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2-D elastic collision circle

  1. Nov 24, 2015 #1
    1. The problem statement, all variables and given/known data

    Hi there! In this exercise, we are supposed to derive this formula for a 2-D elastic with two different masses:
    (x-U*v1)^2 + y^2 = (Uv1)^2 (example, two billiard balls), the second mass is at rest. It's a equation which leads to a circle where all of the possible p2' lie on the circle.

    v1 is the velocity, x and y are the components of p2 vector (the momentum of the second object after collision ),and U= m1*m2/(m1+m2) is the reduced mass.

    2.
    Conversation of momentum in 2-D:

    p1 (vector) = p1' (vector) + p2' (vector) with p2 = 0 (vector)

    Conversation of kinetic energy:

    p1^2/2m1 = p1'^2/2m1 + p2'^2/2m2

    3. Attempt at solution

    Now I know how to get to this point: p1^2/(2m1) = (p1-x)+y^2/(2m1) + (x^2+y^2)/m2 (1)

    What I did, was to put my center of my frame of reference into the second object. There I can you some geometry to get p1' (momentum of object 1 after collision) = (p1-x) + y^2
    And p2' (momentum of object 2 after collision)= x^2+y^2

    I put those 2 into my conversation of kinetic energy equation which leads me to (1)


    Now the problem I face is, that I don't know how to get from equation (1) to the equation from the beginning. I know it is simple algebra, but I've been trying now for 1,5 hour and can't get there. Other sources skip the step from (1) to the equation and just say it is like that.

    Big thanks in advance for help!
     
  2. jcsd
  3. Nov 24, 2015 #2

    haruspex

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    That's not dimensionally consistent. Do you mean
    ((p1-x)^2+y^2)/(2m1)
    ?
     
  4. Nov 24, 2015 #3
    Yes , sorry for that, it's ((p1-x)^2+y^2)/(2m1) like you said
     
  5. Nov 24, 2015 #4

    haruspex

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    You have all those divisions by m1 and m2. In your target equation you have U instead. U involves the product of the two masses. What step does that suggest?
     
  6. Nov 24, 2015 #5
    I suppose that the m1m2 in U comes from adding the two fractions and the m1+m2 comes from dividing a m1+m2. I still don't get it though.
     
  7. Nov 24, 2015 #6

    haruspex

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    To arrive at the target equation, there's another alteration your equ 1 needs. Two of the terms have denominators 2m1, while the third has m2, no factor 2 in that case. That factor of 2 difference must be wrong. Please check your derivation of (1).
    Having corrected that, I asked you what step you should take to get rid of those m denominators. I'm looking for a very simple answer, nothing complicated.
    Another difference is that the target equation has a reference to v1, whereas (1) has a reference to p1. That's easily remedied too.
    Just making incremental steps to approach the form of the target will get you there.
     
  8. Nov 24, 2015 #7
    There is a factor 2 at m2. I don't see where it is wrong. (I guess you corrected it)
    And we'll you can write p1 to m1*v1. Look I tried all steps for hours now and I don't get it. I know that it is just simple steps but by now I would learn much more, if somebody could post the solution.
     
  9. Nov 25, 2015 #8

    haruspex

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    You wrote:
    I can derive the target equation from there if I first change it to ((p1-x)2+y2)/(2m1) + (x2+y2)/(2m2).
    Are you saying you agree with that version, or do you still have (x2+y2)/m2 as the final term?
    I know it's more work for you, but the way these forums operate is that you post your working and we try to spot where it goes wrong. So show us what you tried.
     
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