Question

A proton of mass $m$ collides elastically with a particle of unknown mass at rest. After the collision, the proton and the unknown particle are seen moving at an angle of $90^\circ $ with respect to each other. The mass of unknown particle is:

• A. $\frac{m}{\sqrt{3}}$
• B. $\frac{m}{2}$
• C. $2m$
• D. $m$

Answer 1

Answer: (D)

Apply principle of conservation of momentum,
Along $x-$ direction,
$mu=mv_{1}cos45+Mv_{2}cos45$
$mu=\frac{1}{\sqrt{2}}\left(m v_{1} + M v_{2}\right)...\left(\right.i\left.\right)$
Along $y-$ direction,
$0=mv_{1}sin45-Mv_{2}sin45$
$0=\left(m v_{1} - M v_{2}\right)\frac{1}{\sqrt{2}}...\left(\right.ii\left.\right)$
Coefficient of Resolution, $e=1$
$\Rightarrow \frac{v_{2} - v_{1} cos 90}{u cos 45}=1$
$\Rightarrow \frac{v_{2}}{\frac{u}{\sqrt{2}}}=1$
$\Rightarrow u=\sqrt{2}v_{2}...\left(\right.iii\left.\right)$
Solving equation $\left(\right.i\left.\right),\left(\right.ii\left.\right)\&\left(\right.iii\left.\right)$ , we get
$M=m$