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 :

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 \[M=m\]