Two bodies of masses $2 \ kg$ and $4 \ kg$ are moving with velocities $20 \ m/s$ and $10 \ m/s$ towards each other due to mutual gravitational attraction. What is the velocity of their centre of mass (in $m/s$)?

The figure shows the positions and velocities of two particles. If the particles move under their mutual attraction,then the position of the centre of mass at $t = 1\,s$ is

$A$ mass $m$ is at rest on an inclined plane of mass $M$,which is further resting on a smooth horizontal plane. Now,if the mass $m$ starts moving under gravity,the position of the centre of mass of the system will

Two blocks of masses $10 \; kg$ and $30 \; kg$ are placed on the same straight line with coordinates $(0,0) \; cm$ and $(x, 0) \; cm$ respectively. The block of $10 \; kg$ is moved on the same line through a distance of $6 \; cm$ towards the other block. The distance through which the block of $30 \; kg$ must be moved to keep the position of the centre of mass of the system unchanged 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 blocks,each of mass $m$,are connected to each other by a massless string and placed on two inclined planes as shown in the figure. After releasing the system from rest,find the magnitude of the acceleration of the centre of mass of the two blocks. (Take $g = 10 \, m/s^2$)