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Proof on two dimensional elastic collision.

  1. Feb 29, 2008 #1
    When a ball with mass m collide with another ball with equal mass as m=> m1 at rest, the mathematical proof shows that after the collision, the angle between two masses would be 90'

    I know that when m> m1 , angle <90'
    when m<m1 , angle >90'

    but there is no mathematical proof to show this result....
    does anyone help me solve this problem?
  2. jcsd
  3. Feb 29, 2008 #2

    Doc Al

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    Staff: Mentor

    Can you prove the first result? (For equal masses.)
  4. Feb 29, 2008 #3
    making use of KE conserved and momentum conserved...

    mv=m1v1cosx+mv2cosy- (1)
    0=m1v1sinx-mv2siny - (2)

    v^2 = (v1)^2+(v2)^2 +2v1v(cosxcosy-sinysinx)

    KE conserved
    mv^2 =m1v1^2+mv2^2
    cos(x+y) = 90'
  5. Feb 29, 2008 #4

    Doc Al

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    Staff: Mentor

    Good. Now redo it without canceling the masses. You'll get an expression for cos(x+y) that will depend on m - m1.
  6. Feb 29, 2008 #5
    Thanks ! I've proven that. :)
  7. Apr 26, 2011 #6
    Could you please describe the work in more detail? I understand the entire concept entirely, and even matched up with a lot of the math you did, but I didn't understand the combining you did in the middle and how the angle 90 just appeared out of nowhere.
  8. Apr 26, 2011 #7
    I understand everything except what is in bold. First, you are finding the momentum along the X. Then finding the momentum along the Y. Then somehow combine them in the equation for the conservation of energy. I have all of that already, but I don't understand the combining itself and don't understand the bolded areas
  9. Apr 27, 2011 #8
    NvM. I got it. Law of cosines and sum difference formulas.

    Ø1 + ø2 = 90,

    so it is safe to assume that :

    cos(Ø1 + ø2) = 0

    since cos(90) = 0. Now, for the actual proof, start by using the formula for conservation of kinetic energy:

    K1i + K2i = K1f + K2f.

    After simplifying, you will get:

    V1i^2 = V1f^2 + V2f^2.

    According to law of cosines,

    c^2 = a^2 + b^2 -2(ab)cosC.

    CosC is The same value as cos(ø1 + ø2), so

    c^2 = a^2 + b^2 -2(ab)cos
  10. Apr 27, 2011 #9

    Doc Al

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    Staff: Mentor

    Good! (Sorry I didn't have time to respond earlier.)
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