Two particles $A$ and $B$ initially at rest move towards each other under a mutual force of attraction. At the instant when the velocity of $A$ is $v$ and that of $B$ is $2v$,the velocity of the centre of mass of the system is:

Two bodies of $6 \,kg$ and $4 \,kg$ masses have their velocities $5 \hat{i}-2 \hat{j}+10 \hat{k}$ and $10 \hat{i}-2 \hat{j}+5 \hat{k}$ respectively. Then the velocity of their centre of mass is

Two particles of equal mass have velocities $2\,\hat{i}\,ms^{-1}$ and $2\,\hat{j}\,ms^{-1}$. The first particle has an acceleration $(3\,\hat{i} + 3\,\hat{j})\,ms^{-2}$ while the acceleration of the second particle is zero. The centre of mass of the two particles moves in

Two particles of masses $3 \ kg$ and $6 \ kg$ are approaching towards each other with accelerations $2 \ m/s^2$ and $4 \ m/s^2$ respectively,on a smooth horizontal surface. Then the acceleration of the centre of mass of the system is ..... $m/s^2$.

Two particles of masses $1 \,g$ and $2 \,g$ move towards each other with velocities $10 \,ms^{-1}$ and $20 \,ms^{-1}$ respectively. The velocity of the centre of mass of the system of the two particles is (in $\,ms^{-1}$)

In a system,two particles of masses $m_1 = 3 \text{ kg}$ and $m_2 = 2 \text{ kg}$ are placed at a certain distance from each other. The particle of mass $m_1$ is moved towards the center of mass of the system through a distance of $2 \text{ cm}$. In order to keep the center of mass of the system at the original position,the particle of mass $m_2$ should move towards the center of mass by a distance of . . . . . . $\text{cm}$.