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Chapter 39
453
P39.59
(a)
Since Mary is in the same reference frame,
′
S
, as Ted, she measures the ball to have the
same speed Ted observes, namely
′=
uc
x
0800
..
(b)
∆′=
′
=
×
×
=×
t
L
u
p
x
180 10
750 10
12
3
.
.
m
0.800 3.00 10 m s
s
8
ej
(c)
LL
v
c
c
c
p
=−
=
×
−
1
1 80 10
1
0600
144 10
2
2
12
2
2
12
.
.
.
m
m
af
Since
vc
=
.
and
′=−
x
.
, the velocity Jim measures for the ball is
u
uv
uvc
cc
c
x
x
x
=
′+
+′
=
−+
+−
1
0 800
0 600
1
0 800 0 600
0385
2
.
a
f
a
f
.
(d)
Jim measures the ball and Mary to be initially separated by 1 44 10
12
.
×
m. Mary’s motion at
0.600
c
and the ball’s motion at 0.385
c
nibble into this distance from both ends. The gap closes
at the rate 0 600
0 385
0 985
...
ccc
+=
, so the ball and catcher meet after a time
∆
t
=
×
×
488 10
12
3
.
.
m
0.985 3.00 10 m s
s
8
.
*P39.60
(a)
The charged battery stores energy
Et
==
=
P
120
50
60
3600
.
J
s
m
i
n
sm
i
n
J
bg
a
f
so its mass excess is
∆
m
E
c
×
−
22
14
400 10
J
31
0
m
s
kg
8
(b)
∆
m
m
=
×
×
−
−
−
10
160 10
14
3
12
.
.
kg
25
kg
too small to measure.
P39.61
∆
mc
mc
2
2
4 938 78
3 728 4
100%
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This note was uploaded on 12/14/2011 for the course PHY 203 taught by Professor Staff during the Fall '11 term at Indiana State University .
 Fall '11
 Staff
 Physics

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