Not conserving energy when I should be

  • Thread starter Thread starter msimmons
  • Start date Start date
  • Tags Tags
    Energy
AI Thread Summary
The discussion revolves around a program designed to simulate a chaotic system involving two balls in one dimension under gravity. The user is struggling with energy conservation in their calculations, noticing an unexpected energy loss during the collision between the balls. Initial conditions and calculations for the balls' positions and velocities are provided, leading to a total energy discrepancy. A key mistake identified is the incorrect gravitational acceleration value used, which was later corrected. Participants confirm the importance of recalculating the collision timing based on the balls' movements post-ground impact.
msimmons
Messages
17
Reaction score
0
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 mgh + \frac{1}{2}mv^2

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
t=\frac{x1_0-x2_0}{v2_0-v1_0}
which is derived from
x1_0 + v1_0t + \frac{1}{2}gt^2 = x2_0 + v2_0t + \frac{1}{2}gt^2
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)
x = x_0 + v_0t + \frac{1}{2}gt^2
x1 = .75
x2 = .75
then the velocities before the collision
v = v_0 + gt
v1 = 3.3204
v2 = 5.5340
Total energy = 2*.75*9.8 + .75*9.8 + \frac{1}{2}*2*5.534^2+\frac{1}{2}*3.3204^2 = 58.187

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




Edit
Scratch that.
g = -9.8. not -4.9.
Duh.
 
Last edited:
Physics news on Phys.org
msimmons said:
which is derived from
x1_0 + v1_0t + \frac{1}{2}gt^2 = x2_0 + v2_0t + \frac{1}{2}gt^2
for which I get t = -2/-8.8 = .22588.

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.
 
Volcano said:
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.
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:

msimmons said:
then the velocities before the collision
v = v_0 + gt
v1 = 3.3204
v2 = 5.5340
 
So I know that electrons are fundamental, there's no 'material' that makes them up, it's like talking about a colour itself rather than a car or a flower. Now protons and neutrons and quarks and whatever other stuff is there fundamentally, I want someone to kind of teach me these, I have a lot of questions that books might not give the answer in the way I understand. Thanks
Back
Top