Q 11-05-125JEE MainJEE Main 2022 (25 Jul, Shift 2)Medium
A bag of sand of mass $9.8\ \text{kg}$ is suspended by a rope. A bullet of $200\ \text{g}$ travelling with speed $10\ \text{m s}^{-1}$ gets embedded in it. Then the loss of kinetic energy will be
Answer: (B) $9.8\ \text{J}$
Momentum conservation: $0.2\times10 = 10\,v\Rightarrow v = 0.2\ \text{m s}^{-1}$.
$$\Delta K = \frac12(0.2)(10)^2 - \frac12(10)(0.2)^2 = 10 - 0.2 = 9.8\ \text{J}$$
Solution by Sreeraj P, M.Sc Physics