This preview shows pages 1–3. Sign up to view the full content.
Relativity
379
EXECUTE:
22
3
(1
)
4
mc
mv
γ
−=
.
13
1
4
1
x
x
−
.
2
1
14
x
x
⎛⎞
=+
⎜⎟
−
⎝⎠
.
After a little algebra this becomes
2
91
5
8
0
xx
+−
=
.
( )
2
1
15
(15)
4(9)(8)
18
x
=−
±
+
.
The positive root is
0.425
x
=
.
/
xvc
=
, so
0.652
vx
c
c
==
.
EVALUATE:
The fractional increase of the relativistic expression above the nonrelativistic one increases as
v
increases.
37.46.
The fraction of the initial mass (a) that becomes energy is
3
(4.0015 u)
1
6.382 10 ,
2(2.0136 u)
−
×
and so the energy released
per kilogram is
38
2
1
4
(6.382 10 )(1.00 kg)(3.00 10 m s)
5.74 10 J.
−
××
=
×
(b)
19
4
14
1.0 10 J
1.7 10 kg.
5.74 10 J kg
×
=×
×
37.47.
(a)
2
6
8
2 9
,
(3.8 10
J) (2.998 10 m s)
4.2 10 kg
Em
cmE
c
=
×
×
=
×
.
1 kg is equivalent to 2.2 lbs, so
6
4.6 10
m
tons
(b)
The current mass of the sun is
30
1.99 10
kg,
×
so it would take it
30
9
20
13
(1.99 10
kg) (4.2 10 kg s)
4.7 10 s 1.5 10 years
=
×
=
×
to use up all its mass.
37.48.
IDENTIFY:
Since the final speed is close to the speed of light, there will be a considerable difference between the
relativistic and nonrelativistic results.
SET UP:
The nonrelativistic workenergy theorem is
0
11
F x
mv
mv
Δ=
−
, and the relativistic formula for a
constant force is
2
)
Fx
m
c
−
.
(a)
Using the classical workenergy theorem and solving for
x
Δ
, we obtain
9
8
2
0
6
(
)
(0.100 10
kg)[(0.900)(3.00 10 m s)]
3.65 m.
(
1
.
0
0
1
0
N
)
mv
v
x
F
−
−×
×
=
=
×
(b)
Using the relativistic workenergy theorem for a constant force, we obtain
2
)
.
mc
x
F
−
For the given speed,
2
1
2.29,
10
.
9
0
0
−
thus
98
2
6
(2.29 1)(0.100 10
kg)(3.00 10 m s)
11.6 m.
(1.00 10 N)
x
−
×
=
×
EVALUATE: (c)
The distance obtained from the relativistic treatment is greater. As we have seen, more energy is
required to accelerate an object to speeds close to
c
, so that force must act over a greater distance.
37.49.
(a) IDENTIFY
and
SET UP:
8
0
2.60 10 s
t
−
×
is the proper time, measured in the pion’s frame. The time
measured in the lab must satisfy
,
dc
t
=Δ
where
.
uc
≈
Calculate
t
Δ
and then use Eq.(37.6) to calculate
u
.
EXECUTE:
3
6
8
1.20 10 m
4.003 10 s
2.998 10 m/s
d
t
c
−
×
=
=
×
×
0
1/
t
t
Δ
−
so
1
/
2
0
/
)
t
t
Δ
Δ
and
2
0
/
)
t
t
Δ
Δ
Write
)
Δ
so that
2
(/) (
1 ) 12
12
Δ=−Δ
+
Δ≈−Δ
since
Δ
is small.
Using this in the above gives
2
0
1(
12)
t
t
Δ
−−Δ=
Δ
2
2
8
5
0
6
1
1
2.11 10
4
.
0
0
3
1
0
s
t
t
−
−
−
Δ×
=
=
×
EVALUATE:
An alternative calculation is to say that the length of the tube must contract relative to the moving
pion so that the pion travels that length before decaying. The contracted length must be
88
0
(2.998 10 m/s)(2.60 10 s)
7.79 m.
lc
t
−
=Δ =
×
×
=
0
ll
so
2
0
l
l
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document3710
Chapter 37
Then
(1
)
uc
=−
Δ
gives
2
2
5
3
0
1
1
7.79 m
2.11 10 ,
2
2 1.20 10 m
l
l
−
⎛⎞
Δ=
=
=
×
⎜⎟
×
⎝⎠
which checks.
(b) IDENTIFY
and
SET UP:
2
Em
c
γ
=
(Eq.(37.38).
This is the end of the preview. Sign up
to
access the rest of the document.
 Spring '06
 Buchler
 Physics

Click to edit the document details