P25_045.PDF
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)
Pobierz
Chapter 25 - 25.45
45. (a) The potential energy is
U
=
4
πε
0
d
=
(8
.
99
q
2
×
10
9
N
·
m
2
/
C
2
)(5
.
0
×
10
−
6
C)
2
=0
.
225 J
1
.
00m
relative to the potential energy at infinite separation.
(b) Each sphere repels the other with a force that has magnitude
F
=
4
πε
0
d
2
=
(8
.
99
q
2
×
10
9
N
·
m
2
/
C
2
)(5
.
0
×
10
−
6
C)
2
=0
.
225 N
.
(1
.
00m)
2
According to Newton’s second law the acceleration of each sphere is the force divided by the mass
of the sphere. Let
m
A
and
m
B
be the masses of the spheres. The acceleration of sphere
A
is
a
A
=
F
m
A
=
0
.
225N
10
−
3
kg
=45
.
0m
/
s
2
5
.
0
×
and the acceleration of sphere
B
is
a
B
=
F
m
B
=
0
.
225N
10
−
3
kg
=22
.
5m
/
s
2
.
10
×
(c) Energy is conserved. The initial potential energy is
U
=0
.
225J, as calculated in part (a). The
initial kinetic energy is zero since the spheres start from rest. The final potential energy is zero
since the spheres are then far apart. The final kinetic energy is
2
m
A
v
A
+
2
m
B
v
B
,where
v
A
and
v
B
are the final velocities. Thus,
U
=
1
2
m
A
v
A
+
1
2
m
B
v
B
.
Momentum is also conserved, so
(
m
A
/m
B
)
v
A
,
from the momentum equation into the energy equation, and collecting terms, we obtain
U
=
1
−
2
(
m
A
/m
B
)(
m
A
+
m
B
)
v
A
.Thus,
v
A
=
2
Um
B
m
A
(
m
A
+
m
B
)
=
2(0
.
225J)(10
×
10
−
3
kg)
10
−
3
kg)
=7
.
75 m
/
s
.
(5
.
0
×
10
−
3
kg)(5
.
0
×
10
−
3
kg + 10
×
We thus obtain
v
B
=
−
m
A
m
B
v
A
=
−
5
.
0
×
10
−
3
kg
(7
.
75m
/
s) =
−
3
.
87 m
/
s
.
10
×
10
−
3
kg
0=
m
A
v
A
+
m
B
v
B
.
These equations may be solved simultaneously for
v
A
and
v
B
. Substituting
v
B
=
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