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Com in Perfectly Inelastic Collision

Object A of m...

Question

Object $A$ of mass $m$ , is moving at a velocity $v_{1}$ to the right. It collides and sticks to object $B$ of mass $m_{2}$ moving in the same direction as object $A$ with velocity $v_{2}$ . After collision, both the objects have the same velocity $\frac{1}{2} (v_{1} + v_{2})$ . What is the relationship between $m_{1}$ and $m_{2}$ ? [ $v_{1}$ and $v_{2}$ are not equal]

A

m_{1} = m_{2}

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B

m_{1} = 2 m_{2}

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C

m_{1} = m_{2} / 2

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D

m_{1} = 4 m_{2}

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Solution

The correct option is A $m_{1} = m_{2}$

Here, $v = (\frac{v_{1} + v_{2}}{2})$
Momentum before collision $= m_{1} v_{1} + m_{2} v_{2}$
Momentum after collision $= (m_{1} + m_{2}) \times (\frac{v_{1} + v_{2}}{2})$
There is no external force present. Hence, momentum will be conserved.

From momentum conservation,
$m_{1} v_{1} + m_{2} v_{2} = (m_{1} + m_{2}) \times (\frac{v_{1} + v_{2}}{2})$
$\Rightarrow m_{1} v_{1} + m_{2} v_{2} = \frac{m_{1} v_{1} + m_{1} v_{2} + m_{2} v_{1} + m_{2} v_{2}}{2}$
$\Rightarrow \frac{m_{1} v_{1} + m_{2} v_{2}}{2} = \frac{m_{1} v_{2} + m_{2} v_{1}}{2}$
$m_{1} (v_{1} - v_{2}) = m_{2} (v_{1} - v_{2})$
$(m_{1} - m_{2}) (v_{1} - v_{2}) = 0$
Hence, $m_{1} = m_{2}$ (since $v_{1} \neq v_{2}$ )

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