Nelson Physics 12 Chapter 11 solutions PDF

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Section 11.1: The Special Theory of Relativity Mini Investigation: Understanding Frames of Reference, page 576 Answers may vary. Sample answers: A. The speed of the ball is greater when the person tossing the ball walks forward. The speed appears to be the sum of the speed at which the ball is tossed and the speed of the person tossing the ball. B. The speed of the ball was even faster when the student throwing the ball walked forward more quickly. This supports what we inferred in Question A. C. When the student throwing the ball walked slowly backward, the speed decreased. The decrease occurred because the person tossing the ball was moving away from the catcher. D. Yes, if you know the speed of an object in one inertial reference frame, you can determine its speed in another inertial reference frame, at least at the slow speeds used in this investigation. We can determine the speed of the ball if we know the speed in the catcher’s inertial frame and the speed of the pitcher relative to the catcher. When we change this relative speed between the two inertial frames, we can determine the speed of the ball in the new inertial frame. Section 11.1 Questions, page 579 1. Answers may vary. Sample answers: (a) The three most natural reference frames to use would be: 1) a frame moving alongside the skater at the same velocity; 2) a frame fixed on the deck of the boat; and, 3) a frame fixed on the shoreline. (b) For reference frame 1 in part (a), the student on skates would not be moving along, just moving in place. But the boat would be moving at constant speed, as would the shoreline (unless the skater’s velocity relative to the boat was equal but opposite to the velocity of the boat relative to the shoreline). For reference frame 2, the skater would be moving along, the shoreline would be moving past, but the boat would appear fixed in place. For reference frame 3, the skater would appear to be moving with velocity equal to the vector sum of the boat’s velocity relative to the shoreline and the skater’s velocity relative to the boat. In this inertial frame, the boat would appear to be moving past the shoreline, but the shoreline would appear fixed. 2. (a) An inertial frame of reference moves at constant velocity, whereas a non-inertial frame accelerates (i.e., its velocity changes). (b) Answers may vary. Sample answer: Two examples of an inertial frame of reference being at rest on the ground and being in an airplane cruising at constant altitude, with a fixed direction and fixed speed. Two examples of a non-inertial frame of reference are being in a car accelerating from a traffic light and being on a merry-go-round at a carnival. 3. (a) According to special relativity, the astronaut would measure the speed of light to be c. (b) The speed of the light measured by a person on Earth would equal c. 4. (a) Lutaaq would see Gabor’s ball follow a parabolic arc and Gabor moving along underneath the ball. The ball would be seen to fall directly into Gabor’s hand. (b) Gabor would see the same thing that Lutaaq had seen, that is, Lutaaq’s ball following a parabolic arc and Lutaaq moving along underneath the ball, having the ball land in her hand.

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Chapter 11: Relativity

11.1-1

5. The feature of Einstein’s coil and magnet thought experiment that Einstein found troubling was that in the inertial frame of the coil, an electric field causes a current, but in the inertial frame of the magnet, a magnetic field produces a current. The explanation depended on the inertial frame of the explainer. 6. The two postulates of the special theory of relativity are 1. The laws of physics are the same in all inertial reference frames. 2. For an observer in at least one inertial reference frame, the speed of light in a vacuum is independent of the motion of the light source. 7. The conclusion that results from the combination of Einstein’s two postulates is that all inertial observers, regardless of their motion, will measure the same speed of light in a vacuum, regardless of the motion of the light source. 8. A thought experiment is an imagined experiment that may be possible to do but impractical. The experiments are generally used to test an hypothesis or show a problem with an idea. As an example of the former, if one wonders if a laser beam from Earth could be used to deflect an incoming comet, then one could imagine the experiment, using current knowledge about lasers and comets. The hypothesis might be that the vaporized material causes the comet to change course. In Einstein’s thought experiment concerning the magnet and coil, he was demonstrating a problem with the then-current ideas of electrodynamics. 9. To determine whether your ship is in an inertial frame of reference, you must show that the ship is not accelerating. Qualitatively, if you were fixed to your seat, you would feel any acceleration in your body. For a quantitative measure, the simplest experiment would be to hold a ball out in front of you such that it is not moving relative to you or the ship. Then, carefully let go of the ball without giving it any push or pull. Does the ball move? If so, your frame is non-inertial and you can measure the acceleration. (Unless your spacecraft is as massive as a small planet and you are not at its centre, then the ball should float, free of the influence of gravity.) 10. (a) The ball rolls forward but then suddenly slows down when the train car suddenly accelerates forward. You would have felt this acceleration too. It means your reference frame suddenly became non-inertial. (b) You pushed the ball straight ahead (forward), but it curved to the right because your train car is accelerating to the left. You would sense this acceleration too. It means that your reference frame is non-inertial.

