## College Physics (4th Edition)

(a) The ball's change in momentum was $-5.25~kg~m/s$ (b) The impulse that was applied to the ball was $-5.25~kg~m/s$ (c) The average force exerted on the ball by the catcher's glove is $1840~N$
(a) We can find the change in momentum of the ball: $\Delta p = p_f-p_0$ $\Delta p = 0-m~v_0$ $\Delta p = 0-(0.15~kg)(35~m/s)$ $\Delta p = -5.25~kg~m/s$ The ball's change in momentum was $-5.25~kg~m/s$ (b) The impulse that was applied to the ball is equal to the ball's change in momentum, which was $-5.25~kg~m/s$ (c) We can find the rate of deceleration of the ball: $v_f^2 = v_0^2+2ax$ $a = \frac{v_f^2 - v_0^2}{2x}$ $a = \frac{0 - (35~m/s)^2}{(2)(0.050~m)}$ $a = -12,250~m/s^2$ We can use the magnitude of acceleration to find the average force exerted on the ball by the catcher's glove: $F = ma = (0.15~kg)(12,250~m/s^2) = 1840~N$ The average force exerted on the ball by the catcher's glove is $1840~N$.