A squash ball hits a wall at a 45° angle as shown in Fig. 7–29. What is the direction (a) of the change in momentum of the ball, (b) of the force on the wall?

Given data:

The angle of incidence of the ball is\(\theta = 45^\circ \).

Let \({\vec P_1}\) be the momentum of the ball before the collision of the ball with the wall and \({\vec P_2}\) be the momentum of the ball after the collision of the ball with the wall.

Part (a)

Consider the collision of the ball is purely elastic.

The change in momentum of the ball is \({\vec P_2} - {\vec P_1}\)

You can observe that the change in momentum is perpendicular to the wall and away from the wall. Hence, the direction of the change in momentum of the ball is perpendicular to the wall and away from the wall.

Part (b)

From Newton’s second law, you know that the force on the ball is equal to the rate of change in momentum.

Now, from Newton’s third law, you know that the force on the wall will be equal to the force on the ball but in the opposite direction. Hence, the direction of the force on the wall is along the right perpendicular to the wall at the point of collision. The x-component of the force on the wall will be along the positive x-axis.