A tennis ball of mass 0.050 kg traveling horizontally at 20 m/s strikes a racket and rebounds horizontally in the opposite direction at 20 m/s. If the ball is in contact with the racket for 0.010 s, what is the magnitude of the average force exerted on the ball?
Answer and explanation
Correct answer: A. 200 N
Taking the initial direction as positive, initial momentum is p_i = m v_i = 0.050 kg (+20 m/s) = +1.0 kg·m/s. The rebound momentum is p_f = 0.050 kg * (-20 m/s) = -1.0 kg·m/s. The change in momentum (impulse) is Δp = p_f - p_i = -1.0 - (+1.0) = -2.0 kg·m/s. The average force magnitude is F_avg = |Δp / Δt| = 2.0 kg·m/s / 0.010 s = 200 N.
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