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ChR ISM

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```T, the time between the arrival of
signals on Earth is related to the
proper time interval T0 (the time
between signals in Al\u2019s frame of
reference) through the time-dilation
equation:

( )2201 cv
TT \u2212=
Substituting for T and simplifying
yields:
( )
0
22
Bert 600.1
1
T
cv
f
\u2212=

Substitute numerical values and
evaluate Bertf :

( )
( )( ) 1
2
Bert y 50y 0100.0600.1
600.01 \u2212=\u2212=f
Chapter R

1068

(b) Express the number of signals
N received by Bert in terms of
the number of signals sent by Al:

Al0 tfN \u394=
Use the time dilation equation to
express the elapsed time in Al\u2019s
frame of reference in terms of the
elapsed time in Bert\u2019s frame (the
proper elapsed time):
2
BertAl 1\u394\u394 \u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212=
c
vtt

Substitute for Al\u394t to obtain:

2
Bert0 1\u394 \u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212=
c
vtfN (1)

Find Bert\u394t : y667.6
0.600
y00.4
\u394 Bert =\u22c5= c
ct

Substitute numerical values in
equation (1) and evaluate N: ( )( )
533
600.01y 667.6y100
2
1
=
\u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212= \u2212
c
cN

(c) Express the number of signals N
received by Bert in terms of the
number of signals sent by Al before
he returns:

Al0 TfN \u394=
Because
2
BertAl 1\u394\u394 \u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212=
c
vtt ,
the time in Al\u2019s frame for the
round trip Al\u394T is given by:

2
BertAlAl 1\u3942\u3942\u394 \u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212==
c
vttT

Substituting for Al\u394T in the
expression for N yields:

2
Bert0 1\u3942 \u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212=
c
vtfN

Substitute numerical values and
evaluate N:

( )( )
3
2
1
1007.1
600.01y667.6y1002
×=
\u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212= \u2212
c
cN

Special Relativity

1069
(d) Proceed as in Part (a) to find the
rate at which Al receives signals as
Bert is moving away from him:

1
Al y0.50
\u2212=f

(e) Express the number of signals N

AlAl tfN \u394=
Substitute for Alt\u394 to obtain: 2
BertAl 1\u394 \u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212=
c
vtfN

Substitute numerical values and
evaluate N:

( )( )
267
600.01y667.6y0.50
2
1
=
\u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212= \u2212
c
cN

(f) Express the total number of

return
returnoutboundtot
267 N
NNN
+=
+=

The number, returnN , of signals
received by Al on his return trip is
given by:
2
Bertreturn Al,
Alreturn Al,return
1\u394
\u394
\u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212=
=
c
vtf
tfN

Substitute for returnN to obtain: 2
Bertreturn Al,tot 1\u394267 \u239f\u23a0
\u239e\u239c\u239d
\u239b\u2212+=
c
vtfN

Proceeding as in Part (a), find the
rate at which signals are received by
Al on the return trip:
1
return Al, y200
\u2212=f

Substitute numerical values and evaluate totN :

( )( ) 321tot 1033.11334600.01y667.6y200267 ×==\u239f\u23a0\u239e\u239c\u239d\u239b\u2212+= \u2212 c cN

(g) Their age difference is:

AlBert ttt \u394\u2212\u394=\u394
Chapter R

1070

Substitute numerical values to
obtain:
y2.67
y 100
1067
y 100
1334
\u394 11 =\u2212= \u2212\u2212t
and
Bert.an younger thy 2.67 is Al

Relativistic Energy and Momentum

39 \u2022 Find the ratio of the total energy to the rest energy of a particle of mass
m moving with speed (a) 0.100c, (b) 0.500c, (c) 0.800c, and (d) 0.990c.

Picture the Problem We can use Equation R-15 to find the ratio of the total
energy to the rest energy for the given particle.

From Equation R-15:

2
2
0
2
2
2
11
c
v
E
c
v
mcE
\u2212
=
\u2212
=

Solve for the ratio of E to E0 to
obtain:
2
2
0 1
1
c
vE
E
\u2212
=

(a) Evaluate 0EE for v = 0.100c:
( ) 01.1100.01
1
2
2
100.00
=
\u2212
=\u23a5\u23a6
\u23a4
=
c
cE
E
cv

(b) Evaluate 0EE for v = 0.500c:
( ) 15.1500.01
1
2
2
500.00
=
\u2212
=\u23a5\u23a6
\u23a4
=
c
cE
E
cv

(c) Evaluate 0EE for v = 0.800c:
( ) 67.1800.01
1
2
2
800.00
=
\u2212
=\u23a5\u23a6
\u23a4
=
c
cE
E
cv

(d) Evaluate 0EE for v = 0.990c:
( ) 09.7990.01
1
2
2
999.00
=
\u2212
=\u23a5\u23a6
\u23a4
=
c
cE
E
cv

Special Relativity

1071
40 \u2022 A proton (rest energy 938 MeV) has a total energy of 1400 MeV.
(a) What is its speed? (b) What is its momentum?

