Physics 133 HW8 Spring 2009
1. [HRW 37.05]
2. [HRW 37.08]
3. [HRW 37.14]
5. We solve the time dilation equation for the time elapsed (as measured by Earth
observers):
!
!
t
t
"
#
0
2
1
0 9990
( .
)
where
!
t
0
= 120 y. This yields
!
t
= 2684 y
3
2.68 10 y.
$%
8. Only the “component” of the length in the
x
direction contracts, so its
y
component
stays
sin 30
(1.0 m)(0.50)
0.50m
yy
! "
"
# "
"
!!!
while its
x
component becomes
22
1
(1.0 m)(cos30 ) 1 (0.90)
0.38m.
xx
$
! "
%
"
#
%
"
!!
Therefore, using the Pythagorean theorem, the length measured from
S'
is
&’ &’
2
2
(0.38 m)
(0.50 m)
0.63m.
xy
!
!
!
"
(
"
(
"
!
!
!
14. (a) We solve Eq. 3713 for
v
and then plug in:
2
2
0
1
1
1
0.866.
2
L
L
!
"#
$
%
$
%
$
&’
()
(b) The Lorentz factor in this case is
*+
2
1
2.00
1/
vc
,
$$
%
.
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View Full DocumentPhysics 133 HW8 Spring 2009
4. [HRW 37.15]
5. [HRW 37.76]
22
11
7.09.
1
1 (0.99)
!
"
#
#
#
$$
Thus,
%
t
0
= (26.26 y)/(7.09) = 3.705 y.
15. (a) The speed of the traveler is
v
= 0.99
c
, which may be equivalently expressed as
0.99 ly/y. Let
d
be the distance traveled. Then, the time for the trip as measured in the
frame of Earth is
%
t = d
/
v
= (26 ly)/(0.99 ly/y) = 26.26 y.
(b) The signal, presumed to be a radio wave, travels with speed
c
and so takes 26.0 y to
reach Earth. The total time elapsed, in the frame of Earth, is
26.26 y + 26.0 y = 52.26 y .
(c) The proper time interval is measured by a clock in the spaceship, so
%
t
0
=
%
t
/
. Now
76. (a) The relative contraction is
2
1
2
2
2
0
8
00
12
(1
)

1
1
1
630m/s
1
1
1
1
2
2
2 3.00 10 m/s
2.21 10
.
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 Fall '09
 STEIGMAN
 Physics, Special Relativity, Time Dilation, Spacetime, Lorentz Transformation, HW8 Spring

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