$A$ system consists of two particles of masses $m_1$ and $m_2$ separated by a distance $r$. If the particles move towards each other with velocities $v_1$ and $v_2$ respectively,the velocity of the centre of mass of the system is:

$A$ child sits stationary at one end of a long trolley moving uniformly with a speed $V$ on a smooth horizontal floor. If the child gets up and runs about on the trolley in any manner,what is the speed of the $CM$ of the (trolley + child) system?

$A$ 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:

$A$ 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 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 the centre of mass of the system will be

Two masses $m_1$ and $m_2$ are connected by a massless spring of spring constant $k$ and unstretched length $l$. The masses are placed on a frictionless straight channel,which we consider our $X$-axis. They are initially at rest at $x=0$ and $x=l$,respectively. At $t=0$,a velocity of $v_0$ is suddenly imparted to the first particle. At a later time $t$,the centre of mass of the two masses is at