Two particles of masses 10kg and 30kg are lying on a straight line. The 10kg mass is shifted towards the 30kg mass by a distance of 2cm. By what distance should the 30kg mass be shifted so that the position of their centre of mass does not change?

Consider a point (0, 0) with respect to which the centre of mass will be computed. Assume that the particle is distanced from the reference point by a distance of x cm. In this configuration, the centre of mass will be

${\mathrm{C}}_{\mathrm{x}}=\frac{10\times 0+30\times \mathrm{x}}{10+30}$

Let's assume that the 30 kg mass moves by y cm in the direction of the 10 kg mass if the 10 kg moves by a distance of 2 cm.

Thus the new centre of mass = $\frac{10\times 0+30\left(\mathrm{x}\right)}{10+30}=\frac{10\times 2+30(\mathrm{x}-\mathrm{y})}{10+30}$

$$\begin{gathered} 30\mathrm{x}=20+30(\mathrm{x}-\mathrm{y}) \\ \Rightarrow \mathrm{y}=\frac{2}{3}\mathrm{cm} \end{gathered}$$

Thus the correct answer is 2/3 cm towards 10kg .