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Solving for two unknowns in a 2-D elastic collision. HELP!

  1. Nov 6, 2004 #1
    Hey everyone.

    I am really stuck on a problem that seems simple but I just can't figure it out. The problem goes as follows.

    There is a collision between two pucks on an air-hockey table. Puck A has a mass of 0.039 kg and is moving along the x axis with a velocity of +5.5 m/s. It makes a collision with puck B, which has a mass of 0.064 kg and is initially at rest. After the collision, the two pucks fly apart with the angles of 65 deg for A and 37 deg for B.

    I know this problem is really basic but i just can't figure it out unless I have one of the final speeds. Despite all of my work I will still end up with 2 variables. I have checked a bunch of examples but every single one them gives the final speed for one the pucks and the TA at my help session doesn't really give any hints as to how to solve this. Any help will be greatly appreciated.

    Thank You
     
  2. jcsd
  3. Nov 6, 2004 #2

    Tide

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    Did you apply energy conservation?
     
  4. Nov 6, 2004 #3
    Yes I did let me show you my work:

    X component:
    (M1 x VAi) + (M2 x VBi) = (M1 x VAf cos theta) + (M1 x VBf cos theta)

    Y component is the same except for cos is replaced by sin and negative VBf.

    So I got this for X component:

    (0.039 x 5.5) + (0.064 x 0) = (0.039 x VAf cos 65) + (0.064 x VBf cos 37)

    and this for Y component:

    (0.039 x 5.5) + (0.064 x 0) = (0.039 x VAf sin 65) + (0.064 x - VBf sin 37)

    The intial side of the equation is easily solved but that leaves two unknowns on the final side causing my brain to hurt.
     
  5. Nov 6, 2004 #4

    Pyrrhus

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    You used conservation of momentum, but not conservation of kinetic energy (In elastic collision kinetic energy is conserved).
     
  6. Nov 6, 2004 #5
    I tried using the conservation of kinetic energy but it still leaves me with two unknowns on one side.

    since Ki=Kf

    Ki = (1/2 x M1 x VAi^2) + (1/2 x M1 x VBi^2) = (1/2 x M1 x VAf^2) + (1/2 x M1 x VBf^2) = Kf

    So if I plug in my values it solves the initial side but not the final side because of the two unknown variables.
     
  7. Nov 6, 2004 #6
    Anybody have an idea on how to solve this?
     
  8. Nov 7, 2004 #7
    Hmm... I guess it can't be solved since there are two unknown variables unless I am forgetting something fairly obvious. (or maybe its too hard for all you physics masters :biggrin: )

    Thanks anyway.
     
    Last edited: Nov 8, 2004
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