Two particles $A$ and $B$ (both initially at rest) start moving towards each other under a mutual force of attraction. At the instant, when the speed of $A$ is $v$ and the speed of $B$ is $2v$, the speed of the centre of mass is

Two equal rods joined at one end are kept on a smooth surface as shown and released. The trajectory of the centre of mass of both rods is:

$A$ body falling vertically downwards under gravity breaks into two parts of unequal masses. The centre of mass of the two parts taken together

$A$ uniform rod of mass $M$ and length $L$ is placed on a smooth horizontal surface. $A$ horizontal force $F$ is applied at one end of the rod at an angle $\theta$ with the rod. The acceleration of the centre of mass of the rod is:

In the figure shown,the blocks have equal masses. Friction,mass of the string,and the mass of the pulley are negligible. The magnitude of the acceleration of the centre of mass of the two blocks is (Acceleration due to gravity $= g$).

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