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```second wad of putty
of mass m2 which is initially at rest. Do you expect that after the collision the
combined putty mass will be (a) greater than, (b) less than, (c) the same as

Determine the Concept Because the collision is inelastic, kinetic energy is
converted into mass energy. ( )a is correct.

11 \u2022\u2022 [SSM] Many nuclei of atoms are unstable; for example, 14C, an
isotope of carbon, has a half-life of 5700 years. (By definition, the half-life is the
time it takes for any given number of unstable particles to decay to half that
number of particles.) This fact is used extensively for archeological and
biological dating of old artifacts. Such unstable nuclei decay into several decay
products, each with significant kinetic energy. Which of the following is true?
(a) The mass of the unstable nucleus is larger than the sum of the masses of the
decay products. (b) The mass of the unstable nucleus is smaller than the sum of
the masses of the decay products. (c) The mass of the unstable nucleus is the
same as the sum of the masses of the decay products. Explain your choice.

Determine the Concept Because mass is converted into the kinetic energy of the
fragments, the mass of the unstable nucleus is larger than the sum of the masses of
the decay products. ( )a is correct.

12 \u2022\u2022 Positron emission tomography (PET) scans are common in modern
medicine. In this procedure, positrons (a positron has the same mass, but the
opposite charge as an electron) are emitted by radioactive nuclei that have been
introduced into the body. Assume that an emitted positron, traveling slowly (with
Special Relativity

1045
negligible kinetic energy) collides with an electron traveling at the same slow
speed in the opposite direction. They undergo annihilation and two quanta of light
(photons) are formed. You are in charge of designing detectors to receive these
photons and measure their energy. (a) Explain why you would expect these two
photons to come off in exactly opposite directions. (b) In terms of the electron
mass me , how much energy would each photon have? (1) less than mc2,
(2) greater mc2, (3) exactly mc2. Explain your choice.

Determine the Concept
(a) The linear momentum of the electron-positron pair was zero just before their
annihilation and the emission of the two photons and therefore the linear
momentum of the two photons must be zero.

(b) ( )3 is correct. Each photon has the rest mass energy, mec2, equivalent of an
electron. Because there are two photons and they have the same momentum, they
share equally, each getting half the available initial rest energy of 2mec2.

Estimation and Approximation

13 \u2022\u2022 [SSM] In 1975, an airplane carrying an atomic clock flew back and
forth at low altitude for 15 hours at an average speed of 140 m/s as part of a time-
dilation experiment. The time on the clock was compared to the time on an atomic
clock kept on the ground. What is the time difference between the atomic clock
on the airplane and the atomic clock on the ground? (Ignore any effects that
accelerations of the airplane have on the atomic clock which is on the airplane.
Also assume that the airplane travels at constant speed.)

Picture the Problem We can use the time dilation equation to relate the elapsed
time in the frame of reference of the airborne clock to the elapsed time in the
frame of reference of the clock kept on the ground.

Use the time dilation equation to
relate the elapsed time \u394t according
to the clock on the ground to the
elapsed time \u394t0 according to the
airborne atomic clock:

2
0
1
\u394
\u394
\u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212
=
c
v
tt (1)
Because v << c, we can use the
approximation x
x 2
11
1
1 +\u2248\u2212 to
obtain:

2
2 2
11
1
1 \u239f\u23a0
\u239e\u239c\u239d
\u239b+\u2248
\u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212 c
v
c
v

Chapter R

1046
Substitute in equation (1) to obtain:

0
2
0
0
2
2
1
2
11
t
c
vt
t
c
vt
\u394\u239f\u23a0
\u239e\u239c\u239d
\u239b+\u394=
\u394\u23a5\u23a5\u23a6
\u23a4
\u23a2\u23a2\u23a3
\u23a1 \u239f\u23a0
\u239e\u239c\u239d
\u239b+=\u394
(2)
where the second term represents the
additional time measured by the clock
on the ground.

Evaluate the proper elapsed time
according to the clock on the
airplane:

( ) s1040.5
h
s3600h15 40 ×=\u239f\u23a0
\u239e\u239c\u239d
\u239b=\u394t

Substitute numerical values and
evaluate the second term in equation
(2):
( )
ns9.5s1089.5
s1040.5
m/s10998.2
m/s140
2
1
\u394
9
4
2
8
\u2248×=
×\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
×=
\u2212
t'

14 \u2022\u2022 (a) By making any necessary assumptions and finding certain
stellar distances, estimate the speed at which a spaceship would have to travel for
its passengers to make a trip to the nearest star (not the Sun!) and back to Earth in
1.0 Earth years, as measured by an observer on the ship. Assume the passengers
make the outgoing and return trips at constant speed and ignore any effects due to
the spaceship stopping and starting. (b) How much time would elapse on Earth
during their roundtrip? Include 2.0 Earth years for a low-speed exploration of the
planets in the vicinity of this star.

Picture the Problem We can use the length-contraction equation
( )20 1 cvLL \u2212= and vLt =\u394 to estimate the speed of the spaceship. Note that
there is no time dilation effect during the two-year exploration period. Take the
distance to the nearest sun to be 4.0 c\u22c5y.

(a) According to the travelers, the
elapsed time for the trip is:

( )
v
cvL
v
Lt
2
0 1\u394
\u2212==

Multiplying both sides of the
equation by c yields:

( ) ( )
cv
cvL
v
cvcL
tc
2
0
2
0 11\u394
\u2212=\u2212=

Because c\u394t = 1.0 c\u22c5y and
L0 = 4.0 c\u22c5y: ( )cv
cv 210.4
0.1
\u2212=

Special Relativity

1047
Solve for v to obtain:
ccv 97.0
17
16 ==

(b) Express the elapsed time as the
sum of the travel and exploration
times:

nexploratio wayl1
nexploratiotravel
\u394\u3942
\u394\u394\u394
tt
ttt
+=
+=
(1)
From the Earth frame-of reference:
y 12.40.97
y 0.4
\u394 0 way1 =\u22c5== c
c
v
Lt

Substitute numerical values in
equation (1) and evaluate \u394t: ( ) y 2.10y 0.2y 12.42\u394 =+=t

15 \u2022\u2022 (a) Compare the kinetic energy of a moving car to its rest energy.
(b) Compare the total energy of a moving car to its rest energy. (c) Estimate the
error made in computing the kinetic energy of a moving car using non-relativistic
expressions compared to the relativistically correct expressions. Hint: Use of the
binomial expansion may help.

Picture the Problem We can compare these energies by expressing their ratios.
Assume that the speed of the car is 30 m/s (\u2248 67 mi/h).

(a) Express the ratio of the kinetic
energy of a moving car to its rest
energy:

2
2
2
2
1
2 2
1 \u239f\u23a0
\u239e\u239c\u239d
\u239b==
c
v
mc
mv
mc
K

For a car moving at 30 m/s:

%105
m/s 10998.2
m/s 30
2
1
13
2
82
\u2212×\u2248
\u239f\u23a0
\u239e\u239c\u239d
\u239b
×=mc
K

(b) Express the ratio of the total
energy of a moving car to its rest
energy:
500 000 000 000 000.1
122
2
2
\u2248
+=+=
mc
K
mc
mcK
mc
E

(c) The error made in computing the
kinetic energy of a moving car using
the non-relativistic expression
compared to the relativistically
correct expression is given by: 1
\u394
rel
rel-non
rel
relrel-non
rel
\u2212=
\u2212=
K
K
K
KK
K
K
(1)

From Equation R-14, the relativistic
kinetic energy is given by: 2
2
2
rel 11
2
1
mc
c
vK \u23a5\u23a5\u23a6
\u23a4
\u23a2\u23a2\u23a3
\u23a1
\u2212\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b \u2212=
\u2212
(2)
Chapter R

1048
Expand
2
1
2
2
1
\u2212
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b \u2212
c
v binomially to obtain:

4
4
2
2
4
4
2
2
2
2
8
3
2
11...
8
3
2
111
2
1
c
v
c
v
c
v
c
v
c
v ++\u2248+++=\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b \u2212
\u2212

Substituting in equation (2) and
simplifying yields:
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b +=
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b+=
\u23a5\u23a6
\u23a4\u23a2\u23a3
\u23a1 +=
2
2
rel-non
2
2
22
2
4
4
2
2
rel
4
31
8
3
2
1
8
3
2
1
c
vK
c
vmvmv
mc
c
v
c
vK

Substitute for Krel in equation (1)
and simplify to obtain:
1
4
31
1
4
31
\u394
1
2
2
2
2
rel-non
rel-non
rel
\u2212\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b +=
\u2212
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b +
=
\u2212
c
v
c
vK
K
K
K

Expanding
1
2
2
4
31
\u2212
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b +
c
v binomially yields:

( ) ( )( ) \u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b\u2212\u2248+\u239f\u239f\u23a0```