Two blocks of masses $2 \,kg$ and $1 \,kg$ are tied to the ends of a string which passes over a light frictionless pulley. The blocks are held at rest at the same horizontal level and then released suddenly. The distance traversed by their centre of mass in $2 \,s$ is (acceleration due to gravity $g = 10 \,ms^{-2}$) (in $\,m$)

Two particles $A$ and $B$ initially at rest move towards each other under a mutual force of attraction. At the instant when the velocity of $A$ is $v$ and that of $B$ is $2v$,the velocity of the centre of mass of the system is:

$A$ child is standing at one end of a long trolley moving with a speed $v$ on a smooth horizontal track. If the child starts running towards the other end of the trolley with a speed $u$,the centre of mass of the system (trolley + child) will move with a speed

Two particles of masses $2 \ kg$ and $1 \ kg$ are moving along the same line with speeds of $2 \ m/s$ and $5 \ m/s$ respectively. If both particles are moving in the same direction,the speed of the centre of mass is $v_1$. If both particles are moving in opposite directions,the speed of the centre of mass is $v_2$. Find $v_1$ and $v_2$.

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 $\,m$)

Two particles have masses $m_1$ and $m_2$. If the first particle $(m_1)$ is moved by a distance $d$ towards the center of mass,by what distance should the second particle $(m_2)$ be moved so that the center of mass of the system remains at the same position?