$A$ system consists of two particles of masses $m_1$ and $m_2$. If the particle of mass $m_1$ is moved towards the centre of mass through a distance $d$,then the distance the second particle should be moved,so as to keep the centre of mass at the same position is

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)

$A$ small ball $B$ of mass $m$ is suspended with a light inelastic string of length $L$ from a block $A$ of the same mass $m$,which can move on a smooth horizontal surface as shown in the figure. The ball is displaced by an angle $\theta$ from the equilibrium position and then released. The displacement of the block when the ball reaches the equilibrium position is

Two persons of mass $m_1$ and $m_2$ are standing at the two ends $A$ and $B$ respectively,of a trolley of mass $M$ as shown. Choose the incorrect statement,if $m_1 = m_2 = m$ and both the persons jump one by one,then

$A$ man of mass $m$ is suspended in air by holding the rope of a balloon of mass $M$. As the man climbs up the rope,the balloon

$A$ 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: