# A particle of mass $m$ with an initial velocity

Question:

A particle of mass $m$ with an initial velocity $u \hat{i}$ collides perfectly elastically with a mass $3 \mathrm{~m}$ at rest. It moves with a velocity $v \hat{j}$ after collision, then, $v$ is given by :

1. $\mathrm{v}=\sqrt{\frac{2}{3}} \mathrm{u}$

2. $\mathrm{v}=\frac{1}{\sqrt{6}} \mathrm{u}$

3. $v=\frac{u}{\sqrt{3}}$

4. $\mathrm{v}=\frac{\mathrm{u}}{\sqrt{2}}$

Correct Option: , 4

Solution:

From momentum conservation

$\vec{P}_{i}=\vec{P}_{f}$

$\mathrm{m}(\mathrm{ui})+3 \mathrm{~m}(0)=\mathrm{mvj}+3 \mathrm{~m} \overline{\mathrm{v}}_{1}$

mui $-m v j=3 m \bar{v}_{1}$

$\bar{v}_{1}=\frac{u i-v j}{3}$

or $\left|v_{1}\right|=\frac{\sqrt{u^{2}+v^{2}}}{3}$

or $\mathrm{v}_{1}^{2}=\frac{\mathrm{u}^{2}+\mathrm{v}^{2}}{9} \ldots .(1)$

As collision is perfectely elastic hence

$\mathrm{k}_{\mathrm{i}}=\mathrm{k}_{\mathrm{j}}$

$\frac{1}{2} m u^{2}+\frac{1}{2} 3 m 0^{2}=\frac{1}{2} m v^{2}+\frac{1}{2} 3 m v_{1}^{2}$

$\Rightarrow \mathrm{u}^{2}=\mathrm{v}^{2}+3 \mathrm{v}_{1}^{2}$

$u^{2}=v^{2}+3 \frac{\left(u^{2}+v^{2}\right)}{9}$

$\Rightarrow 3 \mathrm{u}^{2}=3 \mathrm{v}^{2}+\mathrm{u}^{2}+\mathrm{v}^{2}$

$\Rightarrow 2 \mathrm{u}^{2}=4 \mathrm{v}^{2}$

$\mathrm{v}=\frac{\mathrm{u}}{\sqrt{2}}$