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## Main Question or Discussion Point

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

[tex]t=\frac{x1_0-x2_0}{v2_0-v1_0}[/tex]

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?

Edit

Scratch that.

g = -9.8. not -4.9.

Duh.

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

[tex]t=\frac{x1_0-x2_0}{v2_0-v1_0}[/tex]

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?

Edit

Scratch that.

g = -9.8. not -4.9.

Duh.

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