$A$ particle of mass $3m$ is projected from the ground at some angle with the horizontal. The horizontal range is $R$. At the highest point of its path,it breaks into two pieces of mass $m$ and $2m$. The smaller mass comes to rest,and the larger mass finally falls at a distance $x$ from the point of projection. The value of $x$ is:

$A$ man of mass $m$ is suspended in air by holding the rope of a balloon of mass $M$. As the man climbs up the rope,the balloon

In a system,two particles of masses $m_1 = 3 \text{ kg}$ and $m_2 = 2 \text{ kg}$ are placed at a certain distance from each other. The particle of mass $m_1$ is moved towards the center of mass of the system through a distance of $2 \text{ cm}$. In order to keep the center of mass of the system at the original position,the particle of mass $m_2$ should move towards the center of mass by a distance of . . . . . . $\text{cm}$.

$A$ 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 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,

The figure shows a man of mass $m$ standing at the end $A$ of a trolley of mass $M$ placed at rest on a smooth horizontal surface. The man starts moving towards the end $B$ with a velocity $u_{rel}$ with respect to the trolley. The length of the trolley is $L$. As the man walks on the trolley,the centre of mass of the system (man $+$ trolley)