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Conservation of Energy with Metal Spheres

  1. Oct 17, 2011 #1
    1. The problem statement, all variables and given/known data
    Two small metal spheres with masses 2.0g and 4.0g are tied together by a 5.4-cm-long massless string and are at rest on a frictionless surface. Each is charged to +2.1 μC .

    The string is cut. What is the speed of each sphere when they are far apart?

    2. Relevant equations
    U = Kq1q2/r
    F = K q1/q2/r^2
    m1v1 = m2v2

    3. The attempt at a solution
    I'm not entirely sure how to approach this. I calculated that the energy in the system is .74 J and also the tension in the string is 14 N. How can I use that to my advantage?
     
  2. jcsd
  3. Oct 17, 2011 #2
    When they are far apart the potential energy -> 0, so they energy of the system that begins entirely as potential must now be all kinetic. This should give you one equation. To get a 2nd you know that momentum is conserved as well.
     
  4. Oct 17, 2011 #3
    So, .74 = 1/2mv^2. I picked the 2g sphere, so v = .86 m/s. This doesn't give me the right answer :(
     
  5. Oct 17, 2011 #4
    The initial energy is equal to the sum of the kinetic energies of the two spheres at the end. So it should be .73 = .5*m1*v1^2 + .5m2*v2^2. This leaves you with two unknowns (v1 and v2). To get a 2nd equation you can use conservation of momentum.
     
  6. Oct 17, 2011 #5
    Hmm, it's still not quite right. I have the two equations, .74 = .5*m1*v1^2 + .5m2*v2^2 and 2*v1 = 4*v2. I solved for v1, giving me v1 = 2*v2. Plugging into the first equation I end up with v2 = .35 and v1 = .70.

    I understand that all potential energy turns into kinetic energy and that energy is conserved. Am I just using the wrong equation for conservation of energy?
     
  7. Oct 17, 2011 #6
    I get a different answer, maybe check your work? Remember when you plug v1=2v2 into the equation v1^2=4v2^2. Also don't forget mass is in grams.
     
  8. Oct 17, 2011 #7
    Ohh, I forgot to convert grams into kg. Thank you for all the help!
     
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