A body of mass $M$ moving at speed $V_0$ collides elastically with a mass $m$ at rest. After the collision, the two masses move at angles $\theta_1$ and $\theta_2$ with respect to the initial direction of motion of the body of mass $M$. The largest possible value of the ratio $\dfrac Mm$, for which the angles $\theta_1$ and $\theta_2$ will be equal, is:
Answer: (A) $3$
Let $\theta_1 = \theta_2 = \theta$ and final speeds $v_1$ ($M$), $v_2$ ($m$).
Perpendicular momentum: $Mv_1\sin\theta = mv_2\sin\theta \Rightarrow mv_2 = Mv_1$.
Along $V_0$: $MV_0 = (Mv_1 + mv_2)\cos\theta = 2Mv_1\cos\theta \Rightarrow v_1 = \dfrac{V_0}{2\cos\theta}$.
Energy: $MV_0^2 = Mv_1^2 + mv_2^2 = Mv_1^2\left(1 + \dfrac Mm\right)$
$$4\cos^2\theta = 1 + \frac Mm \Rightarrow \frac Mm = 4\cos^2\theta - 1 \le 3$$
Solution by Sreeraj P, M.Sc Physics