A ball of mass 0.15 kg strikes a wall at an initial speed of 12 m/s and bounces back at the same speed. If the average contact force exerted by the wall on the ball is 100 N, find the duration of contact.

0.018 s
0.036 s
0.009 s
0.072 s

Correct answer: 0.036 s

Solution

Taking the initial direction as positive, v_i = +12 m/s and v_f = -12 m/s (bounces back). Change in momentum = m*(v_f - v_i) = 0.15*(-12-12) = 0.15*(-24) = -3.6 N s. The magnitude of impulse = 3.6 N s. F * deltaₜ = 3.6 => deltaₜ = 3.6/100 = 0.036 s.

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