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Since the block does not move significantly during the collision, the elastic and
gravitational potential energies do not change. Thus,
E
is the energy that is transferred.
The ratio is
E
/
E
0
= (6.94 J)/(563 J) = 0.0123 or 1.23%.
69. (a) Assume the bullet becomes embedded and moves with the block before the block
moves a significant distance. Then the momentum of the bulletblock system is
conserved during the collision. Let
m
be the mass of the bullet,
M
be the mass of the
block,
v
0
be the initial speed of the bullet, and
v
be the final speed of the block and bullet.
Conservation of momentum yields
mv
0
=
(
m
+
M
)
v
, so
=
+
=
0.050
150
0.050
+ 4.0
=1.85
.
0
v
mv
mM
kg
m / s
kg
kg
m/s
a
fa
f
When the block is in its initial position the spring and gravitational forces balance, so the
spring is elongated by
Mg
/
k
. After the collision, however, the block oscillates with simple
harmonic motion about the point where the spring and gravitational forces balance with
the bullet embedded. At this point the spring is elongated a distance
A
=+
Mm
gk
af
/,
somewhat different from the initial elongation. Mechanical energy is conserved during
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This note was uploaded on 06/03/2011 for the course PHY 2049 taught by Professor Any during the Spring '08 term at University of Florida.
 Spring '08
 Any
 Physics, Mass, Momentum

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