Answer
(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 $1840~N$
Work Step by Step
(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 $1840~N$.
