Question

A particle of mass $m$ is moving along the x-axis with initial velocity $u\hat{i}$. It collides elastically with a particle of mass $10m$ at rest and then moves with half its initial kinetic energy (see figure). If $\sin {\theta }_0=\sqrt{n}\sin {\theta }_2$, then value of $n$ is

Answer 1

From momentum conservation in perpendicular direction of initial motion.
$m{u}_1\sin {\theta }_1=10m{v}_1\sin {\theta }_2$
It is given that energy of $m$ reduced by half. If $u_1$ be velocity of $m$ after collision, then
$\left(\frac{1}{2}m{u}^2\right)\frac{1}{2}=\frac{1}{2}m{u}_1^2$
$\Rightarrow u_1=\frac{u}{\sqrt{2}}$
If $v_1$ be the velocity of mass 10 m after collision, then
$\frac{1}{2}\times 10m\times v_1^2=\frac{1}{2}m\frac{u^2}{2}\Rightarrow v_1=\frac{u}{\sqrt{20}}$
From equation (i), we have
$\sin {\theta }_1=\sqrt{10}\sin {\theta }_2$
$\Rightarrow n=10$