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Nelson Physics 12 Chapter 11 solutions
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Physics (1401/2)
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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 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., 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.
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.
Solution:
"t s
="t m
1!
v
2
c
2
="t m
1!
(2) 10
8 m/s)
2
(3) 10
8 m/s)
2
="t m
1!
4
5
null$
%
&
'
(
2
="t m
25! 16
25
=(3) 10
! 6 s)
3
5
null$
%
&
'
(
"t s
=2) 10
! 6 s
Statement: At rest, the particle’s lifetime is
2! 10
" 6
s, which is less than the lifetime
of the same particles in a fast-moving beam.
3. Given:
!t
s
=8 s; !t
m
=10 s; c=3" 10
8
m/s
Required: v
Analysis: Use
!t m
!t s
=
1
1 "
v
2
c
2
to solve for v.
!t
m
!t
s
=
1
1 (
v
2
c
2
!t
s
!t
m
= 1 (
v
2
c
2
!t
s
!t
m
"
null$
%
&
'
2
= 1 (
v
2
c
2
v
2
c
2
= 1 (
!t
s
!t
m
"
null$
%
&
'
2
Solution:
v
2
c
2
= 1!
"t s
"t m
null$
%
&
'
(
2
= 1!
4
5
null$
%
&
'
(
2
=
9
25
v
c
=
3
5
v=
3
5
c
=
3
5
null$
%
&
'
(
(3) 10
8 m/s)
v=1) 10
8 m/s
Statement: The spacecraft is moving at
1! 10
8
m/s relative to Earth.
4. (a) Given:
Δt m
=30 h; v=0
Required:
Δt s
Analysis:
!t
m
!t
s
=
1
1 (
v
2
c
2
!t
s
=!t
m
!t
s
!t
m
"
null$
%
&
'
Solution:
!t
s
=!t
m
!t
s
!t
m
"
null$
%
&
'
=!t
m
1 (
v
2
c
2
=!t
m
1 (
(0)
2
c
2
=(30 h)(0)
=21 h (two extra digits carried)
!t
s
=21 h
Statement: The time between the events, as viewed on Earth, is 21 h.
(b) Given:
Δt s
=21 h; v=0
Required: Δt
m
Section 11 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,
8! 10
4 s!
60 s
1 min
!
60 min
1 h
!
24 h
1 day
= 1 day
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:
Δt s
=30 s; v=0
Required: Δt
m
Analysis:
!t
m
!t
s
=
1
1 "
v
2
c
2
!t
m
=
!t
s
1 "
v
2
c
2
Solution:
!t m
=
!t s
1 "
v
2
c
2
=
30 s
1 "
(0)
2
c
2
!t m
= 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:
Δt s
=1 s; v=0
Required: Δt
m
Analysis:
!t
m
!t
s
=
1
1 "
v
2
c
2
!t
m
=
!t
s
1 "
v
2
c
2
Solution:
!t m
=
!t s
1 "
v
2
c
2
=
1 s
1 "
(0)
2
c
2
!t m
=3 s
Statement: The observer on Earth finds that the signals arrive every 3 s.
3. (a) Given: Ls = 2 m; Lm = 2 m; c = 3× 10
8
m/s
Required: v
Analysis:
L
m
L
s
= 1!
v
2
c
2
L
m
L
s
"
null$
%
&
'
2
= 1!
v
2
c
2
v
2
c
2
= 1!
L
m
L
s
"
null$
%
&
'
2
v=c 1!
L
m
L
s
"
null$
%
&
'
2
Solution:
v=c 1!
L
m
L
s
"
null$
%
&
'
2
=(3( 10
8
m/s) 1!
(2 m )
2
(2 m )
2
v=1( 10
8
m/s
Statement: To have contracted from 2 m to 2 m, the car must have moved at
1 × 10
8
m/s.
(b) Given: Ls = 33 m; Lm = 26 m; c = 3× 10
8
m/s
Required: v
Analysis: Same as in part (a) above,
v=c 1!
L
m
L
s
"
null$
%
&
'
2
.
Solution:
v=c 1!
L m
L s
"
null$
%
&
'
2
=(3( 10
8 m/s) 1!
(26 m )
2
(33 m )
2
v=1( 10
8 m/s
Statement: To have contracted from 33 m to 26 m, the rocket must have moved at
1 × 10
8
m/s.
Tutorial 2 Practice, page 596
1. (a) Given:
m=1! 10
" 27 kg; v=0; c=3! 10
8
m/s
Required:
p classical
Analysis:
p classical
=mv
Solution:
p classical
=mv
=m (0)
=(1! 10
" 27 kg)(0)(3! 10
8 m/s)
=4! 10
" 19 kg#m/s (two extra digits carried)
p classical
=4! 10
" 19 kg#m/s
Statement: The proton’s classical momentum is
4! 10
" 19
kg#m/s.
(b) Given:
m=1! 10
" 27
kg; v=0; p
classical
=4! 10
" 19
kg#m/s
Required:
p relativistic
Analysis:
p relativistic
=
mv
1!
v
2
c
2
and
p classical
=mv, so
p relativistic
=
p classical
1!
v
2
c
2
.
Solution:
p relativistic
=
p classical
1!
v
2
c
2
=
4" 10
! 19 kg#m/s
1!
(0)
2
c
2
p relativistic
=8" 10
! 19 kg#m/s
Statement: The proton’s relativistic momentum in the lab frame of reference is
8! 10
" 19
kg#m/s, about twice the classical value.
2. Given:
m=0 kg; v=0; c=3! 10
8
m/s
Required:
p relativistic
Analysis:
p relativistic
=
mv
1!
v
2
c
2
Use
L
m
L
s
= 1!
v
2
c
2
, rearranged to solve for
L m
.
We only need to calculate
L ym
=L ys
1!
v
2
c
2
.
Solution:
V m
=L xm
L ym
L zm
=L xs
L ys
L zs
1!
v
2
c
2
=(0 m)(0 m)(0 m) 1!
(0)
2
c
2
V m
=3" 10
! 4 m
3
V m
=L xs
L ys
L zs
1 !(v/c)
2
=(0 m)(0 m)(0 m) 1!(0)
2
V m
=3" 10
! 4 m
3
Statement: The cube contracts along its direction of motion, resulting in a relativistic
volume of
3! 10
####### " 4
m
####### 3
. This is less than the proper volume V
s
= 1 × 10
–
m
3
(
=L
xs
L
ys
L
zs
).
(c) Given:
V
s
=1! 10
" 3
m
3
; density = 2! 10
4
kg/m
3
; v=0; c=3! 10
8
m/s
Required:
p relativistic
Analysis: First, use the proper volume and density to calculate the rest mass, m. Then,
use
p relativistic
=
mv
1!
v
2
c
2
.
Solution:
p
relativistic
=
mv
1!
v
c
2
2
=
(V
s
"density)(0)
1!
(0)
2
c
2
=
(1# 10
! 3
m
3
)(2# 10
4
kg/m
3
)(0)(3# 10
8
m/s)
0.
p
relativistic
=2# 10
10
kg"m/s
Statement: The cube’s relativistic momentum is
2! 10
10
kg"m/s.
Section 11 .3 Questions, page 597
1. Given:
L m
=475 m; v=0
Required:
L s
Analysis:
L
m
L
s
= 1!
v
2
c
2
, so
L s
=
L m
1!
v
2
c
2
.
Solution:
L
s
=
L
m
1!
v
2
c
2
=
475 m
1!
(0)
2
c
2
L
s
=724 m
Statement: The proper length of spacecraft 2 is 724 m.
2. Given:
L m
=8 ly; v 1
=0; v 2
=0
Required:
L
m
Analysis: Apply the length contraction formula to each astronaut (the proper lengths are
the same).
L
m
L
s
= 1!
v
1
2
c
2
;
L
m
L
s
= 1!
v
2
2
c
2
Divide the relation for astronaut 2 by that for astronaut 1.
L
m
L
m
=
1!
v
2
2
c
2
1!
v
1
2
c
2
, so
L m
=L m
1!
v 2
2
c
2
1!
v 1
2
c
2
Solution:
p
relativistic
=
p
classical
1!
v
2
c
2
=
4" 10
! 19
kg#m/s
1!
(0)
2
c
2
=
4" 10
! 19
kg#m/s
7.
(two extra digits carried)
p
relativistic
=3" 10
! 18
kg#m/s
Statement: The proton’s relativistic momentum is
3! 10
" 18
kg#m/s.
(c) From the intermediate step in part (b) above, the relativistic momentum exceeds that
of the classical value by a ratio of 7 : 1.
5. Given:
p relativistic
/p classical
= 5
Required: v
Analysis:
p relativistic
=
mv
1!
v
2
c
2
and
p classical
=mv, so
p relativistic
=
p classical
1!
v
2
c
2
, or
p relativistic
p classical
=
1
1!
v
2
c
2
. Rewrite this equation in terms of
v
c
.
p relativistic
p classical
=
1
1!
v
2
c
2
p relativistic
p classical
"
null$
%
&
'
2
=
1
1!
v
2
c
2
1!
v
2
c
2
=
1
p relativistic
p classical
"
null$
%
&
'
2
v
2
c
2
= 1!
