Q 11-04-093JEE MainJEE Main 2024 (30 Jan, Shift 1)Medium
A spherical body of mass $100\ \text{g}$ is dropped from a height of $10\ \text{m}$ from the ground. After hitting the ground, the body rebounds to a height of $5\ \text{m}$. The impulse of force imparted by the ground to the body is given by: (given $g = 9.8\ \text{m s}^{-2}$)
Answer: (D) $2.39\ \text{kg m s}^{-1}$
Speed just before impact: $v_1 = \sqrt{2\times9.8\times10} = 14\ \text{m s}^{-1}$ (downward).
Speed just after: $v_2 = \sqrt{2\times9.8\times5} \approx 9.9\ \text{m s}^{-1}$ (upward).
$$J = m(v_1 + v_2) = 0.1\times(14 + 9.9) \approx 2.39\ \text{kg m s}^{-1}$$
Solution by Sreeraj P, M.Sc Physics