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(C) Let the velocity of mass $m$ after collision be $v_1$ and the velocity of mass $10m$ after collision be $v_2$.$1$. Conservation of momentum along

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$10$

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(C) Let the velocity of mass $m$ after collision be $v_1$ and the velocity of mass $10m$ after collision be $v_2$.$1$. Conservation of momentum along the $y$-axis:Since the initial momentum along the $y$-axis is zero,the final momentum components along the $y$-axis must cancel out:$m v_1 \sin 	heta_1 = 10m v_2 \sin 	heta_2$$v_1 \sin 	heta_1 = 10 v_2 \sin 	heta_2$  --- $(i)$$2$. Kinetic energy condition:The final kinetic energy of the particle of mass $m$ is half its initial kinetic energy:$\frac{1}{2} m v_1^2 = \frac{1}{2} (\frac{1}{2} m u^2)$$v_1^2 = \frac{u^2}{2} \Rightarrow v_1 = \frac{u}{\sqrt{2}}$ --- (ii)$3$. Conservation of energy for elastic collision:Initial kinetic energy = Final kinetic energy$\frac{1}{2} m u^2 = \frac{1}{2} m v_1^2 + \frac{1}{2} (10m) v_2^2$$u^2 = v_1^2 + 10 v_2^2$Substitute $v_1^2 = \frac{u^2}{2}$:$u^2 = \frac{u^2}{2} + 10 v_2^2$$\frac{u^2}{2} = 10 v_2^2 \Rightarrow v_2^2 = \frac{u^2}{20} \Rightarrow v_2 = \frac{u}{\sqrt{20}}$ --- (iii)$4$. Substituting (ii) and (iii) into $(i)$:$(\frac{u}{\sqrt{2}}) \sin 	heta_1 = 10 (\frac{u}{\sqrt{20}}) \sin 	heta_2$$\sin 	heta_1 = 10 \cdot \frac{\sqrt{2}}{\sqrt{20}} \sin 	heta_2$$\sin 	heta_1 = 10 \cdot \frac{1}{\sqrt{10}} \sin 	heta_2$$\sin 	heta_1 = \sqrt{10} \sin 	heta_2$Comparing with $\sin 	heta_1 = \sqrt{n} \sin 	heta_2$,we get $n = 10$.

$A$ 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 the value of $n$ is:

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