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 $x$ m. Find the value of $x$.

Answer 1

Answer: 4

Using conservation of momentum,
$m{v}_0=m{v}_2-m{v}_1$

$\frac{1}{2}m{v}_1^2=0.36\times 12m{v}_0^2$
$\Rightarrow v_1=0.6v_0$
The collision is elastic. So,
$\frac{1}{2}M{v}_2^2=0.64\times 12m{v}_0^2$ $[\because$ M = mass of nucleus]
$\Rightarrow V_2=\sqrt{\frac{m}{M}}\times 0.8V_0$
$m{V}_0=\sqrt{mM}\times 0.8V_0-m\times 0.6V_0$
$\Rightarrow 1.6m=0.8\sqrt{mM}$
$\Rightarrow 4m^2=mM$
$\therefore M=4m$