A girl of mass 55 kg throws a ball of mass 0.8 kg against a wall.The ball strikes the wall horizontally with a speed of 25 m/s,and it bounces back with this same speed.The ball is in contact with the wall 0.05 s.What is the average force exerted on the wall by the ball?

Question

Quizplus · Accepted Answer

Explanation:The impulse on the ball is equal to the change in momentum of the ball. The change in momentum is equal to the final momentum minus the initial momentum. The final momentum is equal to the mass of the ball times the final velocity, which is -25 m/s. The initial momentum is equal to the mass of the ball times the initial velocity, which is 25 m/s. Therefore, the change in momentum is equal to:$$\Delta p = m(v_f - v_i) = (0.8 kg)(-25 m/s - 25 m/s) = -40 kg m/s$$The impulse is equal to the average force times the time of contact. Therefore, the average force is equal to the impulse divided by the time of contact. The time of contact is 0.05 s. Therefore, the average force is equal to:$$F = \frac{\Delta p}{t} = \frac{-40 kg m/s}{0.05 s} = -800 N$$The negative sign indicates that the force is in the opposite direction of the change in momentum. Therefore, the average force exerted on the wall by the ball is 800 N.

A Girl of Mass 55 Kg Throws a Ball of Mass