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2D elastic collision

  1. Nov 24, 2011 #1
    1. The problem statement, all variables and given/known data
    Object 1 with velocity v hits object 2, of same mass at velocity 2. Prove that the angle between them after the collision is equal to 90 degrees.


    2. Relevant equations
    PE inital = PE final
    KE initial = KE final


    3. The attempt at a solution

    P x component
    m1 x v1 + 0 = (m1)(v'1)(costheta1) + (m1)(v'2)(costheta2)

    p y comp
    0 = (m1)(v'1)(sintheta1) + (m1)(v'2)(sintheta2)

    kinetic E equation simplifies to
    v1 - v2 = v'2 - v'1

    I can't figure out how to combine the three to a point where they simplify.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 24, 2011 #2

    vela

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    If I recall correctly, this relationship only holds for elastic collisions in one dimension.

    Try squaring each of the two equations you have and then add them together. Then use the fact that the collision is elastic to eliminate some of the terms.
     
  4. Nov 25, 2011 #3
    okay, so i tried that and I'm still stuck. Here's what I did: (' meaning after collision)

    V1x = v1'cos1 + v2'cos2
    V1y = v1'sin1 + v2'sin2

    V^2 = V1x^2 + V2y^2
    (v1'cos1)^2 + 2v1'v2'cos1cos2 + (v2'cos2)^2 + (v1'sin1)^2 + 2v1'v2'sin1sin2 + (v2'sin2)^2

    From kinetic energy:
    V^2 = v1'^2 + v2'^2
    (v1' = v1'cos1 + v1'sin1
    v2' = v2'cos2 + v2sin2)
    (v1'cos1)^2 + 2v1v1cos1sin1 + (v1sin1)^2 + (v2cos2)^2 + 2v2v2cos2sin2 + (v2sin2)^2

    Setting the momentum equation to the kinetic energy equation and simplifying leaves me with:

    v1v2(cos1cos2 + sin1sin2) = v1v1cos1sin1 + v2v2cos2sin2

    using trig identities:
    1. sinusinv = 1/2[cos(u-v) - cos(u+v)] and
    2. cosucosv = 1/2[cos(u-v) + cos (u+v)] results in the left side of the equation simplifying to:
    cos(1-2)v1v2 = v1v1cos1sin1 + v2v2cos2sin2

    i don't know what to do with this equation now.
     
  5. Nov 25, 2011 #4

    vela

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    You're not squaring the velocities correctly. There is no cross term with the x and y components because they are orthogonal. In other words,
    [tex]\vec{v} = v\cos\theta\,\hat{i} + v\sin\theta\,\hat{j},[/tex] so when you square this vector, you get
    [tex]v^2 = \vec{v}\cdot\vec{v} = (v \cos\theta)^2 + (v\sin\theta)^2.[/tex] You don't get [itex](v\cos\theta+v\sin \theta)^2[/itex].
     
  6. Nov 25, 2011 #5
    I'm confused. I think I'm over complicating this problem. So that equation you just gave me, comes from the kinetic energy equation right? But is it v1cos1 or v2cos2? And where does momentum play into this? Do I really need to break it up into the x and y components?
     
  7. Nov 25, 2011 #6

    vela

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    What I'm saying is this part is wrong:
    If an object is moving 3 m/s in the x-direction and 4 m/s in the y-direction, its speed isn't 8 m/s. It's 5 m/s. You don't simply (scalar) add the individual components to get the speed.

    Therefore, the following is wrong
    The cross terms shouldn't be there.
     
  8. Nov 25, 2011 #7
    I now understand why my first approach was wrong, but I don't know how to start this problem.
     
  9. Nov 25, 2011 #8

    vela

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    You're almost there. Just fix your mistake.
     
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