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If momentum is conserved in a perfectly inelastic collision (no lost to heat or defor

  1. Apr 22, 2012 #1
    Heat or deformation cannot contribute to velocity here, as per the view of conservation on momentum. So how is it that momentum is conserved but kinetic energy is not given a perfectly inelastic collision?
    The two masses stick together.
    There is no intrinsic means of expressing lost due to deformation or heating assuming a perfectly inelastic collision considering the view conservation of momentum. Therefore, given the same collision (perfectly inelastic) considering conservation of energy, there can be no transfer of energy via vibrations (or there would have to be as much in the latter view).
    Thus, the only remaining mechanism to account for the decrease of kinetic energy of the body in motion would be an increase in volume of the system. No body with a rest mass can have a zero volume. So, when mass increases volume must also? (holding pressure constant)
    There cannot be deformation in this scenario or heat transfer, because there is no force to oppose the change in velocity of block at rest. As soon as the bullet applies force on the block it accelerates.
    Help?
     
  2. jcsd
  3. Apr 22, 2012 #2
    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    The kinetic energy does go into heat and deformation.
     
  4. Apr 22, 2012 #3

    HallsofIvy

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    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    You say "If momentum is conserved in a perfectly inelastic collision (no lost to heat or defor" but you can't have a perfectly inelastic collision unless there is an increases in heat and/or deformation".
     
  5. Apr 22, 2012 #4
    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    The two masses stick together. There is no intrinsic means of expressing lost due to deformation or heating assuming a perfectly inelastic collision considering the view conservation of momentum. Therefore, given the same collision (perfectly inelastic) considering conservation of energy, there can be no transfer of energy via vibrations (or there would have to be as much in the latter view).

    Thus, the only remaining mechanism to account for the decrease of kinetic energy of the body in motion would be an increase in volume of the system. No body with a rest mass can have a zero volume. So, when mass increases volume must also? (holding pressure constant)
    There cannot be deformation in this scenario or heat transfer, because there is no force to oppose the change in velocity of block at rest. As soon as the bullet applies force on the block it accelerates.


    If the temperature, density, volume, and mass of the bullet and block are equal, by conservation of momentum one would expect v1i=2*vf.

    If total system energy were to remain constant Et=constant, and the Eti=KE1i+ui (KE2i = 0). Where ui is the total system internal energy. I say, given the problem constraints that

    KE1i+ui = KE1f+KE2f+uf, then

    delta u = uf -ui = KE1i-(KE1f+KE2f)

    The difference KE1i-(KE1f+KE2f) is positive, so delta u is positive which means uf>ui.
    Given

    delta u = delta Q + delta W

    in my scenario delta Q = zero, so delta u = delta W (which is what you alluded to earlier)
    I don't have the expertise to know how this delta W is realized, but I would argue (and have) that it is the result of a volume expansion.
    This answer uses the model that permits pi = pf for a perfectly inelastic condition and accounts for the decrease in KE.

    What do you think?
     
  6. Apr 23, 2012 #5
    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    What does this mean?!?

    It seams to me that, because you can't calculate the energy loss to heat and deformation,
    you just deny there is any such loss. This is not a valid argument. The temperature increase can also be easily measured.
     
  7. Apr 23, 2012 #6

    Dale

    Staff: Mentor

    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    In an inelastic collision there is always heat or deformation or other losses. It is self contradictory to posit an inelastic collision without them.
     
  8. Apr 23, 2012 #7
    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    The question posed is entirely theoretical, and I posed it to figure out mathematically how in a perfectly inelastic collision KEi>KEf. The constraints of the problem as I have posed them permit an accurate discussion of the mathematical assumptions, though most contributed to the discussion are missing the point entirely (trying to tell me how things vibrate and heat is exchanged).


    Please excuse me for not being clear, but pi=pf, lets say this scenario refers a bullet m1, and a block m2. So m1 *vi=(m1+m2) *vf, for a perfectly inelastic collision.

    In this model there is no method to account for heating, that is why it is called a perfect collision.

    I'll give you an example, if you will permit.

    m1=10 kg v1i= 10 m/s
    m2=10 kg v2i = 0

    pi = 100 kg m/s = pf = (10+10)*vf -> vf = 5 m/s (perfectly inelastic)

    This represents momentum's point of view. Heat doesn't contribute to the change in velocity (because the collision is perfect).

    Now from Energy's point of view.

    KEi = (1/2)(10)(10)^2 = 500 J

    KEf = (1/2)(20)(5)^2 = 250 J

    so KEi > KEf even in a purely mathematical model with no loss.


    The same collision should mathematically be equivalent no matter how you view it from a momentum point of view or energy point of view. If there is no heat exchange in our m1 *vi=(m1+m2) *vf model, then we can't model heating in our KEi>KEf model. Are you understanding what I am getting at here?

    I am not a physics major, so this topic is relatively new to me and firm believer in proof by mathematics. Most of this physical discussion contributors have attempted to exploit as model flaws, are missing the point that this is a mathematical model, not a physical one.

    So follow me here,

    For simplicity, the temperature, density, volume, and mass of the bullet and block are equal, by conservation of momentum one would expect v1i=2*vf. (Remember the block is hallowed out with a slot where the bullet fits)

    If total system energy were to remain constant Et=constant, and

    Eti = KE1i + ui (KE2i = 0).

