The principle of conservation of linear momentum can be strictly applied during a collision between two particles provided the time of impact is

$A$ bomb of mass $16 \,kg$ explodes into two pieces of masses $4 \,kg$ and $12 \,kg$. The velocity of the $12 \,kg$ mass is $4 \,ms^{-1}$. The kinetic energy of the second piece (the $4 \,kg$ piece) is: (in $\,J$)

$A$ bullet of mass $m$ moving with velocity $v$ is fired into a wooden block of mass $M$. If the bullet remains embedded in the block,the final velocity of the system is:

$A$ 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:

$A$ 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 ............. $m/s$.

$A$ 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