Question

In a billiard table, a ball of mass weighing 200g rolls with a velocity of 2.0 m/s and rebounds back at 2.0 m/s. Solve the following: A. Change in momentum. B. Impulse given to the ball.

Answer 1

A. Change in momentum:

$$\Delta p=p_f-p_i=mv_f-mv_i=m(v_f-v_i),\\ \Delta p=0.2(2-(-2))=0.8\text{ kg-m/s}^2.$$

B. The impulse is the change in momentum:

$$I=\Delta p=0.8\text{ kg-m/s}.$$