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.
Copyright © 2012 Nelson Education Ltd.
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.
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.1-2
Section 11.2: Time Dilation
Tutorial 1 Practice, page 585
1. Given: !ts = 1.00 s; v = 0.60c
Required: !tm " !ts
Analysis: Use
!tm
=
!ts
1
.
v2
1" 2
c
# !t
&
Solution: !tm " !ts = !ts % m " 1(
$ !ts
'
#
&
1
" 1(
= !ts %
2
% 1" v
(
$
'
c2
&
#
1
= !ts %
" 1(
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: !ts
Analysis:
!tm
=
!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
v2
1" 2
c
Solution: !ts = !tm
= !tm 1"
(2.4 # 108 m/s )2
(3.0 # 108 m/s )2
= !tm
$ 4'
1" & )
% 5(
= !tm
25 " 16
25
2
$ 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: !ts = 8.0 s; !tm = 10.0 s; c = 3.0 " 108 m/s
Required: v
Analysis: Use
!tm
=
!ts
!tm
=
!ts
1
1"
2
v
c2
to solve for v.
1
1"
v2
c2
!ts
v2
= 1" 2
!tm
c
2
# !ts &
v2
=
1"
% !t (
c2
$ m'
# !t &
v2
= 1" % s (
2
c
$ !tm '
2
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.2-2
# "t &
v2
Solution: 2 = 1! % s (
c
$ "tm '
# 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 ! 108 m/s relative to Earth.
4. (a) Given: !tm = 30.0 h; v = 0.700c
Required: !ts
Analysis:
!tm
=
!ts
1
v2
c2
# !t &
!ts = !tm % s (
$ !tm '
1"
" !t %
Solution: !ts = !tm $ s '
# !tm &
= !tm 1(
= !tm 1(
v2
c2
(0.700 c )2
c2
= (30.0 h)(0.71414)
= 21.424 h (two extra digits carried)
!ts = 21.4 h
Statement: The time between the events, as viewed on Earth, is 21.4 h.
(b) Given: !ts = 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: !tm =
=
!ts
v2
1" 2
c
21.424 h
1"
(0.950 c )2
c2
!tm = 68.6 h
This problem can also be done without the steps for part (a) by evaluating the ratio of the
!t
ratios m for both speeds. The !ts factor is dropped out.
!ts
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 ! 104 m/s; c = 3.0 ! 108 m/s
!t
Required: m
!ts
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
!
!
= 1 day
8.64 ! 104 s !
1 h
1 day
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: !ts = 30 s; v = 0.85c
Required: Δtm
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.2-5
Analysis:
!tm
=
!ts
!tm =
1
v2
c2
1"
!ts
v2
c2
1"
Solution: !tm =
=
!ts
v2
1" 2
c
30 s
(0.85c ) 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: !ts = 1.0 s; v = 0.95c
Required: Δtm
!t
Analysis: m =
!ts
!tm =
1
v2
1" 2
c
!ts
v2
c2
1"
Solution: !tm =
!ts
1"
=
1"
v2
c2
1.0 s
(0.95c ) 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
Lm
v2
Analysis:
= 1! 2
Ls
c
v2
Lm = Ls 1 ! 2
c
Solution: Lm = Ls 1 !
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.
Copyright © 2012 Nelson Education Ltd.
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
Analysis:
Lm
v2
= 1! 2
Ls
c
2
" Lm %
v2
=
1
!
$ L '
c2
# s&
" Lm %
v2
=
1
!
$ L '
c2
# s&
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 ' .
# Ls &
"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.
Copyright © 2012 Nelson Education Ltd.
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!
(0.85c )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
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.3-3
Solution: prelativistic =
mv
2
v
c2
m(0.30c)
1!
=
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 =
Solution: prelativistic =
=
mv
v2
1! 2
c
mv
2
v
1! 2
c
m(0.750c)
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
Lm
v2
Use
= 1 ! 2 , rearranged to solve for Lm .
