THÔNG TIN TÀI LIỆU
37.1: If
O
sees simultaneous flashes then O will see the
)AA(
flash first since O
would believe that the
A
flash must have traveled longer to reach
O
, and hence started
first.
37.2: a)
.29.2
)9.0(1
1
γ
2
s.105.05s)1020.2()29.2(γ
66
t
b)
km.1.36m101.36s)1005.5()sm1000.3()900.0(
368
vtd
37.3:
2
2
212222
2
1)1(1
c
u
cucu
s.1000.5)(
)sm102(3.00
(3600)hrs)4()sm250(
2
))(11()(
9
0
28
2
2
2
22
0
tt
t
c
u
tcutt
The clock on the plane shows the shorter elapsed time.
37.4:
.79.4
)978.0(1
1
γ
2
ms.0.395s103.95s)104.82()79.4(γ
-46
t
37.5: a)
2
0
2
2
22
0
1
1
t
t
c
u
cu
t
t
2
2
0
42
6.2
11
c
t
t
cu
.998.0 cu
b)
m.126s)10(4.2)sm1000.3()998.0(
78
tux
37.6:
667.1γ
a)
s.300.0
)800.0(
γ
m1020.1
γ
8
0
c
t
t
b)
m.107.20)(0.800s)300.0(
7
c
c)
s.180.0γs300.0
0
t
(This is what the racer measures your clock to read at
that instant.) At your origin you read the original
s.5.0
)sm10(3(0.800)
m1020.1
8
8
Clearly the observes (you and the racer) will not agree on the order of events!
37.7:
yr)(1
sm103.00
sm104.80
11
2
8
6
22
0
tcut
hrs.1.12yr)1028.1()(
4
0
tt
The least time elapses on the rocket’s clock because it had to be in two inertial
frames whereas the earth was only in one.
37.8:
a)
The frame in which the source (the searchlight) is stationary is the spacecraft’s
frame, so 12.0 ms is the proper time. b) To three figures,
.
c
u
Solving Eq. (37.7) for
cu
in terms of
,
γ
.
γ2
1
1)
γ1(1
2
2
c
u
Using
.998.0givesms190ms0.12γ1
0
cutt
37.9:
00.6
)9860.0(1
1
)(1
1
γ
22
cu
a)
km.17.9
6.00
km55
γ
0
l
l
b) In muon’s frame:
%.1.7071.0
17
.
9
651.0
%
km.0.651s)1020.2)(9860.0(
6
h
d
ctud
c) In earth’s frame:
%.1.7
km55.0
km90.3
%
km3.90s)10)(1.32(0.9860
s101.32s)(6.00)102.2(
γ
5
56
0
h
d
ctud
tt
37.10: a)
s.101.51
0.99540
m104.50
4
4
c
t
km.4.31
10.44
km45
γ
10.44
(0.9954)1
1
γ)
2
h
h
b
c)
s;101.44
γ
ands,101.44
0.99540
55
t
c
h
so the results agree but the particle’s
lifetime is dilated in the frame of the earth.
37.11: a)
m3600
0
l
m.3568m)(0.991)(3600
)sm1000.3(
)sm1000.4(
1)m3600(1
28
27
0
2
2
0
l
c
u
ll
)b
.s109.00
sm104.00
m3600
5
7
0
0
u
l
t
c)
.s108.92
sm104.00
m3568
5
7
u
l
t
37.12:
.sm102.86952.0)γ1(1so,3048.01γ
82
ccu
37.13:
22
0
2
2
0
1
1
cu
l
l
c
u
ll
m.92.5
0.600
1
m74.0
2
0
c
c
l
37.14:
Multiplying the last equation of (37.21) by u and adding to the first to eliminate t
gives
,
1
1
2
2
x
c
u
xtux
and multiplying the first by
2
c
u
and adding to the last to eliminate x gives
,
γ
1
1
γ
2
2
2
t
c
u
tx
c
u
t
),(γand)(γso
2
cxutttuxx
which is indeed the same as Eq. (37.21) with the primed coordinates replacing the
unprimed, and a change of sign of u.
