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 $$\sin\theta_1 = \sqrt{n}\sin\theta_2$$, then value of n is ___________.

Initial kinetic energy of mass $$m$$: $$K_i = \frac{1}{2}mu^2$$

Final kinetic energy of mass $$m$$: $$K_f = \frac{1}{2}mv_1^2 = \frac{1}{2}K_i \implies \frac{1}{2}mv_1^2 = \frac{1}{4}mu^2 \implies v_1 = \frac{u}{\sqrt{2}}$$

Conservation of linear momentum along the y-axis:

$$p_{iy} = p_{fy} \implies 0 = m v_1 \sin\theta_1 - 10m v_2 \sin\theta_2$$

$$v_1 \sin\theta_1 = 10 v_2 \sin\theta_2$$

Using the given relation $$\sin\theta_1 = \sqrt{n}\sin\theta_2$$:

$$v_1 \sqrt{n}\sin\theta_2 = 10 v_2 \sin\theta_2 \implies v_1 \sqrt{n} = 10 v_2$$

$$n v_1^2 = 100 v_2^2 \implies v_2^2 = \frac{n}{100}v_1^2$$

Conservation of total kinetic energy:

$$K_i = K_{f1} + K_{f2} \implies \frac{1}{2}mu^2 = \frac{1}{2}mv_1^2 + \frac{1}{2}(10m)v_2^2$$

$$u^2 = v_1^2 + 10v_2^2$$

$$2v_1^2 = v_1^2 + 10\left(\frac{n}{100}v_1^2\right)$$

$$v_1^2 = \frac{n}{10}v_1^2 \implies 1 = \frac{n}{10} \implies n = 10$$