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Final Velocities

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Question

Two particles of masses $m_{1}$ & $m_{2}$ and velocities $u_{1}$ and $(α u_{1}) (α \neq 0)$ 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

A

\frac{m_{2}}{m_{1}} = 9 + 2 \sqrt{2}

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B

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

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C

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

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D

\frac{m_{1}}{m_{2}} = 9 + \sqrt{2}

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Solution

The correct option is C $\frac{m_{1}}{m_{2}} = 3 - 2 \sqrt{2}$
Given,
Two particles of mass $m_{1}$ & $m_{2}$ with velocities $u_{1}$ & $α u_{1}$ respectively.
If the initial kinetic energies of the two particles are equal, and after the collision, $m_{1}$ comes to rest.
$\frac{1}{2} m_{1} u_{1}^{2} = \frac{1}{2} m_{2} (α u_{1})^{2}$
$\Rightarrow \frac{m_{1}}{m_{2}} = α^{2}$ ... (1)

By the conservation of linear momentum,
$m_{1} u_{1} + m_{2} (α u_{1}) = 0 + m_{2} v_{2}$
Here, $v_{2} =$ velocity of $m_{2}$ particle after collision
Using eq. (1)
$m_{2} α^{2} u_{1} + m_{2} α u_{1} = m_{2} v_{2}$
$\Rightarrow v_{2} = u_{1} α (α + 1)$ ... (2)

Collision is elastic, hence the kinetic energy of the particles before and after the collision will be equal.
$\frac{1}{2} m_{1} u_{1}^{2} + \frac{1}{2} m_{2} (α u_{1})^{2} = \frac{1}{2} m_{2} v_{2}^{2}$
On putting $v_{2}$ from equation $(2)$
$m_{1} u_{1}^{2} + m_{2} α^{2} u_{1}^{2} = m_{2} u_{1}^{2} α^{2} (α + 1)^{2}$
$\Rightarrow \frac{m_{1}}{m_{2}} = α^{4} + 2 α^{3}$
From equation $(1)$ ,
$α^{4} + 2 α^{3} = α^{2}$
$α^{2} + 2 α - 1 = 0$
$\Rightarrow α = - 1 + \sqrt{2}$ (taking +ve root)
$\Rightarrow \frac{m_{1}}{m_{2}} = α^{2} = (- 1 + \sqrt{2})^{2}$
Hence, $\frac{m_{1}}{m_{2}} = 3 - 2 \sqrt{2}$

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