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)  2m

C)  3m             

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 \[\frac{1}{2}{{m}_{1}}u_{1}^{2}-\frac{1}{2}{{m}_{1}}v_{1}^{2}=\frac{75}{100}\times \frac{1}{2}{{m}_{1}}u_{1}^{2}\] \[\Rightarrow \] \[u_{1}^{2}-v_{1}^{2}=\frac{3}{4}u_{1}^{2}\] \[\Rightarrow \] \[{{v}_{1}}=\frac{1}{2}{{u}_{1}}\] Also from conservation of momentum, we have \[{{v}_{1}}=\frac{({{m}_{2}}-{{m}_{1}}){{u}_{1}}}{({{m}_{1}}+{{m}_{2}})}\] Thus, \[\frac{1}{2}{{u}_{1}}=\frac{({{m}_{2}}-{{m}_{1}}){{u}_{1}}}{{{m}_{1}}+{{m}_{2}}}\] \[\Rightarrow \] \[{{m}_{2}}=3{{m}_{1}}=3m\]