$A$ body of mass $5 \,kg$ explodes at rest into three fragments with masses in the ratio $1 : 1 : 3$. The fragments with equal masses fly in mutually perpendicular directions with speeds of $21 \,m/s$. The velocity of the heaviest fragment will be

$A$ cannon shell fired breaks into two equal parts at its highest point. If one part retraces its path to the cannon with kinetic energy $E_1$ and the kinetic energy of the second part is $E_2$,then

$A$ ball is projected at $60^{\circ}$ from the horizontal at $200 \, m/s$. At the maximum height during its flight,it explodes into $3$ equal fragments. Out of these,one part travels at $100 \, m/s$ vertically up while another travels at $100 \, m/s$ vertically down. What will be the speed of the third part just after the explosion?

$A$ shell of mass $m$ moving with velocity $v$ suddenly breaks into $2$ pieces. The part having mass $m/3$ remains stationary. The velocity of the other part will be

$A$ bomb at rest explodes into three equal fragments. Two of the fragments move at right angles to each other with velocities of $9 \ m s^{-1}$ and $12 \ m s^{-1}$. The magnitude of the velocity of the third fragment is ....... $m s^{-1}$.

$A$ stationary particle explodes into two particles of masses $x$ and $y$,which move in opposite directions with velocities $v_1$ and $v_2$. The ratio of their kinetic energies $(E_1: E_2)$ is