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An elastic collision in two dimensions

  1. Dec 6, 2008 #1
    1. The problem statement, all variables and given/known data
    http://hypertextbook.com/physics/mechanics/momentum-two-three/an-elastic-collision-in-2d.pdf


    2. Relevant equations
    P = MV
    P= (r,θ)
    p'=mv'
    m1v1 + m2v2=m1v1' +m2v2'

    3. The attempt at a solution
    (2.00kg)(13.42 m/s) + (1.00kg)(12.73 m/s) = (2.00kg)(v1')+ (2.00kg)(21.95 m/s)
    26.84 kgm/s + 12.73 kgm/s = 2.00kg(v1') + 43.9 kgm/s
    39.57 kgm/s = 2.00kg(v1') + 43.9 kgm/s
    -4.33 kgm/s = 2.00kg (v1')
    -2.165 m/s = v1'
     
  2. jcsd
  3. Dec 6, 2008 #2

    alphysicist

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    Hi Woopy,

    I don't believe this is true. It's not that the magnitudes of the momenta that are conserved; you have to find the components of the momenta.

    The total momentum in the x direction is conserved, and separately the total momentum of the y direction is conserved.
     
  4. Dec 6, 2008 #3
    How do I know what the Px and Py are before and after? This seems so overwhelming to me
     
    Last edited: Dec 6, 2008
  5. Dec 6, 2008 #4
    v1
    (2.00kg)(13.42 m/s)(cos 63.43)=12.00 kgm/s
    (2.00kg)(13.42 m/s)(sin 63.43)=24.00 kgm/s

    v2
    (1.00kg)(12.73 m/s)(cos 45)=9.00 kgm/s
    (1.00kg)(12.73 m/s)(sin 45)=9.00 kgm/s
     
  6. Dec 6, 2008 #5

    Doc Al

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    Staff: Mentor

  7. Dec 6, 2008 #6
  8. Dec 6, 2008 #7

    Doc Al

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  9. Dec 6, 2008 #8
    Theres a coordinate system, with v1 and v1' in the top left, v1 is 2.00kg and going 13.42 m/s, and is at 63.43degrees.

    v1' has an unknown velocity and unknown theta. and in the bottom right v2 is 1kg, 12.73 m/s and 45 degrees.

    v2' is also below, at 21.95 m/s and 59.93 degrees

    I have to find px, py, and k of the 2kg, 1kg and total, both before and after collision
     
  10. Dec 6, 2008 #9

    Doc Al

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    I didn't look over your numbers, but the only unknown in the diagram is the final speed and direction of the red particle. By setting up momentum conservation equations for the x and y directions, you should be able to solve for the components of that particle's final velocity.
     
  11. Dec 6, 2008 #10
    so like m1v1 + m2v2 = m1v1' + m2v2', and to get the velocity put cos/sin into equation for each term
     
  12. Dec 6, 2008 #11

    Doc Al

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    This is the initial Px and Py for the red particle. But the sign of Py is incorrect.

    This is the initial Px and Py for the green particle. But the sign of Px is incorrect.

    Compute the final Px & Py of the green particle in a similar manner.

    Then set up your two momentum conservation equations and solve for the final Px & Py for the red particle.
     
  13. Dec 6, 2008 #12
    something tells me that since its in another quadrant, the degrees arent as what they appear?? Like they should be - degrees or something
     
  14. Dec 6, 2008 #13
    and does "sign" mean sin? I don't understand
     
  15. Dec 6, 2008 #14
    oh, now I think I know what your saying. The sign is incorrect because they are in negative directions.
     
  16. Dec 6, 2008 #15

    Doc Al

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    Right.
     
  17. Dec 6, 2008 #16
    (1.00kg)(21.95 m/s)(cos 59.93)=11.00 kgm/s
    (1.00kg)(21.95 m/s)(sin 59.93)= -19.00 kgm/s
     
  18. Dec 6, 2008 #17

    Doc Al

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    OK. That's the final Px & Py for the green particle.
     
  19. Dec 6, 2008 #18
    (2.00kg)(13.42 m/s)(cos 63.43) + (1.00kg)(12.73 m/s)(cos 45) = (2.00kg)(v1')(cosθ) + (1.00kg)(21.95 m/s)(cos 59.93)

    12.00 kgm/s + 9.00 kgm/s = 2.00kg(v1')(cosθ) + 11.00 kgm/s
    21.00 kgm/s = 2.00kg(v1')(cosθ) + 11.00 kgm/s
    10.00 kgm/s = 2.00kg(v1')(cosθ)
    5.00 m/s = v1'(cosθ)
    now what?? theres 2 unknowns
     
  20. Dec 6, 2008 #19

    Doc Al

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    Instead of v1'(cosθ), call it v1'x.
     
  21. Dec 6, 2008 #20
    (2.00kg)(13.42 m/s)(sin 63.43) + (1.00kg)(12.73 m/s)(sin 45) = (2.00kg)(v1')(sinθ) + (1.00kg)(21.95 m/s)(sin 59.93)

    -24.00 kgm/s + -9.00 kgm/s = 2.00kg(v1')(sinθ) + 19.00 kgm/s
    -33.00 kgm/s = 2.00kg(v1')(sinθ) + 19.00 kgm/s
    -52.00 kgm/s = 2.00kg(v1')(sinθ)
    -26.00 kgm/s = (v1')(sinθ)
    2 unknowns!
     
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