$A$ bomb at rest explodes into three parts of equal mass. If the momentum of two parts are $-2 p \hat{i}$ and $p \hat{j}$,find the magnitude of momentum of the third part.

$A$ shell of mass $m$ moving with velocity $v$ suddenly breaks into two pieces. The part having mass $\frac{m}{5}$ remains stationary. The velocity of the other part will be

$A$ block of mass $m$ is moving with a velocity $u$ on a smooth horizontal surface towards a wedge of same mass $m$ initially kept at rest. The wedge is free to move in any direction. Initially, the block moves up the smooth inclined plane of the wedge to a height $h$ and then moves down back to the horizontal plane. In this process, the wedge gains a velocity equal to:

$A$ bomb of $12 \ kg$ divides into two parts whose ratio of masses is $1 : 3$. If the kinetic energy of the smaller part is $216 \ J$,then the momentum of the bigger part in $kg \cdot m/s$ 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 ....... $m s^{-1}$.

$A$ 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