Q 11-05-192JEE MainJEE Main 2019 (12 Jan, Shift 2)Easy
An alpha-particle of mass $m$ suffers 1-dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing $64\%$ of its initial kinetic energy. The mass of the nucleus is
Answer: (B) $4m$
It keeps $36\%$ of its kinetic energy, so its speed becomes $0.6u$, directed backwards:
$$\frac{m - M}{m + M} = -0.6 \Rightarrow m - M = -0.6m - 0.6M \Rightarrow 1.6m = 0.4M \Rightarrow M = 4m$$
Solution by Sreeraj P, M.Sc Physics