1. Given data

• Mass of first particle: $m$
• Initial velocity of first particle: $u\hat i$
• Mass of second particle: $10m$
• Second particle initially at rest
• Collision is elastic
• After collision, the first particle has half its initial kinetic energy.

Let after collision:

• particle $m$ move with speed $v_1$ at angle $\theta_1$
• particle $10m$ move with speed $v_2$ at angle $\theta_2$

We need to find $n$ if $\sin\theta_1=\sqrt n\,\sin\theta_2.$

2. Use the given kinetic-energy condition

Initial kinetic energy of mass $m$: $K_i=\frac12 mu^2$

After collision, first particle has half of this: $K_{1f}=\frac12 K_i=\frac14 mu^2$

But $K_{1f}=\frac12 m v_1^2$

So, $\frac12 m v_1^2=\frac14 mu^2$ $v_1^2=\frac{u^2}{2}$ $v_1=\frac{u}{\sqrt2}$

3. Use conservation of kinetic energy

Since collision is elastic, $\frac12 mu^2=\frac12 m v_1^2+\frac12 (10m)v_2^2$

Substitute $v_1^2=u^2/2$: $\frac12 mu^2=\frac12 m\cdot \frac{u^2}{2}+5m v_2^2$

$\frac12 u^2=\frac14 u^2+5v_2^2$ $\frac14 u^2=5v_2^2$ $v_2^2=\frac{u^2}{20}$ $v_2=\frac{u}{2\sqrt5}$

4. Use conservation of momentum in y-direction

Initially there is no momentum in the $y$-direction. So after collision, $m v_1\sin\theta_1 = 10m v_2\sin\theta_2$

(Careful with sign: the particles go on opposite sides of the x-axis, so magnitudes satisfy this relation.)

Cancel $m$: $v_1\sin\theta_1=10v_2\sin\theta_2$

Hence, $\sin\theta_1=\frac{10v_2}{v_1}\sin\theta_2$

Now substitute $v_1=\frac{u}{\sqrt2}$ and $v_2=\frac{u}{2\sqrt5}$: $\frac{10v_2}{v_1}=10\cdot \frac{u}{2\sqrt5}\cdot \frac{\sqrt2}{u}$ $=5\sqrt{\frac{2}{5}}=\sqrt{10}$

Therefore, $\sin\theta_1=\sqrt{10}\,\sin\theta_2$

Comparing with $\sin\theta_1=\sqrt n\,\sin\theta_2,$ we get $n=10.$

5. Final answer

$\boxed{10}$