$A$ projectile is moving at $20 \, m/s$ at its highest point,where it breaks into two equal parts due to an internal explosion. One part moves vertically up at $30 \, m/s$. Then the other part will move at ............. $m/s$.

$A$ bomb moving with velocity $(40 \hat{i} + 50 \hat{j} - 25 \hat{k}) \text{ m/s}$ explodes into two pieces of mass ratio $1:4$. After the explosion,the smaller piece moves away with velocity $(200 \hat{i} + 70 \hat{j} + 15 \hat{k}) \text{ m/s}$. The velocity of the larger piece after the explosion is:

$A$ 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:

An explosion breaks a rock into three parts in a horizontal plane. Two of them go off at right angles to each other. The first part of mass $1 \, kg$ moves with a speed of $12 \, m s^{-1}$ and the second part of mass $2 \, kg$ moves with $8 \, m s^{-1}$ speed. If the third part flies off with $4 \, m s^{-1}$ speed,then its mass is ......... $kg$.

$A$ 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 of mass $9\,kg$ explodes into $2$ pieces of mass $3\,kg$ and $6\,kg.$ The velocity of the $3\,kg$ mass is $1.6\,m/s.$ The kinetic energy of the $6\,kg$ mass is ............ $J.$