$A$ 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 perpendicular to its length. The acceleration of the centre of mass of the rod is:

Two particles have masses $m_1$ and $m_2$. If the first particle $(m_1)$ is moved by a distance $d$ towards the center of mass,by what distance should the second particle $(m_2)$ be moved so that the center of mass of the system remains at the same position?

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

Consider a system of two identical particles. One of the particles is at rest and the other has an acceleration $\vec{f}$. The centre of mass has an acceleration

$A$ cart of mass $M$ is tied to one end of a massless rope of length $10\, m$. The other end of the rope is held by a man of mass $M$. The entire system is on a smooth horizontal surface. The man is at $x = 0$ and the cart is at $x = 10\, m$. If the man pulls the cart by the rope,the man and the cart will meet at the point:

$A$ 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: