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Homework Help: 2d absolutely elastic vector collision

  1. May 30, 2012 #1
    1. The problem statement, all variables and given/known data
    I have mass of two points ma and mb than their vertical horizontal vector velocity parts before collision vxa0 vya0 vxb0 vyb0 and i need to calculate vector velocities after collision vxa1 vya1 vxb1 vyb1

    I found something like the same vector componets of momemtums will exchange so ma.vxa0<=>mb.vxb0 and of course ma and mb is constant so i got this vxa=(mb.vxb)/ma. Am I right? If no, how to calculate them?
  2. jcsd
  3. May 30, 2012 #2


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    Were the two masses moving before or was one stationary?

    With each component, the momentum will be conserved.

    So you'd have in x-direction: momentum before impact = momentum after impact.

    The same for the y-direction.

    Since it is elastic, you know that KE before = KE after impact.
  4. May 30, 2012 #3
    Both was moving it's collision of two moving balls representated by circles in code but theoreticaly there can one stay on place if three balls colide in one thick (but i don't need it so complex) and I need right equations for counting that four velocities vxa1 vya1 vxb1 vyb1 from this 6 variables m1 m2 vxa0 vya0 vxb0 vyb0.
  5. May 30, 2012 #4


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    Based on the information you have such as values for m1,m2 and the velocities, you can form the equationjs and solve.
  6. Jun 1, 2012 #5
    Apply conservation of mommentum. You get


    Edit: Oops. I mean conservation of energy.

    $$ \left(m_1v_{1i}^2+m_2v_{2i}^2\right)=\left(m_1v_{1f}^2+m_2v_{2f}^2\right) $$
    Last edited: Jun 1, 2012
  7. Jun 1, 2012 #6
    Fine from conservation of momentum in directions i can say that
    from KE conservation I can say that
    I have 3 equations with 4 unkowns what am i missing? Are these right?

    Is this rigth? I found something like the same vector componets of momemtums will exchange so ma.vxa0<=>mb.vxb0 and of course ma and mb is constant so i got this vxa=(mb.vxb)/ma.
  8. Jun 1, 2012 #7
    No, during elastic collisions, kinetic energy is conserved, not momentum.
  9. Jun 1, 2012 #8
    Nor in one direction?
  10. Jun 1, 2012 #9
    I don't get what you are saying?
  11. Jun 1, 2012 #10

    Both momentum and kinetic energy are conserved in elastic collisions....

    In fact, momentum is conserved in both elastic and inelastic collisions. o:)
  12. Jun 1, 2012 #11
    Please can you simply tell me the equations? May be when i will see them i will understand. The four equations for
    with variables of
    Please. I get lost
  13. Jun 1, 2012 #12
    Oh, ok.
  14. Jun 1, 2012 #13
    Yes, these are correct. :approve:

    A little rearrangement in the equations will help. To start with, eliminate 1/2 from the kinetic energy equation, and then bring the terms with the same masses in each equation to the same side and use [itex]a^2-b^2 = (a+b)(a-b)[/itex]. You will get a similar term from the momentum equations which you can substitute to solve further.
  15. Jun 4, 2012 #14
    I hava another question because i still have 3 equations with 4 unknowns could I do something like





    from it

    but i still see 4 things i need to get vxA1 vyA1 vxB1 vyB1 and 3 equations plese which one is fourth?

    Or can somebody solve this to get vxA1 vyA1 vxB1 vyB1 because i know one mathematical rule if i have n unkowns i need n equations.
  16. Jun 4, 2012 #15
    Let (vxA0-vxA1) = p be one variable, and (vxB1-vxB0)=q be another. Do you see how to solve this now?
  17. Jul 3, 2012 #16
    I don't see i think i am missing something about angles and vector redistribution :(
  18. Jul 14, 2012 #17
  19. Nov 30, 2012 #18
    First: Momentum is always conserved.
    Second: In elastic collisions, kinetic energy is also conserved.
    When objects collide elastically in 2 dimensions, the geometry of the collision is also important because the impulse delivered to each object is along the lined connecting the two centers of mass.
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