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Inelastic collision of two particles

  1. May 8, 2014 #1
    1. The problem statement, all variables and given/known data

    After a completely inelastic collision, two particles of the same mass m and same initial speed v are found to move away together with the speed (2/3)v. Find the angle between the initial velocity vectors of the objects.

    3. The attempt at a solution

    It is completely inelastic which means the two objects stick together and share the same final speed. Energy is not conserved but Momentum is so
    mv + mv = (m+m)V

    2mv = 2m(2v/3)
    2mv=4mv/3
    2=4/3

    I definitely did something wrong here
     
  2. jcsd
  3. May 8, 2014 #2
    You need components I am pretty sure, because it is asking for angles. So makes your x components of momentum, and y components of momentum. So what I am saying is that your question asks for an angle, so that they collide at an angle, so you would need two equations one for everything in the y direction, and the other everything in the x direction.
     
    Last edited: May 8, 2014
  4. May 8, 2014 #3
    Should there be two different angles I am looking for here or are they equal? Meaning if I put the final velocity vector of the two particles along the x axis, is the angle in between the x axis and the particle above the x axis equal to the angle between the x axis and the particle below the x axis?
     
  5. May 8, 2014 #4
    So you are kind of on the right track, there is one initial angle θ and for simplicities sake, lets say one of the particles is moving perpendicular to the x axis, and the other is moving at an angle θ from the x axis, and lets say that the new angle formed when they collide is θ'. With this information you should be able to create two different equations, and the mass's and velocities should divide out, then you have two unknowns you could easily solve for.
     
  6. May 9, 2014 #5
    attachment.php?attachmentid=69576&d=1399616444.png

    Is this an accurate representation of what is going on? (my terrible drawing skills aside) the green and red vectors are the two particles before the collision and the blue one with angle θ' is the two of them moving while stuck together.
     

    Attached Files:

  7. May 9, 2014 #6
    I am ending up with an equation for the x components:

    vcosθ = (4/3)vcosθ'

    and for the y components:

    vsin + v = (4/3)vsinθ'
     
  8. May 9, 2014 #7
    for simplicities sake I said, that you can make one of the particles move parallel to the x axis, then this eliminates one of the unknown angles.
     
  9. May 9, 2014 #8
    Wouldnt that have to assume that the angle between the two starting vectors is 90 degrees?
     
  10. May 9, 2014 #9
    No it wouldn't, imagine rotating the x and y axis so that the one particle is moving parallel to the x axis. Instead of two angles, its one.
     
  11. May 9, 2014 #10

    haruspex

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    I think it would be simpler to take the final motion as being along an axis. Given that they have the same mass and initial speed it follows that they made the same angle to the axis on approach,
     
  12. May 11, 2014 #11
    ok so now I am getting

    mvcosθ+mvcos(-θ) = 2/3vm

    and

    mvsinθ+mvsin(-θ) = 0

    I still feel stuck here :[
     
    Last edited: May 11, 2014
  13. May 11, 2014 #12

    Nathanael

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    Their components of velocity in opposite directions (which you made the 'sine component') will balance (or "annihilate") to give zero, but how will the components of velocity parallel to each other affect each other?
    Is that correct? On the left, you have units of momentum, on the right, units of velocity. What's missing?
     
  14. May 11, 2014 #13
    oops! thanks I fixed it. So the m's and v's cancel and it becomes cosθ+cos(-θ) = 2/3
     
  15. May 11, 2014 #14

    Nathanael

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    So that means the mass after they collide (and stick together) will only be m? (That's what your math says, is that correct, though?)

    Also, what is cos(-θ)? Is there another way to write that?
     
    Last edited: May 11, 2014
  16. May 11, 2014 #15
    ok im sorry i was being sloppy there

    mvcosθ+mvcos(-θ) = 2/3v(m+m)

    mvcosθ+mvcos(-θ) = 4/3vm

    cosθ+cos(-θ) = 4/3

    cosθ+cosθ = 4/3

    2cosθ = 4/3

    cosθ = 4/6

    cosθ = 2/3

    θ = cos^-1(2/3)

    therefore the angle between them two starting vectors is double that?
     
  17. May 11, 2014 #16

    haruspex

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    Yes.
     
  18. May 11, 2014 #17
    thanks guys
     
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