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The law of conservation of momentum says that the momentum sum of a closed system before an interaction of its objects remains the same after an interaction of its objects.

If an astronaut in space pushes another astronaut I would expect the momentum of the two together to remain the same after the push (because they're applying force to one another).

It is said that when a resting bomb sets off, the momentum sum of the dispersing shrapnel should add up to 0 (the same prior to the explosion). However, when a resting bomb sets off and shrapnel disperses to different directions, it's not these pieces of shrapnel that applied the necessary force to one another to allow them to disperse, it's the explosion. So why would I expect the momentum sum of the shrapnel to add up to 0? It should rather be the momentum of the shrapnel AND the momentum of the gas particles, because after all they are the ones who passed their momentum unto the shrapnel.

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  • $\begingroup$ "It is said that when a resting bomb sets off, the momentum sum of the dispersing shrapnel should add up to 0 (the same prior to the explosion)." Quoted from where? $\endgroup$ – BioPhysicist Apr 30 at 21:47
  • $\begingroup$ @BioPhysicist Well, I am currently studying Newtonian Mechanics from a book. The provided way to solve the problem assumes that the momentum of all of the shrapnel added together equals the momentum of the bomb before it goes off (i.e zero). $\endgroup$ – RandomUser Apr 30 at 21:56
  • $\begingroup$ " It should rather be the momentum of the shrapnel AND the momentum of the gas particles, because after all they are the ones who passed their momentum unto the shrapnel. " You answered your own question, this is what would be true. The other statement is not completely accurate and is at best an approximation $\endgroup$ – silverrahul May 1 at 8:38
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Well, I am currently studying Newtonian Mechanics from a book. The provided way to solve the problem assumes that the momentum of all of the shrapnel added together equals the momentum of the bomb before it goes off (i.e zero).

The book is probably making a simplifying assumption that the gas isn't shooting off in some direction. In other words, you are correct that we must have $p_\text{shrapnel}+p_\text{gas}=0$, but if we assume that after the explosion $p_\text{gas}=0$, then of course we must also have $p_\text{shrapnel}=0$ as well.

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The law of conservation of momentum says that the momentum sum of a closed system before an interaction of its objects remains the same after an interaction of its objects.

Not quite correct. A closed system, at least in thermodynamics, is one that cannot exchange mass with its surroundings. That doesn't preclude the possibility of external forces acting on the system changing its momentum.

If an astronaut in space pushes another astronaut I would expect the momentum of the two together to remain the same after the push (because they're applying force to one another).

Yes, as long as the astronaut is far enough away from any gravitational forces. Then the momentum of the center of mass of the two astronauts will be unchanged, even if the momentum of each astronaut changes.

It is said that when a resting bomb sets off, the momentum sum of the dispersing shrapnel should add up to 0 (the same prior to the explosion).

That is correct, provided the bomb is set off where there are no external forces in play, e.g, if the bomb is set off in outer space far from the effects of gravity.you ignore the external force of gravity.

However, when a resting bomb sets off and shrapnel disperses to different directions, it's not these pieces of shrapnel that applied the necessary force to one another to allow them to disperse, it's the explosion.

Yes, it is the conversion of the chemical potential energy of the bomb to the kinetic energy of the pieces (along with other forms of energy like heat and sound) that causes the pieces to disperse.

So why would I expect the momentum sum of the shrapnel to add up to 0? It should rather be the momentum of the shrapnel AND the momentum of the gas particles, because after all they are the ones who passed their momentum unto the shrapnel.

It is the momentum of the shrapnel and the gas, as both have mass, that sums to zero. Insofar as conservation of energy is concerned, the only missing part is the increase in temperature of the shrapnel parts and gas which, together with the kinetic energy of the shrapnel and gas, equals the chemical potential energy of the bomb before the explosion.

Hope this helps.

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