Two particles of masses $1\ kg$ and $3\ kg$ move towards each other under their mutual force of attraction. No other force acts on them. When the relative velocity of approach of the two particles is $2\ m/s,$ their centre of mass has a velocity of $0.5\ m/s$. When the relative velocity of approach becomes $3\ m/s,$ the velocity of the centre of mass is ........ $m/s$.

As shown in the figure,two blocks are connected with a light spring. When the spring is at its natural length,velocities are given to them as shown in the figure. Choose the wrong alternative.

$A$ system consists of two particles of masses $m_1$ and $m_2$ separated by a distance $r$. If the particle of mass $m_1$ is moved towards the centre of mass by a distance $d$,then the particle of mass $m_2$ must be moved by a distance $d'$ so that the centre of mass remains at the same position. The value of $d'$ 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 particles of equal masses have velocities $\overrightarrow{v_1} = 2\hat{i} \ m/s$ and $\overrightarrow{v_2} = 2\hat{j} \ m/s$. The first particle has an acceleration $\overrightarrow{a_1} = (3\hat{i} + 3\hat{j}) \ m/s^2$,while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a

Show that the total momentum of a system of particles is equal to the product of the total mass of the system and the velocity of the centre of mass.