$A$ 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

In a system,two particles of masses $m_1 = 3 \text{ kg}$ and $m_2 = 2 \text{ kg}$ are placed at a certain distance from each other. The particle of mass $m_1$ is moved towards the center of mass of the system through a distance of $2 \text{ cm}$. In order to keep the center of mass of the system at the original position,the particle of mass $m_2$ should move towards the center of mass by a distance of . . . . . . $\text{cm}$.

Two blocks of masses $2 \text{ kg}$ and $1 \text{ kg}$ respectively, are tied to the ends of a string which passes over a light frictionless pulley as shown in the figure below. The masses are held at rest at the same horizontal level and then released. The distance traversed by the centre of mass in $2 \text{ s}$ is . . . . . . $\text{m}$. (Take $g = 10 \text{ m/s}^2$)

Two particles $X$ and $Y$,initially at rest,move towards each other under mutual attraction. If at any instant the velocity of particle $X$ is $V$ and the velocity of particle $Y$ is $2V$,then the velocity of their center of mass will be:

$A$ 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)

$A$ $500\,kg$ boat is $9\,m$ long and is floating without motion on still water. $A$ man of mass $100\,kg$ is at one end and if he runs to the other end of the boat and stops,the displacement of the boat is