Discussion
A tennis ball of mass 150 g is moving at 20 ms−1 . A racket strikes it, reversing its direction with a final speed of 30 ms−1 . If the contact time is 0.02 s, then the magnitude of the force (in N) exerted by the racket is
Answer: E
The force is 375 N.
Take the final direction as positive. The change in momentum is
\[ \begin{aligned} \Delta p &= 0.15 \times 30 - 0.15 \times (-20) = 7.5\ \text{kg m/s} \\ F &= \frac{\Delta p}{\Delta t} = \frac{7.5}{0.02} = 375\ \text{N} \end{aligned} \]
Ignoring the reversal (30 − 20 = 10 m/s) gives 75 N, not offered; 150 N and 15 N come from using the mass in grams or dropping a factor of ten, and 3.75 N forgets to divide by the contact time.
The force is large because the contact time is short; this is the impulse-momentum theorem, \(F\,\Delta t = \Delta p\).
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