Q 11-04-005NEETNEET 2025Top questionEasy
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 $g = 9.8\ \text{m/s}^2$)
Answer: (A) $21$ N s
Speed just before impact: $v_1 = \sqrt{2gh_1} = \sqrt{2 \times 9.8 \times 40} = 28\ \text{m s}^{-1}$ (downward).
Speed just after impact: $v_2 = \sqrt{2gh_2} = \sqrt{2 \times 9.8 \times 10} = 14\ \text{m s}^{-1}$ (upward).
Impulse = change in momentum (the velocity reverses direction):
$$J = m(v_2 + v_1) = 0.5 \times (14 + 28) = 21\ \text{N s}$$
Solution by Sreeraj P, M.Sc Physics