Question

A batter hits a foul ball. The 0.140-kg baseball that was approaching him at 40.0 m/s leaves

the bat at 30.0 m/s in a direction perpendicular to the line between the batter and the pitcher.

What is the magnitude of the impulse delivered to the baseball?

7.00 Ns

Answer 1

Since the resultant momentum is at 90° relative to the initial, the change in momentum is

$$\Delta p=\sqrt{(mv_1)^2+(mv_2)^2}=7\text{ Ns}.$$