37.15: a)
c
cc
cvu
uv
v
806.0
)600.0()400.0(1
600.0400.0
1
2
b)
c
cc
cvu
uv
v
974.0
)600.0)(900.0(1
600.0900.0
1
2
c)
.997.0
)600.0)(990.0(1
600.0990.0
1
2
c
cc
cvu
uv
v
37.16:
).)54(if35γ(667.1γ cu
a) In Mavis’s frame the event “light on” has
space-time coordinates
0
x
and
00.5
t
s, so from the result of Exercise 37.14 or
Example 37.7,
)(γ tuxx
and
tttux
c
xu
tt
γ,m1000.2γγ
9
2
.s33.8
b) The 5.00-s interval in Mavis’s frame is the proper time
0
t
in Eq. (37.6), so
,s33.8γ
0
tt
as in part (a).
c)
9
1000.2)800.0()s33.8( c
m, which is the distance
x
found in part (a).
37.17: Eq. (37.18):
22
1 cu
utx
x
Eq. (37.19):
22
1 cuxtux
Equate:
γ
γ
)(
x
tuutx
.
)(1
γ
γ
γ
11
11
1
1
1
γ
γ
1
γ
γ
1
γ
γ
γ
γ
2
2
2
22
2
22
2
2
2
2
cu
cuxt
c
xu
tt
cu
cu
cu
cu
cu
cu
cu
u
x
t
u
x
t
u
x
t
37.18:
Starting from Eq. (37.22),
)1(
)1(
1
2
2
2
2
cvuv
cuvvvuv
uvcuvv
cuv
uv
v
from which Eq. (37.23) follows. This is the same as switching the primed and unprimed
coordinates and changing the sign of
u
.
37.19:
Let the unprimed frame be Tatooine and let the primed frame be the pursuit ship.
We want the velocity
v
of the cruiser knowing the velocity of the primed frame
u
and
the velocity of the cruiser
v
in the unprimed frame (Tatooine).
c
cc
c
uv
uv
v
385.0
)800.0()600.0(1
800.0600.0
1
2
the cruiser is moving toward the pursuit ship at
.385.0 c
37.20: In the frame of one of the particles,
u
and v are both
c9520.0
but with opposite
sign.
.9988.0
)9520.0)(9520.0(1
9520.09520.0
)()(1
)(
2
c
cc
cvu
uv
v
Thus, one particle moves at a speed
c9988.0
toward the other in the other
particle’s frame.
37.21:
.784.0
)650.0)(950.0(1
650.0950.0
1
2
c
cc
c
vu
uv
v
37.22: a) In Eq.(39-24),
)400.0()700.0(1
400.0700.0
1
700.0,400.0
2
cc
cvu
uv
vcvcu
.859.0 c
b)
.s0.31
859
.
0
m1000.8
9
c
v
x
37.23:
uv
c
vuv
v
cuv
uv
v
22
1
c
cc
u
cvv
vv
uvv
c
vv
u
837.0
))920.0)(360.0(1(
920.0360.0
)1(
1
22
moving opposite the rocket, i.e., away from Arrakis.
37.24: Solving Eq. (37.25) for
cu
, (see solution to Exercise 37.25)
,
)(1
)(1
2
0
2
0
ff
ff
c
u
and so (a) if
,0202.0,98.0
0
cuff
the source and observer are moving away
from each other. b) if
,882.0,4
0
cuff
they are moving toward each other.
37.25: a)
2
0
2
0
)()( fucfucf
uc
uc
f
)1λ)λ((
)1λ)λ((
1)(
)1)(()(
2
0
2
0
2
0
2
0
22
0
2
0
2
c
ff
ffc
ff
ffc
u
h.km101.72skm104.77sm1077.4159.0
)1)575675((
)1)575675((
847
2
2
ccu
b)
172$)00.1$(h)km90hkm1072.1(
8
million dollars!
