## Physics for Scientists and Engineers: A Strategic Approach with Modern Physics (3rd Edition)

We can use the graph to find the impulse exerted on the ball. The impulse is equal to the area under the force versus time graph. $J = ~F_x~t$ $J = (500~N)(8.0\times 10^{-3}~s)$ $J = 4.0~N~s$ We can use the impulse to find the final momentum $p_f$. $p_f = p_0+J$ $p_f = m~v_0+J$ $p_f = (0.25~kg)(-10~m/s)+4.0~N~s$ $p_f = 1.5~N~s$ We can use the final momentum to find the rebound velocity $v_{fx}$. $m~v_{fx} = p_f$ $v_{fx} = \frac{p_f}{m}$ $v_{fx} = \frac{1.5~N~s}{0.25~kg}$ $v_{fx} = 6.0~m/s$ The ball's rebound velocity is 6.0 m/s.