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The average force on the 0.057 kg ball in contact with the racquet for 5.0 ms was about:
A6600 N
B660 N
C66 N
D6.6 N
Answer & Solution
Correct answer: B. 660 N
1. Change in momentum is 3.3 kg m/s.
2. Divide by 0.0050 s.
3. That is about 660 N.
_Source: OpenStax College Physics for AP Courses 2e, Chapter 8, Linear Momentum and Collisions (CC BY 4.0)._
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