Two bodies of $6 \,kg$ and $4 \,kg$ masses have their velocities $5 \hat{i}-2 \hat{j}+10 \hat{k}$ and $10 \hat{i}-2 \hat{j}+5 \hat{k}$ respectively. Then the velocity of their centre of mass is

Two particles which are initially at rest,move towards each other under the action of their internal attraction. If their speeds are $v$ and $2v$ at any instant,then the speed of centre of mass of the system will be

$A$ 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 ........ $m$. $(g = 10\, ms^{-2})$

$A$ body of mass $2 \,kg$ is moving towards north with a velocity of $20 \,m \,s^{-1}$ and another body of mass $3 \,kg$ is moving towards east with a velocity of $10 \,m \,s^{-1}$. The magnitude of the velocity of the centre of mass of the system of the two bodies is

Two particles of equal mass have velocities $\overrightarrow{v}_1=4 \hat{i} \text{ ms}^{-1}$ and $\overrightarrow{v}_2=4 \hat{j} \text{ ms}^{-1}$. The first particle has an acceleration $\overrightarrow{a}_1=(5 \hat{i}+5 \hat{j}) \text{ ms}^{-2}$ while the acceleration of the other particle is zero. The centre of mass of the two particles moves in a path of

$A$ 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)