Question

A ball of mass 0.40 kg travels horizontally and strikes a vertical wall with a speed of 5.0 m s. It rebounds horizontally with a speed of 3.0 m s. The ball is in contact with the wall for a time of 0.20 s. Calculate the magnitude of average force exerted on the wall.

Answer 1

To find the magnitude of the average force exerted on the wall by the ball, we can use the impulse-momentum theorem, which relates the impulse exerted on an object to the change in momentum of that object:

Impulse = Change in momentum

We can express the impulse as the product of the average force and the time for which it acts:

Impulse = Average force × Time

Since the ball rebounds horizontally, its initial and final momenta have the same magnitude but opposite directions, so the change in momentum is:

Change in momentum = 2 × (final momentum) = 2 × (0.40 kg × 3.0 m/s) = 2.4 kg m/s

Using the formula for impulse, we can then solve for the average force:

Average force = Impulse / Time = (2.4 kg m/s) / 0.20 s = 12 N

Therefore, the magnitude of the average force exerted on the wall by the ball is 12 N.