Ls
c
We only need to calculate Lym = Lys
v2
1! 2 .
c
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 m 3 . 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 .
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.3-5
Section 11.3 Questions, page 597
1. Given: Lm = 475 m; v = 0.755c
Required: Ls
Analysis:
Lm
v2
= 1 ! 2 , so Ls =
Ls
c
Solution: Ls =
=
Lm
v2
1! 2
c
.
Lm
v2
1! 2
c
475 m
1!
(0.755c )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).
Lm1
v2 L
v2
= 1 ! 12 ; m2 = 1 ! 22
Ls
Ls
c
c
Divide the relation for astronaut 2 by that for astronaut 1.
Lm2
=
Lm1
1!
1!
v22
c 2 , so L = L
m2
m1
v12
c2
1!
1!
v22
c2
v12
c2
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.3-6
Solution: Lm2 = Lm1
1!
1!
v22
c2
v12
c2
1!
= (8.0 ly)
1!
(0.85c )2
c2
(0.55c )2
c2
Lm2 = 5.0 ly
Statement: The second astronaut, who travels at 0.85c, finds the distance to the star to
be 5.0 ly.
3. François does not observe the explosions to occur simultaneously. Assume François is
travelling east relative to Soledad. According to him, Soledad is moving west at a speed
of 0.95c. Now Soledad has arranged the explosions in her moving frame to occur such
that the light from each one reaches her at the same time in her inertial frame of
reference. However, once the light leaves the explosion sites, they travel at speed c. So, if
François had seen the explosions as simultaneous, then Soledad would have detected the
western explosion first, because she was moving toward that explosion. Since she instead
observed the explosions simultaneously, the western explosion must have occurred later
than the eastern explosion. In other words, François sees the explosion next to the front of
his railway car first.
4. (a) Given: m = 1.67 ! 10"27 kg; v = 0.99c; c = 3.0 ! 108 m/s
Required: pclassical
Analysis: pclassical = mv
Solution: pclassical = mv
= m(0.99c)
= (1.67 ! 10"27 kg)(0.99)(3.0 ! 108 m/s)
= 4.960 ! 10"19 kg # m/s (two extra digits carried)
pclassical = 5.0 ! 10"19 kg # m/s
Statement: This proton’s momentum, according to Newton’s definition, is
5.0 ! 10"19 kg # m/s .
(b) Given: pclassical = 4.960 ! 10"19 kg # m/s; v = 0.99c
Required: prelativistic
Analysis: prelativistic =
mv
and pclassical = mv , so prelativistic =
pclassical
v2
v2
1! 2
1! 2
c
c
relativistic factor separately because it will be used in part (c) below.
Copyright © 2012 Nelson Education Ltd.
. Calculate the
Chapter 11: Relativity
11.3-7
Solution: prelativistic =
pclassical
v2
c2
4.960 " 10!19 kg # m/s
1!
=
1!
(0.99 c )2
c2
4.960 " 10!19 kg # m/s
(two extra digits carried)
7.088
= 3.52 " 10!18 kg # m/s
=
prelativistic
Statement: The proton’s relativistic momentum is 3.52 ! 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: prelativistic / pclassical = 5
Required: v
Analysis: prelativistic =
prelativistic
=
pclassical
1
1!
prelativistic
=
pclassical
2
" prelativistic %
$ p
' =
# classical &
1!
2
v
c2
1
1!
mv
v2
1! 2
c
and pclassical = mv , so prelativistic =
. Rewrite this equation in terms of
pclassical
v2
1! 2
c
, or
v
.
c
v2
c2
1
v2
1! 2
c
1
v2
=
2
c2 " p
%
relativistic
$ p
'
# classical &
v2
1
= 1!
2
2
c
" prelativistic %
$ p
'
# classical &
v
1
= 1!
2
c
" prelativistic %
$ p
'
# classical &
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.3-8
Solution:
v
1
= 1!
