135 kg)(0.551 m/s)2 2 Ek(final) 0.0205 J 1 f) Erot(final) I 22 2 1 Erot(final) (8.50 105 kg·m2)(184 rad/s)2 2 Erot(final) 1.43 J 2
[250 kg·m (40 kg)(2.5 m) ](2.0 rad/s) (250 kg·m (40 kg)(1.5 m) ) 2
d a
e) Ek(final) mv22
2
2 2(1.10 m) 0.138 m/s 3.99 s ( )
2 184 rad/s
71. I 250 kg·m2 r 1 2.5 m r 2 1.5 m t-1 2.0 rad/s md 40 kg I 11 I 22 I 11 2 I 2 ( I mdr 12)1 2 ( I mdr 22)
2 2.9 rad/s
t
2
2-i 40 rad/s
2
1 2
b) d v1t at 2 t
m r m r (m r m r ) f
9.8 m/s2 a 8.50 105 kg·m2 1 (0.135 kg)(0.0030 m)2
m r m r (m r m r ) 2 1 1
73. m 0.135 kg I 8.50 105 kg·m2 r 0.0030 m d 1.10 m 1 0 v1 0 g a) a I 1 2 mr
2
g) ETotal(initial) mgh1 ETotal(initial) (0.135 kg)(9.80 m/s2)(1.10 m) ETotal(initial) 1.46 J
Solutions to End-of-chapter Problems
149
ETotal(initial)
9.8 m/s2 8.50 105 kg·m2 1 (0.135 kg)(0.0030 m)2
a
ETotal(initial)
a 0.138 m/s2
1 2
b) d v1t at 2
1 0 at 2 v1t d 2 1 0 (0.138 m/s2)t 2 2 (1.0 m/s)t (1.10 m) 0 (0.0690 m/s2)t 2 (1.0 m/s)t (1.10 m) Use the quadratic formula: t
1.0 m/s (1.0 m /s) 4(0.0 690 m /s )( 1.10 m) 2(0.0690 m/s ) 2
2
2
t 1.03 s (v v1) c) a 2 t v2 v1 at v2 1.0 m/s (0.138 m/s2)(1.03 s) v2 1.14 m/s d) v2 r 2 v 2 2 r
2
1.14 m/s 0.0030 m
2 380 rad/s
1 2 1 Ek(final) (0.135 kg)(1.14 m/s)2 2 Ek(final) 0.088 J 1 f) Erot(final) I 22 2 1 Erot(final) (8.50 105 kg·m2)(380 rad/s)2 2 Erot(final) 6.16 J e) Ek(final) mv22
150
1 1 2 2 (0.135 kg)(9.80 m/s2)(1.10 m) 1 2 (0.135 kg)(1.0 m/s) 2 1 5 kg·m2) (8.50 10 2 1.0 m/s 2 0.0030 m 6.24 J
g) ETotal(initial) mgh1 mv12 I 12
74. m 0.135 kg I 8.50 105 kg·m2 r 0.0030 m d 1.10 m v1 1.0 m/s g a) a I 1 2 mr
Solutions to End-of-chapter Problems
43. The electroscope has an overall positive
Chapter 8 33. Positive signs: protons
Negative signs: electrons 34. a) No charge b) Negative c) Positive d) No charge e) Positive 35. a) Negative b) Positive c) Negative d) Positive 36. a) Negative b) Electrons 37. a) Glass: positive; silk: negative b) Since they have opposite charges, they will be attracted 38. 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 rubbed with wool socks would have a negative charge. 40. a) The electroscope becomes positive because it gives up some electrons to the glass rod to reduce the rod’s deficit of electrons. This is called charging by contact. b) The leaves become positively charged as well. In charging by contact, the charged object receives the same charge as the charging rod. c) Negative charges will enter the leaves if the system is grounded. 41. 1 C 6.25 1018 e, q 15 C q (15 C)(6.25 1018 e/C) q 9.38 1019 e 42. q 1.1 C q 1.1 106 C q (1.1 106 C)(6.25 1018 e/C ) q 6.9 1012 e
charge: q 4.0 1011 e q (4.0 1011 e)(1.602 1019 C/e) q 6.4 108 C 1 44. q (5.4 108 e) 2 1 q (5.4 108 e)(1.602 1019 C/e) 2 q 4.3 1011 C 45. qn 2.4 1012 C (2.4 1012 C)(6.25 1018 e/C) 1.5 107 elementary charges This means that there are 1.5 107 protons in the nucleus, so the neutral atom must have an equal number of electrons: 1.5 107. 46. F e
kqq 2 r
a) F e1
kqq (4r )
F e1
kqq 16r 2
F e1
F e
b) F e2 F e2
2
1 16 k(2q)(2q)
r 2
4kqq r 2
F e2 4 F e
4 16 1 F e2 F e 4 47. Each sphere loses half of its charge to balance with its identical neutral sphere. 1 1 q1 q1, q2 q2 2 2 c) F e3
F e
F e1
kq q r
F e2
kq q r
1 2 2 1 1 2 2
1 1 2 2 k
F e2 F e2
2
q1
q2
r 22
kq q 4r 1 2 2 2
Solutions to End-of-chapter Problems
151
But F e2 F e1 kq q kq q r 4r 1 2 2 2
F e2 F e1
1 2 2 1
1 2
1
4
1 Therefore, r 2 r 1 2 The spheres should be placed one-half their original distance apart to regain their original repulsion. 48. r 100 pm 100 1012 m 1.00 1010 m, q1 q2 1.602 1019 C kq1q2 F e r 2 (9.0 109 N·m2/C2)(1.602 1019 C)2 F e (1.00 1010 m)2
F e2 q2 F e1 3q2
F e2 1 F e1 3
1 The magnitude of F e2 is F e1, and in the 3 opposite direction of F e1. 51. a)
p
F e e
F e 2.3 108 N
F g
49. r 25.0 cm 0.250 m, F e 1.29 104 N, 2 q1 q2 q qo
3
q q
r kq1q2
r 2 F e2 (q)(q) F e1 (q)(3q)
2 r 1 4r 2 r 21 2 r 2
a) F e
kq1q2 2
kqq 2 r
(1.29 10 N)(0.25 m) (9.0 10 N·m /C ) F er 2 k
4
8
q 3.00 10
9
b) q1 1.602 1019 C, q2 1.602 1019 C, m 9.1 1031 kg, g 9.8 m/s2 F g F e kq1q2 mg r 2
r
2
2
2
r
C
2 3 3 qo q 2 3 qo (3.00 108 C) 2 qo 4.5 108 C The type of charge, positive or negative, does not matter as long as they are both the same. (Like charges repel.) 50. q1 q, q2 3q qT q (3q) 2q 2q So q1 q2 q 2
b) q is the original charge on each sphere.
kq1q2 mg
(9.0 109 N·m2/C2)(1.602 1019 C)2 (9.1 1031 kg)(9.8 m/s2)
r 5.1 m 52. q1 2.0 106 C, q2 3.8 106 C, q3 2.3 106 C a) r 1 0.10 m, r 2 0.30 m kq1q3 1 F e3 r 21
1 F e3
9
2
2
6
2
4.14 N (attraction) 1 F e3 4.14 N [right] kq2q3 2 F e3 r 22 1 F e3
2 F e3
9
2
2
6
Solutions to End-of-chapter Problems
6
(9.0 10 N·m /C )(3.8 10 C)(2.3 10 C) (0.30 m) 2
0.87 N (repulsion) 2 F e3 0.87 N [left] 2 F e3
152
6
(9.0 10 N·m /C )( 2.0 10 C)(2.3 10 C) (0.10 m)
eT 4.14 N [right] 0.87 N [left] F eT 3.3 N [right] F b) r 1 0.30 m, r 2 0.10 m kq1q3 1 F e3 r 21
1 F e3
9
2
2
6
6
(9.0 10 N·m /C )( 2.0 10 C)(2.3 10 C) (0.30 m)
2
0.46 N (attraction) 1 F e3 0.46 N [left] kq2q3 2 F e3 r 22 1 F e3
2 F e3
9
2
2
6
6
(9.0 10 N·m /C )(3.8 10 C)(2.3 10 C) (0.10 m) 2
7.86 N (repulsion) 2 F e3 7.86 N [right] eT 0.46 N [left] 7.86 N [right] F eT 7.4 N [right] F c) 1 F e3 4.14 N (attraction) 1 F e3 4.14 N [left] 2 F e3 7.86 N (repulsion) 2 F e3 7.86 N [left] eT 4.14 N [left] 7.86 N [left] F eT 12 N [left] F d) The third charge could only be placed to 2 F e3
Therefore, the charge must be placed 0.53 m to the left of the first charge. The other answer, 0.084 m, would place the charge between the two base charges and therefore is an inappropriate answer. For a charge placement to the right of the two charges, two inappropriate answers are calculated, meaning that the only possible placement for the charge is at 0.53 m to the left of the first charge. 53. The forces on the test charge from the repulsion by the other two charges must equal one another for the test charge to come to rest there. The force of charge 1 on the test charge (1 F qt ) must equal the force of charge 2 on the test charge (2 F qt ). 1 F qt 2 F qt
the left or to the right of the two basic charges for the forces to balance and give a force of 0. For the charge to be placed a distance of r x metres to the left of the first charge: 1 F e3 2 F e3 kq q kq2q3 12 3 r x (0.20 m r x)2 ( 2.0 106 C) (3.8 106 C) r x2 (2.0 101 m r x)2
(3.8 106)r x2 (2.0 106)(4.0 102 4.0 101r x r x2) (3.8 106)r x2 8.0 108 8.0 107r x 2.0 106r x2 Rearranging: 1.8 106r x2 8.0 107r x 8.0 108 0 Solve for r x using the quadratic formula.
r x 8 (8.0 107) (8.0 107)2 4(1.8 106)(8.0 10 ) 6 2(1.8 10 )
kqq k4qq 1 2 t 2
3 3 r
t 2
r
r 2 4r 2 9 4(9) 2 r r 2 Therefore, the net force on the charge would 1 be 0 if it was placed of the distance 3 between the two charges. 54. q2 q1 q3 1.0 104 C, r 1 r 2 r 3 0.40 m 1
1
q
q
2
3
q
q
30° 120° 30°
40 cm
For q1: The force is the vector sum of two e1 and 3 F e1. These two magnitudes forces, 2 F must have the same value.
2 F e1
kqq 2 r
(9.0 109 N·m2/C2)(1.0 104 C)2 2 F e1 (0.40 m)2 2 2 F e1 5.6 10 N 3 F e1 F 2eT 2 F 2e1 3 F 2e1 2(2 F e1)(2 F e1)(cos 120°)
F eT 2(5.6 102 C)2 2(5.6 102 C)2(co s 120° ) 2 F eT 9.7 10 N
So r x 0.53 m or 0.084 m.
Solutions to End-of-chapter Problems
153
From the isosceles triangle with angles of 30°, the total angle is 30° 60° 90°. eT1 9.7 102 N [up] F eT2 9.7 102 N [left 30° down] F eT3 9.7 102 N [right 30° down] F Each force is 9.7 102 N [at 90° from the line connecting the other two charges]. 55. a) l 2.0 102 m, q1 q2 q3 q4 1.0 106 C
56.
now asymmetrical and has its inflection points pushed farther to the right.
