# I need a comprehensive answer,

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
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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?
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Thanks again.

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twits

Homework Helper
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.

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

twits
Thank you.

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

man, that DE is going to be a nightmare!

can you show us what you have?

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.

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

twits said:
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.

v(t) = dr(t)/dt

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

v(t) = dr(t)/dt

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

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

tough!

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

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

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?

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

Homework Helper
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.

faganac
sounds very complicated from here :rofl:

Homework Helper
If you eliminate the medium, considering the two particle 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.

Homework Helper
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).

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Homework Helper
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