Example 7-3 A Collision at the Bowl-a-Rama
The sequence of images in Figure 7-8 shows a bowling ball striking a stationary pin. The second image shows that when
the collision occurs, the ball strikes a bit to the left of the horizontal center of the pin. The third image shows what happens
after the collision: The ball moves on a path to the left of its original direction and the pin shoots off to the right. Just after
one such collision, the ball is moving off horizontally at a 10.0° angle and the pin is moving off horizontally at a 60.0°
angle (measured relative to the original direction of motion of the ball). The mass of the ball is 3.50 times greater than the
mass of the pin. Just after the collision, is the pin moving faster or slower than the ball, and by what factor?
Figure 7-8 ​Hitting a bowling pin off-center After the bowling ball strikes the left-hand side of the pin, the ball deflects to the left
and the pin moves off to the right. How do their speeds compare?
Set Up
The forces that the ball and pin exert on each
other are very strong (imagine what it would
feel like if your finger were between the ball
and pin when they hit!). Compared to these,
we can ignore the external forces exerted on
the system of ball and pin, so the total momentum of this system is conserved. The collision
is two-dimensional, so we must account for the
components of momentum in both the x and
y directions.
Momentum conservation:
We are given the directions of motion of the
ball and pin before and after the collision, and
we want to find the ratio of the pin’s final speed
vPf to the ball’s final speed vBf. We don’t know
the mass of either the ball or the pin, but we do
know the ratio of their masses.
Ratio of the mass of the
ball to the mass of the pin:
mB
= 3.50
mP
Solve
10.0°
ball (B)
sB + p
sP
(7-14)
Ps = p
(B for ball, P for pin) has the
same value just before and
just after the collision
Linear momentum:
s = mv
s
p
The total momentum before the collision
(subscript i) equals the total momentum after
the collision (subscript f). Write this for both
the x component and the y component of
momentum.
Momentum conservation:
Just prior to the collision the ball has no
component of velocity in the y direction (so
vBiy = 0) and the pin is at rest (so vPix = vPiy = 0).
Just after the collision the ball is moving at a
10.0° angle with positive x and y components
of velocity, while the pin is moving at a 60.0°
angle with a positive x velocity and a negative y
velocity. Use these facts to rewrite the equations
for the conservation of x and y momentum.
x equation:
mB vBf cos 10.0°
+ mP vPf cos 60.0° = mB vBi
60.0°
pin (P)
(7-5)
sBf + p
sPf = p
sBi + p
sPi, or
p
mB s
vBf + mP s
vPf = mB s
vBi + mP s
vPi, or
mB vBfx + mP vPfx = mB vBix + mP vPix
(total x momentum is conserved)
mB vBfy + mP vPfy = mB vBiy + mP vPiy
(total y momentum is conserved)
y equation:
mB vBf sin 10.0°
+ (2mP vPf sin 60.0°) = 0
before
PBi
after
y
x
PPi = 0
y
10.0°
60.0°
PPf
PBf
x
Our goal is to find the ratio of the speed of the
pin just after the collision (vPf) to the speed
of the ball just after the collision (vBf). The
equation for x momentum isn’t useful for this
because we aren’t given the value of the ball’s
speed vBi before the collision. Instead, we use
the equation for y momentum to solve for
vPf >vBf , and then substitute the ratio of the
masses of the two objects.
Reflect
mB vBf sin 10.0° = mP vPf sin 60.0°
vPf
mB sin 10.0
0.174
= 13.502
= 0.702
=
vBf
mP sin 60.0
0.866
After the collision, the speed of the pin is 0.702 times that of the ball.
The pin moves relatively slowly because it suffered only a glancing blow from the ball. With a more head-on impact the
pin can end up moving away from the collision at a much faster speed than the ball. We’ll explore this in more detail in
Section 7-5.
The most important direction in this problem is the positive x direction, since that is the direction of the total
­momentum vector. So although we used x and y to label the coordinate axes, it would have been more fitting to have
called these the parallel and perpendicular axes based on their orientation relative to the total momentum vector. The
equation for y momentum is just the statement that the total perpendicular momentum equals zero before and after the
collision, and it was this statement that enabled us to find the pin’s speed in terms of the speed of the ball.