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Elastic collisions

  1. Dec 17, 2012 #1
    1. The problem statement, all variables and given/known data

    A 2.0 kg ball moving with a speed of 3.0 m/s hits, elastically, an identical stationary ball. If the first ball moves away with an angle 30 degrees to the original path, determine the speed of the first ball after the collision, and the speed and direction of the second ball after the collision.

    2. Relevant equations

    v_1 = V_1cos30 + v_2cos(theta) for the movement in the "x" direction

    and

    0 = v_1cos30 + v_2cos(theta) for movement in the "y" directions


    3. The attempt at a solution

    Played with this for hours but to me it does not seem like there is enough information. I feel like I am missing at velocity for after the collision.

    Thanks
     
  2. jcsd
  3. Dec 17, 2012 #2

    tiny-tim

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    welcome to pf!

    hi doub! welcome to pf! :smile:
    an elastic collision is defined as one in which energy is conserved (as well as momentum, which is always conserved in collisions) :wink:
     
  4. Dec 17, 2012 #3
    Right,

    this is the best answer I got however I do not feel anywhere near confident.

    3.0 m/s = v_1'cos30 + v_2'cos(theta) = 2.598 m/s + v_2'cos(theta) = 3.0 m/s - 2.598 m/s
    so v_2'cos(theta) = 0.402 m/s in the "x" direction

    0 = v_1'sin30 + v_2'sin(theta) = 1.299 m/s + v_2'sin(theta). so v_2'sin(theta) = -1.299 m/s

    tan^-1 = v_2'cos(theta)/v_2'sin(theta) = -1.299/0402 = -72 degrees

    and using sqrt(v_2'sin(theta)^2 + v_2'cos(theta)^2 = 1.36 m/s

    am I anywhere in the ballpark at least?
     
  5. Dec 17, 2012 #4

    tiny-tim

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    hi doub! :smile:

    (try using the X2 button just above the Reply box :wink:)
    no, v1' isn't 3.0, it's unknown!

    try the momentum equations again (and you'll need an energy equation also) :smile:
     
  6. Dec 17, 2012 #5
    yeah I'm totally lost now
     
  7. Dec 17, 2012 #6

    tiny-tim

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    start again, with v1' v2' and θ as your three variables

    (you have three equations: x, y, and energy, so that should be solvable :wink:)

    show us what you get :smile:
     
  8. Dec 17, 2012 #7
    Ok,

    The equations I have gotten are

    x --> v_1 = v_1'cos30 + v_2'cos(theta)

    y --> 0 = v_1'sin30 + v_2'sin(theta)

    Energy --> v_1^2 = v_1'^2 + v_2'^2
     
  9. Dec 17, 2012 #8

    tiny-tim

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    fine so far :smile:

    now fiddle about with the first two equations so that cosθ and sinθ are on their own, then use cos2θ + sin2θ = 1 to eliminate θ :wink:
     
  10. Dec 17, 2012 #9
    So,

    cos(theta) = v1 - (v1'cos30)/v2'

    and

    sin(theta) = (-va'sin30)/v2'

    where do the sin2theta come from?
     
  11. Dec 17, 2012 #10

    tiny-tim

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    uhh? :confused:

    square both equations! :smile:
     
  12. Dec 17, 2012 #11
    I am just not seeing this...

    cos2(θ) = (v12 -v12'cos302)/v2'2

    sin2(θ) = (-v12'sin302)/v2'2

    thanks very much for helping btw
     
  13. Dec 17, 2012 #12

    tiny-tim

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    ok, now add

    θ will miraculously disappear! o:)

    :smile: pzzaaam!
     
  14. Dec 17, 2012 #13
    so if we add are we left with;

    (v12 -v12'cos302) + (-v12'sin302) / 2v2'2

    ?
     
  15. Dec 17, 2012 #14

    tiny-tim

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    actually, the RHS of that first equation should be (v1 -v1'cos30)2/v2'2
    how did you get that?

    where has the = sign gone? :confused:
     
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