2.0 cm
q
q
57. The field is similar to the one above, but is
2.0 cm
q
q
2 F e1
kqq 2
2 F e1
(9.0 10 N·m /C )( 1.0 10 C) (2.0 10 m)
r 2
9
2
2
6
2
2
2
22.5 N [left] 4 F e1 22.5 N [up] 2 F e1
58. Parallel plates:
kqq 3 F e1 r 23
+
9
2
2
6
11.25 N [left 45° up]
–
2
(9.0 10 N·m /C )( 1.0 10 C) 2(2.0 10 m)
3 F e1 3 F e1
Coaxial cable:
2
2
From Pythagoras’ theorem: N)2 2 F e1 4 F e1 2(22.5 2 F e1 4 F e1 31.82 N [left 45° up] Therefore, F eT1 (31.82 N 11.25 N) [left 45° up] 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] F Each force is 43.1 N [symmetrically outward from the centre of the square]. b) The force on the fifth charge is 0 N because the forces from each charge are balanced. c) Sign has no effect. If the new fifth charge were either positive or negative, the attractive/repulsive forces would still balance one another.
59. q 2.2 106 C, F e 0.40 N F e q
154
0.40 N 2.2 10 C 6
1.8 105 N/C 60. F e 3.71 N, 170 N/C F q e
q
3.71 N 170 N/C
q 2.2 102 C
Solutions to End-of-chapter Problems
61. q1 4.0 106 C, q2 8.0 106 C, r 2.0 m kq2 kq1 1 2 1 2 r r
2 2 (9.0 10 N·m /C )(8.0 10 C) 2.0 m 2 (9.0 10 N·m /C )(4.0 10 C) 2.0 m 2
9
2
2
6
2
9
2
2
6
2
3.6 104 N/C
Therefore, the field strength is 3.6 104 N/C towards the smaller charge. e 7.5 N [left] 62. a) q 2.0 106 C, F
F e q
7.5 N [left] 2.0 10 C
2
r 1 8.3 102 m, therefore, r 2 1.17 101 m 1.2 101 m 0 at 1.2 101 m from the larger charge, or 8.3 102 m from the smaller charge. 67. q1 q2 q3 q4 1.0 106 C, r 0.5 m
6
0.5 m
6
0.5 m
Take right to be positive. e q F F e (4.9 105 C)(3.8 106 N/C) F e 1.86 102 N The force would be 1.86 102 N [left]. 63. r 0.5 m, q 1.0 102 C kq r 2 (9.0 109 N·m2/C2)(1.0 102 C) 2
q
Since the magnitudes of all four forces are equal, and they are paired with forces in the e2 F e4 and opposite direction ( F e1 F e3), there is no net force. Therefore, F there is no net field strength. 0 N/C 68. q1 q2 2.0 105 C, r 0.50 m
3.6 108 N/C [left] 64. q1 4.0 106 C, q2 1.0 106 C
2 1 (9.0 109 N·m2/C2)(4.0 106 C) 2
P
q 2
P
Take right to be positive.
1
q
(0.5 m)
q
q
3.8 10 N/C [left] b) q2 4.9 105 C
66. r T 0.20 m, q1 1.5 106 C, q2 3.0 106 C r 22 (0.20 r 1)2 1 2 kq1 kq2 2 r 1 r 22 1.5 106 C 3.0 106 C r 21 r 22 r 22 2r 21 Substitute for r 22 and rearrange: 0 r 12 0.4r 1 4.0 102 0.4 (0.4)2 4(4.0 102) r 1
q
(0.40 m)
(9.0 109 N·m2/C2)( 1.0 106 C) (0.30 m)2 3.25 105 N/C [right] 65. r 5.3 1011 m, q 1.602 1019 C
kq r 2
0.50 m
kq 1 r 2
(9.0 109 N·m2/C2)(2.0 105 C) 1 (0.50 m)2 1 7.2 105 N/C 1 2 and T 1 2 Therefore, T 2(1)2 2(1)2(cos 120°) 1.2 106 N/C [at 90° from the line conT necting the other two charges]
(9.0 109 N·m2/C2)(1.602 1019 C) (5.3 1011 m)2
5.1 1011 N/C Solutions to End-of-chapter Problems
155
69. q 0.50 C, ∆V 12 V W q∆V W (0.50 C)(12 V) W 6.0 J 70. W 7.0 102 J, ∆V 6.0 V W q V 7.0 102 J q
W Ee2 Ee1 W 0.18 J 0.045 J W 0.14 J 75. Position in the field has no bearing on the
field strength. 5.0 103 N/C, d 5.0 cm 5.0 102 m V d V (5.0 102 m)(5.0 103 N/C) V 2.5 102 V 76. a) q 1 105 C, 50 N/C F e q F e (1 105 C)(50 N/C) F e 5.0 104 N b) ∆d 1.0 m ∆ Ek W ∆ Ek F e∆d 4 ∆ Ek (5.0 10 N)(1.0 m) 4 ∆ Ek 5.0 10 J c) v 2.5 104 m/s
6.0 V
q 1.2 102 C 71. q 1.5 102 C, F e 7.5 103 N, 2 ∆d 4.50 cm 4.50 10 m W V q F d V e q (7.5 103 N)(4.5 102 m) V 1.5 102 C
V 2.3 104 V 72. 130 N/C, F e 65 N, ∆V 450 V V W q VF e W
m
2
m
2(5.0 10 J) (2.5 10 m/s)
W
W 2.3 102 J 73. d 0.30 m, q 6.4 106 C kq V d (9.0 109 N·m2/C2)(6.4 106 C) V
0.30 m
2
2
6
6
(9.0 10 N·m /C )( 1.0 10 C)( 5.0 10 C) 0.25 m
Ee 0.18 J (repulsion) kq1q2 b) Ee1 r 9
2
2
6
1
Ee kq1q2
1
d2 d1 19 Ee 2.08 10 J
Therefore, the electric potential energy was reduced by 2.08 1019 J, which was transferred to kinetic energy. The energy is spread over both electrons, so the energy for each electron is 1.04 1019 J.
6
(9.0 10 N·m /C )( 1.0 10 C)( 5.0 10 C) 1.00 m
Ee1 0.045 J (repulsion) W ∆ Ee
156
2
m 1.6 1012 kg 77. d1 1.0 109 m, d2 1.0 108 m, q1 q2 1.602 1019 C, m1 m2 9.11 1031 kg Ee E2 E1 kq1q2 kq1q2 Ee d2 d1
Ee1
4
V 1.9 105 V 74. a) q1 1.0 106 C, q2 5.0 106 C, r 0.25 m kq1q2 Ee r 9
v
4
(450 V)(65 N) (130 N/C)
Ee
1 2 2 Ek
Ek mv2
Solutions to End-of-chapter Problems
For one electron: 1 Ek mv2 2 v 2 Ek
ma
d qV a md
2(1.04 10 J) 9.11 10 kg
19
m
v
a
19
31
v 4.78 105 m/s 78. V 2 2V 1 and Ek ∆ Ee qV
With the same charge on each electron, the kinetic energy is also doubled, i.e., Ek2 2 Ek1 Ek2 2 Ek1
Ek1
E k1
1 2 mv2 2 1 2 2 mv1 2 v22 2v21 v2 2v1 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
P (6.25 1018 e/C)
V
number of electrons/s (27 J/s)
1C 1.5 104 J
(6.25 1018 e/C) number of electrons/s 1.1 1016 b) q 1.602 1019 C, m 9.11 1031 kg Accelerating each electron from rest, Ek Ee
1 2
2
mv
v
2 Vq m
2(1.5 104 V)(1.602 1019 C) 9.11 1031 kg
v 7.3 107 m/s 80. a) d 1.2 m, V 7.5 103 V, m 3.3 1026 kg F e q
3
(1.602 10 C)(7.5 10 V) (3.3 10 kg)(1.2 m) 26
a 3.0 1010 m/s2 b) E Vq E (1.602 1019 C)(7.5 103 V) E 1.202 1015 J c) At this speed and energy, relativistic effects
may be witnessed. Although the speed may not be what is predicted by simple mechanics, the total energy should be the same but may be partly contributing to a mass increase of the ion. 81. q1 q2 1.602 1019 C, m1 m2 1.67 1027 kg, v1 v2 2.7 106 m/s ∆ Ek
∆ Ee
The total energy for both ions is: kq q 1 (2)mv2 1 2 r 2
r
kq q mv
r
(9.0 10 N·m /C )(1.602 10 C) (1.67 10 kg)(2.7 10 m/s)
1 2 2
9
2
27
2
19
6
2
2
r 1.9 1014 m 82. a) q 2e, m 6.696 1027 kg, v1v 0 m/s, v1h 6.0 106 m/s, V 500 V, dv 0.03 m, d h 0.15 m
Acceleration is toward the negative plate:
Vq
v
qV
F a e m q a m qV a mdv
2(1.602 1019 C)(500 V) a (6.696 1027 kg)(3.0 102 m)
a 7.97 1011 m/s2
Solutions to End-of-chapter Problems
157
Time between the plates is: d h v h
t
t
0.15 m 6.0 10 m/s 6
t 2.5 108 s
Therefore, 1 d h at 2 2 1 d h (7.97 1011 m/s2)(2.5 108 s)2 2 d h 2.5 104 m d h 0.025 cm The alpha particle is 3.0 cm 0.025 cm 2.975 cm from the negative plate if it enters at the positive plate or 1.475 cm from the negative plate if it enters directly between the two plates. b) v2v v1v at v2v 0 (7.97 1011 m/s2)(2.5 108 s) v2v 2.0 104 m/s
From 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
V d
39.0 V 0.050 m
b) d 3.0 cm 3.0 102 m V d V (3.0 102 m)(2.04 107 N/C) V 6.1 109 V 86. d 0.12 m, V 92 V V d
7.7 102 N/C 87. 3 106 N/C, d 1.0 103 m V d V (1.0 103 m)(3 106 N/C) V 3 103 V Therefore, 3.0 103 V is the maximum poten-
tial difference that can be applied. Exceeding it would cause a spark to occur between the plates. 88. V 50 V, 1 104 N/C V d d
2.04 107 N/C
50 V 1 10 N/C 4
d 5.0 103 m 89. V 120 V, 450 N/C V d d
7.80 102 N/C 84. 2.85 104 N/C, d 6.35 cm 6.35 102 m V d V (6.35 102 m)(2.85 104 N/C) V 1.81 103 V 85. a) m 2mP 2mn 4(1.67 1027 kg), g 9.80 N/kg, q 2e F e F g mg q 4(1.67 1027 kg)(9.80 N/kg) 2(1.602 1019 C)
92 V 0.12 m
120 V 450 N/C
d 2.67 101 m 90. a) m 2.2 1015 kg, d 5.5 103 m, V 280 V, g 9.80 N/kg F e F g q mg qV mg d mgd q V
q
15
3
(2.2 10 kg)(9.80 N/kg)(5.5 10 m) 280 V
q 4.2 1019 C 4.2 1019 C b) N 1.602 1019 e/C
N 2.63 e 3 e
The droplet has three excess electrons.
