Question

A particle of mass m, moving with velocity v collides a stationary particles of mass 2m. As a result of collision, the particle of mass m deviates by ${45}^o$ and has final speed of $\frac{v}{2}$. For this situation mark out the correct statement (s).

• A. The angle of divergence between particles after collision is $\frac{\pi }{2}$
• B. The angle of divergence between particles after collision is less than $\frac{\pi }{2}$
• C. Collision is elastic
• D. Collision is inelastic

Answer 1

Answer: (B), (D)

The correct options are
B The angle of divergence between particles after collision is less than $\frac{\pi }{2}$
D Collision is inelastic

Let velocity of mass 2m is $\overrightarrow{v^{\prime }}$ after collision.;
From conservation of momentum in the xy plane,
$\overrightarrow{p_i}=\overrightarrow{p_f}$;

$mv\hat{i}=\frac{mv}{2}(\cos {45}^0\hat{i}+\sin {45}^0\hat{j})+2m\overrightarrow{v^{\prime }}=\frac{mv}{2\sqrt{2}}(\hat{i}+\hat{j})+2m\overrightarrow{v^{\prime }}$;

$\therefore 2\overrightarrow{v^{\prime }}=\left(\frac{2\sqrt{2}-1}{2\sqrt{2}}\right)v\hat{i}-\frac{v}{2\sqrt{2}}\hat{j}$

$\Rightarrow \overrightarrow{v^{\prime }}=0.32v\hat{i}-0.18v\hat{j}$;

$\therefore \tan \theta =\frac{0.18}{0.32}=0.56$;

$\therefore \theta ={29.35}^0$;

$\left|{\overrightarrow{v}}^{\prime }\right|=\sqrt{(.32{)}^2+(.18{)}^2}v=0.37v$;

Since $(\theta +{45}^0)<{90}^0$, the angle of divergence between particles after collision is less than ${90}^0$;
${KE}_i=\frac{1}{2}m{v}^2$;

${KE}_f=\frac{1}{2}m(\frac{v}{2}{)}^2+\frac{1}{2}2m(0.37v{)}^2$;

$\therefore {KE}_f<{KE}_i$;
$\therefore$ collision is inelastic.