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Inelastic Collisions

  1. Sep 7, 2004 #1
    Hi I have a simple question about inelastic collisions. Clearly energy cannot be conserved because when two things hit, they will create friction, noise, heat and other forms of nonconservative forces, but my question involves momentum. When we solve a problem we conserve momentum, in any collision. Is the reason why we can do this becuase we only conserve it for the instant that they come into contact. because in the case of a balistic pendululm, the thing eventually stops, so obviously momentum cannot be conserved throughout the entire trial of the experiment, at once point the pendulum reaches its max height and stops, so the momentum would be zero. The thing that gives me a little trouble is that energy is conserved immeditaly after the collision, so therefore, mechanical energy cannot be lost during this phase. That only leaves the instant when the two objects collide for energy to be lost. Well if this is the only point in which energy can be lost, how come it would not effect the momentum. It is usually,

    [tex] m_1v_1 = (m_1 + M_2) v_f [/tex] .

    But how come the final momentum is not lost due to the energy losses at that point. Ive kind of been doing circles around this in my reasoning for the last two days. arg.
     
  2. jcsd
  3. Sep 7, 2004 #2
    Momentum is conserved for a system of particles (like an object!) when the only forces are internal forces (forces that operate between the constituent particles). When the resultant 'external force' on the system is not zero the momentum changes as [itex]d \vec{p} /dt= \vec{F}_{ext}[/itex] according to Newtons second law. You might want to look up how the law of conservation of momentum is derived from Newtons laws.

    So this tells you why you can use conservation of momentum in case of collisions and not in the case of the pendulum. The external force there is nonzero, a gravitational force acts 'externally' on the pendulum. It is only when the earth and the pendulum are taken together as a system that momentum is conserved!
     
  4. Sep 7, 2004 #3
    But you can and do use momentum in a balisic pendulum. The reason you do so is becuase you can conserve momentum for the first part as the bullet embeds itself, and you can conserve energyfor the second half, as the pendulum bullet combination raises to some height.
     
  5. Sep 7, 2004 #4

    Doc Al

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    short collision time assumed

    In analyzing the ballistic pendulum, one applies momentum conservation to the collision of the bullet with the block. The simplifying assumption usually made is that the collision time is short enough to ignore any displacement of the block during the collision.
     
  6. Sep 7, 2004 #5
    I talked with my a physics professor and I think (hope) I got the anwser to my question. First of all, I was always given:

    [tex] m_1 v_1 = (M_2 + m_1 ) V_f [/tex],

    the problem that was hanging me up was how to derive this function, the reason being that nowhere in the momentum equation is the two masses combined, its always derived based on the principle that when the two masses hit, they both go off or one goes off for that matter, but they BOTH go somwhere, so you always have the sum of the product [tex] m_1 v_1f + m_2 v_2f [/tex] on the other side of the equation. I think I got the equation correct by saying that if they both move together in the end, then I can think of both masses moving with the same velocity, both magnitude and direction. so then I would get,

    [tex] m_1 v_1_i + m_2 v_2_i = m_1 v_f + m_2 v_f [/tex]. But since they both have the same value for the final velocity, then we can group the masses together, and we get, the basic momentum equation for inelastic collisions.

    [tex] m_1 v_1 + m_2 v_2= (m_1 + M_2) V_f [/tex]. Its kind of obvious and stupidly simple now that I got the equation, but I just wondered how to get it, becuase in my book were just told that " because momentum is conserved it is equal before and after". As a consequence of the way I did it, could I also think of the system as "technically" not being physically connected, meerly they could be side by side, without actually being tied together, and move with the same velocity. They would be in physical contact at some point, forever and ever, but not in the classical textbook sense of a thing of putty being stuck onto some other object.



    My second problem was another basic property of momentum I carelessly forgot. What Kept getting me caught up was the fact that I said to myself, as this bullet goes into the pendulum, its going to continually transfer momentum as long as its moving inside the block. So why can we just say it conserved momentum using that equation I described above, and why would it not be some integral of the momentum vectors from the point of contact until the point of being stuck and stopped inside the block. I think the reason we can just use the conservation of momentum at one point is that if we conserve the total momentum, just limiting ourselves to the case where it strikes the block, it would yeild the same result. I could transfer momentum piece by piece from the point of contact until the point it stops, OR, I could transfer ALL of that momentum in one shot, assuming that happens just at the point of contact. I guess in that sense, it would be as if the bullet hits the front of the block, stops dead, transfers all of its momemtum, and the sytem as a whole moves. Oppositely, I could picture it as what really happens, and that would be that the bullet hits the block, and slows down on the way into the block, and in the process, the block and bullet continually go up to speed with time. In the case where its all at that single instant, I guess it would be like saying right at impact it just goes to that slower speed of the block + bullet system at once. ( is that logic correct?)
     
    Last edited: Sep 7, 2004
  7. Sep 7, 2004 #6
    I guess my only question left now is why: friction, deforming the wood, sound, vibration,etc is not a problem with the momentum equation. I already know the anwser, because its an internal force! But Im trying to get a grip on why this would leave the momentum unaffected. As the bullet is traveling through the block, they will both have equal and opposite reactions to one another for any time interval, but at the same token, throughout that time losses will occur. Hmm any insight?
     
