Question

An alpha-particle of mass $m$ suffers $1-$dimensional elastic collision with a nucleus at rest of unknown mass. It is scattered directly backwards losing, $64\%$ of its initial kinetic energy. The mass of the nucleus is:

• A. $3.5m$
• B. $2m$
• C. $1.5m$
• D. $4m$

Answer 1

The correct option is D $4m$


Using conservation of momentum

$m{V}_0=M{V}_2-m{V}_1---\left(1\right)$

Given that, the mass $m$ loses $64\%$ of its kinetic energy,

$\Rightarrow \frac{1}{2}m{V}_1^2=0.36\times \frac{1}{2}m{V}_0^2$

$\Rightarrow V_1=0.6V_0$

As the collision is elastic, the $64\%$ kinetic energy losy by $m$ is transferred to $M$,

$\Rightarrow \frac{1}{2}M{V}_2^2=0.64\times \frac{1}{2}m{V}_0^2$

$\Rightarrow V_2=\sqrt{\frac{m}{M}}\times 0.8V_0$

Substituting the values of $V_1\text{and}{V}_2$ in $\left(1\right)$

$\Rightarrow m{V}_0=M\sqrt{\frac{m}{M}}\times 0.8V_0-m\times 0.6V_0$
$\Rightarrow 1.6m=0.8\sqrt{mM}$
$\Rightarrow 4m^2=mM$
$\therefore M=4m$

Hence, option $\left(D\right)$ is correct.