The velocity of the centre of mass of the system as shown in the figure is

Two men,of masses $60 \, kg$ and $80 \, kg$,are sitting at the ends of a boat of mass $60 \, kg$ and length $4 \, m$. The boat is stationary. If the men now exchange their positions,then:

$A$ particle of mass $10 \text{ g}$ is moving towards east with a velocity of $10 \text{ ms}^{-1}$ and another particle of mass $15 \text{ g}$ is moving towards north with a velocity of $5 \text{ ms}^{-1}$. The magnitude of the velocity of the centre of mass of the system of the two particles is (in $\text{ ms}^{-1}$)

Two bodies of mass $4 \,kg$ and $5 \,kg$ are moving along east and north directions with velocities $5 \,m/s$ and $3 \,m/s$ respectively. The magnitude of the velocity of the centre of mass of the system is:

Two bodies of masses $2\,kg$ and $4\,kg$ are moving with velocities $2\,m/s$ and $10\,m/s$ respectively towards each other due to mutual gravitational attraction. What is the velocity of their centre of mass,given that the bodies were at rest initially?

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 $g = 10 \text{ m/s}^2$)