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B Transfer of Momentum during a collision

  1. Jun 20, 2017 #1
    For an inelastic impact situation where two bodies travelling in opposite directions (vehicles say) collide and coalesce (perfectly inelastic collision) , one can derive the following expression for equal and opposite transfer of momentum. On the basis that if impact forces are equal and opposite, so too is the transfer of momentum.

    Δp =m1m2(v2-v1) /(m1+m2)

    In passing it's interesting to note that the quantity:

    m1m2 /(m1+m2)

    also occurs in the calculation of reduced mass.

    But that's an aside - my question is can one derive a similar expression for momentum transfer in the case of a simple elastic collision? In terms of the pre-collision parameters.
     
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  3. Jun 20, 2017 #2

    Orodruin

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    For a one-dimensional collision, yes. For a general two- or three-dimensional collision, the momentum transfer will depend on the scattering angle. (There is no scattering angle in a 1D collision.)
     
  4. Jun 20, 2017 #3

    sophiecentaur

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    Whatever the coefficient of restitution happens to be, Momentum is conserved and, in a one dimensional situation, you can use the same sort of calculation for elastic as for inelastic collisions. The resulting motion of the CM of the masses is unchanged, either way. Having done this sort of thing to death in A level Maths at School, I don't feel like doing the sums all over again but it really isn't very hard. Just do a before and after, with the parting velocity equal and opposite to the approaching velocity. Solve a few simultaneous equations and get what you want out of it.
    Just a few extra equations and it will all come out relatively easily if you avoid mistakes in the manipulation. (Always a big risk for me!!)
     
  5. Jun 20, 2017 #4

    Orodruin

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    Not if you do not have the scattering angle or other equivalent information. This is necessary information to find the momentum transfer.
     
  6. Jun 20, 2017 #5

    sophiecentaur

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    Of course. You have know the angles in a two dimensional situation. I though that was taken as read. There is also the issue of angular momentum if the bodies are of significant size. But where do you stop? Starting with billiard balls is a good way to find your feet.
     
  7. Jun 20, 2017 #6

    Orodruin

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    The OP asked if you could find the momentum transfer in an expression using just the in state variables:
    The scattering angle is a post-collision parameter.
     
  8. Jun 20, 2017 #7
    Thanks very much for the answers - I will work at the elastic collision equation and report back!

    Re starting with billiard balls - yes indeed! You could even do the duck meets Jumbo jet problem very easily with the above equation which for m2 >> m1 simplifies to m1(v2 - v1). In effect the momentum transfer is as if the duck flew straight into a stationary Jumbo jet with a velocity of its own plus the flying velocity of the Jumbo jet. Divide that by a very small time interval and the impact force is of the order of a ton weight!
     
  9. Jun 21, 2017 #8
    For a one-dimensional perfectly elastic collision I determine the momentum transfer as:

    Δp =2m1m2(v2-v1) /(m1+m2)

    Does that look right ? Brief check would be bouncing ball with m2 infinite and v2 = 0. Then Δp = -2m1v1 as expected.

    Therefore may we conclude that in a one-dimensional collision between two bodies the transfer of momentum is proportional to the product of reduced mass of the two bodies and their relative velocity ?
     
    Last edited: Jun 21, 2017
  10. Jun 21, 2017 #9

    Orodruin

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    Yes. The easiest (regardless of dimension) is to consider the CoM frame. In this frame, the speeds of the objects remain unchanged (their velocity change though). This gives the momentum transfer ##\mu \Delta v \sqrt{2(1-\cos\theta)}## (assuming I did the math correctly in my head - note that there is a momentum exchange in the original direction as well as in the direction orthogonal to it, the former proportional to ##1-\cos\theta## and the latter to ##\sin\theta##), where ##\mu## is the reduced mass, ##\Delta v## the relative speed, and ##\theta## the scattering angle in the CoM frame.
     
  11. Jun 22, 2017 #10
    Many thanks for the confirmation - "CoM frame" and "scattering angle" are not concepts I am familiar with unfortunately although I would note your equation can be written as:

    2μΔv.sin(θ/2)

    which is similar to what I had except with a collision angle parameter that presumably relates to 'scattering angle'. I will dig up some billiard ball examples and see how this applies. Albeit vector arithmetic also a bit rusty on my side!

    For the record, here are the equations I solved for the elastic collision case. Subtracting p1 from the expression for p3 gives the formula in the post above.
     
    Last edited: Jun 22, 2017
  12. Jun 22, 2017 #11

    Orodruin

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    The CoM (centre of momentum) frame is a frame where the total momentum is zero. It is often preferable to do the computations in this frame. The scattering angle is the angle between the initial direction of the incoming object and the outgoing direction of the same object.
     