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Chapter 11: Relativity

11.1-2

Section 11.2: Time Dilation Tutorial 1 Practice, page 585 1. Given: Δts = 1.00 s; v = 0.60c Required: !tm " !ts Analysis: Use

!tm

1

. v2 1" 2 c " !t % Solution: !tm ( !ts = !ts $ m (1' # !ts & !t s

=

" 1 % = !ts $ (1' 2 $ 1( v ' # & c2 % " 1 (1' = !ts $ 2 ' $ 1( (0.60 c ) ' $ c2 & # % " 1 = !ts $ (1' # 0.64 & = (1.00 s)(0.25) !tm ( !ts = 0.25 s Statement: The proper time interval of 1.00 s of the clock appears to increase by 0.25 s when the clock moves with a speed of 0.60c relative to the observer. 2. Given: !tm = 3.7 # 10"6 s; v = 2.4 # 108 m/s; c = 3.0 # 108 m/s Required: Δts Analysis:

!t m !ts

=

1

v2 c2 " !t % !ts = !tm $ s ' # !tm & 1(

" !t % = !tm $ s ' # !tm & !ts = !tm 1(

v2 c2

Copyright © 2012 Nelson Education Ltd.

Chapter 11: Relativity

11.2-1

Solution: "ts = "tm 1! = "tm 1!

v2 c2 (2.4 )108 m/s )2 (3.0 )108 m/s )2

# 4& = "tm 1! % ( $ 5'

2

25 ! 16 25

= "tm

# 3& = (3.7 )10!6 s)% ( $ 5' "ts = 2.2 )10!6 s Statement: At rest, the particle’s lifetime is 2.2 ! 10"6 s , which is less than the lifetime of the same particles in a fast-moving beam. 3. Given: !t s = 8.0 s; !t m = 10.0 s; c = 3.0 " 10 8 m/s Required: v !t 1 to solve for v. Analysis: Use m = !t s v2 1" 2 c !tm 1 = !ts v2 1( 2 c

!ts v2 = 1( 2 !tm c 2

" !ts % v2 = 1( $# !t '& c2 m " !t % v2 = 1( $ s ' 2 c # !tm &

2

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Chapter 11: Relativity

11.2-2

# "t & Solution: 2 = 1! % s ( $ "tm ' c v2

# 4& = 1! % ( $ 5'

2

2

9 25 v 3 = c 5 3 v= c 5 # 3& = % ( (3.0 )108 m/s) $ 5' =

v = 1.8 ) 108 m/s Statement: The spacecraft is moving at 1.8 ! 10 8 m/s relative to Earth. 4. (a) Given: Δtm = 30.0 h; v = 0.700c Required: Δts Analysis:

!tm = !ts

1

v2 c2 " !t % !t s = !tm $ s ' # !t m & 1(

" !t % Solution: !t s = !t m $ s ' # !t m & = !t m 1( = !t m

1(

v2 c2 (0.700 c )2 c2

= (30.0 h)(0.71414) = 21.424 h (two extra digits carried) !t s = 21.4 h Statement: The time between the events, as viewed on Earth, is 21.4 h. (b) Given: Δts = 21.424 h; v = 0.950c Required: Δtm

Copyright © 2012 Nelson Education Ltd.

