# An alpha-particle of mass m suffers 1-dimensional elastic collision with a nucleus at rest of unknown mass.

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 :

1. (1) $2 \mathrm{~m}$

2. (2) $3.5 \mathrm{~m}$

3. (3) $1.5 \mathrm{~m}$

4. (4) $4 \mathrm{~m}$

Correct Option: , 4

Solution:

(4) Using conservation of momentum,

$\mathrm{mv}_{0}=\mathrm{mv}_{2}-\mathrm{mv}_{1}$

$\frac{1}{2} \mathrm{mv}_{1}^{2}=0.36 \times \frac{1}{2} \mathrm{mv}_{0}^{2}$

$\Rightarrow \mathrm{v}_{1}=0.6 \mathrm{v}_{0}$

The he collision is elastic. So,

$\frac{1}{2} \mathrm{MV}_{2}^{2}=0.64 \times \frac{1}{2} \mathrm{mv}_{0}^{2}[\therefore \mathrm{M}=$ mass of nucleus $]$

$\Rightarrow \mathrm{V}_{2}=\sqrt{\frac{\mathrm{m}}{\mathrm{M}}} \times 0.8 \mathrm{~V}_{0}$

$\mathrm{mV}_{0}=\sqrt{\mathrm{mM}} \times 0.8 \mathrm{~V}_{0}-\mathrm{m} \times 0.6 \mathrm{~V}_{0}$

$\Rightarrow 1.6 \mathrm{~m}=0.8 \sqrt{\mathrm{mM}}$

$\Rightarrow 4 \mathrm{~m}^{2}=\mathrm{mM}$

$\therefore M=4 \mathrm{~m}$