Two particles of masses $m_{1}$ and $m_{2}$ and velocities $u_{1}$ and $u_{2} = {αu}_{1} (α > 0)$ , make an elastic head on collision. If the initial kinetic energies of the two particles are equal and $m_{1}$ comes to rest after collision, then

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$\frac{u_{1}}{u_{2}} = \sqrt{2} - 1$

$\frac{m_{2}}{m_{1}} = 3 - 2 \sqrt{2}$

$\frac{u_{1}}{u_{2}} = \sqrt{2} + 1$

$\frac{m_{2}}{m_{1}} = 3 + 2 \sqrt{2}$

answer is A, C.

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Detailed Solution

Initial K.E’s are equal $\frac{1}{2} m_{1} u_{1}^{2} = \frac{1}{2} m_{2} {(α u_{1})}^{2}$

$m_{1} = α^{2} m_{2} \dots \dots (1)$

After collision first particle velocity is zero

$v_{1} = (\frac{m_{1} - m_{2}}{m_{1} + m_{2}}) u_{1} + \frac{2 m_{2} u_{2}}{m_{1} + m_{2}}$

$∴ m_{1} - m_{2} + 2 m_{2} α = 0$

$α^{2} m_{2} - m_{2} + 2 m_{2} α = 0$

$α^{2} + 2 α - 1 = 0 \Rightarrow α = \sqrt{2} - 1$

$∴ \frac{u_{1}}{u_{2}} = \sqrt{2} + 1, \frac{m_{2}}{m_{1}} = 3 + 2 \sqrt{2}$

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