$A$ body initially at rest breaks up into two pieces of masses $2 M$ and $3 M$ respectively,together having a total kinetic energy $E$. The piece of mass $2 M$,after breaking up,has a kinetic energy:

$A$ body at rest breaks into three fragments. Two fragments of equal mass fly off perpendicular to each other with a speed of $30 \; m/s$. The mass of the third fragment is three times that of the other two fragments. What will be the direction and magnitude of the velocity of the third fragment immediately after the explosion?

$12 \,kg$ bomb at rest explodes into two pieces of $4 \,kg$ and $8 \,kg$. If the momentum of the $4 \,kg$ piece is $20 \,Ns$, what is the kinetic energy of the $8 \,kg$ piece (in $\,J$)?

For a system to follow the law of conservation of linear momentum during a collision, the condition is:
$(1)$ Total external force acting on the system is zero.
$(2)$ Total external force acting on the system is finite and the time of collision is negligible.
$(3)$ Total internal force acting on the system is zero.

Three particles of masses $10\;g, 20\;g$ and $40\;g$ are moving with velocities $10\hat{i}, 10\hat{j}$ and $10\hat{k}\;m/s$ respectively. If due to some mutual interaction,the first particle comes to rest and the velocity of the second particle becomes $(3\hat{i} + 4\hat{j})\;m/s$,then the velocity of the third particle 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: