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 identical particles are moving with velocities $2v$ and $v$ respectively. Find the velocity of the center of mass.

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:

The figure shows a man of mass $m$ standing at the end $A$ of a trolley of mass $M$ placed at rest on a smooth horizontal surface. The man starts moving towards the end $B$ with a velocity $u_{rel}$ with respect to the trolley. The length of the trolley is $L$. The distance moved by the man with respect to ground is

$A$ particle of mass $3m$ is projected from the ground at some angle with the horizontal. The horizontal range is $R$. At the highest point of its path,it breaks into two pieces of mass $m$ and $2m$. The smaller mass comes to rest,and the larger mass finally falls at a distance $x$ from the point of projection. The value of $x$ is:

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