  8. Sep 8, 2004 #7

    Gza

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    If the bullet and the block are your system, all internal forces will cancel yielding a conservation of momentum (in the horizontal direction that is; gravity's influence in the vertical direction causes a change in momentum in that respective direction.) What you call "losses" of momentum, i'll more generally refer to as "changes" in momentum. Finding that there are no forces acting on the (bullet block) system in the horizontal direction means that there is no change of momentum in the horizontal direction. I remember these concepts to be a sticking point for me as well when I was learning classical mechanics. Just be persistent, and continue to think the way you are thinking and it will "click."
     
  9. Sep 8, 2004 #8
    Because of Newtons third law! To find the change in momentum you have to sum all forces, internal and external. But for every internal force (e.g. the force particle one excerts on particle two) there is a force with the same magnitude but in opposite direction (the force particle two excerts on particle one) according to Newtons third law. So these internal forces cancel leaving the momentum of the system in absence of external forces unchanged.

    For external forces the opposite 'third law force' does not act on an element of the system so is excluded, so momentum of the system is not conserved.
     
  10. Sep 8, 2004 #9
    This anwser does not explain internal friction. Heat loss due to friction would not explain for equal and opposite forces when they hit.
     
  11. Sep 8, 2004 #10

    Doc Al

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    Can you restate the exact issue that's troubling you? I'm not seeing it.

    What do you mean by "explain internal friction"? What you call "heat loss" is a transformation of energy from one form to another. The total momentum of the system (bullet + block) remains unaffected.
     
  12. Sep 8, 2004 #11
    Yes I agree with you on that total energy has to be conserved, I have thought about it some more and will try to ask it in a different approach and see if its right or not.
     
  13. Sep 8, 2004 #12
    When that bullet hits the very front edge of that block, it will most likely deliver the most force at that point. As time continues, assuming that the block does not move during the process, the bullet will keep on going inside the block until which point it has stopped. In the process of moving inside the block, I would assume that the force that the bullet imparts to the block will decay until which point the bullet is at rest relative to the inside of the block, and the force is now zero. Now as it is moving inside the block, there has to be tremendous energy losses.(friction, heat, vibration, sound etc.) Is this what causes the bullet to slow down and stop inside the block, or is it the block itself that causes the bullet to stop. Example, say the bullet could enter the block with absoulty NO mechanical energy losses, would it still stop inside the block, or is friction the key to making the bullet stop. I guess Im asking, what is the driving force that makes the bullet stop inside the block?
     
    Last edited: Sep 8, 2004
  14. Sep 9, 2004 #13
    Its the structural bonds of the block. Mechanical energy from the bullet is used to break up the structural bonds of the block which explains why there is a dent or a hole in the block. Pardon me, some of the mechanical energy is also being converted to heat, sound, etc. Net force could also be rewritten as dE/dx, where x is the distance travelled in the direction of the force and E is the energy of the object.

    As for the force, its just a normal reaction force. As observed, any two body in contact would experience a normal reaction force on one another.
     
  15. Sep 9, 2004 #14

    ehild

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    If two bodies collide there is no force between them until they touch. At the very moment as they touch, that is their molecules or atoms start to interact, there are forces, but these are of elastic nature at the beginning and cause deformation. The bodies will stop with respect to each other even these forces are conservative as the kinetic energy transforms into elastic energy. Like the balls do in the picture attached. That is an elastic collision. The blue arrows show the forces acting on both balls. Their kinetic energy gradually transforms into elastic energy and the balls stop with respect to each other for a moment but the forces acting on them start to accelerate they into the opposite direction. In the case of totally inelastic collision (imagine that both balls are made of soft clay), the deformation gets over the elastic limit, the tension gets relaxed, and the mechanical energy transforms into another kinds of energy. In the lack of elastic forces, there is no driving force to accelerate the balls backwards, so they stay in their sticked together state.
     
    Last edited: Jun 29, 2010
  16. Sep 9, 2004 #15

    Gza

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    How did you come up with that? You haven't defined what "energy" you are talking about.
     
  17. Sep 9, 2004 #16
    Maybe a Graph of what im talking about will help. See the funciton of the force of the bullet with time. ( I just made up a curve based on what I thought would actually happen in the situation.) At first the bullet will dilever the most force to the block, but as it slows down and stops, it wil dilever less and less force. But, I could also say it gives some average force during the same time interval. Here is question number one, does this not imply that I could also give some very big force, equal to all of the forces combined, over some infitesmaly short time interval, giving me the purple line on the graph. Question number two, as the grey force line decays with time, is this where we account for the losses due to friction, heat sound energy, vibration etc. Is it the above losses that is a consequence of our grey line decreasing and eventually becoming zero? In other words, is that where energy losses are taken into acount during the evaluation of the situation, and thus the reason why they show up as losses when we show that the ratio of starting energy to final energy is a great deal less in the end, or would that bullet stop and would the grey line slope down and stop even if there were NO energy loss to the system. My intuition tells me that there would have to be friction, becuase if I have a fast enough bullet, it would go right through the block and out the other end in susch short time that the block does not move very much, so I would think in that case that friction was too small to stop the bullet inside the block. Maybe what Im asking is a littler clearer now. Is the energy losses the reason the bullet stops and the force decays to zero.
     
