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Final repulsed velocity of two different charged masses

  1. Oct 24, 2012 #1
    1. Two small metal spheres with masses 2.0g and 4.0g are tied together by a 5cm long massless string and are at rest on a frictionless table. Each sphere has a (mue)c charge.
    The string is cut, what are the velocities of the sphere when they are far apart




    I know two conservative qualities are at play, electric energy, and I'm not sure of the other



    I tried using change under the curve with r being the variable (kQq/r) being equal to work and then said mv^2/2. However I can't articulate how I am sure this is the wrong approach.

    What is the right way/approach?
     
  2. jcsd
  3. Oct 24, 2012 #2
    You have the right idea.

    You want to look at the total energy of the system before and after.

    Before:

    total kinetic energy is zero (both spheres are at rest)
    total potential energy is kqQ/R


    After:

    total kinetic energy is 0.5*m1*v1^2 + 0.5*m2*v2^2
    total potential energy is zero


    So you get this equation as a result:

    total energy before = total energy after
    kqQ/R = 0.5*m1*v1^2 + 0.5*m2*v2^2

    but that's two unknowns (v1, v2) with only one equation.


    You need another bit of information and that is the force each sphere experiences is the same for all time (action / reaction).

    Look at each of the spheres:

    F = m1*a1
    F = m2*a2

    In terms of velocity:

    F*dt = m1*dv1
    F*dt = m2*dv2

    You should be able to integrate that, equate them, and come up with another relationship between v1 and v2.


    It might be worthwhile to check the answer by doing it the hard way, if only for the exercise. F = ma, F=kqQ/R^2, find position of each sphere as function of time or perhaps velocity as a function of x. In either case you take the limit as t or x goes to infinity to find the final velocities.
     
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