Q 11-05-167JEE MainJEE Main 2020 (2 Sep, Shift 1)Medium
A particle of mass $m$ with an initial velocity $u\hat{i}$ collides perfectly elastically with a mass $3m$ at rest. It moves with a velocity $v\hat{j}$ after collision, then, $v$ is given by
Answer: (C) $v = \dfrac{u}{\sqrt{2}}$
Momentum conservation gives the velocity of the $3m$ mass: $3m\vec{V} = mu\hat{i} - mv\hat{j}$, so $V^{2} = \dfrac{u^{2}+v^{2}}{9}$.
Kinetic energy is conserved (elastic collision):
$$\tfrac12 mu^{2} = \tfrac12 mv^{2} + \tfrac12 (3m)\frac{u^{2}+v^{2}}{9}$$
$$3u^{2} = 3v^{2} + u^{2} + v^{2} \;\Rightarrow\; 2u^{2} = 4v^{2} \;\Rightarrow\; v = \frac{u}{\sqrt{2}}$$
Solution by Sreeraj P, M.Sc Physics