Question 119 of 130

A proton of mass $\mathrm{m}$ 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$\frac{\mathrm{m}}{\sqrt{3}}$B$\frac{\mathrm{m}}{2}$C$2 \mathrm{~m}$D$\mathrm{m}$

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Answer: (D) $\mathrm{m}$

Apply principle of conservation of momentum along $x$-direction,
$$\begin{aligned} m u &=m v_1 \cos 45^{\circ}+M v_2 \cos 45^{\circ} \\ m u &=\frac{1}{\sqrt{2}}\left(m v_1+M v_2\right) \end{aligned}$$
Along $y$-direction,
$$\begin{aligned} &o=m v_1 \sin 45^{\circ}-M v_2 \sin 45^{\circ} \\ &o=\left(m v_1-M v_2\right) \frac{1}{\sqrt{2}} \end{aligned}$$


Coefficient of restution $e=1$
$$=\frac{v_2-v_1 \cos 90}{u \cos 45}$$
( $\because$ Collision is elastic)
$$\begin{aligned} &\Rightarrow \frac{v_2}{\frac{u}{\sqrt{2}}}=1 \\ &\Rightarrow u=\sqrt{2} v_2 \end{aligned}$$
Solving eqs (i), (ii), \& (iii), we get mass of unknown particle, $M=m$