A 4.0 kg ball is traveling at 3.0 m/s and strikes a wall. The ball bounces off the wall with a velocity of 4.0 m/s in the opposite direction. If the ball is in contact with the wall for 0.10 seconds, then the magnitude of the average force on the ball during contact is

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Quizplus · Accepted Answer

The change in velocity is the final velocity minus the initial velocity, taking direction into account. Since the ball changes direction, the total change in velocity is $3.0 \, \text{m/s} + 4.0 \, \text{m/s} = 7.0 \, \text{m/s}$. The mass of the ball is $4.0 \, \text{kg}$, so the change in momentum ($\Delta p$) is $4.0 \, \text{kg} \times 7.0 \, \text{m/s} = 28.0 \, \text{kg} \cdot \text{m/s}$. The average force ($F$) is the change in momentum divided by the time interval ($\Delta t$), so $F = \frac{\Delta p}{\Delta t} = \frac{28.0 \, \text{kg} \cdot \text{m/s}}{0.10 \, \text{s}} = 280 \, \text{N}$.

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