Two particles of the same mass are moving towards each other with velocities $2v$ and $v$ respectively. The velocity of the centre of mass of the system is:

$A$ body $A$ of mass $M$ falls vertically downwards under gravity. It breaks into two parts: a body $B$ of mass $1/3 M$ and a body $C$ of mass $2/3 M$. The center of mass of bodies $B$ and $C$ taken together shifts from the direction of $A$ by:

$A$ uniform rod of mass $M$ and length $L$ is placed on a smooth horizontal surface. $A$ horizontal force $F$ is applied at one end of the rod perpendicular to its length. The acceleration of the centre of mass of the rod is:

Two bodies of masses $12 \,kg$ and $6 \,kg$ are projected simultaneously with velocities $15 \,ms^{-1}$ and $20 \,ms^{-1}$ respectively from the top of a tower of height $25 \,m$. The body of mass $12 \,kg$ is projected vertically upwards and the body of mass $6 \,kg$ is projected horizontally. Find the maximum height reached by the centre of mass of the system of two bodies from the ground. $(g = 10 \,ms^{-2})$ (in $\,m$)

Two particles which are initially at rest,move towards each other under the action of their internal attraction. If their speeds are $v$ and $2v$ at any instant,then the speed of centre of mass of the system will be

Two men $A$ and $B$ are standing on a plank. $B$ is at the middle of the plank and $A$ is at the left end of the plank. The bottom surface of the plank is smooth. The system is initially at rest and the masses are as shown in the figure. $A$ and $B$ start moving such that the position of $B$ remains fixed with respect to the ground and $A$ meets $B$. Then the point where $A$ meets $B$ is located at