Two particles of masses $m_1$ and $m_2$ $(m_1 > m_2)$,initially at rest,move towards each other under an inverse square law force of attraction. Pick out the correct statement about the centre of mass $(CM)$ of the system.

$A$ 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?

$A$ 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:

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:

Show that the total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of the centre of mass.

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)