Question

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 $

Answer 1

The Correct Option is C

The correct answer is option (C) : 3m
In elastic collision kinetic energy and momentum are conserved. 
Let \({{u}_{1}}\) be the initial velocity of a particle before the 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 the conservation of momentum, we have 
\({{v}_{1}}=\frac{({{m}_{2}}-{{m}_{1}}){{u}_{1}}}{({{m}_{1}}+{{m}_{2}})}\)
\(Thus,\)\(\frac{1}{2}{{u}_{1}}=\frac{({{m}_{2}}-{{m}_{1}}){{u}_{1}}}{{{m}_{1}}+{{m}_{2}}}\) 
\(\Rightarrow\) \({{m}_{2}}=3{{m}_{1}}=3m\)