A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground, and rebounds to the same height. The

Question

A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground, and rebounds to the same height. The magnitude of impulse imparted to the ball is ($g=10 	ext{ m/s}^2$) nearly:

Accepted Answer

1. **Calculate Impact Velocity:** The velocity of the ball just before hitting the ground is given by kinematic equation $v = \sqrt{2gh}$.
   $$ v = \sqrt{2 	imes 10 	imes 10} = \sqrt{200} = 10\sqrt{2} 	ext{ m/s} $$
   Taking downward direction as negative, $v_i = -10\sqrt{2} 	ext{ m/s}$.
2. **Calculate Rebound Velocity:** Since the ball rebounds to the same height, the magnitude of velocity remains the same, but the direction is reversed (upward).
   $$ v_f = +10\sqrt{2} 	ext{ m/s} $$
3. **Calculate Impulse:** Impulse is defined as the change in momentum ($I = \Delta p = m(v_f - v_i)$).
   $$ I = 0.15 	imes [10\sqrt{2} - (-10\sqrt{2})] $$
   $$ I = 0.15 	imes 20\sqrt{2} $$
   $$ I = 3\sqrt{2} = 3 	imes 1.414 \approx 4.24 	ext{ kg-m/s} $$
   [Source 60, 61]

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