A $1200 \mathrm{~kg}$ car is traveling at $3.0 \mathrm{~m} / \mathrm{s}$ and strikes a wall. The car bounces off the wall with a velocity of 4.0 $\mathrm{m} / \mathrm{s}$ in the opposite direction. If the car is in contact with the wall for 0.10 seconds, then the magnitude of the average force of the wall on the car is

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Quizplus · Accepted Answer

The change in velocity is $3.0 \mathrm{~m/s} + 4.0 \mathrm{~m/s} = 7.0 \mathrm{~m/s}$ (since the car reverses direction, we add the speeds). The change in momentum ($\Delta p$) is mass ($m$) times change in velocity ($\Delta v$), so $\Delta p = 1200 \mathrm{~kg} \times 7.0 \mathrm{~m/s} = 8400 \mathrm{~kg \cdot m/s}$. The average force ($F$) is the change in momentum ($\Delta p$) divided by the time ($\Delta t$), so $F = \frac{\Delta p}{\Delta t} = \frac{8400 \mathrm{~kg \cdot m/s}}{0.10 \mathrm{~s}} = 84000 \mathrm{~N}$.

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A $1200 \mathrm{~kg}$ Car Is Traveling At $3.0 \mathrm{~m} / \mathrm{s}$ And Strikes a Wall