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Physics · LAWS OF MOTION
हिन्दीA rubber ball of mass \( 0.15\text{ kg} \) is dropped from a height of \( 5\text{ m} \) onto a hard floor. It rebounds upwards to a maximum height of \( 3.2\text{ m} \). If the time of collision with the floor is \( 0.03\text{ s} \) and \( g = 10\text{ m/s}^2 \), the average normal force exerted by the floor on the ball during the impact is:
View Answer & Solution
Correct answer · Option D
91.5 N
Explanation
Incident velocity \( v_1 = \sqrt{2(10)(5)} = 10\text{ m/s} \) (downward), rebound velocity \( v_2 = \sqrt{2(10)(3.2)} = 8\text{ m/s} \) (upward). Impulse \( \Delta p = m(v_2 - (-v_1)) = 0.15(8+10) = 2.7\text{ N s} \). Net force \( F_{\text{net}} = \frac{2.7}{0.03} = 90\text{ N} \). Normal force \( N = F_{\text{net}} + mg = 90 + 1.5 = 91.5\text{ N} \).
