10.52.
Model:
Assume an ideal spring that obeys Hooke’s law. There is no friction, hence the mechanical
energy
g
s
K
U
U
+
+
is conserved.
Visualize:
We have chosen to place the origin of the coordinate system on the free end of the spring that is neither stretched
nor compressed, that is, at the equilibrium position of the end of the unstretched spring. The bullet’s mass is
m
and the block’s mass is
M
.
Solve:
(a)
The energy conservation equation
2
s2
g2
1
s1
g1
K
U
U
K
U
U
+
+
=
+
+
becomes
2
2
2
2
e
e
e
e
2
2
2
1
1
1
1
1
1
1
(
)
(
)
(
) (
)
(
)
(
)
(
) (
)
2
2
2
2
m
M v
k y
y
m
M g y
y
m
M v
k y
y
m
M g y
y
+
+
−
+
+
−
=
+
+
−
−
+
−
Noting
2
0 m/s,
v
=
we can rewrite the above equation as
2
2
2
2
2
1
1
1
(
)
2(
) (
)
(
)
(
)
k
y
m
M g
y
y
m
M v
k
y
Δ
+
+
Δ
+ Δ
=
+
+
Δ
Let us express
1
v
in terms of the bullet’s initial speed
B
v
by using the momentum conservation equation
f
i
p
p
=
which is
1
B
block
(
)
.
m
M v
mv
Mv
+
=
+
Since
block
0 m/s,
v
=
we have
1
B
m
v
v
m
M
⎛
⎞
=
⎜
⎟
+
⎝
⎠
We can also find the magnitude of
1
y
from the equilibrium condition
1
e
(
)
.
k y
y
Mg
−
=
1
Mg
y
k
Δ
=
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 Fall '09
 geller
 Conservation Of Energy, Energy, Force, Friction, Mass, Robert Hooke, Energy Conservation Equation, unstretched spring

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