2
c
" prelativistic %
$ p
'
# classical &
= 1!
1
25
v
= 0.980
c
v = 0.980c
Statement: To have increased its momentum by a relativistic factor of 5, the speed of the
particle must be 0.98c.
6. Given: melectron = 9.11 ! 10"31 kg; velectron = 0.999c; mship = 4.38 ! 107 kg;
c = 3.0 ! 108 m/s
Required: vship
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.
m
v
pship = mship vship , pelectron = electron electron , and pship = pelectron , so
(velectron )2
1!
c2
m
v
mship vship = electron electron
(velectron )2
1!
c2
melectron velectron
vship =
(velectron )2
mship 1 !
c2
melectron velectron
Solution: vship =
(v
)2
mship 1! electron
c2
=
(9.11" 10!31 kg )(0.999)(3.0 " 108 m/s)
(4.38 " 107 kg ) 1! (0.999)2
vship = 1.39 " 10!28 m/s
Statement: If the ship has the same momentum as the electron, its speed is
1.39 ! 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.)
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.3-9
7. No, if you were travelling on a spacecraft at 0.99c relative to Earth, you would not feel
compressed in the direction of travel. According to the principle of relativity, experiments
cannot tell us if we are at rest or moving at constant velocity. If we could get the feeling
of being compressed, then we could then find some experiment that would determine this
compression. As no such experiment exists, we cannot feel any changes in our bodies,
including compression.
8. We do not notice the effects of length contraction in our everyday lives because the
speeds that we experience are much less than the speed of light, so we do not notice the
effects of relativistic length contraction. For example, formula-one race cars can slightly
exceed 300 km/h, which is 1/1000 of the speed of light. The length contraction factor
even at this high speed is 1 ! 10!6 , which is about 0.999 999 5. A car that is 5 m long at
rest would shrink by only 5 µm at this speed, a distance we could only detect using a
microscope. This is why cars do not appear shorter when they drive past us at high
speeds.
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.3-10
Section 11.4: Mass–Energy Equivalence
Tutorial 1 Practice, page 602
1. Given: Erest = 2.25 ! 1016 J; c = 3.0 ! 108 m/s
Required: m
Analysis: Rearrange the rest-mass equation by dividing both sides by c2: m =
Solution: m =
=
Erest
Erest
c2
c2
2.25 ! 1016 kg " m 2 /s 2
(3.0 ! 108 )2 m 2 /s 2
m = 0.25 kg
Statement: The cellphone’s rest mass is 0.25 kg.
2. (a) Given: m = 1.67 ! 10"27 kg; v = 0.800c; c = 3.0 ! 108 m/s
Required: Etotal in units of MeV
Analysis: First calculate the rest energy of the proton, Erest = mc 2 , and convert it to units
of MeV. Then, use the equation Etotal =
Erest
v2
1! 2
c
.
Solution: Erest = mc 2
# 1 MeV &
= (1.67 ! 10"27 kg)(3.0 ! 108 m/s)2 %
$ 1.60 ! 10"13 J ('
Erest = 939.38 MeV (two extra digits carried)
Etotal =
939.38 MeV
1"
(0.800 c )
2
c2
Etotal = 1.57 ! 103 MeV
Statement: The given proton has a total energy of 1.57 × 103 MeV.
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.4-1
(b) Given: Erest = 939.38 MeV; v = 0.800c
Required: Ek in units of MeV
Analysis: Use the equations Etotal =
Ek =
Erest
v2
1! 2
c
and Ek = Etotal – Erest to obtain
Erest
! Erest
v2
1! 2
c
"
%
1
Ek = Erest $
! 1'
2
$ 1! v
'
#
&
c2
v
3
4
Use = 0.800 = , making the denominator .
5
c
5
"
%
1
! 1'
Solution: Ek = Erest $
2
$ 1! v
'
#
&
c2
" 1
%
Ek = (939.38 MeV) $
! 1'
" 3%
$$ ' '
# # 5& &
Ek = 6.26 ( 102 MeV
Statement: The proton has a kinetic energy of 6.26 × 102 MeV.
