A 0.2 kg ball hits a wall at 10 m/s and rebounds along the same line at 8 m/s. The contact lasts 0.04 s. What is the average force on the ball?
Correct answer B. 90 N
Explanation
Δp = m(v − u) = 0.2 × (8 − (−10)) = 3.6 kg m/s. F = Δp/Δt = 3.6/0.04 = 90 N. Ignoring the reversal of direction gives only 10 N.
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