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Question on Momentumby Yh Hoo
Tags: momentum 
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#1
Feb1013, 11:55 AM

P: 73

We know that momentum of a closed or isolated system is always conserved .
Now for INELASTIC collision, lets make It simple,consider a system with only two moving particles. If they collide inelastically, we know that energy is surely lost! now the system is no longer a closed system right?? The total momentum of d system must be decreasing right? However we could still use d principle of conservation of momentum equation is because during collision the impulsive force is totally exerted onto both colliding objects and is not to external factor right?? 


#2
Feb1013, 12:04 PM

P: 1,100

Inelastic means that kinetic energy is lost. Energy is still constant if the system is closed. Momentum is also conserved.



#3
Feb1013, 12:05 PM

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#4
Feb1213, 04:42 AM

P: 73

Question on Momentum
Now, I gt another question, for a closed.system, if inelastic collision where kinetic energy is lost occurs between particles, surely the total kinetic energy before must be less than after one right?? the kinetic energy might be lost to the system as other form of energy, bt in this case the total kinetic energy is no longer conserved right? ( even the total energy of the closed system is conserved. ) so accordingly the total momentum should also decreases right??



#5
Feb1213, 04:51 AM

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#6
Feb1213, 05:38 AM

Sci Advisor
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PF Gold
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#7
Feb1213, 10:21 AM

P: 73

Hmm, now there still seems like somethings is absent. It there a theoretical prove that total momentum is conserved also???



#8
Feb1213, 10:30 AM

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PF Gold
P: 12,256

'Proof' by virtue of loads of experimental evidence where results confirm that a model, based on conservation of momentum, works. It's one of the most basic of ideas that hasn't been shaken yet ifaik.
If you were to suggest a World in which momentum is not conserved the you would have to start pretty much from scratch and 'explain' all our observations in a different way. I assume that we are talking Newtonian Physics here? Let's not try to run before we can walk. 


#9
Feb1213, 10:30 AM

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PF Gold
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Zz. 


#10
Feb1213, 10:31 AM

P: 1,100




#11
Feb1213, 10:43 AM

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P: 11,864

Yes, if we assume that the laws of physics are invariant under spatial translation (loosely speaking, that the laws of physics don't depend on location), then Noether's Theorem can be used to prove that total linear momentum is conserved.
This of course begs the question, "how do we prove that the laws of physics are invariant under spatial translation?" In the end, all "proof" in physics is based on experimental observation and is never 100% absolute. That's why we continue to do more and more precise experiments to test the things that we think we know. Sometimes we get a surprise that leads us to new physics! 


#12
Feb1813, 06:42 PM

P: 72

Well, qualitatively one might say that all the molecules of the body collide during inelastic collision. If you could "see" them one by one you would see that momentum is conserved in each collision. Yet, the molecules' momenta are not constrained in the zdirection (direction of motion of the projectiles). Thus they will conserve momentum in all directions. Their collisions are chaotic, thus increasing the temperature of the system> heat is radiated.
One could say that the nr of molecules that go upwards after collision is equal to those going down. The momenta are roughly equal, thus the transverse momentum of the system after the collision is zero. In analogy, if a molecule with momentum p collides with another with p=0, in the final state, p will be distributed between the two. However their vector sum will always be p. Therefore the total momentum parallel to the projectile's motion will be conserved. The above two result in momentum conservation, but mechanical energy nonconservation. 


#13
Feb1813, 07:03 PM

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PF Gold
P: 12,256

That post is confusing momentum conservation with Equipartion of Energy arguments.



#14
Feb1813, 07:26 PM

P: 72

Yes it does!
But if you want "proof" about the conservation of momentum in an inelastic collision, I don't see how you would answer it elseway! In my point of view, the momenta gained after the collisions span in x,y,z , but the net effect is that they are cancelled out leaving the total momentum identical with the initial one. 


#15
Feb1813, 07:42 PM

P: 72

e.g if you assume a frictionless pools table, and the white ball stricking the others while beginning the game, the argument would be as follows:
Assume that the collisions between the balls are elastic. Assume that the white ball immediately stops after collision. The collision is perfectly central. The white ball's momentum is P. The rest of balls are scattered in all directions. However, the total transverse momentum of the balls going up is same as the transverse momentum of the balls going down (why shouldn't it be?). Thus these two components cancel out. What remains is the momentum parallel to the white ball's momentum. Each ball, j, carries some momentum p(j). If you add up these momenta you cannot find more than P , neither less because this would mean that there was kinetic energy lost which cannot be since collisions are assumed to be elastic. thus the p(1)+p(2)+....=P Now, if you take the example in the molecular scale, same goes for the molecules that collide when we have inelastic collisions. Transverse momenta are balanced due to the statistical nature of the effect. However, there is net amount of energy transfered to the transverse directions equal to E(j)=p(j)^2/2m(j), for each molecule. So the system's internal energy increases and there is heat radiated. The p^2 does the trick here. Doesn't this toy model make sense? 


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