Ph1a Solutions 5
Chefung Chan ([email protected]), Fall 2004
Each homework problem is worth 5 points. Please disregard the point values listed on the problem itself.
Use these instead.
5.1
QP14 (5 points)
5.1.a
(1 point)
Use the definition of the center of mass:
y
CM
=
∑
i
m
i
y
i
∑
i
m
i
=
m
1
y
1
+
m
2
y
2
m
1
+
m
2
=
md
+ (2
m
)(0)
m
+ 2
m
=
1
3
d.
5.1.b
(1 point)
Use conservation of energy to find
v
1
i
by equating the energies at the top and bottom of the arc:
E
i
=
mgd
=
1
2
mv
2
1
i
=
E
f
⇒
v
1
i
=
√
2
gd
=
v.
The lower mass
m
2
is at rest, so
v
2
i
= 0.
Momentum is conserved in the collision. Because the collision is elastic the kinetic energy is also conserved.
The initial momenta and energies are
p
1
i
=
mv
0
p
2
i
=
0
p
≡
p
1
i
+
p
2
i
=
mv
0
K
1
i
=
1
2
mv
2
0
K
2
i
=
0
K
≡
K
1
i
+
K
2
i
=
1
2
mv
2
0
The final momentum of the two pendula are given by
p
1
f
=
mv
1
f
and
p
2
f
= 2
mv
2
f
and the final kinetic
energies by
K
1
f
=
1
2
mv
2
1
f
and
K
2
f
=
mv
2
2
f
. Applying the conservations laws:
p
1
f
+
p
2
f
=
p
mv
1
f
+ 2
mv
2
f
=
mv
0
K
1
f
+
K
2
f
=
K
1
2
mv
2
1
f
+
mv
2
2
f
=
1
2
mv
2
0
Solving these equations yields
v
1
f
=

1
3
v
0
and
v
2
f
= +
2
3
v
0
.
5.1.c
(2 points)
The energies of the masses after the collision are
E
1
f
=
1
2
mv
2
1
f
=
1
18
mv
2
0
E
2
f
=
mv
2
2
f
=
4
9
mv
2
0
1
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Apply conservation of energy to each mass, noting that
v
2
0
= 2
gd
:
E
1
f
=
1
18
mv
2
0
=
mgh
1
E
2
f
=
4
9
mv
2
0
= 2
mgh
2
h
1
=
(

1
3
v
)
2
2
g
=
1
9
d
h
2
=
(
2
3
v
)
2
2
g
=
4
9
d.
To find the height of the center of mass, apply the definition again:
y
CM
=
∑
i
m
i
y
i
∑
i
m
i
=
m
1
h
1
+
m
2
h
2
m
1
+
m
2
=
1
9
md
+
4
9
(2
m
)
d
3
m
=
1
3
d.
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 Fall '07
 Goodstein
 Energy, Kinetic Energy, Mass, Work, MV

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