158
Solutions to End-of-chapter Problems
91. V 450 V, me 9.11 1031 kg, e 1.602 1019 C a) ∆ Ee qV 19 ∆ Ee (1.602 10 C)(450 V) 17 ∆ Ee 7.21 10 J
Therefore, Ek Ee
1 2
2
mv
Ee
v v
2 2(7.21 10 J) 9.11 10 kg Ee m
17
31
v 1.26 107 m/s b)
1 2
2
mv
1 3
Ee
v
2 E 3m
v
2(7.21 10 J) 3(9.11 10 kg)
e
17
31
v 7.26 106 m/s 92. k 6.0 103 N/m, d 0.10 m, V 450 V, x 0.01 m V a) d
450 V 0.10 m
4.5 103 N/C
b) The force to deform one spring is: F kx F (6.0 103 N/m)(0.01 m) F 6.0 105 N
The force to deform both springs is: 2(6.0 105 N) 1.2 104 N c) The force on the pith ball must also be 1.2 104 N d) F spring F e F spring q F q spring 1.2 104 N q 4.5 103 N/C q 2.7 108 C
Solutions to End-of-chapter Problems
159
B 1.8 102 T
Chapter 9 22. I 12.5 A B 3.1 105 T I B 2r I r 2 B (4 107 T·m/A)(12.5 A) r 2(3.1 105 T)
NOTE: The solutions to problem 27 are based on a distance between the two conductors of 1 cm. 27. a) F
F
r 8.1 102 m 23. r 12 m I 4.50 103 A I B 2r (4 107 T·m/A)(4.50 103 A) B 2(12 m) 5 B 7.5 10 T 24. I 8.0 A B 1.2 103 T N 1 NI B 2r NI r 2 B (4 107 T·m/A)(1)(8.0 A) r 2(1.2 103 T)
Currents in the same direction wires forced together
Referring to the above diagram, the magnetic fields will cancel each other out because the field from each wire is of the same magnitude but is in the opposite direction.
r 4.2 103 m 25. N 12 r 0.025 m I 0.52 A NI B 2r (4 107 T·m/A)(12)(0.52 A) B
2(0.025 m)
B 1.6 104 T N 35 turns 100 cm 26. L 1 cm 1m N 3500 turns/m L I 4.0 A NI B L N B I L
3500 turns 1 m (4.0 A)
B (4 107 T·m/A)
160
b) F
F x
Currents in opposite directions wires forced apart
I 10 A r 1.0 102 m I B 2r (4 107 T·m/A)(10 A) B 2(1.0 102 m)
B 2.0 104 T
But this field strength (2.0 104 T) is for each of the two wires. Referring to the above diagram, the two fields flow in the same direction when the current in the two wires moves in the opposite direction. The result is that the two fields will add to produce one field with double the strength (4.0 104 T). 28. Coil 1: N 400 L 0.1 m I 0.1 A Coil 2: N 200 L 0.1 m I 0.1 A
Solutions to End-of-chapter Problems
BTotal Bcoil1 Bcoil2 NI NI BTotal L L 7 (4 10 T·m/A)(400)(0.1 A) BTotal
(0.1 m) (4 10 T·m/A)(200)(0.1 A) (0.1 m) 7
BTotal 2.5 104 T
2(0.02 m)
Solenoid: L 2r single loop L 2(0.02 m) L 0.04 15 turns 100 cm N L 1 cm 1m N (1500 turns/m)(0.04 m) N 188
I 4900 A
NI L
b) This current would most likely melt the 7
Bsol
reversed, the magnitude and direction of the resultant force would be 0.57 N [downwards]. 31. a) d(linear density) 0.010 kg/m B 2.0 105 T 90° F dg L F (0.010 kg/m)(9.8 N/kg) L F 9.8 102 N/m (linear weight) L F BIL sin F I BL sin F L I B sin 9.8 102 N/m I (2.0 105 T) sin 90°
29. The single loop: NI Bsingle 2r (4 107 T·m/A)(1) I Bsingle
Bsol
b) If the current through the wire was to be
(4 10 T·m/A)(188)(0.4 A) 0.04
Bsol 7.5 104 T
To cancel the field, the magnitude of the two fields must be equal but opposite in direction. Bsol Bsingle (4 107 T·m/A)(1) I 7.5 104 T
2(0.02 m) (7.5 10 T)2(0.02 m) I 4 10 T·m/A 4
7
I 24 A 30. a) 45° L 6.0 m B 0.03 T I 4.5 A F BIL sin F (0.03 T)(4.5 A)(6.0 m) sin 45° F 0.57 N
The direction of this force is at 90° to the plane described by the direction of the current vector and that of the magnetic field, i.e., upwards.
wire. 32. a) N 60 I 2.2 A B 0.12 T NI B L NI L B (4 107 T·m/A)(60)(2.2 A) L
0.12 T
L 1.38 103 m F BIL sin F (0.12 T)(2.2 A)(1.38 103 m)
(sin 90°) F 3.64 104 N b) F ma F a m 4
a
3.64 10 N 0.025 kg
a 1.46 102 m/s2 33. B 0.02 T v 1.5 107 m/s 90°
Solutions to End-of-chapter Problems
161
q 1.602 1019 C m 9.11 1031 kg F c F B mv2 qvB sin r mv r qB (9.11 1031 kg)(1.5 107 m/s) r (1.602 1019 C)(0.02 T)
According to the right-hand rules #1 and #3, this charge would always be forced towards the wire. 37. a) v 5 107 m/s r 0.05 m I 35 A q 1.602 1019 C F Bqv sin
3
r 4.3 10 m 34. qalpha 2(1.602 1019 C) qalpha 3.204 1019 C v 2 106 m/s B 2.9 105 T malpha 2(protons) 2(neutrons) malpha 4(1.67 1027 kg) malpha 6.68 1027 kg mv r qB (6.68 1027 kg)(2 106 m/s) r (3.204 1019 C)(2.9 105 T)
r 1.4 103 m 35. F g mg F g (9.11 1031 kg)(9.8 N/kg) F g 8.9 1030 N F mag Bqv sin F mag (5.0 105 T)(1.602 1019 C) (2.8 107 m/s) F mag 2.24 1016 N
The magnetic force has considerably more influence on the electron. 36. q 1.5 106 C v 450 m/s r 0.15 m I 1.5 A 90° F Bqv sin B
I 2r
F
Iqv sin 2r
F
(4 10 T·m/A)(1.5 A)(1.5 10 C)(450 m/s)sin 90° 2 (0.15 m)
7
F 1.3 109 N
162
6
F
B
I 2r
F
Iqv sin 2r
7
19
7
(4 10 T·m/A)(35 A)( 1.602 10 C)(5 10 m/s) sin 90° 2 (0.05)
F 1.12 1015 N
According to the right-hand rules #1 and #3, this charge would always be forced toward the wire. b) If the electron moved in the same direction as the current, then it would be forced away from the wire. 38. a) v 2.2 106 m/s r 5.3 1011 m q 1.602 1019 C m 9.11 1031 kg At any given instant, the electron can be considered to be moving in a straight line tangentially around the proton. F mag F c mv2 qvB sin r mv B qr (9.11 1031 kg)(2.2 106 m/s) B (1.602 1019 C)(5.3 1011 m)
B 2.36 105 T
But this field would always be met by a field of the same magnitude but opposite direction when the electron was on the other side of its orbit. Therefore, the net field strength at the proton is zero. b) To keep an electron moving in a circular artificially simulated orbit, the scientist must apply a field strength of 2.36 105 T.
Solutions to End-of-chapter Problems
39. 475 V/m B 0.1 T
The electron experiences no net force because the forces from both the electric and magnetic fields are equal in magnitude but opposite in direction. If all the directions are mutually perpendicu lar, both the electric and magnetic fields will move the electron in the same direction (based on the right-hand rule #3). Therefore, F mag F e qvB q v B v
(475 V/m) (0.1 T)
v 4750 m/s 40. B 5.0 102 T d 0.01 m v 5 106 m/s q 1.602 1019 C F mag F e qvB q V qvB q d V dvB V (0.01 m)(5 106 m/s)(5.0 102 T) V 2500 V 41. r 3.5 m I 1.5 104 A I 1 I 2 L F 2r
F
7
4
2
(4 10 T·m/A)(1.5 10 A) (190 m) 2 (3.5 m)
F 2.44 103 N 42. L 0.65 m I 12 A B 0.20 T F BIL sin F (0.20 T)(12 A)(0.65 m)(sin 90°) F 1.56 N [perpendicular to wire]
43. a) v 5.0 106 m/s r 0.001 m q 1.602 1019 C m 9.11 1031 kg F c F mag mv2 qvB r vm B qr (5.0 106 m/s)(9.11 1031 kg) B ( 1.602 1019 C)(0.001 m)
B 2.8 102 T b) F c mac F c qvB mac qvB qvB ac m ac
19
6
(1.602 10 C)(5.0 10 m/s)(0.028 T) 9.11 10 kg
31
ac 2.5 1016 m/s2 44. a) r 0.22 m B 0.35 T q 1.602 1019 C m 1.67 1027 kg q v m Br qBr v m (1.602 1019 C)(0.35 T)(0.22 m) v (1.67 1027 kg)
v 7.4 106 m/s mv2 b) F c r (1.67 1027 kg)(7.4 106 m/s)2 F c
0.22 m
F c 4.2 1013 N
At the angle shown, the force is: (1.56 N)sin 30° 0.78 N
Solutions to End-of-chapter Problems
163
45.
d) No, the results would be exactly the same.
e 5.7 108 C/kg m B 0.75 T q v m Br mv r Bq d v t 2r v
T
The induced current flow would be in the opposite direction if the poles of the magnet were reversed, but the reduction in speed would be the same. 49. The copper conductor is cutting through the magnetic field lines as it moves, and therefore experiences a force that opposes its motion. The induced and external magnetic fields are in opposite directions, which causes the opposition. Aluminum wire would make no difference as long as it conducts electricity.
T 2r
v
mv Bq v
2
T T
2 q
B m T
2 (0.75 T)(5.7 10 C/kg) 8
T 1.5 108 s 46. m 6.0 108 kg q 7.2 106 C B 3.0 T
1 2
t T
1 2
t
2m Bq
(6.0 108 kg) t (3.0 T)(7.2 106 C)
t 8.7 103 s 47. 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 (down at the south end). c) Yes, the falling magnet would experience a magnetic force that is opposing its motion, as described by Lenz’s law.
164
Solutions to End-of-chapter Problems
1
Chapter 10
T f T 0.77 s/cycle
21. a) 4 m b) A 5 cm c) T 8 s
For 45 rpm: 45 cycles f 60 s f 0.75 Hz 1
1
d) f T f
1
T f T 1.33 s/cycle
8s f 0.1 s1 e) v f v (4 m)(0.1 s1) v 0.4 m/s 22. f f
10 cycles 3.2 s 1
T f
1 3.125 cycles/s
T 0.32 s/cycle 23. f f
1 3
cycles s
f 3.125 cycles/s
T
For 33 rpm: 100 f rpm 3 100 cycles f 180 s f 0.56 Hz 1
cycles s 72 cycles 60 s
f 1.2 cycles/s
T f T 1.8 s/cycle 27. 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 3 x 1 cos rad
4
1
T f T
x 0.71 m d) x A cos x 1 cos (2π rad) x 1 m
1 1.2 cycles/s
T 0.83 s/cycle 24. f 60 Hz
28.
1
T f T 0.017 s/cycle 25. a) f
x
) s / m ( v
1
f
78 cycles 60 s
f 1.3 Hz
360°
180°
360°
θ
At equilibrium ( x 0), v is a maximum (sin 90° 1). When x A, v is a minimum (sin 0° 0).
f 2.5 Hz
26. For 78 rpm:
θ
180°
–A
150 cycles 60 s
b) T f T 0.4 s/cycle
A
) m (
29.
)
2
s / m ( a
θ
180°
360°
The object always accelerates toward equilibrium and slows down as it moves away from equilibrium. Solutions to End-of-chapter Problems
165
a is a maximum when x A (cos 0°); a is a
minimum at equilibrium (cos 90°). The a vector is always directed toward the equilibrium position.
