1. Jul 21, 2005

### twits

I made up this question, but I can not solve it. This bothers me so much. Please try it and explain it to me.

Here it is
--------------
Conditions: 1 proton, 1 electron, 1 meter away from each other.
The medium exerts a force F(Newton)=square root of the velocity of the object(meter/second)
Gravity is negligible
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Question: How long will it take for them to crash? How much energy will be emitted from the collision?
---------------

Thanks again.

Last edited: Jul 22, 2005
2. Jul 21, 2005

### twits

3. Jul 21, 2005

### HallsofIvy

Staff Emeritus
mass times acceleration equals force. You should be able to calculate the total force- it is the force due to their charge minus the friction force which you are given. That gives you a differential equation for the motion of each of the particles- be sure to include the different masses of the particles.

4. Jul 21, 2005

### twits

That's the way I approached the problem, but then I got convoluted with the variables. It's so confusing.

5. Jul 21, 2005

### twits

Thank you.

6. Jul 21, 2005

### twits

Please, anyone? This isn't homework or anything, so if you think you have the answer, just show it !!!
Thanks.

7. Jul 21, 2005

man, that DE is going to be a nightmare!

can you show us what you have?

8. Jul 22, 2005

### twits

Sum of force on the electron= kQ1*Q2/(r(t))^2 - (v(t))^(1/2)
m*v'(t)=kQ1*Q2/(r(t))^2-(v(t))^(1/2)

But then how will I connect r(t) and v(t) ?

I got stuck here.

9. Jul 22, 2005

### twits

Let's just eliminate the medium, will the problem be easier ?

10. Jul 22, 2005

v(t) = dr(t)/dt

yeah, and taking away that drag force will make things easier, i bet.

11. Jul 22, 2005

actually, dr/dt is going to be more difficult than this, since both charges are accelerating towards each but at different rates.

tough!

12. Jul 23, 2005

### sniffer

for twits: do you mean the medium create decelerating force? opposite to the velocity?

13. Jul 23, 2005

### twits

either way is ok. Let's just say it's the opposing force .

14. Jul 23, 2005

### twits

Question: Can we use the known charges, masses of the particles to indicate how far each will go, then just do the problem for 1 particle?

15. Jul 23, 2005

### twits

Can anyone describe how the velocities of the particles will change during their movements(with the dragging force in this case) ?

16. Jul 23, 2005

### Päällikkö

I tried to solve the problem with work done by the field (dW = F dr), but I got infinites.
I think the problem cannot be solved if the distance goes to 0 (well, not with dW = F dr), it can only go somewhere quite close to it (before the particles collide). I suppose there either are some formulas I'm unaware of or I had the wrong approach. The latter is quite likely :).

I did not take relativistic effects into account in my calculations. Those would require some knowledge in general relativity (which I lack) as the particles are accelerating, I suppose.

17. Jul 23, 2005

### faganac

sounds very complicated from here :rofl:

18. Jul 23, 2005

### mukundpa

If you eliminate the medium, considering the two partical system the e force is internal force and so conservation of momentum and conservation of energy will give the velocities of the two particles as a function of distance r between them, their approach velocity is the sum of magnitude of these two velocity.

19. Jul 24, 2005

### Päällikkö

I simplified the problem a bit: No resistance by medium, proton glued (ie. not moving).
With dW = F dr I got the velocity of the electron:
$$v_e = \sqrt{ \frac{-2kq^2(\frac{1}{x}-1)}{m_e} }$$, where x is the "end distance" from the proton. Is this correct?
When x -> 0, the velocity goes infinite (and when x = 1, v = 0, sounds reasonable). I suppose the time will still be finite, though.

as $$v = \frac{dx}{dt}$$, and the v in my equation is dependant on x (v(x), instead of time v(t)) can the time be solved: $$\int dt = \sqrt{ \frac{m_e}{-2kq^2}} \int \frac{dx}{\sqrt{\frac{1}{x}-1}}$$ ?
Well, I can't solve the integral. I'm not too familiar with differential equations anyway.

I do hope I didn't make a fool out of myself, but I'm afraid I did it all totally and utterly wrong :).

EDIT: I put my calculator to solve the integral (the simplified problem) with numerical approximation. I got 0,0693 seconds as the answer.
Sounds logical, as if the force was constantly $$F = k \frac{q^2}{1^2} = kq^2$$ (so r = 1), the trip'd take 0,0888 seconds (provided I typed all the numbers correctly into my calculator).

Last edited: Jul 24, 2005
20. Jul 25, 2005

### mukundpa

The integral is correct. If the proton is not glued then we have to take approach velocity
of the two particles resulting in a factor sqrt[M/(m+M)]
m = mass of electran and M = mass of Proton