37.26: Using
ccu 53600.0
in Eq. (37.25) gives
.2
58
52
531
531
000
ffff
37.27: a)
2/322
2
2
1
2
1
2222
)1(
)2(
11
cv
cvamv
cv
ma
cv
mv
dt
d
dt
dp
F
.1
)1(
)1(
)1(
23
2
2
2322
2322
2222
c
v
m
F
a
cvma
cv
cvcv
ma
b) If the force is perpendicular to velocity then denominator is constant
dt
dp
F
.1
1
22
22
cv
m
F
a
cv
dtdvm
37.28:
The force is found from Eq. (37.32) or Eq. (37.33). (a) Indistinguishable from
N.145.0
maF
b)
.N75.1γ
3
ma
c)
N.7.51γ
3
ma
d)
N,145.0γ
ma
N.03.1,N333.0
37.29: a)
mv
cv
mv
p
2
1
22
.866.0
2
3
4
3
1
4
1
121
22
2
2
22
ccvcv
c
v
cv
b)
c
v
c
v
mamaF
3/2
2
2
3/133
2
1
1
so)2(
γ2γ2γ
.608.021
3/2
37.30: a)
s.m1021.4and140.0so,01.1γ
8
vcv
b) The relativistic
expression is always larger in magnitude than the non-relativistic expression.
37.31: a)
22
22
2
1
mcmc
cv
mc
K
.866.0
4
3
1
4
1
2
1
1
2
2
22
cv
c
v
cv
b)
.986.0
36
35
1
36
1
6
1
1
5
2
2
22
2
ccv
c
v
cv
mcK
37.32:
eV.1088.1J1001.3)sm1000.3(kg)1067.1(22
91028272
mcE
37.33:
2
4
22
8
3
2
1
)1
γ(
c
mv
mvmcK
s.m1089.4163.0
150
4
4
150
2
02.0
8
3
2
1
02.1
722
2
2
4
2
0
ccvcv
v
c
v
mvKKif
37.34: a)
.)1007.4()1γ(
232
f
mcmcKW
b)
( .79.4)γγ
22
if
mcmc
c) The result of part (b) is far larger than that of part (a).
37.35:
a) Your total energy E increases because your potential energy increases;
%103.3)sm10998.2()m30()sm80.9()(
)(so)(
132822
222
cygmm
cymgcEmcmE
y
mg
E
This increase is much, much too small to be noticed.
b)
J36.0m)m)(0.060N1000.2(
22
2
1
2
2
1
kxUE
kg100.4)(
162
cEm
Energy increases so mass increases. The mass increase is much, much too small
to be noticed.
37.36: a)
2
00
cmE
2
0
2
22 cmmcE
0
22
0
0
2
/1
2 m
cv
m
mm
sm1060.2866.0
43
4
3
1
4
1
8
2
2
2
2
cv
cv
c
v
c
v
b)
2
22
0
22
0
1
10
c
cv
m
mccm
sm1098.2995.0
100
99
100
99
100
1
1
8
2
2
2
2
ccv
c
v
c
v
37.37: a)
J1005.400.3means00.4so,
10222
mcKmcEKmcE
b)
22222222
)()(0.15so,00.4;)()( pcmcmcEpcmcE
smkg1094.115
18
mcp
c)
222
1 cvmcE
ccvcvmcE 968.01615and1611gives00.4
222
37.38:
The work that must be done is the kinetic energy of the proton.
a)
2
0
2
0
1
1
1
)1(
2
2
cmcmK
c
v
J1056.7
1
0.011
1
J)1050.1(
1
1
1
)sm10kg)(3.001067.1(
13
10
2
c1.0
2827
c
b)
1J)1050.1(
2
(0.5)1
1
10
K
J1032.2
11
c)
1J)1050.1(
2
(0.9)1
1
10
K
J1094.1
10
37.39:
)smkg102.10kg,1064.6(
1827
pm
a)
222
)()( pcmcE
J.1068.8
10
b)
J.1070.2kg)1064.6(1068.8
10227102
cmcEK
c)
.452.0
kg)10(6.64
J1070.2
227
10
2
cmc
K
37.40:
21
2
2212242
1)(
mc
p
mccpcmE
22
2
2
22
2
2
2
1
22
1
1
mvmc
m
p
mc
cm
p
mc
the sum of the rest mass energy and the classical kinetic energy.
