Question

An $\alpha -$ particle of mass $m$ suffers one-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. $1.5m$
• B. $4m$
• C. $3.5m$
• D. $2m$

Answer 1

The correct option is B $4m$

We have following collision, where mass of $\alpha$ particle $=m$ and mass of nucleus $=M$

Let $\alpha$ particle rebounds with velocity $v_1$, then Given;
final energy of $\alpha =36\%$ of initial energy
$\Rightarrow \frac{1}{2}m{v}_1^2=0.36\times \frac{1}{2}m{v}^2$
$\Rightarrow v_1=0.6v$ ... (i)
As unknown nucleus gained $64\%$ of energy of $\alpha$,
we have
$\frac{1}{2}M{v}_2^2=0.64\times \frac{1}{2}m{v}^2$
$\Rightarrow v_2=\sqrt{\frac{m}{M}}\times 0.8v$ ... (ii)
From momentum conservation, we have
$mv=M{v}_2-m{v}_1$
Substituting values of $v_1$ and $v_2$ from Eqs. (i) and (ii), we have
$mv=M\sqrt{\frac{m}{M}}\times 0.8v-m\times 0.6v$
$\Rightarrow 1.6mv=\sqrt{mM}\times 0.8v$
$\Rightarrow 2m=\sqrt{mM}$
$\Rightarrow 4m^2=mM$
$\Rightarrow M=4m$