Exam

JEE Main

Year

2018

Date

15 April

Shift

Evening Shift

Mode

Online

Subject

Physics

Chapter

Center of Mass, Momentum and Collision

JEE Main 2018 · 15 April · Evening Shift · Online · Physics · Center of Mass, Momentum and Collision

A proton of mass m collides elastically with a particle of unknown...

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:

Step-by-step solution

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,$M=m$