Midterm Solutions
(I)
A bullet of mass
m
moving at horizontal velocity
v
strikes and sticks to the rim of a wheel (a solid disc) of mass
M
, radius
R
, anchored at its center but free to rotate. (i) Which of energy, momentum and angular momentum
is conserved for the bullet+wheel system? Give a few words of explanation. (ii) Find
ω
f
the final angular velocity
of the wheel.
10
(i) Only angular momentum is conserved. The collision is an inelastic collision, so energy cannot be conserved.
Momentum is not conserved because there is an external force acting at the center of the wheel that keeps
the wheelbullet system from moving forward after the collision. Angular momentum is conserved because
there are no external torques acting on the system. (The force acting at the center of the wheel does not
provide a torque because it is acting at the pivot point.)
(ii) To find
ω
f
, we equate the angular momentum of the system just before the bullet hits the wheel with the
final angular momentum of the rotating bulletwheel system.
mvR
=
ω
f
(
I
wheel
+
I
bullet
)
=
ω
f
1
2
MR
2
+
mR
2
ω
f
=
v
R
1
1 +
M
2
m
.
Please note that linear momentum
p
and angular momentum
L
are
not
equivalent. They have separate
conservations laws, different units, and measure fundamentally different quantities. You cannot equate
p
to
L
.
(II)
A block of mass
M
sits on frictionless table
L
meters from the edge. At
t
= 0
bullet of mass
m
and velocity
v
1
penetrates it from left and exits to the right with a speed
v
2
. (i) When will the block fly off the table? (ii) If
the table has a height
h
how far from the edge of the table will it land?
10
(Neglect loss of wood in block due to
bullet penetrating it and the time it takes bullet to traverse block.)
(i) To find when the block will fly off the table, we need to calculate the final velocity of the block by conserving
momentum before and after the collision.
mv
1
=
mv
2
+
Mv
=
⇒
v
=
m
M
(
v
1

v
2
)
There are no external forces on the block, so the block will fly off the table at
t
=
L
v
=
M
m
L
v
1

v
2
.
(ii) To find the distance the block travels in the
x
direction after leaving the table, we first need to find the
total time it spends in the air. We know that the block falls a distance
h
in the
y
direction and has no
initial velocity in
y
. This tells us that
h
=
1
2
gt
2
=
⇒
t
=
2
h
g
(clearly, the negative solution for
t
is not physical for this problem). There are no forces in the
x
direction,
so the total distance the block travels once it leaves the table is
x
=
vt
= (
v
1

v
2
)
m
M
2
h
g
.
(III)
Consider the force
F
=
i
2
xy
3
+
j
3
x
2
y
2
.
(i) Show that it is conservative.
(ii) What is the potential energy
U
(
x, y
)
associated with it ? (iii) What is the work done by the force along a path
y
=
x
123456789
joining
(0
,
0)
to
(1
,
1)
?
10
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2
(i) A force given by
F
=
i
F
x
+
j
F
y
is conservative if
∂F
x
∂y
=
∂F
y
∂x
. This is because in order for a force to be
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 Fall '08
 RAMAMURTISHANKAR
 Physics, Angular Momentum, Energy, Force, Mass, Momentum

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