2.1 m 2 9.8 m/s
30. a) T 2 L g T
T 0.711 s/cycle
0.40 kg 2 20 N/m
b) T 2
2
0.15 m 2 9.8 m/s
c) T 2 L g
T 0.889 s/cycle
0.21 kg 2 200 N/m
c) T 2
2
L g
T 0.204 s/cycle
T 2
Moon
2.1 m 1.6 m/s 2
T 7.2 s/cycle L g
ii) T 2
Moon
80 m 1.6 m/s 2
T 44 s/cycle
0.15 m 2 1.6 m/s
iii) T 2
L g Moon
T
2
T 1.9 s/cycle
2.1 m 2 24.6 m/s
T 2
L g Jupiter
T
2
T 1.8 s/cycle
80 m 2 24.6 m/s
ii) T 2
L g Jupiter
T
2
33. a) f
1
k
2 m k 42 f 2m k 42(12 Hz)2(0.402 kg) k 2.3 103 N/m
b) F kx F (2.3 103 N/m)(0.35 m) F 8.0 102 N 34. a) v f v f 3.00 108 m/s f 6.50 107 m
f 4.62 1014 Hz v b) f 3.00 108 m/s f 6.00 107 m
f 5.00 1014 Hz v c) f 3.00 108 m/s f 5.80 107 m
f 5.17 1014 Hz
T 11 s/cycle 166
m k
T
T 0.78 s/cycle
T 2
m k
T
T 18 s/cycle
T 2
m k
T
80 m 2 9.8 m/s
T
2
0.30 kg 2 23.4 N/m
32. a) T 2
b) T 2 L g
b) i)
T
2
T
L g Jupiter
T 0.49 s/cycle
T 2.9 s/cycle
31. a) i)
0.15 m 2 24.6 m/s
iii) T 2
Solutions to End-of-chapter Problems
v d) f 8
f
3.00 10 m/s 5.20 10 m 7
f 5.77 1014 Hz v e) f 3.00 108 m/s f 4.75 107 m
f 6.32 1014 Hz v f) f 3.00 108 m/s f 4.00 107 m
f 7.50 1014 Hz d 35. a) t v 1.49 1011 m t 3.00 108 m/s t 497 s t 8.28 min t 0.138 h d b) t v 3.8 108 m t 3.00 108 m/s t 1.3 s t 2.1 102 min t 3.5 104 h d c) t v 5.8 1012 m t 3.00 108 m/s t 1.9 104 s t 3.2 102 min t 5.4 h d d) t v 9.1 1010 m t 3.00 108 m/s t 3.03 102 s t 5.1 min t 8.4 102 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 d t v t 3.15 109 s t 100 a d 38. t v t
160 m 3.00 10 m/s 8
t 5.33 107 s 39. r Earth 6.38 106 m, c Earth 2π(6.38 106 m) 4.01 107 m d t v 4.01 107 m t 3.00 108 m/s t 0.134 s 40. For the minimum frequency, 4 107 m v f 3.00 108 m/s f 4 107 m
f 8 1014 Hz
For the maximum frequency, 8 108 m v f
3.00 108 m/s f 8 108 m
f 4 1015 Hz
Thus, the range is 8 1014 Hz to 4 1015 Hz. 41. For the car: ∆t car (50 h)(3600 s/h) ∆t car 180 000 s For light: d t light v
4.00 106 m t light 3.00 108 m/s t light 0.01 s Comparing the two, 180 000 s t car 1.8 107 times 0.01 s t light
Solutions to End-of-chapter Problems
167
42. a) b) c) d) e) f) g) 43. a) b) c) d) e)
sin 30° 0.5 sin 60° 0.866 sin 45° 0.707 sin 12.6° 0.218 sin 74.4° 0.963 sin 0° 0 sin 90° 1 sin1 (0.342) 20° sin1 (0.643) 40° sin1 (0.700) 44.4° sin1 (0.333) 19.5° sin1 (1.00) 90°
c c) v n 8
v
v 2.26 108 m/s c d) v n 3.00 108 m/s v
1.30
c 44. v n 8
v
3.00 10 m/s 0.90
v 3.3 108 m/s
This speed is impossible, since it is greater than the speed of light. 45. n1 1.00, n2 1.98, 1 22 n1 sin 1 n2 sin 2 sin 22 1.98 sin 2 2 sin 2 cos 2 1.98 sin 2 1.98 cos 2 2 2 8.1° 46. 1.5 sin 30° n2 sin 50° n2 0.98 As in problem 44, this value is impossible. 47.
light ray
3.00 10 m/s 1.33
wavefronts
v 2.31 108 m/s n 49. Use 1 with n1 1.00: n2 a) 0.413 b) 0.658 c) 0.752 d) 0.769 c 50. v n 3.00 108 m/s v
1.33
v 2.26 108 m/s d t v t
1200 m 2.26 10 m/s
t 5.31 105 s 51. a) tan 1 tan B n tan 1 2 n1 tan 1 1.42 1 54.8° b) n1 sin 1 n2 sin
glass
2 sin1 1.00·
sin 54.8° 1.42
52. Polaroid glasses are most effective when the 8
3.00 10 m/s 2.42
v 1.24 108 m/s c b) v n 3.00 108 m/s v
1.52
v 1.97 108 m/s
168
2
2 35.2° c) 54.8° (i r)
c 48. a) v n v
8
light is most polarized. The light is 100% polarized at Brewster’s angle, B. n2 n1
tan B
tan B
1.33 1.00
B 53.1° elevation 90° 53.1° elevation 36.9°
Solutions to End-of-chapter Problems
53. a) I 2 0.5 I o cos2 I 2 0.5 I o cos2 30° I 2 0.375 I o I 2 37.5% I o I b) 2 20.7% I o I c) 2 5.85% I o 54. When viewing something through a doubly
refracting crystal, two images are seen, since two rays of polarized light are produced. If another crystal was laid over-top and rotated, nothing would be seen, since the two crystals now act as a polarizeranalyzer pair, with an angle of 90° between their axes. 55. Use an analyzer (another polarizer, rotated). 56. tan
B
n2 n1
B tan1
1.33 1.00
B 53°
57. a) tan
B
n2 n1
B tan1
1.33 1.00
B 53.1° n b) tan B 2 n1 B tan1
1.50 1.00
B 56.3° n c) tan B 2 n1 B tan1
1.33 1.50
n2 tan 60° n2 1.73 59. The first Polaroid will remove exactly one
component (50%) of the incident light. The third Polaroid, having been placed at any angle but 90° to the first one, will remove a fraction of the remaining light, allowing one component of the light to pass through. The second Polaroid will then remove only a single component of the residual light. Thus, a fraction of the incident light passes through all three Polaroids. 60. a) I 2 0.5 I o cos2 I 2 0.5 I o cos2 10° I 2 0.485 I o I 2 48.5% I o I b) 2 37.5% I o I c) 2 5.85% I o I d) 2 0.380% I o 61. I 2 0.4 I o I 2 0.5 I o cos2 0.4 I o 0.5 I o cos2 cos1
0.4 0.5
26.6° 62. I 2 0.5 I o cos2 1 and I 3 I 2 cos2 2 I 3 0.5 I o cos2 1 cos2 2 I 3 0.5 I o cos2 60° cos2 10° I 3 0.121 I o I 3 12.1% I o
B 41.6° n d) tan B 2 n1 B tan1
1.33 1.30
B 45.7° n 58. tan B 2 n1 n2 tan 60°
1.00
Solutions to End-of-chapter Problems
169
Chapter 11 23. a) b) c) d) 24.
b) d
constructive constructive partial destructive
1.0 m 10 500 slits
d 9.52 105 m m sin m d (2)(5.50 107m) sin 2 9.52 105 m
1
sin 2 0.01155 2 0.662° 27. L 1.0 m
S1 0
m
dxm L
xm
m L d
x2
2(5.50 10 m)(1.0 m) 2.0 10 m
S1
7
1
25.
6
x2 0.55 m
2
28.
1
1
0
2
S1
1
S2
1
2
3
3
Minima numbers
2
26. m 2 550 nm a) d 2.0 106 m m sin m d
1
sin 2 0.55 2 33.4°
2
2
(2)(5.50 107 m) sin 2 2.0 106 m
1
29. 560 nm d 4.5 106 m a) m 1 m sin m d
(1)(5.60 107 m) sin 1 4.5 106 m
sin 1 0.12444 1 7.14°
170
Maxima numbers
Solutions to End-of-chapter Problems
b) sin
d 1 m 2
3 (5.60 107 m) 2 4.5 106 m
32. 585 nm L 1.25 m m9 x 3.0 cm
sin
d
sin 0.18667 10.8°
(3)(5.60 107 m) sin 3 4.5 106 m
sin 3 0.37333 3 21.9° 1 m 2 d) sin m
d
7 7 m) (5.60 10 2 sin 3 4.5 106 m sin 3 0.43556 3 25.8° 30. 610 nm m2 2 23o d
m sin
19 (5.85 107 m)(1.25 m) 2 3.0 102 m
d
m c) sin m d
x 1 2
m L
d 2.3 104 m d 0.23 mm 33. 630 nm d 3.0 105 m d sin m
For maximization, sin 1. d m 5
m
3.0 10 m 6.30 10 m 7
m 4.76 1011 34. a) The light now travels an extra
twice 4 between the original and the second positions. This produces an extra shift
m
(2)(6.10 107 m) d sin 23° d 3.12 106 m d 3.12 m 31. d 0.15 mm m2 x2 7.7 m L 1.2 m
dxm
(1.5 10 m)(7.7 m) (2)(1.2 m)
mL
4
4.81 104 m 481 m
of . The observer therefore sees a dark 2 band and the fringe pattern moves by half a band. b) The light now travels an extra
twice 2 between the original and the second positions. This produces an extra shift of . The observer therefore sees a bright band and the fringe pattern moves by a full band. 3 c) The light now travels an extra twice 4 between the original and the second positions. This produces an extra shift 3 of . The observer therefore sees neither 2 a bright nor a dark band and the fringe 3 pattern moves by of a band. 2
Solutions to End-of-chapter Problems
171
d) The light now travels an extra twice
between the original and the second positions. This produces an extra shift of 2. The observer therefore sees a bright band and the fringe pattern moves by two full bands. 35. ∆ PD 4 n 1.42 600 nm PD t 2(n 1) (4)(6.00 107 m) t
m
g
1 2(3.64 107 m) (3.64 107 m) 2 m 3.64 107 m
m 2.5
The interference is destructive. 1 3 39. a) 2 2 2 2 destructive 1 b) 2 4 2 constructive 5 1 c) 2 3 4 2 constructive 7 1 15 d) 2 2 2 2 destructive 40. ng 1.40 560 nm t 4.80 106 m
2(0.42)
t 2.857 106 m t 2.86 m 36. ∆ PD 12 t 3.60 microns 640 nm PD 1 n 2t (12)(6.40 107 m) n 1 2(3.60 106 m)
n 2.07 37. ∆ PD 10 vm 1.54 108 m/s t 2.80 microns c nm vm 3.0 108 m/s nm 1.54 108 m/s nm 1.948 nm 1.95 2t (n 1) PD 2(2.80 106 m)(0.948)
10
5.309 107 m 531 nm 38. t 364 nm 510 nm ng 1.40 g ng
Because there is a half-phase shift between air and gas, 1 2t g 2
ng
g
g
5.60 10 m 1.40
7
g 4.00 107 m
Because there is a half-phase shift between air and gas, 1 2t g 2 m
g
1 2(4.80 106 m) (4.00 107 m) 2 m 4.00 107 m
m 24.5
The interference is destructive and a dark area will result.
