$A$ system consists of two particles of masses $m_1$ and $m_2$. If the particle of mass $m_1$ is moved towards the centre of mass through a distance $d$,then the distance the second particle should be moved,so as to keep the centre of mass at the same position is

Three blocks $A$,$B$ and $C$ are arranged as shown in the figure such that the distance between two successive blocks is $10 \ m$. Block $A$ has a mass of $10 \ kg$,block $B$ has a mass of $25 \ kg$,and block $C$ has a mass of $15 \ kg$. Block $A$ is displaced towards block $B$ by $2 \ m$ and block $C$ is displaced towards block $B$ by $3 \ m$. The distance through which the block $B$ should be moved so that the centre of mass of the system does not change is

Two particles of masses in the ratio $1:2$ are placed along a vertical line. The lighter particle is raised through a height of $9 \ cm$. To raise the centre of mass of the system by $2 \ cm$,the heavier particle should be

$A$ small ball $B$ of mass $m$ is suspended with a light inelastic string of length $L$ from a block $A$ of the same mass $m$,which can move on a smooth horizontal surface as shown in the figure. The ball is displaced by an angle $\theta$ from the equilibrium position and then released. The displacement of the block when the ball reaches the equilibrium position is

Two bodies of masses $2\,kg$ and $4\,kg$ are moving with velocities $2\,m/s$ and $10\,m/s$ respectively towards each other due to mutual gravitational attraction. What is the velocity of their centre of mass,given that the bodies were at rest initially?

$A$ system of two particles has masses $m_1$ and $m_2$. If the particle of mass $m_1$ is pushed towards the centre of mass of the particles through a distance $d$,by what distance should the particle of mass $m_2$ be moved to keep the centre of mass of the particles at the original position?