  13. Jun 22, 2017 #12

    sophiecentaur

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    Unfortunately, Scattering Angle is not very useful if you want to work out 'what happens' in a collision, given the initial geometry. Unless the bodies are spherical and frictionless, it will soon get pretty much impossible to calculate. A glancing contact can cause a lot of the Energy to be transferred to rotational of both bodies. That's the advantage of billiard balls. All the forces can be considered to act along the line of centres. Very well suited to A level..
     
  14. Jun 22, 2017 #13

    Orodruin

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    Of course not, but that was not the question. The question was "can you determine the momentum transfer from the pre-collision parameters?", which I took to mean the very same parameters as quoted by the OP. The answer to this question is: No. If you have more data you might be able to in classical mechanics.

    In addition, scattering angle is very useful in the description of what happens in a collision at the quantum level. I know that this particular case is not QM or QFT based, but I think calling the scattering angle "not very useful" is a sweeping generic statement that is not universally true.
     
  15. Jun 22, 2017 #14

    sophiecentaur

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    What I wrote was:
    That was not "sweeping" but very specific. The OP was about "e.g. vehicles". Given that, how can you object to my comment? Problems involving Scattering Angle are rare in everyday life, I think. It's a bit back to front for most situations.
     
  16. Jun 22, 2017 #15
    Can we use your formula to solve this problem which cropped up elsewhere in the forum:

    Consider an elastic collision (ignoring friction and rotational motion).
    A queue ball initially moving at 2.6 m/s strikes a stationary eight ball of the same size
    and mass. After the collision, the queue ball’s final speed is 1.2 m/s.
    Find the queue ball’s angle (theta) with respect to its original line of motion.
    Answer in units of ◦.

    I tried but didn't seem to get far - I would guess I am misunderstanding how the formula should be applied.
     
  17. Jun 22, 2017 #16

    sophiecentaur

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    Ideal case: Draw a line along the line of centres. Initial and final velocity components of both balls will be unchanged at right angles to that line. The difference in velocity components along the line will be multiplied by the coefficient of restitution (less than or equal to 1). You can do a (one dimensional) momentum calculation along the line of centres to get the resulting velocity components. You then add the components of velocity for each ball to find the resulting velocities of the two balls.
    This link deals with equal weight balls but you can carry on with your own calculation using different masses.
     
  18. Jun 22, 2017 #17
    I am getting correct answers using the formula 2μΔv.sin(θ) rather than 2μΔv.sin(θ/2) for momentum transfer! So I am not sure where the half angle came in ? I solved:

    2μΔv.sin(θ) = m√(2.6^2-1.2^2). Right hand side based on conservation of kinetic energy in elastic collision.

    Given that the masses were all equal, this simplified to: Δv.sin(θ)=√(2.6^2-1.2^2). With Δv = 2.6.

    Eureka (I think - well at least it seems to give the same answers as per the forum example here.)!
     
  19. Jun 23, 2017 #18
    I think this is why your formula has a half angle in it , whereas mine does not. In effect the reference frame I am using to solve the problem above is one in which the second object is stationary. I would guess the CoM reference would apportion half of the cue-ball momentum to the 8-ball. It makes sense since then you only have to deal with Δp as per formula. Just have to deal with the mechanics of changing reference frame.
     
  20. Jun 23, 2017 #19
    Well now you won't need to any more - all you have to do is use the formula: Δp = (1+e).μ.Δv and Bob's your uncle.
     
  21. Jul 3, 2017 #20
    A useful addition to the above is to calculate energy of two masses following a collision by solving (in the case of an elastic collision):

    Conservation of energy: e1 + e2 = e3 + e4
    Conservation of momentum: √(2m1*e1) + √(2m2*e2) = √(2m1*e3) + √(2m2*e4) which simplifies on removal of √2:
    √(m1*e1) + √(m2*e2) = √(m1*e3) + √(m2*e4)

    Send this off to the computational engine and we obtain these solutions. Note I have used the letter "a" for energy rather than e since the engine choked on e (perhaps confusing with natural log base e ?).

    Well they look algebraically complex but in the case of a small mass m1 striking a larger one m2 at rest, the answers simplify considerably:

    e3 = e1 * [(m2 - m1)/(m2+m1)]^2 and e4 = e1 * (1 - [(m2 - m1)/(m2+m1)]^2)

    It will be noted than in the case of m2 being infinitely large (perfectly elastic bouncing ball), e3 = e1 and e4 = 0 (no energy transfer).
     
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