$A$ child is standing at one end of a long trolley moving with a speed $v$ on a smooth horizontal track. If the child starts running towards the other end of the trolley with a speed $u$,the centre of mass of the system (trolley + child) will move with a speed

$A$ 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 $X$ and $Y$,initially at rest,move towards each other under mutual attraction. If at any instant the velocity of particle $X$ is $V$ and the velocity of particle $Y$ is $2V$,then the velocity of their center of mass will be:

Three blocks $A$,$B$ and $C$ are arranged as shown in the figure such that the distance between two successive blocks is $10 \ m$. Block $A$ has a mass of $10 \ kg$,block $B$ has a mass of $25 \ kg$,and block $C$ has a mass of $15 \ kg$. Block $A$ is displaced towards block $B$ by $2 \ m$ and block $C$ is displaced towards block $B$ by $3 \ m$. The distance through which the block $B$ should be moved so that the centre of mass of the system does not change is

Two particles of masses in the ratio $1:2$ are placed along a vertical line. The lighter particle is raised through a height of $9 \ cm$. To raise the centre of mass of the system by $2 \ cm$,the heavier particle should be

Two particles $A$ and $B$ (both initially at rest) start moving towards each other under a mutual force of attraction. At the instant, when the speed of $A$ is $v$ and the speed of $B$ is $2v$, the speed of the centre of mass is