Answer to Question #124676 in Physics for Dr. Horus

Question

A 0.25kg ball is released from a height of 15m. The ball hits the floor and bounces in the opposite direction with 80% of its velocity. Calculate the impulse imparted by the floor on the ball.

Answer 1

The velocity of the ball before collision with floor

$$v_i=\sqrt{2gh}=\sqrt{2\times 9.8\times 15}=17.1\:\rm m/s$$

The velocity of the ball after collision with floor

$$v_f=0.8v_i=0.8\times 17.1=13.7\:\rm m/s$$

Hence, the impulse imparted by the floor on the ball

$$\Delta p=m|{\bf v}_f-{\bf v}_i|=m(v_f+v_i)\\ =0.25\times(13.7+17.1)=7.7\:\rm kg\cdot m/s$$