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Not conserving energy when I should be

  1. Mar 16, 2008 #1
    I'm writing a program to demonstrate the chaotic system of two balls in one dimension with gravity. Before I can get even remotely close to that, however, it would help if my energy was conserved, so I'm apparently doing something wrong but I can't figure it out.

    There is no damping forces.
    energy is calculated with [tex]mgh + \frac{1}{2}mv^2[/tex]

    ball 1 is the BOTTOM ball.
    ball 2 is the TOP ball
    x1 corresponds to the position of the bottom ball, etc.
    initial conditions:
    x1 = 1
    x2 = 3
    v1 = v2 = 0
    m1 = 1
    m2 = 2
    (total system energy = 68.6 )

    I'll walk through until my loss of energy, not far away.
    First I calculate when the bottom ball hits the ground, they've both accelerated to the same speed and they've both dropped one meter, so the conditions are now
    x1 = 0
    x2 = 2
    v1 = 4.4272 (positive, after the elastic collision with the ground which ain't moving)
    v2 = -4.4272
    (total system energy = 68.8 )

    Next collision will be between the two balls. This is where I spontaneously lose energy
    I first calculate the time it takes with
    which is derived from
    [tex]x1_0 + v1_0t + \frac{1}{2}gt^2 = x2_0 + v2_0t + \frac{1}{2}gt^2[/tex]
    for which I get t = -2/-8.8 = .22588.

    next with elementary equations we calculate the position of each ball at that time (or just assume that they're at the same place, which they are, but I'll calculate just in case I am doing something wrong)
    [tex]x = x_0 + v_0t + \frac{1}{2}gt^2[/tex]
    x1 = .75
    x2 = .75
    then the velocities before the collision
    [tex]v = v_0 + gt[/tex]
    v1 = 3.3204
    v2 = 5.5340
    Total energy = [tex]2*.75*9.8 + .75*9.8 + \frac{1}{2}*2*5.534^2+\frac{1}{2}*3.3204^2 = 58.187[/tex]

    And that's it. Anyone see what I did wrong?

    Scratch that.
    g = -9.8. not -4.9.
    Last edited: Mar 16, 2008
  2. jcsd
  3. Mar 17, 2008 #2
    I think, above equation is wrong. Honestly, could not undertand the meaning too.

    The balls collides after first ball return from ground. You must calculate the time AFTER that time. BTW, you may calculate the previous time too(before first ball arrive ground). Maybe couldn't understand right.
  4. Mar 17, 2008 #3

    Doc Al

    User Avatar

    Staff: Mentor

    That part (the calculation of the time) is correct.

    Realize that msimmons already spotted his mistake (see his edit at the bottom of the post), which was here:

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