Chapter 9 Problem 25

Problem

A tennis ball of mass $m$ = 0.060 kg and speed $v$ = 25m/s strikes a wall at a 45° angle and rebounds with the same speed at 45° . What is the impulse (magnitude and direction) given to the ball?

Solution

The impulse given the ball is the change in the ball’s momentum. From the symmetry of the problem, the vertical momentum of the ball does not change, and so there is no vertical impulse. Call the direction AWAY from the wall the positive direction for momentum perpendicular to the wall.

$\Delta P_{\perp }=mv_{\perp ,f}-mv_{\perp ,i}$ $=m\left(v\sin(45^{\circ })-v\sin(-45^{\circ })\right)$ $=2mv\sin(45^{\circ })$ $=2\left(6.0\times 10^{-2}{\text{kg}}\right)\left(25{\text{m/s}}\right)\sin(45^{\circ })$

$=2.1{\text{kg.m/s}}$ to the left