An artillery piece of mass $M_1$ fires a shell of mass $\mathrm{M}_2$ horizontally. Instantaneously after the firing, the ratio of kinetic energy of the artillery and that of the shell is:

A spacecraft of mass $M$ moves with velocity $V$ in free space at first, then it explodes breaking into two pieces. If after explosion a piece of mass $m$ comes to rest, the other piece of spacecraft will have a velocity

You are on a frictionless horizontal plane. How can you get off if no horizontal force is exerted by pushing against the surface

A body of mass $M$ at rest explodes into three pieces, in the ratio of masses $1: 1: 2$. Two smaller pieces fly off perpendicular to each other with velocities of $30 \,ms ^{-1}$ and $40 \,ms ^{-1}$ respectively. The velocity of the third piece will be ............... $\,ms ^{-1}$

If final momentum is equal to initial momentum of the system then

A shell is fired from a cannon with velocity $v m/sec$ at an angle $\theta $ with the horizontal direction. At the highest point in its path it explodes into two pieces of equal mass. One of the pieces retraces its path to the cannon and the speed in $m/sec $ of the other piece immediately after the explosion is