Q 11-05-112JEE MainJEE Main 2023 (8 Apr, Shift 2)Medium
A bullet of mass $0.1$ kg moving horizontally with speed $400\ \text{m s}^{-1}$ hits a wooden block of mass $3.9$ kg kept on a horizontal rough surface. The bullet gets embedded into the block and moves $20$ m before coming to rest. The coefficient of friction between the block and the surface is ______.
Answer: (D) 0.25
Momentum: $v=\dfrac{0.1\times400}{4}=10\ \text{m s}^{-1}$. Then $v^2=2\mu gs\Rightarrow100=2\mu\times10\times20\Rightarrow\mu=0.25$.
Solution by Sreeraj P, M.Sc Physics