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

Answer 1

The correct option is A $4m$

$m{v}_0=m{v}_2-m{v}_1$
$\frac{1}{2}m{V}_1^2=0.36\times \frac{1}{2}m{V}_0^2$
$v_1=0.6v_0$
$\frac{1}{2}M{V}_2^2=0.64\times \frac{1}{2}m{V}_0^2$
$V_2=\sqrt{\frac{m}{M}}\times 0.8V_0$
$m{V}_0=\sqrt{mM}\times 0.8V_0-m\times 0.6V_0$
$⟹1.6m=0.8\sqrt{mM}$
$4m^2=mM$