3. (a) Given: m = 23 kg; c = 3.0 ! 108 m/s
Required: Erest
Analysis: Use the rest-mass equation, Erest = mc2.
Solution: Erest = mc 2
= (23 kg) (3.0 ! 108 m/s)2
Erest = 2.1 ! 1018 J
Statement: The typical CANDU fuel bundle has a rest-mass energy of 2.1 × 1018 J.
(b) Dividing the result in (a) by the average daily home energy use of 3.6 × 1010 J/day
gives 5.8 × 107 days, which is nearly 160 000 years.
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.4-2
4. Given: m = 2500 kg; Ek = 1.5 ! 1020 J; c = 3.0 ! 108 m/s
Required: v
Analysis: Rearrange the equation for Ek.
1
Ek = mc 2 (
! 1)
v2
1! 2
c
Ek
1
+1=
2
mc
v2
1! 2
c
v2
1
1! 2 =
Ek
c
+1
mc 2
v2
1
1! 2 =
2
c
" Ek
%
$# mc 2 + 1'&
1
v
= 1!
2
c
" Ek
%
$# mc 2 + 1'&
Solution:
1
v
= 1!
2
c
" Ek
%
$# mc 2 + 1'&
= 1!
= 1!
1
"
%
1.5 ( 1020
$ (2.5 ( 103 )(3.0 ( 108 )2 + 1'
#
&
2
1
"2
%
$# 3 + 1'&
2
v 4
=
c 5
= 0.80
v = 0.80c
Statement: The asteroid has a speed of 0.80c.
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.4-3
Section 11.4 Questions, page 603
1. Given: m = 1 kg; v = 0.95c
Required: mrelativistic
Analysis: mrelativistic =
m
1!
Solution: mrelativistic =
v2
c2
m
1!
=
1!
v2
c2
1 kg
(0.95c ) 2
c2
mrelativistic = 3.2 kg
Statement: The relativistic mass of the 1 kg object is 3.2 kg.
2. The famous equation E = mc2 tells us that the rest mass of an object and its energy are
equivalent, or Erest = mc 2 . This equation shows that even at rest, and with no other
sources of potential energy, a mass nevertheless has energy. It suggests the possibility,
now well established, that mass can be converted to energy and vice versa, and thus the
conservation of relativistic energy is really a conservation of mass and energy. In
particular, the rest mass m can be converted to another form of energy in the amount of
Erest = mc2.
3. Given: Erest = 4.20 ! 106 J; c = 3.0 ! 108 m/s
Required: m
Analysis: Erest = mc 2 , so m =
Solution: m =
=
Erest
c2
Erest
c2
.
4.20 ! 106 kg " m 2 /s 2
(3.0 ! 108 )2 m 2 /s 2
m = 4.67 ! 10#11 kg
Statement: A rest mass of about 4.67 × 10-11 kg has as much rest energy as 1 kg of
TNT explosive.
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.4-4
4. Given: m = 1.67 ! 10"27 kg; c = 3.0 ! 108 m/s
Required: Erest for two masses m that annihilate
Analysis: Erest = mc 2 . Use the rest mass energy of one proton and multiply by two.
Solution: Erest = 2mc 2
= 2(1.67 ! 10"27 kg)(3.0 ! 108 m/s)2
Erest = 3.01 ! 10"10 J
Statement: When a proton and anti-proton annihilate each other, the energy released is
the sum of their rest-mass energies, which equals 3.01 × 10–10 J.
5. (a) Given: Erest = 1.28 MeV; c = 3.0 ! 108 m/s
Required: m
Analysis: Erest = mc 2 , so m =
Erest
c2
. Change the energy units to joules.
