$A$ 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 $m$)

In a system of two particles of masses $m_{1}$ and $m_{2}$,the first particle is moved by a distance $d$ towards the centre of mass. To keep the centre of mass unchanged,the second particle will have to be moved by a distance:

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

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 $L$. As the man walks on the trolley,the centre of mass of the system (man $+$ trolley)

$A$ body falling vertically downwards under gravity breaks into two parts of unequal masses. The centre of mass of the two parts taken together

Discuss the explosion of a projectile.