Explanation

(b) Initial position of centre of mass $r_{cm} = \frac{m_{1}x_{1} + m_{2}x_{2}}{m_{1} + m_{2}}$ ...(i)
If the particles of mass $m_{1}$ is pushed towards the centre of mass of the system through distance $d$ and to keep the centre of mass at the original position let second particle be displaced through distance $d'$ away from the centre of mass
Now $r_{cm} = \frac{m_{1}\left( x_{1} + d \right) + m_{2}(x_{2} + d^{'})}{m_{1} + m_{2}}$ …(ii)
Equating (i) and (ii)
$\frac{m_{1}x_{1} + m_{2}x_{2}}{m_{1} + m_{2}} = \frac{m_{1}\left( x_{1} + d \right) + m_{2}\left( x_{2} + d^{'} \right)}{m_{1} + m_{2}}$
By solving $d^{'} = - \frac{m_{1}}{m_{2}}d$
Negative sign shows that particle $m_{2}$ should be displaced towards the centre of mass of the system