Conservation of momentum and energy in an explosion One simple problem is physics is to determine the mechanical energy difference after an explosion. To do this, you must assume that momentum is conserved because in a explosion you have internal forces, so, using Newton's third law, you get that the total momentum is conserved.
How true is that? I mean, when a explosion happens you have LIGHT and SOUND being propagated. And they carry energy and momentum. It's clear to me that if you only consider the momentum of the parts of the body + the momentum of the sound wave + the momentum of the light wave this should be equal to zero. However, it's not so clear if the momentum of the body should be equal of the sum of the momentum of its fragments.
I have some conjectures, but I didn't find nothing related to that:
I) The waves propagate symmetrically, so, because momentum is a vector, it should be zero. However, the explosion may not be total symmetric, so I don't know if this argument is valid.
II) The momentum is not conserved for the body, but the momentum of the wave is negligible. 
These are the two possibilities I considered, so I don't know if they are valid.
 A: Introductory physics problems often limit the momentum economy to the motion of large particles or fragments (collisions and explosions) for simplicity of calculations. In reality, the momentum transferred to any surrounding gas (air) should ideally be part of the conservation. These introductory problems are constructed so that compression waves and huge amounts of EM radiation are negligibly small. Even in collision experiments we don't initially account for the sound produced by masses hitting each other. Later, we mention that the sound should be considered as momentum and energy lost from the colliding masses.
In real explosions of large bombs, the sound and EM are not negligible components. Compressions waves are often the most destructive part of a bomb; the air definitely receives energy and momentum and the resulting small particles may or may not carry much. Some bombs are designed to be concussive (huge amplitude pressure waves) and others are fragmentary (scattering massive particles).
Depending on the geometry and strength of the casing the distribution of momentum and energy may or may not be symmetric.  If you go to a good fireworks show you will see a variety of distributions of particles and sounds.
The extra mechanical energy comes from the chemical or nuclear potential energies converted in the reaction, as @Natasha mentioned.
A: In the case of an explosion, before the explosion the momentum of the bomb is zero, so according to law of conservation of momentum, the momentum after explosion should also be zero. 
So, momentum of the bomb before collision = momentum of the bomb after collision. 
As for sound and light energy, I think that it is the chemical energy of the bomb that is converted into this energy. In case of an atomic bomb, it will be atomic energy converted into light and sound energy. Also, according to law of conservation of energy, this energy will not be destroyed with the explosion, but converted.
A: Before explosion the bomb is at rest. Its total momentum is zero. As it explodes, it breaks into many parts of masses $m_1,m_2,m_3$ etc which fly of in different directions with velocities $v_1,v_2, v_3$ etc. these diff parts have different momenta $m_1v_1,m_2v_2, m_3v_3$,etc. For eg,- 
If the bomb explodes in two parts then both of them fly into opposite direction with equal momentum, so the final momentum is also zero. Hence the explosion of bomb obeys the law of conservation of momentum. 
