ChR ISM
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ChR ISM


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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 
m1 + m2. Explain your answer. 
 
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