7
g
5.10 10 m 1.40
g 3.64 107 m
172
Solutions to End-of-chapter Problems
b) Because of the phase shift and constructive
41. 500 nm a) nf 1.44 f nf
interference, 1 m s 2 t 2
7
f
5.00 10 m 1.44
f 3.47 107 m
Because there is a half-phase shift, 1 2t f 2 m
f
t
1 2
m f
1 1 (4.36 107 m) 2 t 2 t 1.09 107 m t 109 nm v 43. a) f
2
350 m/s 250 Hz
1.40 m v b) f
1 1 (3.47 107 m) 2 t 2 t 8.675 107 m t 86.8 nm b) nf 1.23
8
2.50 10 m/s 4.81 10 Hz 14
5.20 107 m 520 nm v c) f
f nf
5.00 107 m f 1.23 f 4.07 107 m Because the shifts cancel,
8
3.0 10 m/s 1.2 10 Hz 8
2.5 m v 44. a) f
m t f
2 (1)(4.07 107 m) t 2 t 2.03 107 m t 203 nm 42. 580 nm ns 1.33
3.0 108 m/s f 2.0 1012 m
f 1.5 1020 Hz b) v
14 km 1h 1000 m 1h 3600 s 1 km
v 3.889 m/s v f
s ns 7
s
5.80 10 m 1.33
s 4.36 107 m a) Because of the phase shift and destructive
f
3.889 m/s 1.2 m
f 3.2 Hz
interference, m t s
2 (1)(4.36 107 m) t 2 t 2.18 107 m t 218 nm
Solutions to End-of-chapter Problems
173
Maximum:
46. a) m 2 580 nm w 2.2 105 m
x
1 2 m
sin
w
1 2 (5.80 107 m) 2 sin 2.2 105 m sin 0.0659 3.78° b) m 2 550 nm w 2.2 105 m sin m
m w
(2)(5.50 107 m) sin 2 (2.2 105 m)
sin 2 0.05 2 2.87° 47. w 1.2 102 mm m1 1 4° w sin m
(1.2 10 m) sin 4° 1 7
8.37 10 m 837 nm 48. L 1.0 m m2 837 nm w 1.2 102 mm
x 0.174 m x 174 mm 49. w 1.1 105 m 620 nm m2 a) Minimum: m sin m w
(2)(6.20 107 m) sin 2 1.1 105 m
sin 2 0.113 2 6.47° b) Maximum:
1 2 w
1 2 (6.20 107 m) 2 sin 1.1 105 m sin 0.141 8.10° 50. Width of central maximum 6.6° 400 nm w
sin
4.00 107 m w sin 3.3° w 6.949 106 m w 6.95 m 51. 585 nm w 1.23 103 cm L 1.2 m a) m 3
Minimum: m L w
(2)(8.37 107 m)(1.0 m) x2 1.2 105 m
x2 0.1395 m x2 140 mm
1 2 (8.37 107 m)(1.0 m) 2 x 1.2 105 m
sin
m
xm
w
m
5
1 2
m L
xm
m L w
x3
(3)(5.85 10 m)(1.2 m) 1.23 10 m
x3 0.171 m x3 171 mm 174
Solutions to End-of-chapter Problems
7
5
w d N 1 102 m d
b) m 2
1 2 m L
x
10 000
w
1 2 (5.85 107 m)(1.2 m) 2 x 1.23 105 m
sin 2
m
d
d
0.0427
53. 493 nm w 5.65 104 m L 3.5 m m1 m L 1 a) xm w 2
m1 57. m 2 7
(1)(4.93 10 m)(3.5 m) (5.65 10 m)
4
m b) sin w 2 7
(1)(4.93 10 m) (5.65 10 m)
8.73 104
0.10°
54. 450 nm 55. 530 nm N 10 000 slits w 1 cm m1
2000
x 6.1 103 m x 6.1 mm
w N 1 102 m
6
d 5.00 106 m d sin m 1 m 2 6 (5.00 10 m) sin 11.25° 1 m 7 (6.50 10 m) 2
4.90°
sin 2 sin 2
(1)(5.30 10 m) 1 10 m
sin 1 0.53 1 32° 56. 650 nm N 2000 slits w 1 cm m 11.25°
sin w 2 (1)(4.70 107 m) sin 2 1.10 105 m
x 2
7
sin 1
x 0.1426 m x 143 mm 52. w 1.10 103 cm 470 nm m1
d 1 106 m m sin m d
4
d
1 2.3 10 slits/mm 4
d 4.35 105 mm L 0.95 m 610 nm m L xm d (2)(6.10 107 m)(0.95 m) x2 (4.35 108 m)
x2 27 m 58. N 10 000 slits w 1.2 cm w d N 1.2 102 m d
10 000
d 1.2 106 m
Solutions to End-of-chapter Problems
175
a) 600 nm d m 1.2 106 m m 6.00 107 m
m2 b) 440 nm d m 1.2 106 m m 4.40 107 m
m 2.7 m2 1 102 m 59. d
d 1 105 m d m400 1 105 m m400 4.00 107 m
m400 25 d m700
m 2.3 m2 61. 1 589 nm 2 589.59 nm w 2.5 cm N 104 w d N 2.5 102 m d 4
d 2.5 106 m m2 m1 sin1 sin1 d d 1
sin
sin1
1 105 m m700 7.00 107 m
m700 14.3 m700 14
Orders needed: 25 14 11 60. d 1.0 microns a) 610 nm
5.8959 10 m 2.5 10 m 5.89 10 m 2.5 10 m
6
1.0 10 m 6.10 10 m 7
m 1.6 m1 b) 575 nm d m 1.0 106 m m 5.75 107 m
7
6
13.641° 13.627° 1.39 102° 62. ∆ 2 1 7 7 ∆ 5.8959 10 m 5.89 10 m 10 ∆ 5.9 10 m 2 avg 1 avg
6
2 (5.89 10 m) (5.8959 10 m) 2
avg 5.8930 107 m m2 N avg m (5.8930 107 m) N (5.9 1010 m)(2)
N 500 63. N 106 w 2.5 cm 520 nm w d N 2.5 102 m d 6
10
d 2.5 108 m 176
7
7
d m
m 1.7 m1
10
1000
m
c) 430 nm d m 1.0 106 m m 4.30 107 m
Solutions to End-of-chapter Problems
7
d m 8
m
2.5 10 m 5.20 10 m 7
m 0.0481 m0 64. N 4000 m1 7 7 ∆ (6.5648 10 m) (6.5630 10 m) 10 ∆ 1.8 10 m avg
7
7
(6.5648 10 m) (6.5630 10 m) 2
avg 6.5639 107 m avg R 6.5639 107 m R 1.8 1010 m
R 3647 Nm (4000)(1) Nm 4000 R Nm, therefore it will not be resolved. 65. 0.55 nm d
1m 2.5 10 6
d 4.0 107 m m1 m sin d (1)(5.5 1010 m) sin 4.0 107 m
sin 1.375 103 7.9 102° Diffraction is not apparent. 66. d 0.40 nm 0.20 nm m3 sin
m
2d (3)(2.0 1010 m) sin 2(4.0 1010 m)
sin 0.75 49°
Solutions to End-of-chapter Problems
177
Chapter 12
23. E 4.5 eV, W 0(gold) 5.37 eV E W 0. The gold will absorb all of the
19. max 597 nm 5.97 107 m
The temperature can be found using Wien’s law: 2.898 103 max
T 2.898 10 T 3
max
2.898 103 T 5.97 107 m
T 4854.27 K T 4854.27 273°C T 4581.27°C 20. T 2.7 K max can be found using Wien’s law: 2.898 103 max T 2.898 103 max 2.7 K max 1.07 103 m 21. T 125 K max can be found using Wien’s law: 2.898 103 max T 2.898 103 max
125 K
max 2.32 105 m
The peak wavelength of Jupiter’s cloud is 2.32 105 m. It belongs to the infrared part of the electromagnetic spectrum. 22. P 2 W, 632.4 nm 6.324 107 m We are to find the number of photons leaving the laser tube per second. Let us symbolize this quantity by N . Using Planck’s equation, we can express the energy for a single photon: hc E
The number of photons leaving the tube can be found as follows: P N E P N hc N
energy of the incident photons, hence there will be no photoelectric effect observed (see Figure 12.13). 24. 440 nm 4.4 107 m, W 0(nickel) 5.15 eV First, we shall calculate the energy of the incident photons. Using Planck’s equation: hc E
(6.626 1034 J·s)(3.0 108 m/s) E 4.4 107 m
E 4.52 1019 J 4.52 1019 J E 1.6 1019 C
E 2.82 eV Since E W 0, the photoelectric effect will
not be exhibited (see Figure 12.13). 25. P 30 W, 540 nm 5.4 107 m We are to find the number of photons radiated by the headlight per second. Let us symbolize this quantity by N . Using Planck’s equation, we can express the energy for a single photon: hc E
The number of photons radiated by the head light can be found as follows: P N E P N hc N
34
8
N 8.15 1019 photons/s 26. W 0 3 eV 4.8 1019 J, 219 nm 2.19 107 m a) The energy of photons with cut-off fre-
quency is equal to the work function of the metal. Hence, E W 0 4.8 1019 J 7
(2 W)(6.324 10 m) (6.626 10 J·s)(3.0 10 m/s)
34
8
N 6.36 1018 photons/s
178
7
(30 W)(5.4 10 m) (6.626 10 J·s)(3.0 10 m/s)
Solutions to End-of-chapter Problems
The frequency can be found using Planck’s equation: E hf E f h
4.8 1019 J f 6.626 1034 J·s
f 7.24 1014 Hz b) The maximum energy of the ejected pho-
tons can be found using the equation: Ekmax E W 0 hc Ekmax W 0 34
8
(6.626 10 J·s)(3.0 10 m/s) 2.19 10 m
Ekmax
7
19
4.8 10 J Ekmax 4.28 1019 J 27. a) To avoid unwanted electrical currents and change in bonding structure of the material of the satellite, the number of electrons ejected from the material should be minimal. The greater the work function of the metal, the more photon energy it will absorb and the fewer electrons will be ejected. Hence, the material selected should have a relatively high work function. b) The longest wavelength of the photons that
could affect this satellite would have an energy equal to the work function of the material, i.e., E W 0
hc
Using Planck’s equation E , max
hc
(if W 0 is in Joules)
W 0 hc max (if W 0 is in eV) W 0e 28. W 0(platinum) 5.65 eV 9.04 1019 J
From problem 27, we know that: hc max W 0 max
34
8
(6.626 10 J·s)(3.0 10 m/s) 9.04 10 J
19
max 2.2 107 m
29. a) For a material with a work function
greater than zero, the typical photoelectric effect graph has a positive x intercept. If the graph passes through the origin, the work function of the material is zero, which means that the photoelectric effect would be observed with incident photons having any wavelength. b) If the graph has a positive y intercept, we would observe the photoelectric effect without the presence of incident photons. 30. 400 pm 4.0 1010 m a) The frequency of the photon can be found using the wave equation: c f 8
f
3.0 10 m/s 4.0 10 m 10
f 7.5 1017 Hz b) The momentum of the photon can be com-
puted using de Broglie’s equation: h p 34
p
6.626 10 J·s 4.0 10 m 10
p 1.66 1024 N·s c) The mass equivalence can be found using
de Broglie’s equation: p mv p m c 24
m
1.66 10 N·s 3.0 10 m/s 8
m 5.53 1033 kg 31. mproton 1.673 1027 kg
First, we have to express the rest energy of the proton. It can be found using: Eproton mc 2
The energy of the photon, which is equal to the rest energy of the proton, can be expressed using Planck’s equation: hc E
