$A$ ball of mass $M$ moving with a speed of $2 \ m \ s^{-1}$ hits another ball of mass $1 \ kg$ moving in the same direction with a speed of $1 \ m \ s^{-1}$. If the kinetic energy of the center of mass is $4/3 \ J$,then the magnitude of $M$ is: (in $kg$)

The equation $M\vec{A} = \vec{F}_{ext}$ denotes what?

Two blocks of masses $10 \; kg$ and $30 \; kg$ are placed on the same straight line with coordinates $(0,0) \; cm$ and $(x, 0) \; cm$ respectively. The block of $10 \; kg$ is moved on the same line through a distance of $6 \; cm$ towards the other block. The distance through which the block of $30 \; kg$ must be moved to keep the position of the centre of mass of the system unchanged is

$A$ 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:

$A$ 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:

$A$ body of mass $20 \ kg$ is moving with a velocity of $2v$ and another body of mass $10 \ kg$ is moving with velocity $v$ along the same direction. The velocity of their centre of mass is