1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    rock.freak667

    User Avatar
    Homework Helper

    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

    rock.freak667

    User Avatar
    Homework Helper

    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

    $$m_1v_{1i}+m_2v_{2i}=m_1v_{1f}+m_2v_{2f}$$

    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
    vxA0*mA+vxB0*mB=vxA1*mA+vxB1*mB
    vyA0*mA+vyB0*mB=vyA1*mA+vyB1*mB
    from KE conservation I can say that
    mA*(vxA0^2+vyA0^2)/2+mB*(vxB0^2+vyB0^2)/2=mA*(vxA^2+vyA^2)/2+mB*(vxB^2+vyB^2)/2
    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
    :uhh:

    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
    vxa=
    vya=
    vxb=
    vyb=
    with variables of
    ma
    mb
    vxa0
    vya0
    vxb0
    vyb0
    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

    mA*(vxA0^2+vyA0^2)/2+mB*(vxB0^2+vyB0^2)/2=mA*(vxA^2+vyA^2)/2+mB*(vxB^2+vyB^2)/2

    mA*(vxA0^2+vyA0^2)+mB*(vxB0^2+vyB0)=mA*(vxA^2+vyA^2)+mB*(vxB^2+vyB^2)

    fine

    mA*(vxA0^2+vyA0^2)-mA*(vxA^2+vyA^2)=mB*(vxB^2+vyB^2)-mB*(vxB0^2+vyB0)

    vxA0*mA-vxA1*mA=vxB1*mB-vxB0*mB
    vyA0*mA-vyA1*mA=vyB1*mB-vyB0*mB
    from it
    mA*(vxA0-vxA1)=mb*(vxB1-vxB0)
    mA*(vyA0-vyA1)=mb*(vyB1-vyB0)

    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: 2d absolutely elastic vector collision
  1. 2D elastic collision (Replies: 7)

  2. 2D Elastic Collision (Replies: 9)

  3. 2D Elastic Collisions? (Replies: 1)

  4. 2D Elastic Collisions (Replies: 1)

Loading...