37.41: a)
0376.1
1
1
γsm108
22
7
cv
v
p
mm J1034.5
2
1
122
0
mvK
.06.1 J1065.5)1γ(
0
122
K
K
mcK
b)
3.203γsm1085.2
8
v
.88.4J1031.31)(γ
J1078.6
2
1
0
102
112
0
KKmcK
mvK
37.42:
MeV.3105J1097.4)sm10kg)(3.001052.5(
102827
37.43: a)
VeVqK
V1006.2
MeV2.06J10295.3025.41
1
1
6
132
22
2
eKV
mc
cv
mcK
b) From part (a),
MeV2.06J1030.3
13
K
37.44:
a) According to Eq. 37.38 and conservation of mass-energy
.292.1
)7.16(2
75.9
1
2
122
222
M
m
McmcMc
Note that since
,
22
1
1
cv
we have that
.6331.0
)292.1(
1
1
1
1
22
c
v
b) According to Eq. 37.36, the kinetic energy of each proton is
MeV.274
J101.60
MeV00.1
)sm10kg)(3.001067.1)(1292.1()1(
13
28272
McK
c) The rest energy of
0
is
J101.60
MeV00.1
28282
13
)sm10kg)(3.001075.9(mc
MeV.548
d) The kinetic energy lost by the protons is the energy that produces the
,
0
MeV).2(274MeV548
37.45: a)
eV.104.20MeV420.0
5
E
b)
eVJ106.1
)sm1000.3(kg)1011.9(
eV1020.4
19
2831
52
mcKE
eV.109.32eV105.11eV1020.4
555
c)
2
2
22
2
1
1
E
mc
cv
cv
mc
E
.sm1051.2836.0
eV109.32
eV1011.5
1
8
2
5
5
cc
d) Nonrel:
kg109.11
)eVJ10eV)(1.61020.4(22
2
1
31
915
2
m
K
vmvK
.sm1084.3
8
37.46:
a) The fraction of the initial mass that becomes energy is
,10382.6
u)2(2.0136
u)0015.4(
1
3
and so the energy released per kilogram is
J.1074.5)sm10kg)(3.0000.1)(10382.6(
14283
b)
kg.107.1
kgJ105.74
J100.1
4
14
19
[...]... 10 8 s 2.20 10 8 m s 37. 61: x 2 c 2 t 2 ( x ut ) 2 2 c 2 2 t ux c 2 2 x ut c(t ux c 2 ) u 1 x1 x(u c) t (u c) x ct c c x 2 c 2t 2 37. 62: a) From Eq (37. 37), (3.00 10 4 m s) 4 1 3 v4 3 K mv 2 m 2 (90.0 kg) 304 J 2 8 c 8 (3.00 10 8 m s) 2 2 (3 8) mv 4 c 2 3 v 7.50 10 9 b) 2 (1 2)mv 4c 37. 63: dv F (1 v 2 c... m mc E c, so E mc 2 magnitude of momentum, E c b) v mc 37. 58: a) v 37. 59: Speed in glass v 1 1 v2 c2 c c 1.97 10 8 m s n 1.52 1.326 K ( 1)mc 2 (0.326)(0.511 MeV) 0.167 MeV 1.67 10 5 eV 37. 60: a) 80.0 m s is non-relativistic, and K 1 2 mv 186 J 2 b) ( 1)mc 2 1.31 1015 J c) In Eq (37. 23), v 2.20 10 8 m s , u 1.80 10 8 m s , and so v ... t 12 2.11 10 5 37. 50: One dimension of the cube appears contracted by a factor of 1 , so the volume in γ S is a 3 γ a 3 1 (u c) 2 37. 51: Need a b l 0 a, l b l b b 1 u 2 c2 l0 a 1.40b 2 2 b 1 u c 1 c 1 0.700c a 1.40 2.10 108 m s 37. 52: The change in the astronaut’s biological age is t 0 in Eq (37. 6), and t is the distance to... Tachyons seem to violate causality 37. 67: Longer wavelength (redshift) implies recession (The emitting atoms are moving (λ λ) 2 1 away.) Using the result of Ex 37. 26: u c 0 (λ 0 λ) 1 (656.3 953.4) 2 1 8 u c 0.3570c 1.071 10 m s 2 (656.3 953.4) 1 37. 