1
p relativistic
p classical
"
null$
%
&
'
2
v
c
= 1!
1
p relativistic
p classical
"
null$
%
&
'
2
Solution:
v
c
= 1!
1
p relativistic
p classical
"
null$
%
&
'
2
= 1!
1
25
v
c
=0.
v=0
Statement: To have increased its momentum by a relativistic factor of 5, the speed of the
particle must be 0.
6. Given:
m
electron
=9! 10
" 31
kg; v
electron
=0; m
ship
=4! 10
7
kg;
c=3! 10
8
m/s
Required:
v
ship
Analysis: The speed of the ship will be much less than that of the electron, so its
relativistic momentum will be very close to its classical momentum. This is not the case
for the electron, so we will use the relativistic expression for just the electron.
p
ship
=m
ship
v
ship
,
p electron
=
m electron
v electron
1!
(v electron
)
2
c
2
, and
p
ship
=p
electron
, so
m
ship
v
ship
=
m
electron
v
electron
1!
(v
electron
)
2
c
2
v
ship
=
m
electron
v
electron
m
ship
1!
(v
electron
)
2
c
2
Solution:
v
ship
=
m
electron
v
electron
m
ship
1!
(v
electron
)
2
c
2
=
(9" 10
! 31
kg)(0)(3" 10
8
m/s)
(4" 10
7
kg) 1!(0)
2
v
ship
=1" 10
! 28
m/s
Statement: If the ship has the same momentum as the electron, its speed is
1! 10
" 28
m/s. (This speed is so low that the ship would have to travel for nearly a
million million years before travelling the distance of one hydrogen atom.)
Section 11 .4 : Mass–Energy Equivalence
Tutorial 1 Practice, page 602
1. Given:
E
rest
=2× 10
16
J; c=3× 10
8
m/s
Required: m
Analysis: Rearrange the rest-mass equation by dividing both sides by c
2
:
m=
E
rest
c
2
Solution:
m=
E rest
c
2
=
2! 10
16 kg"m
2 /s
2
(3! 10
8 )
2 m
2 /s
2
m=0 kg
Statement: The cellphone’s rest mass is 0 kg.
2. (a) Given:
m=1! 10
" 27 kg; v=0; c=3! 10
8
m/s
Required:
E total
in units of MeV
Analysis: First calculate the rest energy of the proton,
E
rest
=mc
2
, and convert it to units
of MeV. Then, use the equation
E total
=
E rest
1!
v
2
c
2
.
Solution:
E rest
=mc
2
=(1( 10
! 27 kg)(3( 10
8 m/s)
2 1 MeV
1( 10
! 13 J
"
null$
%
&
'
E rest
=939 MeV (two extra digits carried)
E total
=
939 MeV
1!
(0)
c
2
2
E total
=1( 10
3 MeV
Statement: The given proton has a total energy of 1× 10
3
MeV.
(b) Given:
E rest
=939 MeV; v=0
Required:
E k
in units of MeV
Analysis: Use the equations
E total
=
E rest
1!
v
2
c
2
and
E k
=E total
- E rest
to obtain
E k
=
E rest
1!
v
2
c
2
!E rest
E k
=E rest
1
1!
v
2
c
2
! 1
"
null$
$
%
&
'
'
Use
v
c
=
0 =
4
5
, making the denominator
3
5
.
Solution:
E k
=E rest
1
1!
v
2
c
2
! 1
"
null$
$
%
&
'
'
E k
=(939 MeV)
1
3
5
"
null$
%
&
'
! 1
"
null$
$
%
&
'
'
E k
=6( 10
2 MeV
Statement: The proton has a kinetic energy of 6 × 10
2
MeV.
3. (a) Given:
m=23 kg; c=3× 10
8
m/s
Required: E
rest
Analysis: Use the rest-mass equation, E
rest
= mc
2
.
Solution:
E rest
=mc
2
=(23 kg) (3× 10
8 m/s)
2
E rest
=2× 10
18 J
Statement: The typical CANDU fuel bundle has a rest-mass energy of 2 × 10
18
J.
(b) Dividing the result in (a) by the average daily home energy use of 3 × 10
10
J/day
gives 5 × 10
7
days, which is nearly 160 000 years.
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Nelson Physics 12 Chapter 11 solutions
Course: Physics (1401/2)
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Copyright © 2012 Nelson Education Ltd. Chapter 11: Relativity 11.1-1
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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