I invite with due regards Mr. Hardik Kothadia of IIT as one of my follower or any other Researchers to answer it with proper explanation for higher momentum after explosion, very soon.
Hello U. Dreher
I can not recomend your andwer. Because before explosion every particle in the bomb or cracker was at rest/zero momentum and after explosin every particle moves with considerable large momentiim. Here my question is on the isolated bomb what is the force tjat causes the particles to eject out with lar ge momentum. If you half enough idea please answer it.
With best tegards.
Sorry to object: when (vector!) adding up all individual momentums, the resulting momentum after the explosion is the same as the one prior the explosion. Zero if the cracker/bomb was at rest before.
A simplified example (only 2 "particles") involving some explosion: The firing of some weapon: while the bullet exits with considerable speed (moment), there is the recoil exerted on the weapon.
mBullet * vBullet = mWeapon * vWeapon.
Going back to the bomb: you get fragments flying in all directions (some maybe burying into the ground). If the bomb ws at rest before, the vector sum of all momentums results in 0.
Convinced now?
Hello U. Dreher
I agree with the total vector sum before and after blasting to zero. But my point of the question was actually that the rate of change of momentum of each particle is a force. Before blasting all the particles were at rest and there was no inter-particle collision, However after blasting each particle experiences some force and undergoes momentum change. Here the point of my question is to identify that force on each particle. Please if you have some Idea, share it.
Hello U. Dreher
Still I encourage you that if you find the answer anywhere, please communicate with that very soon.
The exact question is still not made clear.
If it is about the initial acceleration force: this is the pressure building up from the explosion: the pressure rises to extreme levels - until the containment disintegrates (and the pressure thus rapidly drops). Be aware that the containment's "wall" is driven outwards even before disintegration. Thus we have the force (pressure) by way product that equals the "work" (aka energy) the particles have when flying. The momentum transfer is similar as the pressure buildup is only possible by exerting the pressure omnidirectionally on the containment.
Question answered now?
Hello U. Dreher
I am ending the further question on clearing my thought. My thought is that the explosion of a bomb or a cracker involves some chemical change and releases a large quantity of internal energy whose consumption performs some work on the product materials which then gain the momentum. This momentum gain due consumption of own internal energy will disturb the applicability of general momentum balance equation, rather is applied n modified form.
Initial momentum + momentum gain from internal energy consumption
= final momentum
We already agreed that the vector sum of the whole "thing" remains constant. Thus I insist in the momentum balance equation being applicable - without any modification. Selecting a single particle to judge on this is not possible.
Getting back to the weapon (the very simple case): do you think it is feasible to solely concentrate on the bullet - ignoring the weapon's recoil?
Hello U. Dreher
Sorry for a long gap touch with you. Now I am agree with your answer. Still I think a set of momentum balance equations can be written for individual particles on conserving the total momentum. I request the equations for your bullet- weapon system.
Practically I am very much interested on single particle analysis because I could not accept the plugging of flow of a fluid through a tube even beyond the sonic condition if there is consumption of internal source of energy in gain of momentum in the original direction of flow. If you have any idea in this respect please share with me.
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