    Where ui is the total system internal energy.

    I say, given the problem constraints that

    KE1i+ui = KE1f+KE2f+uf, then

    delta u = uf -ui = KE1i-(KE1f+KE2f)

    The difference KE1i-(KE1f+KE2f) is positive, so delta u is positive which means uf>ui.

    Given

    delta u = delta Q + delta W (thermodynamics)

    in my scenario delta Q = zero, so delta u = delta W

    I don't have the expertise to know how this delta W is realized, but I would argue (and have) that it is the result of a volume expansion.

    Now is it more clear?
     
  9. Apr 23, 2012 #8
    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    Your scenario is not physical.
    The problem is that mathematical proof only works if you postulate some physical principles. Here, the principles are that energy and momentum are conserved. From that, you mathematically show that in an inelastic collision, there must be heat/vibrations transferred.

    If you postulate that no heat is transferred, you find that momentum is not conserved, but that has nothing to do with physics. It will not happen. Math and physics are two different fields.

    You are missing their points: your constraints do NOT permit a discussion of physics.
     
  10. Apr 23, 2012 #9
    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    Would you agree that if there were any vibration/heat/deformation in the collision under discussion that vf would be less than predicted given m1vi=(m1+m2)vf?
     
  11. Apr 23, 2012 #10
    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    There will be vib/heat/def and the final velocity of the combined mass will be less. Correct. The part that is not physical is requiring that there is no vib/heat/def.
     
  12. Apr 23, 2012 #11
    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    So for a mathematical model, how can you defend supporting a perfect inelastic collision from the point of view of momentum, but not from the point of view of Energy?

    m1*vi=(m1+m2)*vf only works perfectly if there is no heat/vibration and is a purely mathematical model.

    Given the same collision from the point of Energy, I am saying we have to make the same assumptions no heat/vibration.

    What say you?
     
  13. Apr 23, 2012 #12

    Doc Al

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    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    A 'perfectly inelastic collision' just means that the colliding bodies stick together. (Sure, in any real collision there will be some sound or light, which will may remove momentum or energy from the system. But that can be a small correction.) So momentum conservation is a very reasonable assumption. Using that alone you can easily see that mechanical energy is not conserved.

    Assuming no heat/vibration has nothing whatsoever to do with it being a perfectly inelastic collision. As already pointed out, that would create an impossible situation.
     
  14. Apr 23, 2012 #13
    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    So given
    m1=10 kg v1i= 10 m/s
    m2=10 kg v2i = 0

    pi = 100 kg m/s = pf = (10+10)*vf -> vf = 5 m/s (perfectly inelastic)

    Now from Energy's point of view.

    KEi = (1/2)(10)(10)^2 = 500 J

    KEf = (1/2)(20)(5)^2 = 250 J

    so KEi > KEf

    You are saying that the decrease in KE is totally due to heat?
     
  15. Apr 23, 2012 #14

    Doc Al

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    Staff: Mentor

    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    I would say that the only way that two such objects can collide and stick together is if some of their translational KE is transformed into internal energy--in this case random thermal energy and deformation.

    Another way to think of a perfectly inelastic collision is to view things from the center of mass frame. After the collision, the total translational KE in the center of mass frame is zero. (A perfectly inelastic collision is one in which the maximum amount of translational KE is transformed into internal energy.)
     
  16. Apr 23, 2012 #15
    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    Would you mind elaborating on the center of mass frame?

    BTW, Doc Al, I have been discussing this topic for nearly 24 hours online in different forums. Your explanation is the most satisfying thus far.
     
    Last edited: Apr 23, 2012
  17. Apr 23, 2012 #16

    Doc Al

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    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    View the collision from a frame of reference in which the center of mass is at rest (one that moves along with the center of mass). That is also called the center of momentum frame; it's a frame in which the momentum is zero.

    In your example, the center of mass moves with speed +5 m/s. So, in that frame, m1 moves with speed +5 m/s and m2 moves with speed -5 m/s. After the collision nothing is moving.
     
  18. Apr 23, 2012 #17
    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    The kinetic energy lost in an inelastic collision does not have to solely end up as heat, it can also end up as potential energy (or an increased mass of the system according to Einstein). Consider two large blocks flying at each other on a frictionless surface, and one has a large spring on its side situated so as to be in between the two blocks when they collide. There is also a latch that falls down from the top of one block and locks them together once they get close enough. The blocks collide, the spring compresses, the latch falls, so they remained joined. This is a perfectly inelastic collision. A large part of the lost kinetic energy now resides as potential energy in the spring. Lift the latch and the spring shoots the blocks out, changing the potential energy back to kinetic.
     
  19. Apr 23, 2012 #18

    Dale

    Staff: Mentor

    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    Your calculations are all correct. The decrease in KE doesn't all have to go to heat, it can in principle go to any form of internal energy, but most of those other forms will themselves eventually go to heat.
     
  20. Apr 23, 2012 #19
    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    I think one of the issues is how we're allowed to assume there is no deformations with an elastic collision, yet with an inelastic we are not allowed to assume no deformation.
     
  21. May 1, 2012 #20
    Re: If momentum is conserved in a perfectly inelastic collision (no lost to heat or d

    In an inelastic collision, KE is always lost as heat and the body gets deformed.
     
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