Question: 10. A ball of mass 0.5 kg is dropped from a height of 40 m. The ball hits the ground and rises to a height of 10 m. The impulse imparted to the ball during its collision with the ground is (Take \mathrm{g}=9.8\,\mathrm{m/s^{2}})
(1) 21 Ns
(2) 7 Ns
(3) 0
(4) 84 Ns
Answer: Option (1)
Explanation:
Before the collision, the ball falls freely from a height of 40 m. Its speed just before impact is
v_1=\sqrt{2gh}=\sqrt{2\times 9.8\times 40}=\sqrt{784}=28\ \text{m/s}.This velocity is downward, so take it as negative:
v_1=-28\ \text{m/s}.After the collision, the ball rises to a height of 10 m. Its speed just after impact is
v_2=\sqrt{2gh}=\sqrt{2\times 9.8\times 10}=\sqrt{196}=14\ \text{m/s}.This velocity is upward, so it is positive:
v_2=14\ \text{m/s}.Impulse is the change in momentum:
J = m(v_2 - v_1).Substitute values:
J = 0.5\,(14 - (-28)) = 0.5\,(42) = 21\ \text{Ns}.Thus, the impulse imparted to the ball is 21\ \text{Ns}.