An object of mass $m_{1}$ collides with another object of mass $m_{2}$ , which is at rest. After the collision, the objects move at equal speeds in opposite directions. The ratio of the masses $m_{2} : m_{1}$ is:

$2 : 1$

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$1 : 1$

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$1 : 2$

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$3 : 1$

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Solution

The correct option is D
$3 : 1$

Step 1. Given data
$m a s s e s = m_{1} a n d m_{2}$
Step 2: Find the ratio between $m_{1} a n d m_{2}$ :
From the conservation of linear momentum
$P_{i} = P_{f}$
$\Rightarrow m_{1} u_{1} + m_{2} (0) = m_{1} (- v) + m_{2} v$
$\Rightarrow m_{1} u_{1} = v (m_{2} - m_{1}) \dots (i)$
Since, $e = 1 = (V s e p / V a p p) = v - (- v) / u_{1} = 2 v / u_{1}$
$\Rightarrow u_{1} = 2 v \dots (i i)$
from $(i)$ & $(i i)$
$m_{1} (2 v) = v (m_{2} - m_{1})$
$\Rightarrow 2 m_{1} = m_{2} - m_{1}$
$\Rightarrow 3 m_{1} = m_{2} \Rightarrow m_{2} / m_{1} = 3 / 1$
Hence ,(D) is the correct option.

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