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Velocities of colliding particles (Kinetic and Potential Enegies)

  1. Feb 25, 2013 #1
    1. The problem statement, all variables and given/known data

    Two 2.0mm diameter beads, C and D, are 9.0mm apart, measured between their centers. Bead C has mass 1.0g and charge 2.5nC . Bead D has mass 1.5g and charge -1.0nC .
    K= 9*10^9 Nm^2/C^2
    Vinitials =0
    mass of C= 1 g
    mass of D= 1.5 g
    charge of C = 2.5nC
    charge of D = -1nC
    distance r between them= 9mm

    2. Relevant equations

    a) If the beads are released from rest, what is the speed vc at the instant the beads collide?
    b) If the beads are released from rest, what is the speed vd at the instant the beads collide?

    3. The attempt at a solution
    I start by setting up my diagram of the two beads lying on the x-axis. They are 9mm apart from the center, We know initial velocities of both are zero. So I decided to use the conservation of energies formula. Kintial +Uinitial = Kfinal + Ufinal
    C: 1/2mvi^2 + (K*C*D)/rinitial = 1/2mvf^2 + (K*C*D)/rfinal

    Both Initial Kinetic is zero since they start as rest, as well as Final potential because the distance between them would be zero

    So I have (K*C*D)/rinitial = 1/2(mvf^2)
    But we have to take into account the kinetic energy of Bead D as well. When doing a similar problem I was told that both Final Kinetic energies must be added. But I would have two unknowns. Both vfinals. I dont know where to go from here. Any hint?
    Thanks
     
  2. jcsd
  3. Feb 25, 2013 #2

    Simon Bridge

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    Lets see if I understand you:
    ... I thought the potential changed as 1/d (d is the distance between their centers) ... but 1/0 is undefined.

    But surely the beads stop before r=0?

    Apart from that:
    PE before the motion = PE after the motion + total kinetic energy.
    But you also know something else that is conserved.
     
    Last edited: Feb 25, 2013
  4. Mar 1, 2013 #3
    ok thanks for the reply/ so are you saying that the I should use the distance between the centers as rfinal for the PEfinal?
     
  5. Mar 1, 2013 #4

    Simon Bridge

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    You can answer that question for yourself by looking at how the Coulomb force works.

    For two spheres touching - only the charges immediately next to each other have zero separation - the charges on the rest of the two spheres are further apart. But - classically - the area of the contact is zero so the charge in the contact area is also zero.

    If you crunch the numbers you'll find that for r>R (where R is the radius of the bead/sphere), the force acts as if the entire charge is concentrated in the center at r=0. But for r<R, the actual charge distribution matters. Fortunately, two hard beads of charge won't get their centers that close together.
     
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