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Chapter 37
p.1
CHAPTER 37 – Special Theory of Relativity
1. (
a
) [1 – (
v
/
c
)
2
]
1/2
= {1 – [(20,000 m/s)/(3.00
×
10
8
m/s)]
2
}
1/2
=
1.00
.
(
b
) [1 – (
v
/
c
)
2
]
1/2
= [1 – (0.0100)
2
]
1/2
=
0.99995
.
(
c
) [1 – (
v
/
c
)
2
]
1/2
= [1 – (0.100)
2
]
1/2
=
0.995
.
(
d
) [1 – (
v
/
c
)
2
]
1/2
= [1 – (0.900)
2
]
1/2
=
0.436
.
(
e
) [1 – (
v
/
c
)
2
]
1/2
= [1 – (0.990)
2
]
1/2
=
0.141
.
(
f
)
[1 – (
v
/
c
)
2
]
1/2
= [1 – (0.999)
2
]
1/2
=
0.0447
.
2.
You measure the contracted length.
We find the rest length from
L
=
L
0
[1 – (
v
/
c
)
2
]
1/2
;
28.2 m =
L
0
[1 – (0.750)
2
]
1/2
, which gives
L
0
=
42.6 m
.
3.
We find the lifetime at rest from
?
t
= ?
t
0
/[1 – (
v
2
/
c
2
)]
1/2
;
4.76
×
10
–6
s = ?
t
0
/{1 – [(2.70
×
10
8
m/s)/(3.00
×
10
8
m/s)]
2
}
1/2
, which gives ?
t
0
=
2.07
×
10
–6
s
.
4.
You measure the contracted length:
L
=
L
0
[1 – (
v
/
c
)
2
]
1/2
= (100 ly){1 – [(2.50
×
10
8
m/s)/(3.00
×
10
8
m/s)]
2
}
1/2
=
55.3 ly
.
5.
We determine the speed from the time dilation:
?
t
= ?
t
0
/[1 – (
v
2
/
c
2
)]
1/2
;
4.10
×
10
–8
s = (2.60
×
10
–8
s)/[1 – (
v
/
c
)
2
]
1/2
, which gives
v
=
0.773
c
.
6.
We determine the speed from the length contraction:
L
=
L
0
[1 – (
v
/
c
)
2
]
1/2
;
25 ly = (75 ly)[1 – (
v
/
c
)
2
]
1/2
, which gives
v
=
0.94
c
.
7.
For a 1.00 per cent change, the factor in the expressions for time dilation and length contraction must
equal 1 – 0.0100 = 0.9900:
[1 – (
v
/
c
)
2
]
1/2
= 0.9900, which gives
v
=
0.141
c
.
8.
In the Earth frame, the clock on the
Enterprise
will run slower.
(
a
) We find the elapsed time on the ship from
?
t
= ?
t
0
/[1 – (
v
2
/
c
2
)]
1/2
;
5.0 yr = ?
t
0
/[1 – (0.84)
2
]
1/2
, which gives ?
t
0
=
2.7 yr
.
(
b
) We find the elapsed time on the Earth from
?
t
= ?
t
0
/[1 – (
v
2
/
c
2
)]
1/2
= (5.0 yr)/[1 – (0.84)
2
]
1/2
=
9.2 yr
.
9. (
a
) To an observer on Earth, 95.0 ly is the rest length, so the time will be
t
Earth
=
L
0
/
v
=
(95.0 ly)/0.960
c
=
99.0 yr
.
(
b
) We find the dilated time on the spacecraft from
?
t
= ?
t
0
/[1 – (
v
2
/
c
2
)]
1/2
;
99.0 yr = ?
t
0
/[1 – (0.960)
2
]
1/2
, which gives ?
t
0
=
27.7 yr
.
(
c
) To the spacecraft observer, the distance to the star is contracted:
L
=
L
0
[1 – (
v
/
c
)
2
]
1/2
= (95.0 ly)[1 – (0.960)
2
]
1/2
=
26.6 ly
.
(
d
) To the spacecraft observer, the speed of the spacecraft is
v
=
L
/?
t
= (26.6 ly)/27.7 yr =
0.960
c
,
as expected.
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View Full Document Chapter 37
p.2
10. (
a
) You measure the contracted length.
We find the rest length from
L
=
L
0
[1 – (
v
/
c
)
2
]
1/2
;
4.80 m =
L
0
[1 – (0.660)
2
]
1/2
, which gives
L
0
=
6.39 m
.
Distances perpendicular to the motion do not change, so the rest height is
1.25 m
.
(
b
) We find the dilated time in the sports vehicle from
?
t
= ?
t
0
/[1 – (
v
2
/
c
2
)]
1/2
;
20.0 s = ?
t
0
/[1 – (0.660)
2
]
1/2
, which gives ?
t
0
=
15.0 s
.
(
c
) To your friend, you moved at the same relative speed:
0.660
c
.
(
d
) She would measure the same time dilation:
15.0 s
.
11. In the Earth frame, the average lifetime of the pion will be dilated:
?
t
= ?
t
0
/[1 – (
v
2
/
c
2
)]
1/2
.
The speed as a fraction of the speed of light is
v
/
c
=
d
/
c
?
t
=
d
[1 – (
v
2
/
c
2
)]
1/2
/
c
?
t
0
;
v
/
c
= (15 m)[1 – (
v
2
/
c
2
)]
1/2
/(3.00
×
10
8
m/s)(2.6
×
10
–8
s),
which gives
v
=
0.89
c
= 2.7
×
10
8
m/s
.
12. With the standard orientation of the reference frames, the Galilean transformation is
x
=
x
′
+
vt
,
y
=
y
′
,
z
=
z
′
.
(
a
) For the given data we have
x
= 25 m + (30 m/s)(2.5 s) = 100 m;
y
= 20 m;
z
= 0.
Thus the coordinates of the person are
(100 m, 20 m, 0).
(
b
) For the given data we have
x
= 25 m + (30 m/s)(10.0 s) = 325 m;
y
= 20 m;
z
= 0.
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This note was uploaded on 02/05/2010 for the course PHYSICS 37,38,39 taught by Professor Fielding during the Spring '09 term at Los Angeles Valley College.
 Spring '09
 Fielding
 Physics, Theory Of Relativity

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