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Physics 140  Discussion 11  Solutions
8.47.
I
DENTIFY
:
When the spring is compressed the maximum amount the two blocks aren’t
moving relative to each other and have the same velocity
V
r
relative to the surface. Apply
conservation of momentum to find
V
and conservation of energy to find the energy stored in
the spring. Since the collision is elastic, Eqs. 8.24 and 8.25 give the final velocity of each block
after the collision.
S
ET
U
P
:
Let
x
+
be the direction of the initial motion of
A
.
E
XECUTE
:
(a)
Momentum conservation gives
(2 00 kg)(2 00 m/s)
(12 0 kg)
V
.
.
=
.
and
0 333 m/s.
V
= .
Both blocks are moving at 0.333 m/s, in the direction of the initial motion of block
A
.
Conservation of energy says the initial kinetic energy of
A
equals the total kinetic energy at
maximum compression plus the potential energy
b
U
stored in the bumpers:
2
2
1
1
b
2
2
(2 00 kg)(2 00 m/s)
(12 0 kg)(0 333 m/s)
U
.
.
=
+
.
.
and
b
3.33 J.
U
=
(b)
2
1
2 00 kg 10 0 kg
(2 00 m/s)
1 33 m/s.
12 0 kg
A
B
A x
A x
A
B
m
m
v
v
m
m

.

.
=
=
.
=  .
+
.
Block
A
is moving in the
x

direction at 1.33 m/s.
2
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This note was uploaded on 02/15/2012 for the course PHYS 140 taught by Professor Uher during the Winter '10 term at University of Michigan.
 Winter '10
 Uher
 Momentum

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