A particle of mass m is moving along the x-axis with initial velocity  $u\widehat{i}$. It collides elastically with a particle of mass 10m at rest and then moves with half its initial kinetic energy, (see figure). If $\sin {\theta }_1=\sqrt{n}\sin {\theta }_2$  then value of n is

$$\begin{gathered} Given\to \frac{1}{2}M{V}_1^2=\frac{1}{2}\left(\frac{1}{2}M{u}^2\right) \\ \Rightarrow V_1=\frac{u}{\sqrt{2}} \end{gathered}$$

From conservation of momentum in y – direction
 $V_1\sin {\theta }_1=10V_2\sin {\theta }_2\to \left(1\right)$
  As collision is elastic KE remains conserved
 $\frac{1}{2}M{u}^2=\frac{1}{2}M{V}_1^2+\frac{1}{2}\left(10M\right)V_2^2$
$\Rightarrow \frac{1}{2}M{u}^2=\frac{1}{2}\left(\frac{1}{2}M{u}^2\right)+\frac{1}{2}\left(10M\right)V_2^2$ 
$\Rightarrow \frac{1}{4}M{u}^2=\frac{10}{2}M{V}_2^2$ 
$$\begin{gathered} \Rightarrow \frac{u^2}{20}=V_2^2 \\ \Rightarrow V_2=\frac{u}{\sqrt{20}} \end{gathered}$$ 
Put in equation (1)
$\Rightarrow \frac{u}{\sqrt{2}}\sin {\theta }_1=10\frac{u}{\sqrt{20}}\sin {\theta }_2$ 
 $\Rightarrow \sin {\theta }_1=\sqrt{10}\sin {\theta }_2$