Example 7-4 A Tricky Billiards Shot
On a billiards table, the 7-ball and the 8-ball are initially at rest and touching each other. You hit the cue ball in such a way
that it acquires a speed of 1.7 m>s before hitting the 7-ball and 8-ball simultaneously. After the collision, the cue ball is at
rest and the other two balls are each moving at 45° from the direction that the cue ball was moving. What are the speeds
of the 7-ball and 8-ball immediately after the collision? Each billiard ball has a mass of 0.16 kg.
Set Up
This is a more complicated collision than in the
previous two examples, but the fundamental
principle is the same: Total momentum is
conserved. Note that all three balls have the
same mass m = 0.16 kg.
Initially all of the momentum is in the
cue ball and points along the direction of its
motion. After the collision, the 7-ball and
8-ball must travel on opposite sides of the cue
ball’s initial path as shown. (If they were on
the same side, there would be a nonzero total
momentum perpendicular to the cue ball’s
initial path, and momentum would not be
conserved.)
As in Example 7-2, we know the directions
of motion of the balls before and after the collision. We know the initial speed vcue, i = 1.7 m>s
and final speed vcue, f = 0 of the cue ball, and we
want to find the final speeds v7f and v8f of the
7-ball and 8-ball.
Solve
Write in component form the momentum
conservation equation, which says that the total
momentum before the collision (subscript i)
equals the total momentum after the collision
(subscript f). Take the positive x axis to be
in the direction that the cue ball was moving
before the collision.
Momentum conservation:
scue + p
s7 + p
s8
Ps = p
before
vcue,i = 1.7 m/s
(7-14)
8
(cue ball, 7-ball, and 8-ball)
has the same value just before and
just after the collision
after
v7f = ?
Linear momentum:
s = mv
s
p
7
7
(7-5)
45°
45°
8
v8f = ?
Momentum conservation:
scue, f + p
s7f + p
s8f = p
scue, i + p
s7i + p
s8i
p
The 7-ball and 8-ball are not moving before the collision,
s7i = p
s8i = 0, and the cue ball is not moving after the
so p
scue, f = 0. So the momentum conservation
collision, so p
equation becomes
s7f + p
s8f = p
scue, i
p
s7f + mv
s8f = mv
scue, i
mv
or in component form
mv7fx + mv8fx = mvcue, ix (total x momentum is conserved)
mv7fy + mv8fy = mvcue, iy (total y momentum is conserved)
The cue ball’s initial velocity had zero y
component (so vcue, iy = 0). Of the 7-ball and
8-ball, one has positive y velocity and the
other negative y velocity in order to satisfy the
condition that total y momentum is conserved.
Use these observations to simplify the
momentum equations.
x equation:
mv7f cos 45°
+ mv8f cos 45° = mvcue, i
y equation:
mv7f sin 45°
+ (2mv8f sin 45°) = 0
before
pcue,i
y
p7i = p8i = 0
after
y
x
p7f
45°
x
pcue,f = 0
45°
p8f
Solve the momentum conservation equations
for the final speeds of the 7-ball and 8-ball.
From the y equation,
mv7f sin 45° = mv8f sin 45° so
v7f = v8f
After the collision, the 7-ball and the 8-ball move with the same
speed.
To find this common speed, replace v8f in the x equation with v7f
and then solve for v7f:
mv7f cos 45° + mv7f cos 45° = mvcue, i
2mv7f cos 45° = mvcue, i
v7f =
vcue, i
2 cos 45
=
1.7 m>s
2 cos 45
= 1.2 m>s
After the collision, both the 7-ball and the 8-ball are moving at
1.2 m>s.
Reflect
Note that we didn’t need the value of the billiard ball mass. That’s because all three balls have the same mass m, so
m canceled out in the equations. That didn’t happen in Example 7-3 because the bowling ball and pin had different
masses.
There were three objects in this collision rather than two as in Examples 7-2 and 7-3. But the same ideas of momentum
conservation apply no matter how many objects are involved in the collision.