Erest
c2
# 1.60 ! 10"13 J &
1.28 MeV
=
('
(3.0 ! 108 )2 m 2 /s 2 %$ 1 MeV
Solution: m =
m = 2.28 ! 10"30 kg
Statement: The subatomic particle’s rest mass is 2.28 × 10–30 kg.
(b) Given: Erest = 1.28 MeV; Etotal = 1.72 MeV
Required: Ek
Analysis: The total energy is the sum of the rest energy and the kinetic energy. Thus, the
kinetic energy is the difference between the two given energies.
Solution: Ek = Etotal ! Erest
= 1.72 MeV ! 1.28 MeV
Ek = 0.44 MeV
Statement: The relativistic kinetic energy of the subatomic particle is 0.44 MeV.
(c) Given: Erest = 1.28 MeV; Etotal = 1.72 MeV; c = 3.0 ! 108 m/s
Required: v
Analysis: Use the expression for the relativistic total energy and solve for v/c.
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.4-5
Etotal =
Erest
1!
2
" Etotal %
$E ' =
# rest &
v2
c2
1
v2
1! 2
c
1
v2
1! 2 =
2
c
" Etotal %
$E '
# rest &
1
v2
= 1!
2
2
c
" Etotal %
$E '
# rest &
1
v
= 1!
2
c
" Etotal %
$E '
# rest &
Solution:
1
v
= 1!
2
c
" Etotal %
$E '
# rest &
= 1!
1
" 1.72 MeV %
$
'
# 1.28 MeV &
2
v
= 0.6680 (two extra digits carried)
c
v = 2.00 ( 108 m/s
Statement: The particle’s speed in the laboratory is 2.00 × 108 m/s (about 2/3 the speed
of light).
6. Given: m = 7.35 ! 1022 kg; v = 1.02 ! 103 m/s; c = 3.0 ! 108 m/s
Required: mass that, when converted to energy, would produce the observed kinetic
energy of the Moon
Analysis: The speed of the Moon is more than 5 orders of magnitude less than the speed
of light. Hence, the kinetic energy of the Moon equals the classical kinetic energy to at
least 10 decimal places. So, we calculate the classical kinetic energy of the Moon and
determine the mass equivalent using Ek-classical = (mass equivalent)c2.
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.4-6
Solution:
Ek-classical = (mass equivalent)c 2
Ek-classical
c2
1 2
mv
2
=
c2
1
(7.35 ! 1022 kg)(1.02 ! 103 m/s )2
= 2
(3.0 ! 108 m/s )2
mass equivalent =
mass equivalent = 4.25 ! 1011 kg
Statement: A conversion of 4.25 × 1011 kg to energy would be enough to accelerate the
Moon from rest to its present orbital speed.
7. Given: Erest T = 2809.4 MeV; Erest P = 938.3 MeV; Erest N = 939.6 MeV
Required: energy released, Ereleased, when two neutrons combine with one proton to form
one tritium nucleus
Analysis: Use conservation of mass–energy, Etotal(before) = Etotal(after), and the fact that
the neutrons and protons are initially at rest. After the nuclear reaction, we have only the
rest energy of the tritium plus Ereleased, where Ereleased may include any allowed
combination of tritium kinetic energy and radiative energy of gamma radiation.
Solution:
Etotal (before) = Etotal (after)
Erest P + 2Erest N = Erest T + Ereleased
(938.3 MeV) + 2(939.6 MeV) = (2809.4 MeV) + Ereleased
Ereleased = 8.1 MeV
Statement: The energy released when the proton combines with the two neutrons to form
a tritium nucleus is 8.1 MeV (which may contribute to the tritium’s kinetic energy and
produce gamma radiation).
8. Given: Ek = Erest
Required: v
mc 2
Analysis: Substitute Erest = mc 2 into Etotal =
and then substitute into
v2
1! 2
c
Erest
Ek = Etotal – Erest to obtain Ek =
! Erest . Set Ek = Erest and solve for v.
v2
1! 2
c
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.4-7
Erest
Ek =
! Erest
v2
1! 2
c
"
%
1
Ek = Erest $
– 1'
2
$ 1! v
'
#
&
c2
Ek
+1=
Erest
1
1!
v2
c2
Set Ek = Erest , so
1
2=
1!
