The position of the centre of mass for a two-particle system is given by $x_{cm} = \frac{m_1 x_1 + m_2 x_2}{m_1 + m_2}$. Since the centre of mass remains at its original position, the shift in the centre of mass is zero, i.e., $\Delta x_{cm} = 0$. Therefore, $\frac{m_1 \Delta x_1 + m_2 \Delta x_2}{m_1 + m_2} = 0$, which implies $m_1 \Delta x_1 + m_2 \Delta x_2 = 0$. Considering only the magnitudes of the displacements (as both must move towards the centre of mass to keep it stationary), we have: $m_1 d = m_2 d'$ Where $d$ is the distance moved by $m_1$ and $d'$ is the distance moved by $m_2$. Thus, the distance moved by the second particle is $d' = \frac{m_1}{m_2}d$.