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Elastic collision of the rigid balls

  1. Nov 22, 2014 #1
    1. The problem statement, all variables and given/known data
    I can't believe it how my brains stopped cooperating today.
    We have the first ball with ##m_1 ## and ##v_1## and of course the second one with ##m_2## and velocity ##v_2##. Covariance matrix before the collision is ##M=
    \begin{bmatrix}
    \sigma _1^2 & 0\\
    0& \sigma _2^2
    \end{bmatrix}## . Calculate the covariance matrix after the elastic collision

    2. Relevant equations
    ##{\vec x }'=\Phi {\vec x }## if ##\Phi ## is the transformational matrix and if ##{(xyz)}'## indicates the values after the collision.

    3. The attempt at a solution
    Now my idea was to move to the center of mass system but due to my brain blockage I am not able to find or to work with any given equation. -.-

    This is wrong, and I can't find out why:

    Lets use notation ##u## for the velocities in the center of mass system and let ##v_{CMS}## be the center of mass velocity. Than I guess ##u_1=v_1-v_{CMS}## and ##u_2=-v_2-v_{CMS}##.

    Of course in center of mass system ##m_1u_1-m_2u_2=0## which leaves me with the most stupid thing ever, saying that ##v_{CMS}=\frac{m_1v_1+m_2v_2}{m_1-m_2}##.

    Could somebody please help me a bit?
     
  2. jcsd
  3. Nov 22, 2014 #2

    gneill

    User Avatar

    Staff: Mentor

    If you assume that v1 and v2 are signed values, then the total momentum of the initial system is just:

    P = m1*v1 + m2*v2

    You can then find a velocity VCMS (the relative velocity of the center of momentum frame) to add to both v1 and v2 such that the momentum becomes zero. That is,

    0 = m1*(v1 + VCMS) + m2*(v2 + VCMS)

    Solve for VCMS
     
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