The figure shows the positions and velocities of two particles. If the particles move under their mutual attraction,then the position of the centre of mass at $t = 1\,s$ is

If a bomb is thrown at a certain angle with the horizontal and after exploding on the way the different fragments move in different directions,then the centre of mass

$A$ 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 centre of mass of the $A+B$ system until the string becomes vertical is

$A$ balloon of mass $M$ with a light rope and a monkey of mass $m$ are at rest in mid-air. If the monkey climbs up the rope and reaches the top of the rope,the distance by which the balloon descends will be (Total length of the rope is $L$):

$A$ 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

Two bodies of masses $12 \,kg$ and $6 \,kg$ are projected simultaneously with velocities $15 \,ms^{-1}$ and $20 \,ms^{-1}$ respectively from the top of a tower of height $25 \,m$. The body of mass $12 \,kg$ is projected vertically upwards and the body of mass $6 \,kg$ is projected horizontally. Find the maximum height reached by the centre of mass of the system of two bodies from the ground. $(g = 10 \,ms^{-2})$ (in $\,m$)