A ball weighing 10 grams hits a hard surface vertically with a speed of 5 m/s and rebounds with the same speed. The ball remains in contact with the surface for (0.01) sec. the average impulsive force exerted by the surface on the ball is
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The average impulsive force is calculated using the impulse-momentum theorem: . The change in momentum (Δp) is m(v_final - v_initial). The ball rebounds with the same speed, so its velocity changes from -5 m/s (downward) to +5 m/s (upward). Therefore, Δv = 5 - (-5) = 10 m/s. The mass is 10 g = 0.01 kg. Δp = (0.01 kg)(10 m/s) = 0.1 kg·m/s. The time Δt is 0.01 s. Thus, F_avg = (0.1 kg·m/s) / (0.01 s) = 10 N.
Final answer: 10 N