Ek
= 1.
Erest
v2
1! 2
c
v2 1
=
c2 4
v2 3
=
c2 4
3
v
=
2
c
v=
c 3
2
c 3
2
= (3.0 ! 108 m/s)(0.8660)
Solution: v =
v = 2.60 ! 108 m/s
Statement: The particle’s speed is 2.60 × 108 m/s.
9. Given: m = 2500 kg; Ek = 2.0 ! 1019 J; c = 3.0 ! 108 m/s
Required: v
Analysis: Rearrange the equation for Ek.
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.4-8
"
%
1
! 1'
Ek = mc 2 $
2
$ 1! v
'
#
&
c2
Ek
+1=
mc 2
1
v2
1! 2
c
v2
1
=
2
Ek
c
+1
mc 2
v2
1
1! 2 =
2
c
" Ek
%
$# 2 + 1'&
mc
1!
1
v
= 1!
2
c
" Ek
%
$# 2 + 1'&
mc
Solution:
1
v
= 1!
2
c
" Ek
%
$# 2 + 1'&
mc
= 1!
1
"
%
2.0 ( 1019
+
1
3
8
2
$# (2.5 ( 10 )(3.0 ( 10 )
'&
2
v
= 0.40
c
v = 0.40c
Statement: The spacecraft travels at 0.40 times the speed of light.
10. Given: Erest = 512 keV; the classical kinetic energy Ek-class = 30.0 keV
Required: the relativistic kinetic energy, Ek
1
Analysis: Rewrite the classical equation for kinetic energy, Ek-class = mv 2 , as
2
2
2Ek-class v
= 2 . Then, substitute that into the relativistic kinetic energy equation. Use
mc 2
c
2
Erest = mc to calculate.
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.4-9
"
%
1
! 1'
Ek = mc 2 $
2
$ 1! v
'
&
#
c2
"
%
1
Ek = mc 2 $
! 1'
$ 1! 2Ek-class
'
#
&
mc 2
"
%
1
Solution: Ek = mc 2 $
! 1'
$ 1! 2Ek-class
'
#
&
mc 2
"
%
1
! 1'
= (512 keV) $
$ 1! 2(30.0 keV ) '
$#
'&
512 keV
Ek = 32.9 keV
Statement: The actual kinetic energy of the electron is 32.9 keV.
11. Given: mfuel = 100.0 mg; c = 3.0 × 108 m/s; car mileage = 30 km from 1.0 × 108 J
of fuel
Required: number of kilometres travelled on mfuel , d
Analysis: First, using Erest = mc 2 , calculate the rest energy in 100.0 mg of fuel, then
multiply by the mileage ratio, 3.0 × 10–7 km/J.
Solution: Erest = mc 2
= (1.000 ! 10"4 kg)(3.0 ! 108 m/s)2
Erest = 9.0 ! 1012 J
30.0 km
1.0 ! 108 J
30.0 km
= 9.0 ! 1012 J !
1.0 ! 108 J
d = Erest !
d = 2.7 ! 106 km
Statement: The car could hypothetically drive 2.7 × 106 km using the energy contained
in the rest mass of 100.0 mg of fuel.
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.4-10
12. Given: reduction in uranium mass, Δm = 100.000 kg – 99.312 kg = 0.688 kg;
c = 2.997 99 × 108 m/s
Required: Erest from !mc 2
Analysis: Erest = !mc 2
Solution: Erest = !mc 2
= (0.688 kg)(2.99799 " 108 m/s)2
Erest = 6.1837 " 1016 J
Statement: The amount of energy released during the four years was 6.1837 × 1016 J.
Copyright © 2012 Nelson Education Ltd.
Chapter 11: Relativity
11.4-11