    Last edited: Dec 26, 2005
  18. Sep 9, 2004 #17
    I think I found the flaw in my thinking, and a new understanding. If you look at the graph on my last post, you can see that clearly the force must be zero at the point in which it stops. I was thinking if that we increase the friction, then the bullet must stop at some point which is smaller than where it stopped initially. But the problem in my logic was that I assumed that the bullet would start with the same force value on every instant (x=0), and this would mean that friction played a role in momentum ( since less time would mean small x, and if F(x=0) were to be the same for every intial x=0 point of contact, and the force has to decay to zero until some point where it stops, then total momentum would change in each instance and depend on how much energy was lost due to the systems friction, because now the areas under the curves would not all be equal, they would be shorter and shoter as friction went up); however, we know this is not the case. So I guess this gives me a little insight into materials properties. Since momentum is conserved, the area under the force curve must be constant. If we are to assume that the internal forces in a material are uniform, then we can clearly see that for the area to be equal, materials with HIGH internal friciton forces will have overall HIGH forces at the instant of impact and to the point where it stops, meanwhile materials with Low internal forces will have smaller forces values during impact, but they will be stretched over longer periods of time. I was thinking that a moving bullet would give the same initial impact force no matter what it hits, since it has a constant mass/ area, but this is not what determines the impact force at that first instant x=0. Maybe now the question Ive been asking makes more sense to you, it was hard for me to explain it, since I wasent quite sure what I was asking also :-) sorry about that. I attached a graph of what im refering to. As you can see, if we started with the same F(x=0) force value in every case, but gradually increase friction so that the stopping distance shortened, the areay under the curve, and thus the momentum transfered would change, and be dependent on friction, this was what keept confusing me. But this should really look something more like the second graph attached, where friction increases, then the force values at F(x=0) increase as well.
     
    Last edited: Dec 26, 2005
  19. Sep 9, 2004 #18

    ehild

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    In the case of a bullet penetrating into a block and staying inside, the answer is "yes". This is a completely inelastic collision. The velocity of the colliding bodies is equal after collision. According to the conservation of momentum, [tex]
    mv_1=(M+m) v_f \rightarrow v_f=v_1\frac{m}{M+m}\mbox.[/tex] There is always some loss of mechanical energy in this case, as [tex]KE_1=0.5mv_1^2 \mbox{, and }
    KE_2=0.5(M+m)v_f^2=0.5mv_1^2\frac{m}{M+m}<KE_1\mbox. [/tex]

    In general, all colliding bodies stop with respect to each other for a very short time (except when they fly through each other). The bodies interact just during this short period, and the forces of their interaction can be either conservative or non-conservative. In the first case, the mechanical energy is conserved and the collision is said elastic. When the bodies stop moving with respect to each other their mechanical energy is stored as elastic energy, a kind of potential energy. This potential energy transforms back to KE. In the other case, when there are non-conservative forces during the interaction, some mechanical energy is lost and the overall KE of the bodies is less after the collision than before it.

    Imagine that the block is made of such a hard material that the bullet can not penetrate but bounces back from it. There is interaction for a very short time between the bullet and block, it might happen that there is no energy loss; it might happen that there is some. But there is a moment when the bullet touches the block, so it is stopped with respect to it even in the case when the mechanical energy is conserved.
    Imagine that the bullet is made of some very soft material, bubble gum or something like that. It will stick to the block, and move together with it. This is a completely inelastic collision, we have energy loss, and it is due to the deformation of the bullet.
    Now the third case when the bullet flies through the block and comes out at the other side. This is the only case when it never comes to rest with respect to the block. Some energy certainly is lost during its travel inside the block, and we don't know what will be the final velocity of the bullet and that of the block.

    ehild
     
  20. Sep 9, 2004 #19
    But is my assertion about the force that the bullet imparts to the material correct. For a given bullet colliding with a block, the force that the bullet exerts initially will depend on the material of the block. So say I shoot a mass of lead, it will stop inside the lead (assuming it does) say 2 inches, but the force will be very very very high during the time it traveled that 2 inches. Now, say we shot a bullet into an equal mass of wood, and it traveled 10 inches inside. Upon impact with the wood, the force at x=0 should be much much less than the case with the lead. Similarly, the force as it decays during that time interval should be shorter overall as well. But in each case, the sum of all the forces are equal. They both have the same area under the curve. I guess this would mean the bullet shot into lead would be smushed up at the tip due to the high forces, whereas in the wood, if we spilt the wood in half, we might be able to retreieve a nearly perfect bullet with little damage to it.
     
  21. Sep 10, 2004 #20

    ehild

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    We don't know much about the exact time dependence of the force of interaction. Its time integral is equal to the momentum exchanged. If this happened in a shorter time, the average interaction force was bigger.

    ehild
     
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