Question 23 of 130

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$1.5m$B$4m$C$3.5m$D$5m$

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Answer: (B) $4m$

From momentum conservation,
$mv+0=-m{v}_{1}+M{v}_{2}...(1)$

Using kinetic energy conservation,

$\dfrac{1}{2}m{v}^{2}=\dfrac{1}{2}m{v}_{1}^{2}+\dfrac{1}{2}M{v}_{2}^{2} \Rightarrow \dfrac{1}{2}m{v}^{2}=0.36\times \dfrac{1}{2}m{v}^{2}+\dfrac{1}{2}M{v}_{2}^{2} \Rightarrow 0.64m{v}^{2}=M{v}_{2}^{2} \Rightarrow {v}_{2}=0.8v\sqrt{\dfrac{m}{M}}$

As it retain $36$,$\Rightarrow 0.36\times \left(\dfrac{1}{2}m{v}^{2}\right)=\dfrac{1}{2}m{v}_{1}^{2}\Rightarrow 0.36{v}^{2}={v}_{1}^{2}$ Hence, ${v}_{1}=0.6v$
From $(1)$, $mv=-m\left(0.6v\right)+M\left(0.8v\sqrt{\dfrac{m}{M}}\right)$
$\Rightarrow 1=-0.6+0.8\sqrt{\dfrac{M}{m}}$
On solving, we get $M=4m$