An artillery piece of mass $M_1$ fires a shell of mass $M_2$ horizontally. Instantaneously after the firing,the ratio of kinetic energy of the artillery and that of the shell is:

$A$ bomb of mass $9 \,kg$ explodes into two pieces of masses $3 \,kg$ and $6 \,kg$. The velocity of mass $3 \,kg$ is $16 \,m/s$. The kinetic energy of mass $6 \,kg$ in joule is

$A$ cannon shell fired breaks into two equal parts at its highest point. One part retraces the path to the cannon with kinetic energy $E_1$ and the kinetic energy of the second part is $E_2$. The relation between $E_1$ and $E_2$ is:

$A$ man of mass $m$ is standing on a cart of mass $2m$. Initially, the cart is at rest. The man jumps horizontally with a relative velocity $u$ with respect to the cart. The work done by the internal forces of the man during the process of jumping will be:

At a certain height,a body at rest explodes into two equal fragments with one fragment receiving a horizontal velocity of $10\sqrt{3} \text{ m/s}$. The horizontal distance between the two fragments,when their displacement vectors are inclined at $60^{\circ}$ relative to each other,is (take $g = 10 \text{ m/s}^2$):

$A$ cannon ball is fired with a velocity $200 \, m/s$ at an angle of $60^{\circ}$ with the horizontal. At the highest point of its flight,it explodes into $3$ equal fragments. One fragment goes vertically upwards with a velocity $100 \, m/s$,and the second one falls vertically downwards with a velocity $100 \, m/s$. The third fragment will be moving with a velocity: