question_answer

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: [JEE Online 15-04-2018 (II)]

A) \[\frac{m}{\sqrt{3}}\]    

B) \[\frac{m}{2}\]

C) \[2m\]

D) \[m\]

Correct Answer: D

Solution :

[d] Apply principle of conservation of momentum along x-direction,

\[mu=m{{v}_{1}}\cos 45+M{{v}_{2}}\cos 45\]

\[mu=\frac{1}{\sqrt{2}}(m{{v}_{1}}+M{{v}_{2}})\]...(1)

Along \[y-direction,\]

\[o=m{{v}_{1}}\sin 45-M{{v}_{2}}\sin 45\]

\[o=(m{{v}_{1}}-M{{v}_{2}})\frac{1}{\sqrt{2}}.....(2)\]

Coefficient of \[e=1=\frac{{{v}_{2}}-{{v}_{1}}\cos 90}{u\cos 45}\]

Restution

\[\Rightarrow \frac{{{v}_{2}}}{\frac{u}{\sqrt{2}}}=1\]

\[\Rightarrow u=\sqrt{2}{{v}_{2}}.....(3)\]

Solving eqn. (1), (2), & (3) we get

\[M=m\]