Chapter 11: Relativity

11.2-3

Analysis:

!tm = !ts

!tm =

1 v2 c2

1"

!ts v2 c2

1" Solution: !t m =

!t s v2 c2 21.424 h

1" =

1"

(0.950 c ) 2 c2

!t m = 68.6 h This problem can also be done without the steps for part (a) by evaluating the ratio of the Δt ratios m for both speeds. The Δts factor is dropped out. Δts Statement: By going faster, the crew measured the time between events to increase from 30.0 h to 68.6 h. 5. (a) Given: v = 1.1 ! 10 4 m/s; c = 3.0 ! 10 8 m/s Δt Required: m Δts Analysis: Use

Solution:

!tm = !ts

!tm = !ts

1 v2 1" 2 c

.

1 1"

v2 c2 1

= 1"

(1.1 # 104 )2 (3.0 # 108 )2

= 1 + 0.67 # 10"9 !tm = 1.000 000 001 !ts Statement: The time dilation factor is 1.000 000 001. (b) For objects going much slower than the speed of light, measuring the effects of time dilation requires extremely high accuracy.

Copyright © 2012 Nelson Education Ltd.

Chapter 11: Relativity

11.2-4

Section 11.2 Questions, page 587 1. For observer 2 to measure the same time for the light pulse on the light clock that observer 1 measures, she would have to be in the same inertial frame as observer 1. That is, she would have to be moving at the same velocity (same speed and direction) as the railway car. 2. (a) The process will always take longer for the observer who is moving relative to the process. (b) The observer who is at rest with respect to the process measures the proper time of the process. 3. (a) The clocks do not remain synchronized. As in the case of the Hafele-Keating experiment (atomic clocks on aircraft), the clock that orbits Earth will be observed by a person on Earth to run slow. So when the clock returns to Earth, it will have recorded less elapsed time. That is, the clocks will no longer be synchronized. (b) Although the clock that orbited will have less elapsed time after returning to Earth, it will thereafter run at the same rate as other clocks that are stationary on Earth. It will not run slow. (c) As described in part (a), the time elapsed on the clock that orbited will be less. The times will be different. (d) We are assuming that the clocks are ideal; they run as designed, keeping perfect time. So, the stationary clock did not have the wrong time. Any differences between the stationary and the orbiting clock are due to the nature of time as revealed by special relativity. (e) The orbiting clock was also assumed to be ideal. Its time is correct, and all differences with the stationary clock were due to the nature of time, not the clock. It did not have the wrong time. 4. Notice that 1) the aircraft is travelling in the opposite direction to that of the ground on the spinning Earth, and 2) the aircraft took exactly one day to travel around the world, 60 s 60 min 24 h ! 8.64 ! 104 s ! ! = 1 day 1 day 1 h 1 min Therefore, for an observer at rest with respect to the centre of Earth, the plane was stationary and the clock at the airport was moving east (the speed will depend on the latitude, but near the equator, the speed would exceed 400 m/s). Thus, as in the case of the Hafele-Keating experiment, the airport clock would show less elapsed time. The airport clock would have run slower. 5. The accuracy of a GPS system depends on correcting satellite clocks for special relativity because the GPS satellites, which send the signals, are moving rapidly with respect to the GPS receivers, which are stationary on Earth. Thus, according to the receiver, the clocks on the satellites run slow. Even if the time dilation effect is small, differences in elapsed time continue to build. To ensure that that both the receiver and satellite agree on the elapsed time, the GPS system should take into account time dilation from special relativity. 6. (a) Roger, not Mia, moves in Roger’s inertial reference frame. Thus, Roger, not Mia, measures Roger’s proper time. (b) Given: Δts = 30 s; v = 0.85c Required: Δtm

Copyright © 2012 Nelson Education Ltd.

Chapter 11: Relativity

11.2-5

Analysis:

!tm = !ts

!tm =

1 1"

!ts 1"

Solution: !tm =

v2 c2 v2 c2

!ts v2 c2 30 s

1" =

(0.85 c ) 2

1"

c2

!tm = 57 s Statement: Roger observes that over a period of 30 s in his reference frame, Mia’s watch has elapsed 57 s. 7. Given: Δts = 1.0 s; v = 0.95c Required: Δtm !t Analysis: m = !ts

!tm =

1 1"

!ts 1"

Solution: !tm =

v2 c2 v2 c2

!ts 1"

= 1"

v2 c2 1.0 s (0.95 c ) 2 c2

!tm = 3.2 s Statement: The observer on Earth finds that the signals arrive every 3.2 s.