The maximum wavelength of the photon that could generate the photoelectric effect on the platinum surface is 2.2 107 m. Solutions to End-of-chapter Problems
179
Then,
To find the Compton shift,
Eproton E hc mc 2 h mc
f i 1 109 m 9.9552 1010 m 4.48 1012 m The Compton shift is 4.48 1012 m. 34. 180°, vf 7.12 105 m/s
Using de Broglie’s equation:
From the conservation of energy,
h p
Ei Ef Ek hc hc 1 mvf 2 2 i f
Hence, p mc p (1.673 1027 kg)(3.0 108 m/s) p 5.02 1019 N·s 32. 10 m 1 105 m
Using de Broglie’s equation:
From the conservation of momentum, pi pf pe h h mvf i f
(eq. 2)
(The negative sign signifies a scatter angle equal to 180°.) Multiplying equation 2 by c and adding the result to equation 1, 2hc 1 mvf 2 cmvf 2 i
h p 34
p
(eq. 1)
6.626 10 J·s 1 10 m 5
p 6.63 1029 N·s 33. f 1 nm 1 109 m
i
Consider the following diagram:
2hc 1
2
m
y
2
vf
cvf
34
i
31
(9.11 10
e x f C
kg)
2 (7.12
5
2
10 m/s) (3.0 108 m/s)(7.12 105 m/s)
i 2.04 109 m 35. i 18 pm 1.8 1011 m, energy loss is 67%
θ
43°
8
2(6.626 10 J·s)(3.0 10 m/s) 1
x
The initial energy of the photon can be computed using Planck’s equation:
x i
hc Ei i
From the conservation of energy, Ei Ef Ek hc hc 1 mvf 2 2 i f
Ei
(eq. 1)
From the conservation of momentum, pi pf pe
In the direction of the x axis: h h cos 43° mvf cos (eq. 2) i f In the direction of the y axis: h sin 43° mvf sin (eq. 3) i Using math software to solve the system of equations that consists of equations 1, 2, and 3, the value for i 9.9552 1010 m. 180
34
8
(6.626 10 J·s)(3.0 10 m/s) 1.8 10 m
11
Ei 1.1 1014 J
Since 67% of the energy is lost, the final energy of the photon is: Ef 0.33 Ei Ef 0.33(1.1 1014 J) Ef 3.64 1015 J The final wavelength can be calculated using Planck’s equation: hc f Ef
(6.626 1034 J·s)(3.0 108 m/s) f 3.64 1015 J
f 5.45 1011 m
Solutions to End-of-chapter Problems
The Compton shift as a percentage is: 5.45 1011 m f 100% 1.8 1011 m i
v
f 302% i The wavelength of a photon increases by 302%. 36. m 45 g 0.045 kg, v 50 m/s Using de Broglie’s equation: h mv 34
6.626 10 J·s (0.045 kg)(50 m/s)
The wavelength associated with this ball is 2.9 1034 m. 37. mn 1.68 1027 kg, 0.117 nm 1.17 1010 m Using de Broglie’s equation: h mv h m
6.626 10 J·s (1.68 10 kg)(1.17 10 m) 27
10
v 3371 m/s
The velocity of the neutron is 3371 m/s. 38. mp 1.67 1027 kg, 2.9 1034 m Using de Broglie’s equation:
h
mv h v m
31
v 4.19 106 m/s Now can be found using de Broglie’s
equation:
h
6.626 10 J·s (9.11 10 kg)(4.19 10 m/s)
mv
34
31
6
b) The Bohr radius is 5.29 1011 m. The
wavelength associated with an electron is longer than a hydrogen atom. 40. The photon transfers from n 5 to n 2. The energy at level n is given by: 13.6 eV En
n 2
E E5 E2 13.6 eV 13.6 eV E 2 2
5
2
E 2.86 eV E 4.58 1019 J
To compute the wavelength: hc E
(6.626 1034 J·s)(3.0 108 m/s) 4.58 1019 J
34
v
18
The energy released when the photon transfers from n 5 to n 2 is: 34
v
Ek m
1.73 1010 m
2.9 1034 m
v
v
2 2(8 10 J) 9.11 10 kg
6.626 10 J·s (1.67 10 kg)(2.9 10 m) 27
34
v 1.37 1027 m/s
The speed of the proton would have to be 1.37 1027 m/s. Since v is much greater than c , this speed is impossible. 39. Ek 50 eV 8 1018 J, me 9.11 1031 kg a) We shall first compute the velocity using the kinetic energy value: 1 Ek mv2 2
4.34 107 m
The wavelength released when the photon transfers from n 5 to n 2 is 4.34 107 m. It is in the visual spectrum and it would appear as violet. 41. a) The electron transfers from n 1 to n 4. The energy of the electron is given by: 13.6 eV En
n 2
The energy needed to transfer the electron from n 1 to n 4 is: E E4 E1 13.6 eV 13.6 eV E 2 2
4
1
E 12.75 eV Solutions to End-of-chapter Problems
181
b) The electron transfers from n 2 to n 4.
Similarly, the energy needed to transfer the electron from n 2 to n 4 is: E E4 E2 13.6 eV 13.6 eV E 2 2
2
4
42. 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 r n (5.29 1011 m)n2 The difference in radii is: ∆r r 3
r 2 11 ∆r (5.29 10 m)(3)2 (5.29 1011 m)(2)2 10 ∆r 2.64 10 m 43. n 1
The radius of the first energy level can be found using: r n (5.29 1011 m)n2 r n (5.29 1011 m)(1)2 r n 5.29 1011 m The centripetal force is equal to the electrostatic force of attraction:
9
2
2
19
2
11
2
F 8.22 108 N
The centripetal force acting on the electron to keep it in the first energy level is 8.22 108 N. 44. F 8.22 108 N, r 5.29 1011 m F m4 2rf 2 1 F f 2 mr
f 3.08 1015 Hz
The frequency of the photon is 3.08 1015 Hz, or one-half the number of cycles per second completed by the electron in problem 44. 46. Bohr predicted a certain value for energy at a given energy level. From the quantization of energy, there can be only specific values for velocity, v, and radius, r . Thus, the path of the orbiting electron can attain a specific path (orbit) around the nucleus, which is an orbital. 48. v 1000 m/s, m 9.11 1031 kg ∆py ∆ y ≥
–h
∆p
m ∆v –h y mv
1 8.22 10 N 2 (9.11 10 kg)(5.29 10 m)
31
8
11
y 1.16 107 m
Hence, the position is uncertain to 1.16 107 m. 49. ∆ y 1 104 m The molecular mass of oxygen is 32 mol. The mass of one oxygen molecule is 32 mol 5.32 1026 kg 23 6.02 10 mol/g
From ∆py∆ y ≥ –h and ∆p m∆v, the maximum speed is: v
h mv
v
1.0546 10 J·s (5.32 10 kg)(1 10 m)
f 6.56 1015 Hz
–
34
The electron is orbiting the nucleus 6.56 1015 times per second.
182
2.04 1018 J 6.626 1034 J·s
(8.99 10 N·m /C )(1.6 10 C) (5.29 10 m)
f
1.0546 1034 J·s y (9.11 1031 kg)(1000 m/s)
ke2 F r 2
f
the energy released is equal to 12.75 eV 2.04 1018 J. The frequency is then equal to: E f h
E 2.55 eV
F
45. Consider an electron transferring from n 4 to n 1. As computed in problem 41,
26
v 1.98 105 m/s
Solutions to End-of-chapter Problems
4
Chapter 13 28. a) 3 cm/a
3 cm 1m 1a 1a 100 cm 365.25 d
1d 86400 s 9.5 1010 m/s 9.5 1010 m/s 3.16 1018 3.0 108 m/s 0.1 mm 1m b) 0.1 mm/s 1s 1000 mm 1.0 104 m/s 1.0 104 m/s 3.3 1013 8 3.0 10 m/s 10.8 m/s c) 3.6 108 8 3.0 10 m/s d) Mach 6.54 6.54 332 m/s Mach 6.54 2171.28 m/s 2171.28 m/s 7.24 107 8 3.0 10 m/s 2.2 106 m/s e) 7.33 103 8 3 10 m/s 29. a) Snoopy must fly 50 km/h [N then E]. Let y resultant ground speed y (130 km/h)2 (50 km/h )2 y 120 km/h b) The Baron going west has a ground speed of:
d)
t S t B
t S t B
b vw w vg bvg 130 km/h 50 km/h bvg 180 km/h [W]
vw
vw vw v vw vw
2
S
t B
2
v
t S t B t S t B
v w v 2
2
2
w2 v
1 2
30. In our rest frame, we observe the contracted
length:
v2 c L (1.0 m)1 (0.080)2 L 0.6 m L L0 1
2
1 3
31. L L0
v2 L L0 1 c 2
While going east, bvg 130 km/h 50 km/h bvg 80 km/h [E] c) The time for Snoopy: 200 km 3600 s 6000 s 120 km/h 1h Time for the Baron: 100 km 100 km 3600 s 180 km/h 80 km/h 1h
t v w
bvg
2
(v w)(v w) t S t B v(v w)(v w) t (v w)(v w) S t B v
2
100 100 200 ) ( )( 200 ( )() v w
200 v w
6500 s
1 v2 1 3 c 2 v2 1 1 c 2 9 v2 8 2 c 9
89
v c
v 0.943c v 2.83 108 m/s
Therefore, Snoopy wins the race by 500 s or 0.139 h.
Solutions to End-of-chapter Problems
183
32. Length is contracted for the moving stop-
watch. The time it measures is:
L t v
t
v2 v2 1 1 2 2c 2 c v2 L0 1 1 2 2c t
v2 1 2 c v
L0
Using the low-speed approximation when v c :
(180 m)1 (0 .7)2 t 0.7(3 108 m/s) t 6.12 107 s 33. We observe the dilated half-life of the muon:
35 m/s
t
t
t 0 v2
35 000 m t
(35 m/s) 2(3.0 10 m/s) 2
35 m/s
t 6.81 1012 s
c 2 2.6 108 s
and its rest half-life: t 0 2.2 106 s
1 (0 .998) 2
t 4.11 107 s
t
The distance travelled is:
2.2 1 2 .8
2
35. The time the girlfriend measures is: L0 t f
35 m/s
The time Henry measures is: d t h v
35 m/s
Their time difference is: t
2
c
L 35 m/s 35 m/s 0
v2 L0 1 c 2
v
1.856 108 m/s v d circumference d 2r d vt vt r 2
(1.856 108 m/s)(2.8 106 s) 2 r 82.7 m 37. Only the component of L0 in the direction of travel is contracted: Lx L0 cos 30° The contracted length seen by Tanya in the direction of travel ( x) is: r
v2 L0 1 c 2
1 1
2.2 v2 2.8 c 2 v2 2.2 2 1 c 2 2.8
34. Katrina measures a contracted distance: v L L0 1 c 2 L (7.83 1010 m)1 (0.25)2 L 7.58 1010 m
t 0 v2 c 2
d vt d (0.998c )(4.11 107 s) d 123 m
t h
2
36. Given the muon’s dilated half-life: t 2.8 106 s
1
8
cos 30° 1
v2 Lx Lx 1 c 2 Lx L0
v2 c 2
The perpendicular length, Ly, is L0 sin 30° for both Katrina and Tanya.
