$A$ body falling vertically downwards under gravity breaks into two parts of unequal masses. The centre of mass of the two parts taken together

$A$ 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 $kg$)

$A$ body of mass $m_1 = 4 \text{ kg}$ moves at $5 \hat{i} \text{ m/s}$ and another body of mass $m_2 = 2 \text{ kg}$ moves at $10 \hat{i} \text{ m/s}$. The kinetic energy of the centre of mass is:

Two particles which are initially at rest move towards each other under the action of their mutual attraction. If their speeds are $v$ and $2v$ at any instant,then the speed of the center of mass of the system is,

$A$ firecracker is thrown with a velocity of $30 \, ms^{-1}$ in a direction that makes an angle of $75^{\circ}$ with the vertical axis. At some point on its trajectory,the firecracker splits into two identical pieces in such a way that one piece falls $27 \, m$ far from the shooting point. Assuming that all trajectories are contained in the same plane,how far will the other piece fall from the shooting point? (Take $g = 10 \, ms^{-2}$ and neglect air resistance)

Three particles $A, B$ and $C$ of masses $m, 2m$ and $3m$ are moving towards north, south and east respectively. If the velocities of the particles $A, B$ and $C$ are $6 \,ms^{-1}, 12 \,ms^{-1}$ and $8 \,ms^{-1}$ respectively, then the velocity of the centre of mass of the system of particles is (in $\,ms^{-1}$)