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:

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3.5 m

1.5 m

4 m

answer is D.

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Detailed Solution

Using conservation of momentum.
$\begin{array}{l} {mv}_{0} = M v_{2} - {mv}_{1} \\ \frac{1}{2} {mv}_{1}^{2} = 0 .36 \times \frac{1}{2} {mv}_{0}^{2} \end{array}$
After
collision
$\Rightarrow v_{1} = 0 .6 v_{0}$

The collision is elastic. So,
$\begin{array}{l} \frac{1}{2} {MV}_{2}^{2} = 0 .64 \times \frac{1}{2} {mv}_{0}^{2} \\ [∴ M = mass of nucleus) \\ \Rightarrow V_{2} = \sqrt{\frac{m}{M}} \times 0 .8 V_{0} \\ m V_{o} = \sqrt{mM} \times 0 .8 V_{0} - m \times 0 .6 V_{0} \\ \Rightarrow 1 .6 m = 0 .8 \sqrt{mM} \Rightarrow 4 m^{2} = mM \\ ∴ M = 4 m \end{array}$

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