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)

Two particles of equal mass have velocities $2\,\hat{i}\,ms^{-1}$ and $2\,\hat{j}\,ms^{-1}$. The first particle has an acceleration $(3\,\hat{i} + 3\,\hat{j})\,ms^{-2}$ while the acceleration of the second particle is zero. The centre of mass of the two particles moves in

Two bodies of masses $m_1$ and $m_2$ are connected by a massless string passing over a massless,frictionless pulley as shown in the figure. Given $m_2 < m_1$. The acceleration of the center of mass of the system is:

Two spherical objects have masses $M$ and $5M$ and radii $R$ and $2R$ respectively. They are released in free space with their centers at a distance of $12R$. If they attract each other only due to gravitational force,the distance traveled by the smaller object before collision is: (in $R$)

$A$ 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)

Two particles of masses $1\ kg$ and $3\ kg$ move towards each other under their mutual force of attraction. No other force acts on them. When the relative velocity of approach of the two particles is $2\ m/s,$ their centre of mass has a velocity of $0.5\ m/s$. When the relative velocity of approach becomes $3\ m/s,$ the velocity of the centre of mass is ........ $m/s$.