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A ball of mass $0.5\text{ kg}$ is dropped from a height of $40\text{ m}$. The ball hits the ground and rises to a height of $10\text{ m}$. The impulse imparted to the ball during its collision with the ground is (Take $g = 9.8\text{ m/s}^2$):
A
21 Ns
B
7 Ns
C
0
D
84 Ns
Explanation
Impulse is $J = m(v_f + v_i) = 0.5(14 + 28) = 21\text{ N s}$.
Detailed Solution
Velocity just before hitting the ground: $v_i = \sqrt{2gh_1} = \sqrt{2 \times 9.8 \times 40} = \sqrt{784} = 28\text{ m/s}$ (downward). Rebound velocity to reach $10\text{ m}$: $v_f = \sqrt{2gh_2} = \sqrt{2 \times 9.8 \times 10} = \sqrt{196} = 14\text{ m/s}$ (upward). Taking upward as positive: Impulse $J = \Delta p = m(v_f - (-v_i)) = m(v_f + v_i) = 0.5 \times (14 + 28) = 0.5 \times 42 = 21\text{ N s}$.