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(C) The average force $F$ is given by the rate of change of momentum: $F = \frac{\Delta p}{\Delta t}$.Given mass $m = 140\, g = 0.140\, kg$.The initia

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$9100$

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(C) The average force $F$ is given by the rate of change of momentum: $F = \frac{\Delta p}{\Delta t}$.Given mass $m = 140\, g = 0.140\, kg$.The initial velocity $v_1 = 39.0\, m/s$ (let this be in the positive direction).The final velocity $v_2 = -39.0\, m/s$ (opposite direction).The change in momentum $\Delta p = m(v_2 - v_1) = 0.140 	imes (-39.0 - 39.0) = 0.140 	imes (-78.0) = -10.92\, kg \cdot m/s$.The impact time $\Delta t = 1.20\, ms = 1.20 	imes 10^{-3}\, s$.The magnitude of the average force is $F = \frac{|\Delta p|}{\Delta t} = \frac{10.92}{1.20 	imes 10^{-3}} = \frac{10.92}{0.0012} = 9100\, N$.

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