Question

A particle of mass $m$ is moving along the x-axis with an 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 $\sin \left({\theta }_1\right)=\sqrt{n}\sin \left({\theta }_2\right)$ then value of $n$ is

Answer 1

Step 1: Given data and assumptions.

Mass of the particle, $A$ $m_A=m$

Initial velocity of particle $A,u_1$ $=u\hat{i}$

Mass of the other particle, $B$, $m_B=10m$

Step2: Finding the value of final velocity, $v_1$ of the particle, $A$ of mass $m$

For particle of mass $m$, final kinetic energy $K_f$ is equals to half of the initial kinetic energy, $K_i$

$K_f=\frac{1}{2}{K}_i$

$$\begin{gathered} \frac{1}{2}{m}_A{v}_1^2=\frac{1}{2}\times \left(\frac{1}{2}{m}_A{u}_1^2\right) \\ {v}_1^2=\frac{u_1^2}{2} \\ {v}_1=\frac{u_1}{\sqrt{2}} \\ {v}_1=\frac{u}{\sqrt{2}}[u_1=u] \end{gathered}$$

Step 3. Finding the value of $n$

As we know collision is elastic KE remains conserved then,

$\Rightarrow$ $\frac{1}{2}{m}_B{v}_2^2=\frac{1}{2}{m}_A{v}_1^2$ [ $v_2$ is the final velocity of particle $B$]

$\Rightarrow$$\frac{1}{2}\times 10m\times v_2^2=\left(\frac{1}{2}m\frac{u^2}{2}\right)$ $\left[\because u_1=\frac{u}{\sqrt{2}}\right]$

$\Rightarrow$ $v_2=\frac{u}{\sqrt{20}}$

Now, momentum in the y-direction will remain conserved then,

$m_A{v}_1\sin \left({\theta }_1\right)=m_B{v}_2\sin \left({\theta }_2\right)$

$\Rightarrow$$m\times \frac{u}{\sqrt{2}}\sin \left({\theta }_1\right)=10m\times \frac{u}{\sqrt{20}}\sin \left({\theta }_2\right)$

$\Rightarrow$ $\sin \left({\theta }_1\right)=\sqrt{10}\sin \left({\theta }_2\right)$

Hence, the value of $n$ is $10$ as compared given equation with $\sin \left({\theta }_1\right)=\sqrt{10}\sin \left({\theta }_2\right)$.