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Velocity of spheres after collision

  1. May 13, 2013 #1
    1. The problem statement, all variables and given/known data
    Three spheres, each of mass m, can slide freely on a frictionless, horizontal surface. Spheres A and B are attached to an inextensible inelastic cord of length / and are at rest in the position shown when sphere B is struck directly by sphere C which is moving to the right with a velocity ##v_o##. Knowing that the cord is taut when sphere B is struck by sphere C and assuming perfectly elastic impact between B and C, determine the velocity of each sphere immediately after impact.
    attachment.php?attachmentid=58700&stc=1&d=1368454112.png

    2. Relevant equations



    3. The attempt at a solution
    Let the velocity of A, B and C after collision be ##v_A##, ##v_B## and ##v_C## respectively.

    Conserving linear momentum,
    ##mv_0=mv_A+mv_B+mv_C##

    As the string is inextensible, the components of velocity of A and B along the string should be equal i.e ##v_A\sin\theta=v_B\sin\theta \Rightarrow v_A=v_B## but this is wrong because as per the answer key the velocities of A and B are different.
     

    Attached Files:

  2. jcsd
  3. May 13, 2013 #2
    Hmm interesting question I think that the problem lies in the directions you have decided for the velocities of A and B. How can the velocity of A have a component perpendicular to the string if the tension can only provide impulse to it? Also for B the tension will give an impulse and the velocity vector wont be parallel to Vo. You have to use vector equations for conservation of linear momentum.
     
  4. May 13, 2013 #3

    haruspex

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    It's a string, not a rod.
     
  5. May 13, 2013 #4
    Let the velocity of B in horizontal and vertical direction be ##v_{Bx}## and ##v_{By}## and similarly for A, ##v_{Ax}## and ##v_{Ay}##

    Conserving momentum in the horizontal direction,
    ##mv_o=mv_{Ax}+mv_{Bx}+mv_C##

    Conserving momentum in the vertical direction,
    ##0=mv_{Ay}+mv_{By} \Rightarrow v_{By}=-v_{Ay}##

    I still need more equations to solve the problem. I can only think of conserving energy.
    [tex]\frac{1}{2}mv_o^2=\frac{1}{2}mv_C^2+\frac{1}{2}m(v_{Ax}^2+v_{Ay}^2)+ \frac {1}{2}m(v_{Bx}^2+v_{By}^2)[/tex]
    [tex]\Rightarrow v_o^2=v_C^2+v_{Ax}^2+2v_{Ay}^2+v_{Bx}^2[/tex]

    Is it right to assume that C moves only in horizontal direction even after the collision?
    As the collision is elastic, ##v_o=v_{Bx}-v_{C}##.

    Am I doing this right? I still need more equations.
     
  6. May 13, 2013 #5
    One equation is to equate the velocities along the string. As it is an elastic collision don't use conservation of energy just use coefficient of restitution as energy conservation gives complicated equations and the same result as coefficient of restitution.

    EDIT:

    And yes C will move in horizontal direction only as no vertical impulse acts on it. We have 5 unknowns and 5 equations but I don't think we will get the answer as conserving energy isn't a "unique equation" and using coefficient of restitution is the same thing.
     
    Last edited: May 13, 2013
  7. May 13, 2013 #6
    I think I cannot do that. Check haruspex's post.
     
  8. May 13, 2013 #7
    Well I thought that since the string remains tight after the collision the components along it would be equal. As for how it remains tight I don't have any mathematical reason but I cannot imagine practically how the the string would become loose after the collision.
     
  9. May 14, 2013 #8

    haruspex

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    Ive come around to your view. Initially, at least, it will remain taut. And since it is taut before the collision, you can treat the collision as work conserving.
    No, that ignores the role of A. Consider KE of whole system.
     
  10. May 14, 2013 #9
    Does this mean I can conserve energy in this problem?

    And can I equate the components of velocity of A and B along the string?
     
  11. May 14, 2013 #10

    haruspex

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    Yes, I think you can justify both of those.
     
