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Um
v
m
v
AA
BB
=+
1
2
1
2
22
.
Momentum is also conserved, so
0
=
+
mv
mv
AA
BB
.
These equations may be solved simultaneously for
v
A
and
v
B
. Substituting
(/)
BA
B
A
vm
m
v
=−
, from the momentum equation into the energy equation, and collecting
terms, we obtain
Um
m
m
m
v
AB A
B
A
=+
1
2
2
(/)
(
)
.
Thus,
3
333
2
2(0.225 J)(10 10 kg)
7.75 m/s.
(
)
(5.0 10 kg)(5.0 10 kg 10 10 kg)
B
A
AA B
Um
v
mm m
−
−−−
×
==
=
+×
×
+
×
51. (a) The potential energy is
U
q
d
==
×⋅
×
=
−
2
96
4
899 10
50 10
100
0 225
π
0
ε
..
.
.
Nm C
C
m
J
22
2
ch
c
h
relative to the potential energy at infinite separation.
(b) Each sphere repels the other with a force that has magnitude
F
q
d
==
×⋅
×
=
−
2
2
96
4
899 10
50 10
0 225
π
0
ε
..
.
Nm C
C
1.00 m
N.
22
2
2
ch
c
h
bg
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
F
m
A
A
==
×
=
0 225
450
3
.
.
N
5.0 10 kg
ms
2
and the acceleration of sphere
B
is
a
F
m
B
B
==
×
=
−
0 225
22 5
3
.
..
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This note was uploaded on 12/13/2010 for the course PHY 211 taught by Professor Ld during the Spring '10 term at NE Texas CC.
 Spring '10
 LD
 Physics, Energy, Force, Potential Energy

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