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Distance of closest approach of like charge

  1. Jul 24, 2014 #1
    1. The problem statement, all variables and given/known data
    A proton and an α -particle are projected with velocity v=√[itex]\frac{e^{2}}{4∏εmL}[/itex] each from infinity as shown.The perpendicular distance between their initial velocities is L .Let the distance of their closest approach be D .

    Given .Find (8D/L - 5)^2

    Details and assumptions

    1)mass of proton=m,charge=+e

    2)mass of α-particle=4m,charge=2e



    2. Relevant equations



    3. The attempt at a solution
    As the particles are located at infinite distance from each other so I think dist.L will not matter much.
    I tried to use reduced mass problem here.
    Here reduced mass(M) =4m/5 let it be at infinite distance from origin O.
    What should be the initial velocity of reduced mass here? Is it 2v.(I used 2v here)
    Lat it be at a distance D from O after some time.
    At D all its kinetic energy wiil be converted to potential energy so
    Mv^2/2=k2e^2/R
    On putting values and after doing some calcilations I got
    4D=5L
    I got the value of 25.Am I right?
    I think I am not.

    This is the link of figure.
    d3pq38zxuosm5i.cloudfront.net/solvable/2676130fa9.92b5dbd0d6.O7DvUt.png
     
    Last edited: Jul 24, 2014
  2. jcsd
  3. Jul 24, 2014 #2

    BiGyElLoWhAt

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    I'm not following. What do you mean by the perpendicular distance between their velocity? You've got misspellings and random punctuation out the ***.
    I would suggest cleaning this post up a bit.
     
  4. Jul 25, 2014 #3
    A proton and an α -particle are projected with velocity v=√e^2/4∏εmL each from infinity to each other as shown in figure.The perpendicular distance between their initial velocities is L .Let the distance of their closest approach be D .


    The direction of velocities of two particles are opposite to each other but the particles do not lie on the same line.The perpendicular distance between their initial velocities means the perpendicular distance between the direction of their initial velocities.
    I think it will be clear from the figure here. http://d3pq38zxuosm5i.cloudfront.net/solvable/2676130fa9.92b5dbd0d6.O7DvUt.png
    Sorry for the inconvenience.
     
  5. Jul 25, 2014 #4

    ehild

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    Do you mean [itex]v=\sqrt{\frac{e^{2}}{4\pi εmL}}[/itex] ?

    It does :tongue: There is angular momentum of the system, and you have to take its conservation into account.

    No. the radial component of the velocity is zero, but but not the whole velocity.


    ehild
     

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  6. Jul 26, 2014 #5
    Do you want me to apply law of conservation of momentum.

    I did not have knowledge about angular momentum and torque, so I explored about them yesterday.
    I found that these quantities are associated with rotational motion of system.
    In rotational motion the axis of system does not move and other particles of the system move around it.

    Are concepts of rotational motion related to this question? If yes,then please explain how.

    You want me to use angular momentum here but angular momentum is calculated w.r.t a fixed point.
    Where is this 'fixed point' in this question?

    I don't understand why the kinetic energy is not zero at the distance of closest approach of particles.As particles stops momentarily at the distance of closest approach so Kinetic energy should be 0 at that point.
    Thank you ehild.
     
  7. Jul 26, 2014 #6
    Yes. I didn't get the symbol of 'pi'
     
  8. Jul 26, 2014 #7

    ehild

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    No, I want you to use conservation of angular momentum. It is conserved in a central force field.
    You can define angular momentum for a particle with respect to a fixed origin.
    It is ##\vec L= \vec r \times \vec p ##, the cross product of the position vector with the linear momentum. The time derivative of the angular momentum is equal to the torque. A central force has zero torque, so the angular momentum is conserved in a central force field. Your virtual particle moves in a central force field, where the force acts away from a repulsive centre.

    If you have the real particles you can choose the fixed point at the CM when they are at the closest approach. In the reduced mass approach, you have a fixed centre of force, determine the angular momentum with respect to it.

    The kinetic energy becomes zero at closest approach if the particles move along the same line. Here they move along curved lines which do not intersect. See picture. If they stopped somewhere, they would move along a straight line away from the CM and each other and the angular momentum of the system would be zero.
     

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  9. Jul 27, 2014 #8
    I have made the figure ehild.Please see that.

    I have some problem in finding the initial angular momentum.
    L=mvrsinθ.Here r=∞.

    But from the figure
    cosA=∞/∞=1
    so A=0.
    So the angle between velocity vector and position vector=180(joining both vector from tail to tail)
    but sin180=0
    So initial angular momentum is 0.Is it right.
     
  10. Jul 27, 2014 #9
    physics figure..png .
    This is the figure.
     
  11. Jul 27, 2014 #10

    ehild

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    You can not calculate with infinity like that. ∞/∞ is not 1. Think about x/x2, both the numerator and the denominator goes to infinity, what about the ratio?
    You never reach infinity. It is only a limit - you can increase something more and more but it is still finite.

    Initially the point is very far away but still at a finite distance R. Rsin(A)=L.
    So the angular momentum of the virtual particle is μ(2v)L.



    ehild
     
  12. Jul 28, 2014 #11
    Ok I get that.
    For calculating final angular momentum.
    I have made a figure.

    L=mvrsinθ(θ in this case should be equal to 90 I think)
    Lf=μv'r.
    Is it right?

    physics figure..png
     
  13. Jul 28, 2014 #12

    ehild

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    Yes. The initial angular momentum is μVL, (V=2vo) and the final one is μv'r, r is the distance of closest approach (the problem text uses D for it). The initial angular momentum is equal to that at the closest approach.


    ehild
     
  14. Jul 28, 2014 #13
    μ2vL=μv'D
    or 2vL=v'D.
    If I use law of conservation of energy and substitute v'=2vL/D.Will I get the answer?
     
  15. Jul 28, 2014 #14

    ehild

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    I hope. :smile:

    ehild
     
  16. Jul 28, 2014 #15
    v'=[itex]\frac{2vL}{D}[/itex]

    [itex]\frac{4μv^{2}}{2}[/itex]=[itex]\frac{k2e^{2}}{D}[/itex]+[itex]\frac{4μv^{2}L^{2}}{2D^{2}}[/itex]

    On simplifying and putting the values of v,μ=4m/5 and k.
    I got

    4D[itex]^{2}[/itex]=5DL+4L[itex]^{2}[/itex].

    I am not able to find (8D/L - 5)^2
     
  17. Jul 28, 2014 #16

    ehild

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    Divide the whole equation with L2. You get a quadratic equation for D/L. Write the quadratic formula, but do not evaluate. What do you get?

    ehild
     
  18. Jul 28, 2014 #17
    Is it 89?
     
  19. Jul 28, 2014 #18

    ehild

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    It looks all right.

    ehild
     
  20. Jul 29, 2014 #19
    Thank you ehild.You are the best.
     
  21. Jul 29, 2014 #20

    ehild

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    Thank you, but it is an overstatement:shy:

    ehild
     
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