  12. May 14, 2013 #11
    I wrote the equation for conservation of energy in post #4.

    (I have assumed that x component of velocity both A and B is in the right direction and y-component in vertically upward direction)
    Equating the components of velocity along the string,
    [tex]v_{By}\cos \theta+v_{Bx}\sin \theta=v_{Ax}\sin \theta+v_{Ay}\cos \theta[/tex]
    [tex](v_{Bx}-v_{Ax})\sin \theta=2v_{Ay}\cos \theta[/tex]

    From the figure, ##\sin \theta=1/2## and ##\cos \theta=\sqrt{3}/{2}##
    [tex]\Rightarrow v_{Bx}-v_{Ax}=2\sqrt{3}v_{Ay}[/tex]

    That's going to be really dirty. :yuck:

    Can you give me a few ideas how should I begin to solve this? I have got enough equations I think but substituting expressions in the energy equation gives so many squared terms. :uhh:
     
    Last edited: May 14, 2013
  13. May 14, 2013 #12

    haruspex

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    When you substitute in the energy equation, you should find a common factor (a velocity) that cancels out, reducing it to a linear form.
     
  14. May 14, 2013 #13
    I had ##v_o=v_{Ax}+v_{Bx}+v_C \Rightarrow v_C=v_o-(v_{Ax}+v_{Bx})##
    Also ##v_{Bx}-v_{Ax}/2\sqrt{3}=v_{Ay}##

    Substituting them in the energy equation,
    [tex]v_o^2=v_o^2+v_{Ax}^2+v_{Bx}^2-2v_o(v_{Ax}+v_{Bx})+\frac{v_{Bx}^2+v_{Ax}^2-2v_{Ax}v_{Bx}}{6}+v_{Bx}^2[/tex]
    [tex]\Rightarrow 12v_o(v_{Ax}+v_{Bx})=7v_{Ax}^2+19v_{Bx}^2-2v_{Ax}v_{Bx}[/tex]
    I don't see how can I simplify that. :confused:
     
  15. May 14, 2013 #14

    ehild

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    One equation is still missing: conservation of angular momentum.

    ehild
     
  16. May 14, 2013 #15
    About what point should I do this? :confused:

    About A?
     
  17. May 14, 2013 #16

    ehild

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    I would do it around the point above C, in the horizontal line between A and B. But A is good, too.
     
    Last edited: May 14, 2013
  18. May 14, 2013 #17
    I am assuming the point X shown in attachment.
    attachment.php?attachmentid=58713&stc=1&d=1368523091.png
    About that point,
    [tex]0=\frac{mv_{By}\ell}{2}+\frac{mv_{Ax}\sqrt{3}\ell}{2}[/tex]
    [tex]\Rightarrow v_{By}=-\sqrt{3}v_{Ax}[/tex]

    Correct?
     

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  19. May 14, 2013 #18

    ehild

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    correct
     
  20. May 14, 2013 #19
    I further solved it but I am not able to reach the right answer.

    I had ##v_{Bx}-v_{Ax}x=2\sqrt{3}v_{Ay}##. As ##v_{By}=-\sqrt{3}v_{Ax}=-v_{Ay}##

    [tex]v_{Bx}-v_{Ax}=6v_{Ax} \Rightarrow v_{Bx}=7v_{Ax}[/tex]

    From the energy equation:
    [tex]12v_o(v_{Ax}+v_{Bx})=7v_{Ax}^2+13v_{Bx}^2-2v_{Ax}v_{Bx}[/tex]
    [tex]\Rightarrow 12v_o(8v_{Ax})=(7+13\times 49-14)v_{Ax}^2[/tex]
    [tex]\Rightarrow v_{Ax}=\frac{16v_o}{105}[/tex]

    [tex]v_A=\sqrt{v_{Ax}^2+v_{Ay}^2}=2v_{Ax}=\frac{32v_o}{105}[/tex]

    But this is wrong. :confused:
     
  21. May 14, 2013 #20

    ehild

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    You might have an error in the energy equation. Check.

    ehild
     
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