A girl of mass 55 kg throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of $25 \mathrm {~m} / \mathrm { s } \text {, }$ and it bounces back with this same speed. The ball is in contact with the wall $0.050 \mathrm {~s}$ What is the magnitude of the average force exerted on the wall by the ball?

Question

Quizplus · Accepted Answer

The impulse on the ball is equal to the change in its momentum:$$J = \Delta p = m_b v_f - m_b v_i = m_b (v_f - v_i)$$where $m_b$ is the mass of the ball, $v_i$ is the initial velocity of the ball, and $v_f$ is the final velocity of the ball.The average force exerted on the wall by the ball is equal to the impulse divided by the time of contact:$$F_{avg} = \frac{J}{t} = \frac{m_b (v_f - v_i)}{t}$$Substituting the given values, we get:$$F_{avg} = \frac{(0.80 \mathrm{~kg})(25 \mathrm{~m/s} - (-25 \mathrm{~m/s}))}{0.050 \mathrm{~s}} = 800 \mathrm{~N}$$Therefore, the magnitude of the average force exerted on the wall by the ball is 800 N.

A Girl of Mass 55 Kg Throws a Ball of Mass