184
Solutions to End-of-chapter Problems
39. 1 ca vt 1 ca (3.0 108 m/s) (365.25 24 60 60 s) 1 ca 9.47 1015 m 40. Using spacetime invariance: (∆s2) c 2(∆t J)2 (∆ x J)2
Ly tan 45° Lx Ly 1 Lx Ly L0 sin 30° v2 Lx L0 cos 30° 1 c 2
Therefore: sin 30° 1 cos 30° 1 1 tan 30° 1 1 3 L0
v2 c 2
L0
and: (∆s2) c 2(∆t T)2 (∆ xT)2 For Ted, the distance between events is: c 2(1.0 106 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 m 41. Ted’s length, L, has contracted relative to Jane’s length, L0:
v2 c
2
v2 c v2 1 1 2 c 3 v2 2 2 c 3 v 0.816c v 2.45 108 m/s
2
520 m
300 m/s
t 0 1.336 105 s
For the clocks on Earth, use the low-speed approximation for v c :
t
t 0 v2
1 1 2 c 2 v2 c 2
t 0
The difference in the flying clocks compared to the ones on Earth is: t t t 0
t
v
(300 m/s) 10 s) 2(3.0 10 m/s) 2
5
t 6.68 108 s
t
t 0 v2
1 1 c 2
4.0 s v2 5.0 s c 2 v2 16 1 c 2 25 3 v c 5 The distance travelled in the 5.0 s is: 3 5 d 9.0 108 m 43. See problem 42: 3 v c 5 v 1.8 108 m/s
d (3.0 108 m/s)(5.0 s)
v2 1 1 2c 2
t (1.336
v2 c
2
d vt
v2 t t 0 1 t 0 c 2
2r
v2 169 1 c 2 225 v2 56 2 c 225 v 0.499c v 1.50 108 m/s 42. The dilated time of the stationary observer is:
38. The time to travel a circumference is: 2r t v 2(6.38 106 m) t 0
t
(600 m) 1
v2 L L0 1 c 2
8
2
Solutions to End-of-chapter Problems
185
44. Trevor’s time is: d t 0 v
46. The centripetal force is provided by the elec-
trical coulomb force: 2
mv r
v2 c
v 2 L0 1 2
t 0
The time difference is: t 1 a
v2 c
v
∆m
v2 c
2
tv v
1 2
2
v2 c 2
m m0
m0
m m0
v2 c 2
m0 v2 c 2
m
4
2
8
2
m 3.0 107 kg
0
2
qB 1 c 2 [9.11 1031 kg][0.8(3.0 108 m/s)]
48. Use the high-speed approximation: v2 2 1 v 1 2 c c
m
(1.602 10 C)(1.5 T) 1 (0 .8) 2
m
m0 v2
1 c 2 m0
v 21 c
9.11 1031 kg m 2(1 0.999 999 999 67) 9.11 1031 kg m 10 2(3.3 10 ) m 3.55 1026 kg
186
2
8
2
m m0
m
mv v
r 1.52 103 m
c 2
2(1 1 (0 .95c ) )
19
19
1 2 1 1 2 2 (60 kg) (3.0 10 m/s) 2 (3.0 10 m/s)
c 2 (1 a)(0.95c )
r
Use the low-speed binomial approximation when v c : 1 v2 1 v2 2c 2 1
L0 0.691 ca 45. q 1.6 1019 C v 0.8c B 1.5 T m0 m v2 1 2 c mv r qB r
2
31
masses is:
2 1
L0
47. The difference between the dilated and rest
L0
2
r 6.26 1015 m
2 L0 1
9
2 L0 1 2
2 L t 0 1 v
ke mv2
(9.0 10 N·m C )(1.602 10 C) 1 (0 .6) r (9.11 10 kg)[0.6(3.0 10 m/s)]
v
v
r
v2 ke2 1 c 2 r m0v2
t
kQq r 2 2
His sister’s time is: 2 L0 t
Solutions to End-of-chapter Problems
2
49. For a charge moving perpendicular to a mag-
51. Using the relativistic equation of velocity addi-
netic field, the centripetal force equals the magnetic force:
tion, the velocity of the light relative to the duck is:
mv2 Bqv r
lvd
mv v 0
lvd
2
31
19
8
2
B 2.26 102 T
density
m
lvd
c
av b
xyz
1 2
c v2 x0 1 yz c 2 0 v2 1 2 c
2
1 2 c
v2 1 1 2 c 2 2 v 1 2 c 2 v 0.7071c v 2.1 108 m/s
a E
E b
a E E b 2
1 c
av b
0.2c 0.3c 1 (0.2)(0.3)
0.472c 8 av b 1.42 10 m/s
53. The speed of rocket A relative to Earth is: av b bvE avE av b bvE 1 c 2 0.8c 0.7c avE 1 (0.8)(0.7) avE 0.962c 8 avE 2.88 10 m/s 54. The speed of the positron relative to the elec-
When the density of an object is dilated twice as much as its density at rest, 20 : 0
v v v v
v
av b
m0 v2
20
1.2c 1.2
lvd
tion, the velocity of star A relative to star B is:
where x, y, and z are the rectangular dimensions. Contraction occurs only in the direction of motion, so density is:
c 0.2c (c )(0.2c )
52. Using the relativistic equation of velocity addi-
mass volume
50. density
1
c
(9.1 10 kg)(3.0 10 m/s)(0.999 999 986) (1.602 10 C)(450 m) 1 (0 .999 9 99 986 )
c d
l c c d 2
1 2
qr 1 c 2
B
l c
c
Due to mass dilation, the magnetic field is: B
v v vv
tron is: pve
v v vv p g
g e
p g g e 2
1 c
pve
0.95c 0.85c 1 (0.95)(0.85)
0.996c 8 pve 2.988 10 m/s pve
55. Bob’s velocity relative to Earth, bvE 0.3c ;
Nicole’s velocity relative to Earth, nvE 0.9c pvE, the phaser bullet’s velocity relative to Earth.
Solutions to End-of-chapter Problems
187
The velocity of the phaser bullet relative to Bob, pv b, is: pv b
v v vv p E
E b
p E E b 1 2
c
pv b
0.9c 0.3c 1 (0.9)(0.3)
0.822c 8 pv b 2.47 10 m/s pv b
56. Kirk’s velocity relative to Earth: kvE X the module’s velocity relative to Kirk: mvk X
the module’s velocity relative to Earth: mvE 0.8c mvk kvE mvE mvkkvE 1 c 2 X X 0.8c X 2 1 c 2 0.8 X 2 0.8c 2 X c 0.8 X 2 2cX 0.8c 2 0 2 X 2 5cX 2c 2 0 (2 X c )( X 2c ) 0 The speed of the Enterprise is: c 8 1.5 10 m/s
2 57. The mass, m, equivalent to the chemical energy released is: E mc 2 m
60. E mc 2 E (m0c 2 Ek)
The work done in increasing an electron’s speed is: Ek Ek Ek Ek (mc 2 m0c 2) (mc 2 m0c 2) Ek (mc 2) mc 2
Ek m0c 2
c 2
m 3.56 10 kg 58. The mass, m, equivalent to the chemical
energy released is: E mc 2 10
9.2 10 J c
2
1 2 c
2
1 2 c
For v 0.5c to v 0.9c : 1 1 Ek m0c 2 2 1 (0 1 (0 .9) .5)2 Ek 1.139m0c 2 For v 0.9c to v 0.95c : 1 1 Ek m0c 2 2 1 (0 1 (0 .95) .9)2 Ek 0.908m0c 2 It takes more work to increase from 0.5c to 0.9c . 61. To find the equivalent mass of the particle:
8.19 1014 J m (3.0 108 m/s)2 m 9.1 1031 kg m the mass of an electron 62. To find the difference between the dilated relativistic and the classical momentum,
p p p0 p mv m0v
2
m 1.02 106 kg 59.To find the energy equivalent of 1.0 kg of
p m0v
bananas:
1 1 v2 1 2
2
E mc E (1.0 kg)(3.0 108 m/s)2 E 9.0 1016 J 9.0 1016 J E 3.6 106 J/kWh E 2.5 1010 kWh
188
1 1 v v
E mc 2
3.2 104 J 13
m
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 billion Conversely, the rate of relativistic “banana” power is: $1.29 $0.000 000 000 052/kWh 2.5 1010 kWh
Solutions to End-of-chapter Problems
c
Since v 75 103 m/s, or v c , use the low-speed binomial approximation: v2 1 1 2 2c v2 1 2
c
v2 p m0v 1 1 2c 2 3
p
mv 2c
p
(125 kg)(75 000 m/s) 2(3.0 10 m/s)
0
65. Accelerating the electron of mass, m0 9.1 1031 kg, and charge, q 1.6 1019 C, from rest, through a potential of V results in a new total energy: E m0c 2 Vq E mpc 2 mpc 2 m0c 2 Vq m c 2 m0c 2 V p q
2
3
8
V
2
p 0.29 kg·m/s 63. Since v c , use the low-speed approxima-
tion: 1
v 1 2 2c 2 v
1 c 2
The work done, ∆ Ek, in speeding Mercury from rest is given by: Ek mc 2 m0c 2
1 1 v2 1 2 c
v2 Ek m0c 2 1 1 2c 2
2
mv 2
Ek
(3.28 10 kg)(4.78 10 m/s) 2
23
4
2
Ek 3.75 1032 J
The mass equivalent, ∆m, to this amount of energy is: m
Ek c 2
27
31
19
V 9.38 108 V V 938 MV
v mvc (where E mc 2 m0c 2 + Ek) c mc 2 v 20 J c 29 J v 0.69c
For 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 J To find the velocity of B,
E 64. m 2 c qV m c 2 19
8
(1.602 10 C)(1.35 10 V ) (3.0 10 m/s) 8
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 J To find the velocity of A,
v mvc c mc 2 v 18.9 J c 29 J v 0.65c
m 2.4 1028 kg
3.75 1032 J m (3.0 108 m/s)2 m 4.16 1015 kg
m
2
Ek
0
66. Using the energy triangle, E2 (mvc )2 (m0c 2)2 E2 (m0c 2 Ek)2
2
Ek m0c 2
8
(3.0 10 m/s) [(1.67 10 kg) (9.1 10 kg)] 1.60 10 C
2
Particle A has the greater speed. 67. E2 (mvc )2 (m0c 2)2 E2 (mvc )2 (938.3 MeV)2 E2 (0.996mc 2)2 (938.3 MeV)2 E2(1 0.9962) 8.804 105 MeV2 E 1.05 104 MeV
Solutions to End-of-chapter Problems
189
68. E mc 2 m0c 2 Ek mc 2 (0.511 MeV) (3.1 103 MeV) mc 2 3100.511 MeV
From the energy triangle: cos
m0c E
cos
0.511 MeV 3100.5 MeV
89.990557° mvc sin mc 2 v sin c 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: mvc sin mc 2 mvc v 2 mc c mvc tan m0c 2
For particle A: mvc (4 108 N·s)(3 108 m/s) m0c 2 20 J
mvc 0.60 m c 0
2
tan 0.60 30.96° v c v 0.514c
sin (30.96°)
For particle B: mvc (5 108 N·s)(3 108 m/s) mc 2 30 J
mvc 0.50 mc 2 v 0.50c
Particle A is faster.
190
Solutions to End-of-chapter Problems
52. Assuming the uranium nucleus is fixed at rest
Chapter 14 43. a) Cl b) Rn c) Be d) U e) Md 44. For A Z X , Z is the number of protons and A Z
is the number of neutrons: a) 17 protons, 18 neutrons b) 86 protons, 136 neutrons c) 4 protons, 5 neutrons d) 92 protons, 146 neutrons e) 101 protons, 155 neutrons 45. Since 1 u 931.5 MeV/c 2, then 18.998 u 931.5 MeV/c 2/u 17 697 MeV/c 2. 106 MeV/c 2 0.114 u. 46. Conversely, 931.5 MeV/c 2/u 47. To find the weighted average of the two isotopes: 0.69(62.9296 u) 0.31(64.9278 u) 63.55 u This is closest to the mean atomic mass of Cu. 48. B [ Zm(1H) Nmn m(146 C)]c 2 B [6(938.78) 8(939.57) (14.003 242 u)(931.5)] MeV B 105.22 MeV B 105.22 MeV 7.5 MeV/nucleon A 14 nucleons
49. Since 146C → 147N 01 e v, the
N ratio Z
8 7 4 changes from to or from to the 6 7 3 1 more stable . 1 50. The binding energy is: B [m(3He) mn m(4He)]c 2 B [3.0160 u 1.008 665 u 4.002 60 u]c 2 931.5 MeV/c 2/u B 20.55 MeV 228 4 51. Since 232 92U → 90Th 2He Ek, Ek [mU mTh m]c 2 Ek [232.037 131 u 228.028 716 u 4.002 603 u]c 2 931.5 MeV/c 2/u Ek 5.41 MeV
and the kinetic energy of the alpha particle becomes electrical potential, Ek
kq q r
r
kq q E
1 2
1 2 k
r
9
2
19
2
(8.99 10 J m/C )(1.6 10 C) (2)(92) (5.3 10 eV)(1.6 10 J/eV)
·
6
19
r 5.0 1014 m 0 231 53. 231 v 90Th → 91Pa 1e 235 231 4 92U → 90Th 2He 54. The mass difference is: ∆m mn (mp me) 2 ∆m [939.57 938.27 0.511] MeV/c 2 ∆m 0.789 MeV/c 55. From problem 54, the energy equivalent of 0.789 MeV/c 2 is 0.789 MeV.