68: The baseball had better be moving non-relativistically, so the Doppler shift formula (Eq (37. 25)) becomes f f 0 (1 (u... Ft m So as t , v c v 1 ( Ft mc) 2 Ft v2 m 2 37. 64: Setting x 0 in Eq (37. 21), the first equation becomes x ut and the last, upon multiplication by c, becomes ct ct Squaring and subtracting gives c 2 t 2 x 2 γ 2 (c 2 t 2 u 2 t 2 ) c 2 t 2 , or x c t 2 t 2 4.53 10 8 m 37. 65: a) Want t t 2 t1 x1 ( x1 ut1 ) γ x2 ( x2 ... yr 24.7 yr γu γ(0.9910) 37. 53: a) E mc 2 and 10 1 1 (v c ) 2 v c v 99 γ2 1 c 2 c γ 100 0.995 v 2 2 b) ( pc) m v γ c , E m c γ 1 c 2 2 1 1 E ( pc) 0.01 1% 2 2 1 (10 (0.995)) 2 E 2 v 1 c 2 2 2 2 2 2 2 4 37. 54: a) Note that the initial velocity is parallel to the x-axis Thus, according to Eqn 37. 30, 3 2 3 Fx v (3.00... does not enter the front of the barn until the later time t 3 37. 72: In Eq (37. 23), u V , v (c n), and so ( c n) V (c n ) V v cV 1 (V nc) 1 2 nc For V non-relativistic, this is v ((c n) V )(1 (V nc)) (c n) V (V n 2 ) (V 2 nc) c 1 1 2 V n n 1 so k 1 2 For water, n 1.333 and k 0. 437 n dv dt (dt udx c 2 ) dt dv vu u dv dv... s)]2 x 1.81 m 2F 2(4.20 10 4 N) b) Using the relativistic work-energy theorem for a constant force (Eq 37. 35) we obtain ( 1)mc 2 x F 1 2.55, thus For the given speed, 2 1 0.920 (2.55 1)(2.00 10 12 kg)(3.00 10 8 m s) 2 6.65 m (4.20 10 4 N) c) According to Eq 37. 30, x 3 3 3 v2 2 (4.20 10 4 N) v 2 2 F v2 2 1 2 (2.10 1016 m s 2 )1 2 , a 1... to Eqn 37. 33, 2 1 1 Fy v 2 2 (5.00 10 12 N) 2 2 2 1 2 ay (1 0.900 ) 1.31 1015 m s 27 c m (1.67 10 kg) b) The angle between the force and acceleration is given by cos θ Fx a x Fy a y Fa ( 3.001012 N)( 1.491014 m s 2 ) ( 5.001012 N)(1.311015 m s 2 ) ( 3.001012 N)2 ( 5.001012 N) 2 ( 1.491014 m s 2 ) 2 (1.311015 m s 2 ) 2 θ 24.5 37. 55:... 1.44 10 8 s c) Part (b): t c c x 1 37. 66: a) (100 s)(0.600)(3.00 10 8 m s) 1.80 1010 m b) In Sebulbas frame, the relative speed of the tachyons and the ship is 3.40c, and so the time t 2 100 s 1.80 1010 m 118 s At t 2 Sebulba measures that Watto is a distance from him of 3.4c (118 s)(0.600)(3.00 10 8 m s) 2.12 1010 m c) From Eq (37. 23), with v 4.00c and u 0.600c, v . a)
m3600
0
l
m.3568m)(0.991)(3600
)sm1000.3(
)sm1000.4(
1)m3600(1
28
27
0
2
2
0
l
c
u
ll
)b
.s109.00
sm104.00
m3600
5
7
0
0
u
l
t
c)
.s108.92
sm104.00
m3568
5
7
u
l
t
37. 12:
.sm102.86952.0)γ1(1so,3048.01γ
82
ccu
37. 13:
22
0
2
2
0
1
1
cu
l
l
c
u
ll
m.92.5
0.600
1
m74.0
2
0
c
c
l
37. 14:
Multiplying. in Eq. (37. 6), so
,s33.8γ
0
tt
as in part (a).
c)
9
1000.2)800.0()s33.8( c
m, which is the distance
x
found in part (a).
37. 17: Eq. (37. 18):
22
1
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