A 140,g ball,in horizontal flight with a spee | vedclass

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Question

(C) Given: Mass $m = 140\,g = 0.140\,kg$. Initial velocity $v_i = 39.0\,m/s$. Final velocity $v_f = -39.0\,m/s$ (opposite direction). Impact time $\De

Vedclass · Accepted Answer

$9100$

Vedclass · Answer

(C) Given: Mass $m = 140\,g = 0.140\,kg$. Initial velocity $v_i = 39.0\,m/s$. Final velocity $v_f = -39.0\,m/s$ (opposite direction). Impact time $\Delta t = 1.20\,ms = 1.20 	imes 10^{-3}\,s$. The change in momentum is $\Delta p = m(v_f - v_i)$. $\Delta p = 0.140 	imes (-39.0 - 39.0) = 0.140 	imes (-78.0) = -10.92\,kg\cdot m/s$. The average force $F$ is given by $F = \frac{\Delta p}{\Delta t}$. $F = \frac{-10.92}{1.20 	imes 10^{-3}} = -9100\,N$. The magnitude of the average force acting on the ball is $9100\,N$.

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