Consider a two-particle system with particles having masses $m_1$ and $m_2$. If the first particle is pushed towards the center of mass through a distance $d$,by what distance should the second particle be moved so as to keep the center of mass at the same position?

The center of mass of three particles of masses $10 \ kg, 20 \ kg$,and $30 \ kg$ is at $(0, 0, 0)$. Where should a fourth particle of mass $40 \ kg$ be placed so that the center of mass of the system shifts to $(3, 3, 3)$?

Three identical spheres,each of mass $1 \ kg$,are placed touching each other with their centres on a straight line. Their centres are marked $K, L$,and $M$ respectively. The distance of the centre of mass of the system from $K$ is

The position vectors of three particles of masses $1\, kg, 2\, kg$,and $3\, kg$ are $\vec{r_1} = (\hat{i} + 4\hat{j} + \hat{k})\,m$,$\vec{r_2} = (\hat{i} + \hat{j} + \hat{k})\,m$,and $\vec{r_3} = (2\hat{i} - \hat{j} - 2\hat{k})\,m$ respectively. The position vector of their centre of mass is:

On a horizontal frictionless frozen lake,a girl of mass $36 \,kg$ and a box of mass $9 \,kg$ are connected to each other by means of a rope. Initially,they are $20 \,m$ apart. The girl exerts a horizontal force on the box,pulling it towards her. How far has the girl travelled when she meets the box?

$A$ point mass $m_A$ is connected to a point mass $m_B$ by a massless rod of length $l$ as shown in the figure. It is observed that the ratio of the moment of inertia of the system about the two axes $AA$ and $BB$,which are parallel to each other and perpendicular to the rod,is $\frac{I_{BB}}{I_{AA}} = 3$. The distance of the centre of mass of the system from the mass $m_A$ is