$A$ body of mass $20 \ kg$ is moving with a velocity of $2v$ and another body of mass $10 \ kg$ is moving with velocity $v$ along the same direction. The velocity of their centre of mass is

$A$ piece of wood of mass $0.03\, kg$ is dropped from the top of a $100\, m$ height building. At the same time,a bullet of mass $0.02\, kg$ is fired vertically upward,with a velocity $100\, ms^{-1}$,from the ground. The bullet gets embedded in the wood. Then the maximum height to which the combined system reaches above the top of the building before falling below is ........ $m$. $(g = 10\, ms^{-2})$

Three blocks $A$,$B$ and $C$ are arranged as shown in the figure such that the distance between two successive blocks is $10 \ m$. Block $A$ has a mass of $10 \ kg$,block $B$ has a mass of $25 \ kg$,and block $C$ has a mass of $15 \ kg$. Block $A$ is displaced towards block $B$ by $2 \ m$ and block $C$ is displaced towards block $B$ by $3 \ m$. The distance through which the block $B$ should be moved so that the centre of mass of the system does not change is

Two particles of equal mass have velocities $2\,\hat{i}\,ms^{-1}$ and $2\,\hat{j}\,ms^{-1}$. The first particle has an acceleration $(3\,\hat{i} + 3\,\hat{j})\,ms^{-2}$ while the acceleration of the second particle is zero. The centre of mass of the two particles moves in

$A$ 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 $\text{ ms}^{-1}$)

In the system of two particles of masses $m_1$ and $m_2$,the first particle is moved by a distance $d$ towards the centre of mass. To keep the centre of mass unchanged,the second particle will have to be moved by a distance