Answer to Question #211408 in Physics for Anonymous student

Question

A 500g-ball drops vertically onto the ground, hitting with speed 14 m/s. It rebounds to height 5 m. The ball is in contact with the ground for 15 ms. (a) Determine the magnitude of the impulse acting on the ball. Sketch the impulse vector. (b) Calculate the average force acting on the ball. What is the origin of this average force?

Answer 1

The speed of a ball after rebound

$$v_2=\sqrt{2gh_2}=\sqrt{2*9.8*5}=9.9\:\rm m/s$$

(a) The magnitude of impulse acting on the ball

$$\Delta p=m|v_2+v_1|=0.5*|9.9+14|=12\:\rm kg\cdot m/s$$

(b) The average force acting on the ball

$$F=\frac{\Delta p}{\Delta t}=\frac{12}{0.015}=797\:\rm N$$

This force is the force of elasticity.