Copyright © 2012 Nelson Education Ltd.

Chapter 11: Relativity

11.2-6

Section 11.3: Length Contraction, Simultaneity, and Relativistic Momentum Tutorial 1 Practice, page 591 1. Given: Ls = 5.0 m; Lm = 4.5 m; c = 3.0 ×108 m/s Required: v Analysis:

Lm v2 = 1! 2 Ls c 2

" Lm % v2 = 1 ! $L ' c2 # s& "L % v2 = 1! $ m ' 2 c # Ls &

2

"L % v = c 1! $ m ' # Ls & "L % Solution: v = c 1 ! $ m ' # Ls &

2

2

= (3.0 ( 108 m/s) 1 !

(4.5 m )2 (5.0 m )2

v = 1.3 ( 108 m/s Statement: To have had a relativistic contraction to 4.5 m, the 5.0 m long object must have moved at 1.3 × 108 m/s. 2. Given: Ls = 120 m; v = 0.80c Required: Lm

Analysis:

Lm v2 = 1! 2 Ls c Lm = Ls 1 !

Solution: Lm = Ls 1 !

v2 c2 v2 c2

= (120 m) 1 !

(0.80 c )2 c2

Lm = 72 m Statement: The relativistic contraction reduces the length of the spacecraft to 72 m.

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Chapter 11: Relativity

11.3-1

3. (a) Given: Ls = 2.5 m; Lm = 2.2 m; c = 3.0 ×108 m/s Required: v Lm v2 Analysis: = 1! 2 Ls c 2

" Lm % v2 = 1 ! $L ' c2 # s& "L % v2 = 1! $ m ' 2 c # Ls &

2

"L % v = c 1! $ m ' # Ls & "L % Solution: v = c 1 ! $ m ' # Ls &

2

2

= (3.0 ( 108 m/s) 1 !

(2.2 m )2 (2.5 m )2

v = 1.4 ( 108 m/s Statement: To have contracted from 2.5 m to 2.2 m, the car must have moved at 1.4 × 108 m/s. (b) Given: Ls = 33 m; Lm = 26 m; c = 3.0 ×108 m/s Required: v 2

"L % Analysis: Same as in part (a) above, v = c 1 ! $ m ' . #L & s

"L % Solution: v = c 1 ! $ m ' # Ls &

2

= (3.0 ( 108 m/s) 1 !

(26 m )2 (33 m )2

v = 1.8 ( 108 m/s Statement: To have contracted from 33 m to 26 m, the rocket must have moved at 1.8 × 108 m/s.

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Chapter 11: Relativity

11.3-2

Tutorial 2 Practice, page 596 1. (a) Given: m = 1.67 ! 10"27 kg; v = 0.85c; c = 3.0 ! 108 m/s Required: pclassical Analysis: pclassical = mv Solution: pclassical = mv = m (0.85c) = (1.67 ! 10"27 kg)(0.85)(3.0 ! 108 m/s) = 4.259 ! 10"19 kg # m/s (two extra digits carried) pclassical =4.3 ! 10"19 kg # m/s Statement: The proton’s classical momentum is 4.3 ! 10"19 kg # m/s . (b) Given: m = 1.67 ! 10"27 kg; v = 0.85c; pclassical = 4.259 ! 10"19 kg # m/s Required: prelativistic Analysis: prelativistic =

Solution: prelativistic =

mv v2 1! 2 c pclassical

and pclassical = mv , so prelativistic =

pclassical v2 1! 2 c

.

v2 c2 4.259 " 10!19 kg # m/s 1!

=

1!

(0.85 c )2 c2

prelativistic = 8.1 " 10!19 kg # m/s Statement: The proton’s relativistic momentum in the lab frame of reference is 8.1 ! 10"19 kg # m/s , about twice the classical value.

