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

Answer: (C)

3m

Answer 2

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}}}
$\Rightarrow$ ${{m}_{2}}=3{{m}_{1}}=3m$