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Elastic collision of three balls

  1. Nov 25, 2014 #1
    1. The problem statement, all variables and given/known data
    The first ball collides with ##v_1## collides with the second two as shown in the picture. All balls are identical and the second balls have no speed at the beginning.
    Find the transform matrix and find the covariation matrix after the collision if it is ##
    \begin{bmatrix}
    \sigma ^2 & 0\\
    0 & 0
    \end{bmatrix}## before the collision.
    Captureg.PNG

    2. Relevant equations


    3. The attempt at a solution
    The momentum is conserved: $$mv_1=2m{v_2}'cos\varphi$$ where luckily the balls are identical, therfore ##cos\varphi =\frac{1}{\sqrt 2}##. And also kinetic energy is conserved $$\frac 1 2 v_1^2={v_2}'^2$$ This brings me to transform matrix $$
    \begin{bmatrix}
    {v_1}'\\
    {v_2}'
    \end{bmatrix}=A\begin{bmatrix}
    {v_1}\\
    {v_2}
    \end{bmatrix}=\begin{bmatrix}
    0 & 0\\
    0 & \frac{1}{\sqrt 2}
    \end{bmatrix}\begin{bmatrix}
    {v_1}\\
    {v_2}
    \end{bmatrix}$$ and finally also to covariance matrix $${M}'=AM_0A^T=
    \begin{bmatrix}
    0 &0 \\
    0&0
    \end{bmatrix}$$ The last step was done using mathematica. And the most confusing part here is that all the components are completely correlated - I am having some troubles to interpret this absolute correlation. Could somebody help me with that? It is also possible that my result is completely wrong in that case, please let me know.
     
  2. jcsd
  3. Nov 25, 2014 #2

    mfb

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    Staff: Mentor

    Why?
    And why do you think ball 1 stops?

    How could they be independent (and independent of what)?
     
  4. Nov 25, 2014 #3
    How would I then find ##\varphi ##?

    If ball 1 doesn't stop, than from the conservation of momentum $$v_1=2{v_2}'cos\varphi + {v_1}'$$and from conservation of kinetic energy$$v_1^2=2{v_2}'^2+{v_1}'^2$$

    Now I have two equations for three unknown parameters ##{v_1}'##,##{v_2}'## and ##\varphi ##.
     
    Last edited: Nov 25, 2014
  5. Nov 25, 2014 #4
    Should I work in the CMS system?

    ##u_1=v_1+v_{CMS}## and ##u_2=v_2-v_{CMS}=-v_{CMS}##. Conservation of momentum than brings me to ##mu_1-mu_2=0## meaning ##v_{CMS}=-\frac{v_1}{2}## and conservation of energy to ##u_1^2=2{u_2}'^2+{u_1}'^2=3{u_1}'^2##.

    Now ##{u_1}'=\sqrt 3({v_1}'+v_{CMS})=u_1=v_1+_{CMS}## mening $${v_1}'=v_1\frac{1+\sqrt 3}{2\sqrt 3}$$ and similar for $${v_2}'=v_1\frac{\sqrt 3 -1}{2\sqrt 3}$$.

    Would this be ok?
     
  6. Nov 25, 2014 #5

    mfb

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    Momentum transfer is along the line "contact point <-> center of balls".

    You can get the angle from the geometry of the problem.

    I would not go to the cms. Also, your formulas look wrong there.
     
  7. Nov 25, 2014 #6
    Hmmm, from the geometry?

    So, since the balls are identical, I can say that ##\varphi = \pi /4## ?
     
  8. Nov 25, 2014 #7

    mfb

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    Why pi/4? Where does that number come from?

    Yes from geometry, see the first part of my post.
     
  9. Nov 25, 2014 #8
    Blaaah, that's a mistake, I mean't to write ##\varphi =\pi /3##
     
  10. Nov 25, 2014 #9

    mfb

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    Keeping guessing numbers does not make it better. Sure eventually you'll hit the right one (not yet), but that is not a solution.
     
  11. Nov 25, 2014 #10
    I wasn't guessing, the geometry is clear. Equilateral triangle with all three sides ##2R## if ##R## is the radius of each ball.

    The angles in the equilateral triangle are ##\pi /3##. However, you are right, If I wouldn't be so careless I would see that the angle is actually half of that, therefore ##\pi /6##.
     
  12. Nov 25, 2014 #11

    mfb

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    Right.
     
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