2. Given: m = 0.1 kg; v = 0.30c; c = 3.0 ! 108 m/s Required: prelativistic Analysis: prelativistic =

mv 1!

v2 c2

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Chapter 11: Relativity

11.3-3

Solution: prelativistic =

=

mv 2

v 1! 2 c m(0.30c) 1!

=

(0.30 c )2 c2

(0.1 kg)(0.30)(3.0 " 108 m/s)

0.91 prelativistic = 9.4 " 10 kg # m/s Statement: The projectile’s relativistic momentum with respect to Earth is 9.4 ! 106 kg " m/s . 6

3. Given: m = 1.67 ! 10"27 kg; v = 0.750c; c = 3.0 ! 108 m/s Required: prelativistic Analysis: prelativistic =

mv v2 c2 mv

1! Solution: prelativistic =

2

v c2 m(0.750c)

1! =

1! =

(0.750 c )2 c2

(1.67 " 10–27 kg)(0.750)(3.0 " 108 m/s)

0.4375 prelativistic = 5.68 " 10 kg # m/s Statement: The proton’s relativistic momentum in the lab frame of reference is 5.68 ! 10"19 kg # m/s . 4. (a) The motion affects only the component of length along the direction of motion. So, only direction y is affected. (b) Given: Lxs = 0.100 m; Lys = 0.100 m; Lzs = 0.100 m (proper lengths of the cube); !19

v y = 0.950c (speed along y-axis); c = 3.0 ! 108 m/s Required: relativistic volume of the cube, Vm Analysis: Vm = Lxm Lym Lzm . We know that x- and z-directions are unaffected by the motion, so Lxm = Lxs and Lzm = Lzs .

Copyright © 2012 Nelson Education Ltd.

Chapter 11: Relativity

11.3-4

Use

Lm v2 = 1 ! 2 , rearranged to solve for Lm. Ls c

We only need to calculate Lym = Lys 1 !

v2 . c2

Solution: Vm = Lxm Lym Lzm = Lxs Lys Lzs 1 !

v2 c2

= (0.100 m)(0.100 m)(0.100 m) 1 !

(0.950 c )2 c2

Vm = 3.12 " 10!4 m 3 Vm = Lxs Lys Lzs

1 ! (v / c)2

= (0.100 m)(0.100 m)(0.100 m) 1 ! (0.950)2 Vm = 3.12 " 10!4 m 3 Statement: The cube contracts along its direction of motion, resulting in a relativistic volume of 3.12 ! 10"4 m3 . This is less than the proper volume Vs = 1.00 × 10–3 m3 ( = Lxs Lys Lzs ).

(c) Given: Vs = 1.00 ! 10"3 m 3 ; density = 2.26 ! 104 kg/m 3 ; v = 0.950c; c = 3.0 ! 108 m/s Required: prelativistic Analysis: First, use the proper volume and density to calculate the rest mass, m. Then, mv . use prelativistic = v2 1! 2 c mv Solution: prelativistic = 2 v 1! 2 c (V "density)(0.950c) = s (0.950 c ) 2 1! c2

(1.00 #10!3 m 3 )(2.26 #104 kg/m 3 )(0.950)(3.0 # 108 m/s) 0.3122 10 = 2.06 #10 kg " m/s =

prelativistic

Statement: The cube’s relativistic momentum is 2.06 ! 1010 kg " m/s .

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Chapter 11: Relativity

11.3-5

Section 11.3 Questions, page 597 1. Given: Lm = 475 m; v = 0.755c Required: Ls

Lm v2 = 1 ! , so Ls = Analysis: Ls c2

Solution: Ls =

Lm v2 1! 2 c

.

Lm 1!

= 1!

v2 c2 475 m (0.755 c )2 c2

Ls = 724 m Statement: The proper length of spacecraft 2 is 724 m. 2. Given: Lm1 = 8.0 ly; v1 = 0.55c; v2 = 0.85c Required: Lm2 Analysis: Apply the length contraction formula to each astronaut (the proper lengths are the same).

v12 Lm2 v22 Lm1 = 1! 2 ; = 1! 2 Ls Ls c c Divide the relation for astronaut 2 by that for astronaut 1. Lm2 = Lm1

1!

v22

1!
...