Two balls are thrown in the air. The second ball is thrown $2 \ s$ after the first ball. What is the acceleration of the center of mass of the two balls while they are in the air?

An object $A$ of mass $M$ is falling vertically downwards under the influence of gravity. It breaks into two parts: part $B$ of mass $\frac{1}{3}M$ and part $C$ of mass $\frac{2}{3}M$. The center of mass of the combined parts $B$ and $C$ will be displaced relative to object $A$ in which direction?

$A$ small ball $B$ of mass $m$ is suspended with a light inelastic string of length $L$ from a block $A$ of the same mass $m$,which can move on a smooth horizontal surface as shown in the figure. The ball is displaced by an angle $\theta$ from the equilibrium position and then released. The displacement of the block when the ball reaches the equilibrium position 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

Two blocks of masses $10\,kg$ and $4\,kg$ are connected by a spring of negligible mass and placed on a frictionless horizontal surface. An impulse gives a velocity of $14\,m/s$ to the heavier block in the direction of the lighter block. The velocity of the centre of mass is ........ $m/s$.

$A$ 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: