6 CONSERVATION OF MOMENTUM 6.7 Collisions in 2 - dimensions
3
p
p
By definition, the net momentum change is equal to the impulse imparted by the
bat, so
I = ∆p = 11.55 N s.
The average force exerted by the bat on the ball is simply the net impulse
divided by the time interval over which the ball is in contact with the bat. Hence,
f ̄^ =
I
t
11.55
=
0.01
= 1155.0 N.
Worked example 6.3: Skater and medicine ball
Question: A skater of mass M = 120 kg is skating across a pond with uniform
velocity v = 8 m/s. One of the skater’s friends, who is standing at the edge of
the pond, throws a medicine ball of mass m = 20 kg with velocity u = 3 m/s to
the skater, who catches it. The direction of motion of the ball is perpendicular
to the initial direction of motion of the skater. What is the final speed of the
skater? What is the final direction of motion of the skater relative to his/her
initial direction of motion? Assume that the skater moves without friction.
p
2
1
x
Answer: Suppose that the skater is initially moving along the x-axis, whereas the
initial direction of motion of the medicine ball is along the y-axis. The skater’s
initial momentum is
p 1 = (M v, 0) = (120 × 8, 0) = (960, 0) N s.
Likewise, the initial momentum of the medicine ball is
p 2 = (0, m u) = (0, 20 × 3) = (0, 60) N s.
y