Physics 1311
Spring 2010
PS #4 Solutions
Page 1 of 10
Physics 1311
Problem Set #4 Solutions
Problem 1:
A
6
.
20
g bullet is fired horizontally at two blocks resting on a smooth tabletop, as
shown in the top figure. The bullet passes through the first block, with mass
1
.
20
kg, and embeds
itself in the second, with mass
1
.
80
kg. Speeds of
0
.
430
m/s and
1
.
06
m/s, respectively, are
thereby imparted to the blocks, as shown in the bottom figure. (a) Neglecting the mass removed
from the first block by the bullet, find the speed of the bullet immediately after it emerges from
the first block. (b) Find the bullet’s original speed.
m
1
m
2
1
v
v
2
v
0
Throughout this set of collisions, the total momentum of
the system will be constant.
Initially only the bullet is
moving at some unknown speed, so we write the initial
momentum as
p
i
=
m
b
v
0
.
After the bullet passes through the first block, the block
is moving at a known speed
v
1
and the bullet is moving at
some speed
v
b
that is slower than
v
0
. The total momentum
now is
p
f
1
=
m
b
v
b
+
m
1
v
1
.
Finally, when the bullet lodges itself in the second block, the total momentum consists of
these two objects moving together at
v
2
, plus the first block still moving at
v
1
:
p
f
2
=
m
1
v
1
+ (
m
b
+
m
2
)
v
2
.
Note that we have all the quantities for
p
f
2
, so we can use this and the equation for
p
f
1
to
solve for the unknown
v
b
:
p
f
1
=
p
f
2
=
⇒
m
b
v
b
+
m
1
v
1
=
m
1
v
1
+ (
m
b
+
m
2
)
v
2
=
⇒
m
b
v
b
= (
m
b
+
m
2
)
v
2
v
b
=
m
b
+
m
2
m
b
v
2
=
1 +
m
2
m
b
v
2
.
Numerically, this is
v
b
=
1 +
1
.
80
0
.
00620
(1
.
06 m/s) = 308
.
8 m/s
.
=
309 m/s
.
Now, using the fact that momentum is conserved in the first collision, we can solve for the
bullet’s initial speed:
p
i
=
p
f
1
=
⇒
m
b
v
0
=
m
b
v
b
+
m
1
v
1
=
⇒
v
0
=
v
b
+
m
1
m
b
v
1
.
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Physics 1311
Spring 2010
PS #4 Solutions
Page 2 of 10
Using our previous result and the given numbers,
v
0
= (308
.
8 m/s) +
1
.
20
0
.
00620
(0
.
430 m/s) =
392 m/s
.
Problem 2:
Two pucks collide on an air hockey table.
Puck A has a mass of
18
g and is
initially traveling in the
+
x
direction at
7
.
1
m/s. Puck B has a mass of
54
g and is initially at
rest. After the pucks collide, puck A moves away at an angle of
60
◦
above the
+
x
axis, while
puck B travels at an angle of
25
◦
below the
+
x
axis. (a) Calculate puck A’s final speed. (b)
Calculate puck B’s final speed.
We begin by setting up the equations for this problem. In
this 2D collision, momentum is conserved
in each coordinate
direction
. In the
x
direction, the conservation of momentum
equation can be written as
p
ix
=
p
fx
=
⇒
m
A
v
i
=
m
A
v
A
cos
θ
A
+
m
B
v
B
cos
θ
B
.
In the
y
direction, there was no momentum to start with, so
there’s no final momentum either:
p
ix
=
p
fx
=
⇒
0 =
m
A
v
A
sin
θ
A

m
B
v
B
sin
θ
B
.
We write the second term with a negative sign because
puck B heads o
ff
below the
x
axis.
θ
A
θ
B
v
v
v
B
A
i
A
B
A
B
(a) We have two momentum conservation equations, and two unknowns. To find
v
A
, we
first solve the
y
equation for
v
B
(actually for
m
B
v
B
):
m
B
v
B
sin
θ
B
=
m
A
v
A
sin
θ
A
=
⇒
m
B
v
B
=
m
A
v
A
sin
θ
A
sin
θ
B
.
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 Spring '10
 Wiegert
 Physics, Force, Momentum, Solutions Page

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