In a two-particle system with particle masses $m_{1}$ and $m_{2}$ , the first particle is pushed towards the centre of mass through a distance d, the distance through which second particle must be moved to keep the centre of mass at the same position is

\frac{m_{2} d}{m_{1}}

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d

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\frac{m_{1} d}{(m_{1} + m_{2})}

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\frac{(m_{1} + m_{2}) d}{m_{1}}

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\frac{m_{1} d}{m_{2}}

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Solution

The correct option is E $\frac{m_{1} d}{m_{2}}$
given masses are $m_{1}$ and $m_{2}$
let x and y be the distance of $m_{1}$ and $m_{2}$ from the centre of mass respectively, then
$m_{1} x = m_{2} y$
If the mass $m_{1}$ is moved by a distance $d_{1}$ , and mass $m_{2}$ be moved by a distance $d_{2}$ , then

$m_{1} (x - d_{1}) = m_{2} (y - d_{2})$
$m_{1} x - m_{1} d_{1} = m_{2} y - m_{2} d_{2}$
$m_{1} d_{1} = m_{2} d_{2}$
$d_{2} = \frac{m_{1}}{m_{2}} d_{1}$ $(∵ c o n s t a n t s)$
$d_{2} = \frac{m_{1}}{m_{2}} d$ $(∵ d_{1} = d)$

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