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2D inelastic collision

  1. Apr 29, 2014 #1
    1. The problem statement, all variables and given/known data
    https://scontent-a-sjc.xx.fbcdn.net/hphotos-prn1/t1.0-9/603671_1409914319284787_3441607163537875560_n.jpg


    2. Relevant equations

    conservation of total momentum

    M1V1+M2V2 = MtotVf

    3. The attempt at a solution

    I tried using the COTM

    as such:

    x: M1V1iCOSΘ1+M2V2SINΘ2=MtotVf

    y: M1V1iSINΘ1+M2V2COSΘ2=MtotVf


    then dividing 1/2:

    COTΘ1 + TANΘ2 = 1

    Θ2 = TAN-1(1-COTΘ1)

    After plugging in the values im getting Θ2 = -10.85°

    What am I doing wrong?
     
  2. jcsd
  3. Apr 29, 2014 #2
    I think the main issue is in setting both the x and y equations equal to ##M_{tot}V_{f}##, since these are only component equations, so neither will equal the entire momentum. Instead, you'll want to use the components of the total momentum, so ##M_{tot}v_{f,x}## and ##M_{tot}v_{f,y}##
     
  4. Apr 29, 2014 #3
    I cant see that working because the ending vector is directly along the x-axis, there is no sin or cos involved with that vector.
     
  5. Apr 29, 2014 #4
    No, it works. The angle from the vertical will just be 0 degrees, so you'll have something like:

    ##P_{x} = P_{tot} \cdot cos(0)## and
    ##P_{y} = P_{tot} \cdot sin(0)##
     
  6. Apr 29, 2014 #5
    so the equation becomes TANΘ1+COTΘ2=0

    Θ2=COT-1(-TANθ1)

    Θ2= -.025°
     
  7. Apr 29, 2014 #6
    Well, you can't divide the equations like that.
    $$\frac{a + b}{c + d} \neq \frac{a}{c} + \frac{b}{d}$$
     
  8. Apr 29, 2014 #7
    Ok so before dividing I have

    x: M1V1iCOSθ1+M2V2iSINθ2=MtotVfCOS(0)

    y: M1V1iSINθ1+M2V2iCOSθ2=MtotVfSIN(0)

    look legit?
     
  9. Apr 30, 2014 #8
    Yes, that looks good. I think the easiest way to solve would be to move the components with ##M_{1}v_{1}## over, and then you could divide the equations.
     
  10. Apr 30, 2014 #9
    Hey there!

    $$V_{total} = \sqrt{V_{x}^2+V_{y}^2}$$

    and

    $$\theta = tan^{-1}(\frac{V_{y}}{V_{x}})$$

    These seem to be the only two equations that you are forgetting about. As long as you can solve for both component of your resultant velocities, you should be good!
     
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