Question

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___

Answer 1

Answer: 10.00

By momentum conservation along y :
$m_1{u}_1\sin {\theta }_1=m_2{u}_2\sin {\theta }_2$
$\text{ i.e. }m{u}_1\sin {\theta }_1=10m{u}_2\sin {\theta }_2$
$\Rightarrow u_1\sin {\theta }_1=10u_2\sin {\theta }_2.....(i)$
$k{f}_{m_1}=\frac{1}{2}k{i}_{m_1}$ i.e. $\frac{1}{2}m{u}_1^2=\frac{1}{2}\times \frac{1}{2}m{u}^2$
i.e. $u_1=\frac{u}{\sqrt{2}}....(ii)$
Also collision is elastic $:k_i=k_f$
$\frac{1}{2}m{u}^2=\frac{1}{2}m{u}_1^2+\frac{1}{2}.10m.u_2^2$
$\frac{1}{2}m{u}^2=\frac{1}{2}\times \frac{1}{2}m{u}^2+\frac{1}{2}\times 10m.u_2^2$
$\frac{1}{4}m{u}^2=\frac{1}{2}\times 10\times m{u}_2^2$
$u_2=\frac{u}{\sqrt{20}}....(iii)$
Putting $(ii)$ & $(iii)$ in $(i)$
$\frac{u}{\sqrt{2}}\sin {\theta }_1=10.\frac{u}{\sqrt{20}}\sin {\theta }_2$
$\sin {\theta }_1=\sqrt{10}\sin {\theta }_2$