Picture the Problem The rest energy E0 is equal to mc2. We are given E0 and E,
where E0 = 938 MeV and the total energy E = 1400 MeV. (The total energy is the
rest energy plus the kinetic energy). We can find the momentum p of the proton
using 2 2 2 2 4E p c m c= + (Equation R-17), and once we have p we can solve for the
speed v using v/c = pc/E (Equation R-16).

(b) Use Equation R-17 to relate the
momentum to the total energy and
the rest energy:

42222 cmcpE += and 20 mcE =
so
2
0
222 EcpE += \u21d2
c
EE
p
2
0
2 \u2212=

Substitute numerical values and
evaluate p: ( ) ( )
c
c
c
p
MeV/ 1040
MeV/ 1039
Mev 938Mev 1400 22
=
=
\u2212=

(a) Use Equation R-16 to express the
speed of the proton: E
pc
c
v = \u21d2 c
E
pcv \u239f\u23a0
\u239e\u239c\u239d
\u239b=

Substitute numerical values and
evaluate v:
( ) cccv 742.0
MeV 1400
MeV/ 1039 =\u239f\u23a0
\u239e\u239c\u239d
\u239b=

41 \u2022 [SSM] How much energy would be required to accelerate a particle
of mass m from rest to (a) 0.500c, (b) 0.900c, and (c) 0.990c? Express your
answers as multiples of the rest energy, mc2.

Picture the Problem We can use Equation R-14 to find the energy required to
accelerate this particle from rest to the given speeds.

From Equation R-14 we have:
( )
( )
2
2
2
2
2
1
1
1
1
mc
cv
mc
cv
mcK
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
=
\u2212
\u2212
=

Chapter R

1072
(a)Substitute numerical values and
evaluate K(0.500c):

( ) ( )
2
2
2
155.0
1
500.01
1500.0
mc
mc
cc
cK
=
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
=

(b) Substitute numerical values and
evaluate K(0.900c):

( ) ( )
2
2
2
29.1
1
900.01
1900.0
mc
mc
cc
cK
=
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
=

(c) Substitute numerical values and
evaluate K(0.990c): ( ) ( )
2
2
2
09.6
1
990.01
1990.0
mc
mc
cc
cK
=
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
=

42 \u2022 If the kinetic energy of a particle equals its rest energy, what
percentage error is made by using p = mv for its momentum? Is the non-
relativistic expression always low or high compared to the relativistically correct
expression for momentum?

Picture the Problem We can use Equations R-10 and R-14 to express the error
made in using p = mv for the momentum of the particle when K = E0.

The error in using p = mv for the
momentum of the particle is given
by:

relrel
rel 1
p
p
p
pp \u2212=\u2212 (1)
From Equation R-14 we have:
( )
( )
2
2
2
2
2
1
1
1
1
mc
cv
mc
cv
mcK
\u239f\u239f\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
=
\u2212
\u2212
=

For K = E0:
( ) KcvK \u239f
\u239f
\u23a0
\u239e
\u239c\u239c\u239d
\u239b
\u2212
\u2212
= 1
1
1
2
or
( ) 21
1
2
=
\u2212 cv

Special Relativity

1073
From Equation R-10, the relativistic
momentum of the particle is:

( ) mvcv
mvp 2
1 2
rel =\u2212
=
Substitute in equation (1) and
simplify to obtain:
%50
2
11
rel
=\u2212=\u2212
mv
mv
p
p

The ratio of the non-relativistic
momentum of a particle to its
relativistic momentum is given by:

22
22
rel
1
1
cv
cv
p
p
p
p \u2212=
\u2212
=
Because 1 1 22 <\u2212 cv , the non-relativistic expression is always low compared to
the relativistically correct expression for momentum.

43 \u2022 What is the total energy of a proton whose momentum is 3mc?

Picture the Problem We can use Equation R-17 to find the total energy of any
proton whose momentum is given. See Problem 42 for the rest energy of a proton.

The total energy, momentum, and
rest energy of the proton are related
by Equation R-17:

( )22222 mccpE +=
Substitute for the momentum of the
proton:

( ) ( )
424242
22222
109
3
cmcmcm
mccmcE
=+=
+=

Solving for E yields: 210mcE =

Substitute for m and evaluate E: ( ) GeV97.2MeV/93810 22 == ccE

44 \u2022\u2022 Using a spreadsheet program or graphing calculator, make a graph of
the kinetic energy of a particle with rest energy of 100 MeV for speeds between 0
and c. On the same graph, plot 221 mv by way of comparison. Using the graph,
estimate at about what speed the non-relativistic expression is no longer a good
approximation to the kinetic energy. As a suggestion, plot the energy in units of
MeV and the speed in the dimensionless form v/c.

Picture the Problem We can create a spreadsheet program to plot both the
classical and relativistic kinetic energy of the particle.

Chapter R

1074```