Conversion of EPE to Kinetic Energy in Magnetism

  • Thread starter lluke9
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So suppose we have two objects, both with positive charge Q Coulombs and X meters apart.
And now suppose Object 1 has a mass of m1 and Object 2 has a mass of m2.
In other words, the charges are identical, but masses are not.

So each of these will have an Electric Potential Energy of -QEΔX, because EPE = -QEΔd.
Easy peasy, right? EPE = KE, just like PE = KE for a falling object! Just like mgh = (1/2)mv^2, -QEX = (1/2)mv^2. So:

EPE = KE, so for Object 1 it's
-QEΔX = (1/2)m1v2.

and for Object 2 the formula is
-QEΔX = (1/2)m2v2.

Apparently, that's wrong and I was actually supposed to work it out this way:
-QEΔX = (1/2)m1v2 + (1/2)m2v2

It does make sense... but it doesn't. I mean, each object has each of their own potential energy, right? And each object loses that EPE and changes it 100% into KE, right? Isn't that what the potential energy is?

I mean, I don't remember doing PE = KE1 + KE2 for a falling object!
As far as I know, I've always done:
mgh = (1/2)mv2.

I don't remember ever making the mass of the earth share the potential energy.
If I apply the same concept to gravity, it would be like:
mgh = (1/2)mEarthv2 + (1/2)mobjectv2.
What!!? I don't remember ever doing that!

Help me out here, guys! I am greatly befuddled! :cry:
 

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  • #2
kuruman
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I mean, each object has each of their own potential energy, right?
Wrong. See my response (post #2) here.
It's probably not your fault that you are befuddled. To solve problems of this kind, where you start with an initial configuration and speeds of two objects and you want to know (say) the speeds of the two objects in their final configuration, you need to use two conservation laws, mechanical energy and momentum. That's because you have two unknowns, namely the speeds, and energy conservation is not enough. In most textbooks energy conservation comes before momentum conservation. Why? Because in the case of a small object falling near the surface of the Earth, for every 1 Joule of energy gained by the Earth, the object gains energy of order ##\mathrm{10^{24}}##.
Here is the correct calculation using the conservation laws for a rock of mass ##m## released from rest by ##h##. Initially, the rock and stone are at rest relative to each other. we are looking for the final speeds of the two objects. We get them by using the two conservation laws. Subscript ##e## stands for "Earth".

1. Momentum conservation, ##P_{before}=P_{after}##.
$$0+0=M_eV_e+mv=0~\rightarrow~V_e=-\frac{m}{M_e}v$$
2. Mechanical energy conservation, ##ME_{before}=ME_{after}##.
$$mgh=\frac{1}{2}M_eV_e^2+\frac{1}{2}mv^2=\frac{1}{2}M_e \left( -\frac{m}{M_e}v \right)^2+\frac{1}{2}mv^2=\frac{1}{2}\left(1+\frac{m}{M_e}\right) mv^2$$Now you see that for a mass of order 1 kg, the ratio of the masses is ##m/M_e=10^{-24}## which means that to an extremely super good approximation ##mgh\approx \frac{1}{2}mv^2##, i.e. all of the potential energy of the two-body system goes into kinetic energy of the rock. Clearly that's not the case when the masses are of the same magnitude; you have a 50-50 split when the masses are equal. But let's find the final speeds anyway. We have $$v=\sqrt{\frac{2gh}{1+m/M_e}}\approx \sqrt{2gh}$$and $$V_e=\frac{m}{M_e}\sqrt{\frac{2gh}{1+m/M_e}} \approx 0.$$ No wonder you are befuddled. Once you learned about momentum conservation, it would have been a good idea for your instructor to have shown you what's under the hood.
 
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