$A$ ball of mass $100 \ g$ is dropped at time $t=0$. $A$ second ball of mass $200 \ g$ is dropped from the same point at $t=0.2 \ s$. The distance between the center of mass of the two balls and the release point at $t=0.4 \ s$ is: (Assume $g=10 \ m \ s^{-2}$) (in $m$)

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 $L$. When the man reaches the end $B$,the distance moved by the trolley with respect to ground is

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

$A$ $500 \ kg$ boat is $9 \ m$ long and is floating without motion on still water. $A$ man of mass $100 \ kg$ is at one end. If he runs to the other end of the boat and stops,the displacement of the boat is:

On a smooth horizontal plane,a uniform string of mass $M$ and length $L$ is lying in a state of rest. $A$ man of the same mass $M$ is standing next to one end of the string. Now,the man starts collecting the string. Finally,the man collects all the string and puts it in his pocket. What is the displacement of the man with respect to the Earth in the process of collection?

Two particles $A$ and $B$,initially at rest,move towards each other under a mutual force of attraction. At an instance when the speed of $A$ is $v$ and the speed of $B$ is $2v$,the speed of the centre of mass $(CM)$ is: