Two particles of masses $1 \,g$ and $2 \,g$ move towards each other with velocities $10 \,ms^{-1}$ and $20 \,ms^{-1}$ respectively. The velocity of the centre of mass of the system of the two particles is (in $\,ms^{-1}$)

$A$ 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:

Two bodies of masses $m_1$ and $m_2$ are connected by a massless string passing over a massless,frictionless pulley as shown in the figure. Given $m_2 < m_1$. The acceleration 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)

$A$ ball of mass $100 \ g$ is dropped at time $t=0$. $A$ second ball of mass $200 \ g$ is dropped from the same point at $t=0.2 \ s$. The distance between the center of mass of the two balls and the release point at $t=0.4 \ s$ is: (Assume $g=10 \ m \ s^{-2}$) (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: