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

Answer 1

Answer: (B)

$m$

Answer 2

In figure (i) before collision, $m '$ is mass of unknown particle; $m$ is mass of proton; $v_{i}$ is initial velocity.

Now, in figure (ii), $v_{1}$ is final velocity of unknown particle and $v_{2}$ is final velocity of proton.

By conservation of momentum, we have
Momentum before collision = Momentum after collision
Consider $x$ -component, we have
$m v_{1}+m' .0 =m' v_{1} \cos 45^{\circ}+m v_{2} \cos 45^{\circ}$
$m v_{i} =\frac{1}{\sqrt{2}}\left(m' v_{1}+m v_{2}\right)$
Consider $y$ -component, we have
$0=m' v_{1} \sin 45^{\circ}-m v_{2} \sin 45^{\circ}$
$\frac{1}{\sqrt{2}}\left(m' v_{i}-m v_{2}\right)=0$
$\Rightarrow m' v_{1}=m v_{2}$
Substitute E (2) in E (1), we get
$m v_{i}=\frac{1}{\sqrt{2}}\left(m v_{2}+m v_{2}\right)=\sqrt{2} m v_{2}$
$\Rightarrow v_{i}=\sqrt{2} v_{2}$
Using Eqs. (2) and (3) in (1), we get
$m v_{i} =\frac{1}{\sqrt{2}}\left(m v_{2}+m v_{2}\right)=\sqrt{2} m v_{2}$
$\Rightarrow v_{i} =\sqrt{2} v_{2}$