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Homework Help: I need a comprehensive answer, please help

  1. Jul 21, 2005 #1
    I need a comprehensive answer, please help !!!

    I need a comprehensive answer,please help !!!
    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
    ---------------
    Question: How long will it take for them to crash? How much energy will be emitted from the collision?
    ---------------


    All answers will be appreciated.
    Thanks again.
     
    Last edited: Jul 22, 2005
  2. jcsd
  3. Jul 21, 2005 #2
    Anyone ??? Please help
     
  4. Jul 21, 2005 #3

    HallsofIvy

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    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.
     
  5. Jul 21, 2005 #4
    That's the way I approached the problem, but then I got convoluted with the variables. It's so confusing.
    Please be more specific !!!!!
     
  6. Jul 21, 2005 #5
    Thank you.
     
  7. Jul 21, 2005 #6
    Please, anyone? This isn't homework or anything, so if you think you have the answer, just show it !!!
    Thanks.
     
  8. Jul 21, 2005 #7
    man, that DE is going to be a nightmare!

    can you show us what you have?
     
  9. Jul 22, 2005 #8
    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.
     
  10. Jul 22, 2005 #9
    Let's just eliminate the medium, will the problem be easier ?
     
  11. Jul 22, 2005 #10
    v(t) = dr(t)/dt

    yeah, and taking away that drag force will make things easier, i bet.
     
  12. Jul 22, 2005 #11
    actually, dr/dt is going to be more difficult than this, since both charges are accelerating towards each but at different rates.

    tough!
     
  13. Jul 23, 2005 #12
    for twits: do you mean the medium create decelerating force? opposite to the velocity?
     
  14. Jul 23, 2005 #13
    either way is ok. Let's just say it's the opposing force .
     
  15. Jul 23, 2005 #14
    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?
     
  16. Jul 23, 2005 #15
    Can anyone describe how the velocities of the particles will change during their movements(with the dragging force in this case) ?
     
  17. Jul 23, 2005 #16

    Päällikkö

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    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.
     
  18. Jul 23, 2005 #17
    sounds very complicated from here :rofl:
     
  19. Jul 23, 2005 #18

    mukundpa

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    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.
     
  20. Jul 24, 2005 #19

    Päällikkö

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    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:
    [tex]v_e = \sqrt{ \frac{-2kq^2(\frac{1}{x}-1)}{m_e} }[/tex], 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 [tex]v = \frac{dx}{dt}[/tex], and the v in my equation is dependant on x (v(x), instead of time v(t)) can the time be solved: [tex]\int dt = \sqrt{ \frac{m_e}{-2kq^2}} \int \frac{dx}{\sqrt{\frac{1}{x}-1}}[/tex] ?
    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 [tex]F = k \frac{q^2}{1^2} = kq^2[/tex] (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
  21. Jul 25, 2005 #20

    mukundpa

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    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
     
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