Question

A particle of mass $m_0,$ travelling at speed $v_0,$ strikes a stationary particle of mass $2m_0.$ As a result, the particle of mass $m_0$ is deflected through ${45}^o$ and has a final speed of $\frac{v_0}{\sqrt{2}}$. Then the speed of the particle of mass $2m_0$ after this collision is

• A. $\frac{v_0}{2}$
• B. $\frac{v_0}{2\sqrt{2}}$
• C. $\sqrt{2}{v}_0$
• D. $\frac{v_0}{\sqrt{2}}$

Answer 1

The correct option is B $\frac{v_0}{2\sqrt{2}}$

A particle of mass $m_0,$
Before collision

$m{v}_0=2mvcos\theta +\frac{m{v}_0}{2}..........\left(i\right)$

$\theta =\frac{m{v}_0}{2}-2mvsin\theta ............\left(ii\right)$

By (i) & (ii), $\theta ={45}^0$
Now again momentum conservation in y direction

$\frac{m{v}_0}{2}=2mv.\frac{1}{\sqrt{2}}\Rightarrow v=\frac{v_0}{2\sqrt{2}}$