Consider a system having two masses $m_{1}$ and $m_{2}$ in which first mass is pushed towards the center of mass by a distance $a$ . The distance by which the second should be moved to keep the center of mass at same position is

\frac{m_{1}}{m_{2}} a

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

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\frac{m_{2}}{m_{1}} a

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

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Solution

The correct option is A $\frac{m_{1}}{m_{2}} a$
Given,
The $2$ 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.
now, $m_{1} x = m_{2} y$
the mass $m_{1}$ is moved by a distance $d$ .
Let, the mass $m_{2}$ be moved by a distance $D$ .
Therefore,
$m_{1} (x - d) = m_{2} (Y - D)$
$m_{1} x - m_{1} d = m_{2} y - m_{2} D$
by equation $(i)$ $m_{1} x = m_{2} y$
$= - m_{1} d = - m_{2} D$
$= D = \frac{m_{1}}{m_{2}} d$ .

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