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Distance of Closest Approach

  1. Oct 28, 2014 #1
    1. The problem statement, all variables and given/known data

    A proton and an alpha particle (q= +2e, m=4u) are fired directly at one another from far way, each with an initial velocity of 0.01c. What is their distance of closest approach, as measured between their centres?

    2. Relevant equations

    mivi = mf vf
    and KEi +Ui = KEf + Uf

    U = (q1 x q2 ) / (4επr)

    KE= 0.5mv^2

    Mass of proton: 1.67E-27
    Charge of Proton: 1.692E-19
    c = 3E8


    3. The attempt at a solution

    I'm sorry this question has been asked before but I didn't understand the explanations in those threads.

    I tried using the conservation of momentum:
    (1.67E-27 * 0.01c) - (4 * 1.67E-27 * 0.01c) = -(1.67E-27 * velocity of proton) + (4* 1.67E-27 * velocity of alpha particle)
    ∴ -9000000= -velocity of proton + 4*velocity of alpha particle
    but then I don't understand how to find the velocities of the particles individually or how to get this equation into the KE equation.

    When I tried using the conservation of energy I got:

    ½*1.67*10^-27*(0.01c)^2 + ½*4*1.67*10^-27*(0.01c)^2 = (3*1.602*10^-19) /4επr

    ∴ r = 115008.95

    which is so far off, as the answer in the back of the book says: 1.93x10^-14

    Which equation am I using wrong and how can I fix it?
     
  2. jcsd
  3. Oct 28, 2014 #2

    rude man

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    Using energy arguments, equate the initial k.e. of both particles to the potential energy of the particles when all their k.e. is gone.
     
  4. Oct 28, 2014 #3

    rude man

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    Use the correct units. What is 0.01c in SI units? What is the formula for potential energy of two charges a distance r apart? Nothing like (3*1.602*10^-19) /4επr for charges e and 2e.
     
  5. Oct 28, 2014 #4

    gneill

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    Staff: Mentor

    I think you'll want to do the calculation in the center of momentum frame of reference. Because the masses are not equal, for observers in any other frame the closest approach will happen when the system still has kinetic energy.
     
  6. Oct 29, 2014 #5

    Orodruin

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    The absolutely easiest way of solving this problem is to ask the following questions:
    What is the velocity of the particles at the time of closest approach?
    What is the kinetic energy at this point?
    What must the potential energy be?
    At what distance is this fulfilled?

    You can go to the CoM frame but it is not necessary. I also suggest you keep the symbolic representations of your quantities and only insert them at the very end.
     
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