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

Using conservation of momentum,
$mv_{0} =mv_{2} -mv_{1}$

$\frac{1}{2} mv^{2}_{1} = 0.36 \times {1}{2} mv^{2}_{0}$
$\Rightarrow v_{1}=0.6\,v_{0}$
The collision is elastic. So,
$\frac{1}{2} M v^{2}_{2}=0.64\times{1}{2} mv^{2}_{0}$ $[\because$ M = mass of nucleus]
$\Rightarrow V_{2} = \sqrt{\frac{m}{M}}\times0.8 V_{0}$
$mV_{0}=\sqrt{mM}\times0.8V_{0}-m\times0.6V_{0}$
$\Rightarrow 1.6\,m=0.8 \sqrt{mM}$
$\Rightarrow 4m^{2}=mM$
$\therefore M=4m$