2 Thus (0.789 MeV) 0.526 MeV. 3 56. Since the total momentum before decay is equal to the total momentum after decay, and p 0 p, the three momentum vectors must form a right-angle triangle. From Pythagoras’ theorem: pC2 pe2 p2 21 2 pC (2.64 1021)2 (4.76 10 ) 21 pC 5.44 10 N·s p2 57. Using Ek , the recoiling carbon nucleus 2m
will have 21
Ek
2
(5.44 10 N·s) 2(12.011 u)(1.6605 10 kg/u) 27
Ek 7.42 1016 J 58. For a fixed gold nucleus at rest, the kinetic
energy of the 449-MeV alpha particle is converted to electrical potential. Thus, for the radius, Ek
kq q r
r
kq q E
1 2
1 2 k
r
9
2
19
2
(8.99 10 J·m/C )(1.6 10 C) (2)(79) (449 10 eV)(1.6 10 J/eV)
6
19
r 5.07 1016 m
Solutions to End-of-chapter Problems
191
59.
62. If the amount of radioactive material is 23%
% of Original Dose still Radioactive vs. Time
100
of the original amount after 30 d, then, 1 t N N 0 T 2 30 d 1 0.23 N 0 N 0 T 2 30 d 1 log (0.23) log T 2
80 e v i t 60 c a o i d a r 40 %
1
2
1
2
1 (30 d) log 2 1 2
20
0
4
8
12 t (h)
16
When t 8 h, 39.7% of the original dose is still radioactive. 60. For carbon-14, T 12 5730 a. Comparing the relative amount, N R, of a 2000-a relic with the amount, N S, in a shroud suspected of being 2002 a 1350 a 650 a, yields:
1 2
1 2 N R 1 N S 2
T 12
20
t
R
T 1
log (0.23)
T 12 14 d
63. The molar amount of 235U is
5.12 0.0218 235
3.4 2 and of 207Pb is 0.0165. The 207 original molar amount of 235U was 0.0218 0.0165 0.0383. Using the decay formula where T 7.1 108 a, 1 2
2
t
R
N N 0
T 1
2
1 2
t
T 1
2
2000 a 650 a N R 5730 a N S N R 0.85 N S
0.0218 0.0383 log
2
0.0383 1 log 2 0.0218
log t
T 1
8
1
t
t 0.0218 1 log 0.0383 7.1 10 a 2
61. The half-life of Po-210 is: T 12 138 d 198 720 min The half-life of Po-218 is T 12 3.1 min
After 7.0 min, there will be: 1 t 1 210 Po: N N 0 T 2 2 1 log N (3.5 105)log 2 N 100% t 1 1 218 T Po: N N 0 2 2 1 log N (2.26)log 2 N 20.9% There will be a total of: 1(1 g) 0.209(1 g) 1.21 g Therefore, 1.21 106 g of radioactive Po remains.
1 2
(7.1 108 a)
7.0 min 198 720 min
t 5.78 108 a
2
1
2
192
7.0 min 3.1 min
64. Using the activity decay formula where T 12 5730 a for 14C decay,
t 1 T N N 0 2 1 750 900 2 t 750 1 log log 900 5730 a 2 5 log (5730 a) 6 t 1 log 2 t 1507 a 1
2
t 5730 a
Solutions to End-of-chapter Problems
65. For an isotope to be doubly stable, its values for both Z and N A Z must be “magic”
nuclear shell numbers, where the numbers are 2, 8, 20, 28, 50, 82, and 126. The other 48 doubly stable isotopes are 42He, 168O, 40 20Ca, 20Ca, 78 132 28Ni, and 50 Sn. 137 0 66. 137 v Ek. To determine the 55Cs → 56Ba 1 e maximum Ek available per disintegration, find the mass difference of the parent nucleon and the daughter plus the electron. Ek [136.9071 u (136.9058 u 0.000 549 u)]c 2 931.5 MeV/c 2/u Ek 0.6996 MeV 67. Dose
Dose
activity time energy percentage absorbed mass
6
19
(3700 Bq)(365 24 60 60 s)(1.0 10 eV)(1.6 10 J/eV)(5%) 70 kg
Dose 0.013 mGy 68. The pilots fly for 52 weeks 20 h/week 1040 h per year. Thus, their exposure is: (7.0 106 Sv/h)(1040 h/a) 7.28 103 Sv/a. Compared with the average of 2 mSv/a, this 7.28 value is about 3.64 times greater. 2 238 206 69. Since 92U → 82Pb, 238 206 32 nucleons are lost through alpha decay in groups of 4 nucleons per decay. 32 Thus, there are 8 alpha particles 4 emitted. The number of beta decays is equal to the number of neutrons changed into protons. N protons in Pb protons left after alpha decay N 82 (92 8 2) N 6 beta particles emitted 70. Four beta decays means that four neutrons 208 were changed into protons, or 208 82Pb 824 X . Six alpha decays means that 208 78 X came from 20864 232 7862Y 90Y . From the periodic table, this element is thorium-232 or 232Th.
71. The separation distance of an alpha particle ( A 4) and a nitrogen nucleus ( AN 14)
is given by: r s r r N 3 3 r s 1.2 A 1.2 AN 3 3 r s 1.2 4 1.2 14 r s 4.8 fm 72. Considering the nitrogen nuclei to be fixed at rest, the Ek of the incoming alpha particle
is converted to electrical potential, or Ek
kq q , where q r
Ek
(9.0 10 J·m/C )(1.6 10 C) (2)(7) (4.8 10 m)(1.6 10 J/MeV)
1 2
1
9
2e and q2 7e
2
19
15
2
13
Ek 4.2 MeV 73. The half-life of hassium-269 is T 12 9.3 s.
The original amount of hassium is N
mass of hassium Avogadro’s number mass per mole
Using the activity equation: 0.693N Activity
T 1 2
(1.0 10 g)(6.022 10 mol ) 269 g/mol
0.693 Activity
3
23
1
9.3 s
Activity 1.67 1017 Bq Using the decay formula for a time of 1 s: 1 t N N 0 T 2 1 N 2 N 92.82% If 92.82% remains after 1 s, then 100% 92.82% 7.18% has decayed. This activity equals:
1
2
1s 9.3 s
(1.0 10 g)(6.022 10 mol ) 269 g/mol
Activity (7.18%)
3
23
1
Activity 1.61 1017 Bq 74. The energy released is equivalent to the energy of the mass difference: E [m(1H) m(2H) m(3He)]c 2 E [1.007 825 u 2.014 102 u 3.016 029 u]c 2 931.5 MeV/c 2/u E 5.49 MeV
Solutions to End-of-chapter Problems
193
75. One mole of 235U releases 23 500 GJ of energy. mo 2(fuel used) mo 2(moles of U used)(mass/mol) power time mo 2 (0.235 kg/mol) energy mol mo
(0.7 GW)(2)(3.1536 10 s) 2 23 500 GJ/mol
7
(0.235 kg/mol) mo 883 kg electrical energy produced 76. % E fission energy released (electrical power) time % E mass of U (energy/mol) mo lar mass (0.7 GW)(86400 s) % E 2.5 kg (23 500 GJ/mol) 0.235 kg/mol % E 0.242 About 24.2% of the fission energy is transformed into electrical energy. 77. Since a mole of 235U releases 23 500 GJ of energy, the 50 kg releases (50 kg)(23 500 GJ/mol) 5 106 GJ 0.235 kg/mol 5 1015 J 78. Since the electron keeps only 10% of its kinetic energy with each collision, the energy remaining after x collisions is given by: Ex Eo(0.1) x 0.05 eV (5.0 106 eV)(0.1) x 0.05 eV x log (0.1) log 5.0 106 eV log (108) x log (101)
x 8 collisions 79. The incoming speed of a neutron with
3.5 MeV of kinetic energy is: 2 Ek v v
2(3.5 10 eV)(1.602 10 J/eV) (1.008 665 u)(1.6605 10 kg/u) m
6
19
27
v1
(2.5876 107 m/s) v1 1.782 104 m/s A 1 141 1 80. For the reaction 235 92U 0n → 56Ba Z Y 30n, conservation of atomic mass number for the reaction yields 235 1 141 A 3(1), or A 92. Conservation of atomic number yields 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/c 2/u) m0 1.928 105 MeV/c 2 Its total energy is: E m0c 2 Ek E 1.928 105 MeV 7.000 106 MeV E 7.1928 TeV
At relativistic speeds, use Einstein’s energy triangle: (mvc )2 E2 (m0c 2)2 (mvc )2 (7.1928 1012 eV)2 (1.928 1011 eV)2 mvc 7.1902 1012 eV Rearranging for v, v 7.1902 1012 eV
82. The de Broglie wavelength is: h mv hc mvc (6.626 1034 J·s)(3.0 108 m/s) (7.19 1012 eV)(1.6 1019 J/eV)
1.73 1019 m
For head-on elastic collisions,
, where mn mx v mn mx
v is the recoil
velocity of the neutron.
194
c mc 2 v 7.1902 1012 eV c 7.1928 1012 eV v 0.999639c v 2.9989 108 m/s
v 2.5876 107 m/s v
1.008 665 u 1.007 276 u 1.008 665 u 1.007 276 u
Solutions to End-of-chapter Problems
83. At relativistic speeds, the mass becomes
dilated: m
m0 v2
1 c 2
m
27
1.673 53 10 kg 1 0. 75 2
h
6.626 10 J·s (2.53 10 kg)(0.75c )
mv
34
27
1.16 1015 m 1.16 fm qB 84. f 2m 2mf B q 2(2.53 1027 kg)(23 106 Hz) B (1.6 1019 C)
B 2.28 T 85. Electrons and protons with the same
de Broglie wavelength have the same h mv
. Using Einstein’s
energy triangle and MeV units, for the electron: (mvc )2 (m0c 2 Ek)2 (m0c 2)2 (mvc )2 (0.511 MeV 9 103 MeV)2 (0.511 MeV)2 mvc 9.0005 GeV The 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 (938.27 MeV)2 Ek 8951.47 MeV 938.27 MeV Ek 8.1 GeV 86. Using the energy equation mc 2 m0c 2 Ek to find the dilated mass of the proton,
Ek m m0 c 2 m 938.27 MeV/c 2
B
2mf q
B
2 (2.3856 10 kg)(20 10 Hz) 1.6 10 C
27
6
19
400 MeV c
2 1 1 3 3 3 2 1 b) ud 1 3 3 1 1 c) db 0 3 3 2 2 d) cc 0 3 3 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 antiu d d. neutron is 90. The mass of the top quark is 176 103 MeV/c 2 188.94 u. The element 931.5 MeV/c 2/u with the closest atomic mass is osmium (Os), with an atomic mass of 190.2 u. 91. The pion has a quark combination of ud and a charge of e. Conversely, a pion has u d, and its charge is the combination 2 1 e e e. 3 3 87. a) uds 0
The de Broglie wavelength is:
qB , yields: 2m
B 1.87 T
m 2.53 1027 kg
momentum
The cyclotron frequency, f
92. t
d
v
2.4 1015 m t 3 108 m/s t 8 1024 s 93. a) Two protons approach and exchange a virtual meson, then recoil from each other. b) An atom sits at rest, then one of its electrons drops to a lower energy level and emits a photon, so the atom is pushed in the opposite direction. c) A pion decays into a muon and a muon neutrino.
2
m 1338.27 MeV/c 2 m 2.3856 1027 kg Solutions to End-of-chapter Problems
195
