Question

A particle of mass m moving with velocity 1 m/s collides perfectly elastically with another stationary particle of mass 2m. If the incident particle is deflected by 90${}^o$, the heavy mass will make and angle $\theta$ with the initial direction of m equal to :

• A. ${60}^o$
• B. ${45}^o$
• C. ${15}^o$
• D. ${30}^o$

Answer 1

The correct option is D ${30}^o$

Let initial velocity of mass $m$ be $u$.

Also let the velocities of mass $m$ and $2m$ be $v$ and $V$ respectively after the collision.

Applying conservation of momentum in x direction :

$mu=0+\left(2m\right)Vcos\theta$ (Given $u=1$ $m{s}^{-1}$)

$⟹Vcos\theta =\frac{1}{2}$ ..............(1)

Applying conservation of momentum in y direction :

$0=mv-\left(2m\right)Vsin\theta$ $⟹Vsin\theta =\frac{v}{2}$ ..............(2)

Solving (1) and (2), we get $V^2=\frac{1+v^2}{4}$ and $tan\theta =v$

Loss in kinetic energy in elastic collision is zero i.e $K.E_{initial}=K.E_{final}$

$\frac{1}{2}m{u}^2=\frac{1}{2}m{v}^2=\frac{1}{2}\left(2m\right)V^2$ (Given $u=1$ $m{s}^{-1}$)

$⟹2V^2=1-v^2$

$\therefore 2\times \frac{1+v^2}{4}=1-v^2$ $⟹v=\frac{1}{\sqrt{3}}$

Thus $tan\theta =v=\frac{1}{\sqrt{3}}$ $⟹\theta ={30}^o$