question_answer

An \[\alpha -\]particle of mass m suffers one dimensional elastic collision with a nucleus of unknown mass. After the collision the \[\alpha -\]particle is scattered directly backwards losing \[75%\] of its kinetic energy. The mass of the unknown nucleus is:

A) \[m\]                                    

B) \[2\,\,m\]

C) \[3\,\,m\]           

D) \[\frac{3}{2}m\]

Correct Answer: C

Solution :

Key Idea: In elastic collision kinetic energy and momentum are conserved. Let \[{{u}_{1}}\] be the initial velocity of a particle before collision and \[{{v}_{1}}\] the final velocity after collision, then change in kinetic energy is given by                 \[\frac{1}{2}{{m}_{1}}u_{1}^{2}-\frac{1}{2}{{m}_{1}}v_{1}^{2}=\frac{75}{100}\times \frac{1}{2}{{m}_{1}}v_{1}^{2}\] \[\Rightarrow \]               \[u_{1}^{2}-v_{1}^{2}=\frac{3}{4}u_{1}^{2}\] \[\Rightarrow \]               \[{{v}_{1}}=\frac{1}{1}{{u}_{1}}\] Also from conservation of momentum, we have                 \[{{v}_{1}}=\frac{({{m}_{2}}-{{m}_{1}}){{u}_{1}}}{({{m}_{1}}+{{m}_{2}})}\] Thus,     \[\frac{1}{2}{{u}_{1}}=\frac{({{m}_{2}}-{{m}_{1}}){{u}_{1}}}{{{m}_{1}}+{{m}_{2}}}\] \[\Rightarrow \]               \[{{m}_{2}}=3{{m}_{1}}=3m\]