Consider a two-particle system with the particles having masses $m_{1}$ and $m_{2}$ . If the first particle is pushed towards the centre of mass through a distance d, by what distance should the second particle be moved so as to keep the centre of mass at the same position?

()d

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()d

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Solution

The correct option is C
d

Again, as we aleady know ,
$△$ $X_{C M}$ = $\frac{m_{1} △ X_{1} + m_{2} △ X_{2}}{m_{1} + m_{2}}$
$△$ $X_{C M}$ = 0
Assuming leftards as negative and rightwards as potive
$0 = \frac{m_{1} d + m_{2} d^{'}}{m_{1} + m_{2}}$
d' = $\frac{- m_{1} d}{m_{2}}$
The negative sign indicates that $m_{2}$ should be shifted by
( $\frac{m_{1}}{m_{2}}$ ) d leftwards in order to keep the COM stationary.

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