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Perfectly Elastic Collision

Particle A ma...

Question

Particle $A$ makes a perfectly elastic collision with another particle $B$ at rest. After collision, they fly apart in opposite directions with equal speeds. The ratio of masses $\frac{m_{A}}{m_{B}}$ is

1 : 2

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1 : 3

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1 : 4

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1 : \sqrt{3}

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Solution

The correct option is B $1 : 3$
Assume $u_{1} =$ initial speed of particle $A$
$v_{1} =$ speed of particle $A$ & $B$ after collision.
There is no external force, hence from linear momentum conservation :
$m_{1} u_{1} + 0 = m_{1} (- v_{1}) + m_{2} (v_{1})$
$m_{1} u_{1} = (- m_{1} + m_{2}) v_{1}$ ... $(1)$

As we know, collision is elastic.

$\Rightarrow e = 1 = \frac{v_{1} - (- v_{1})}{u_{1} - 0} = 1$
or $2 v_{1} = u_{1}$ ... (2)
From equation $(1)$ and $(2)$ ,
$m_{1} (2 v_{1}) = (- m_{1} + m_{2}) v_{1}$
$2 m_{1} = - m_{1} + m_{2}$
or $\frac{m_{1}}{m_{2}} = \frac{m_{A}}{m_{B}} = \frac{1}{3}$

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