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Simultaneous equation - momentum and kinetic energy

  1. Oct 4, 2009 #1
    I'm trying to work out elastic collision equations for some small game development.

    In any body, p is momentum, m mass, KE kinetic energy:

    Code (Text):
    p = mv
    KE = m v² /2
    Therefore

    Code (Text):
    KE = p² / 2m  (also confirmed in many textbooks and resources)
    So, two colliding bodies, A and B, where A.m is the mass of A, B.p the momentum of B and so on, T is total, so T.E being total energy and T.p being total momentum:

    (after collision)

    A.E + B.E = T.E

    So

    Code (Text):
    A.p²     B.p²
    ----  +  ----  =  T.E
    2A.m     2B.m
    Since A.p + B.p = T.p ; A.p = T.p - B.p:

    Code (Text):
    A.p²     (T.p-A.p)²  
    ----  +  ----------  =  T.E
    2A.m        2B.m
    Added fractions:

    Code (Text):

    A.p² 2B.m      (T.p-A.p)² 2A.m
    ----------  +  ----------------  =  T.E
    2A.m 2B.m      2B.m       2A.m


    A.p² 2B.m   +  2A.m  (T.p-A.p)²
    ------------------------------- = T.E
                4A.m B.m
    Expanded the bracket:

    Code (Text):
    A.p² 2B.m   +  2A.m  (T.p² - 2A.p T.p + A.p²)
    ---------------------------------------------- = T.E
                    4A.m B.m
    Multiplied by the denominator:

    Code (Text):
    T.E * 4 A.m B.m = A.p² 2B.m   +  2A.m  (T.p² - 2A.p T.p + A.p²)
    Expanded the remaining bracket:

    Code (Text):
    T.E * 4 A.m B.m = A.p² 2B.m   +  2A.m T.p² - 4A.m A.p T.p + 2A.m A.p²
    Moved all products without A.p to the left, and added some pre-collection brackets:

    Code (Text):
    T.E * 4 A.m B.m - 2A.m T.p²  =  A.p² (2B.m) - A.p(4A.m T.p) + A.p² (2A.m)
    Collect up:

    Code (Text):
    T.E * 4 A.m B.m - 2A.m T.p²  =  A.p² (2B.m+2A.m) - A.p (4A.m T.p)
    Make equal to zero:

    Code (Text):
    0  =  A.p² (2B.m+2A.m) + A.p (-4A.m T.p) - (T.E * 4 A.m B.m - 2A.m T.p²)
    Now we should be able to apply the quadratic equation, since 0 = ax² + bx + c. However, calculating the final momentum and E of A by this method and calculating the remaining momentum and E for B, the final energy is always many times greater than the initial input energy. Help >__<

    I've taken

    Code (Text):
    a = (2B.m+2A.m) + A.p (-4A.m T.p)
    b =  -4A.m T.p
    c = - (T.E * 4 A.m B.m - 2A.m T.p²)

    Used x = -b ± sqrt(b²-4ac), both plus and minus come out vastly incorrect
    Thanks :)
     
  2. jcsd
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