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 ____

From momentum conservation in perpendicular direction of initial motion 

$m{u}_1\sin {\theta }_1=10m{v}_1\sin {\theta }_2\mathrm{......}\left(i\right)$

It is given that energy of m reduced half . If u1 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 v1 be the velocity of mass 10m after collision, then $\frac{1}{2}\times 10m\times v_1^2=\frac{1}{2^m}\frac{u^2}{2}\times v_1=\frac{u}{\sqrt{20}}$

For equation (i), we have $\sin {\theta }_1 =\sqrt{10} \sin {\theta }_2$