Two men,of masses $60 \, kg$ and $80 \, kg$,are sitting at the ends of a boat of mass $60 \, kg$ and length $4 \, m$. The boat is stationary. If the men now exchange their positions,then:

Three particles $A, B$ and $C$ of masses $m, 2m$ and $3m$ are moving towards north, south and east respectively. If the velocities of the particles $A, B$ and $C$ are $6 \,ms^{-1}, 12 \,ms^{-1}$ and $8 \,ms^{-1}$ respectively, then the velocity of the centre of mass of the system of particles is (in $\,ms^{-1}$)

$A$ projectile of mass $M$ is fired so that the horizontal range is $4\, km$. At the highest point,the projectile explodes into two parts of masses $M/4$ and $3M/4$ respectively. The heavier part starts falling down vertically with zero initial speed. The horizontal range (distance from the point of firing) of the lighter part is .................. $km$.

$M$ mass of a man stands at one end of a plank of length $L$ which lies at rest on a frictionless surface. The man walks to the other end of the plank. If the mass of the plank is $\frac{M}{3}$,then the distance that the man moves relative to the ground is:

Consider a two-particle system with particles having masses $m_1$ and $m_2$. If the first particle is pushed towards the centre of mass through a distance $d$,by what distance should the second particle be moved,so as to keep the centre of mass at the same position?

$A$ 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)