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Electric Potential Energy Question With two Similarly charged Objects

  1. Dec 21, 2011 #1
    1. The problem statement, all variables and given/known data
    Two frictionless pucks are placed on a level surface as shown, at an initial distance of 20m.

    Puck 1 has a mass of 0.8 kg and a charge of + 3x10^-4 while puck 2 has a mass of 0.4 kg and a charge of +3 x10^-4.

    The initial velocity of puck 1 is 12 m/s [E] and the initial velocity of puck 2 is 8 m/s [W]. Find the minimum separation of the two pucks; the min distance between the two pucks

    2. Relevant equations

    Pt=Pt'
    M1V1+M2V2=(M1+M2)V
    Ek=1/2mv2


    3. The attempt at a solution

    Conservation of momentum is used to find the velocity of closest approach, since it must be constant for both.

    Pt=Pt'
    M1V1+M2V2=(M1+M2)V
    (0.8)(12)+(0.4)(8)=(1.2)V (Should I set East or west as +, and therefore should 1 velocity be negative?)

    V=10.7 m/s when pucks reach point of closest approach

    Ek1+Ek2= Ee +Ek
    1/2M1v12 +1/2M2v22=(kq1q2/r) + 1/2(M1+M2)V

    1/2(0.8)(12)2+1/2(0.4)(8)2= [(9x10^9)*(3x10-4)2/r] + [1/2(1.2)*(10.7)]

    (70.4-6.42)r=(9x109)(3x10-4)2
    r= 12.66 m

    SO my question is, is 12.7 m the closest they reach or would it be 20-12.7?


    I think I put all the variables into the equations correctly and accounted for conservation of energy and momentum, any help would be great. Im quite confused.

    Thanks everyone and merry Christmas, happy holidays and a happy winter solstice
     
  2. jcsd
  3. Dec 21, 2011 #2

    SammyS

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    The pucks travel in opposite directions, so, yes, their initial velocities should have opposite sign. You pick what direction is positive. That is often East, but it also makes sense to have the sign of the one with greater speed. In this case, that's also East.

    It's also possible to work this out in the center of mass reference frame, but that's not necessary.

    Added in Edit:

    The initial potential energy is not zero. you need to include it of the left side of your equation:
    (1/2)M1v12 +(1/2)M2v22 = (kq1q2/r) + (1/2)(M1+M2)V2
     
    Last edited: Dec 21, 2011
  4. Dec 23, 2011 #3
    DO you have any insight as to whether im close with my answer of 12.7 m?

    I cant help but wonder if it should be 20-12.7m but I can't really explain why.
     
  5. Dec 23, 2011 #4
    I have another thought about the question ...
    if we work in the frame of COM of the pucks ... it has no external force acting so it is in inertial frame ...

    So simply using energy conservation in that case will give a solution because final velocities will be zero of both pucks ... (i guess :biggrin: correct me if i'm wrong) ...
     
  6. Jan 18, 2012 #5
    Im still working at this question, 12.7 M seems like way to big of a distance to be the minmum distance these pucks come between each other. Am I doing something wrong in my calculations/equations?
     
  7. Jan 18, 2012 #6
    I am also wondering is there electric potential energy at 20 m away so it should be

    EE + EΔK=EE

    This gives me an answer of 7.3 m, a little better but that still seems huge.. I also get 11.5 m sometimes as well.

    Im really confused here.
     
  8. Jan 18, 2012 #7

    SammyS

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    How much force do the pucks exert on each other at those distances ?

    Added in Edit:

    I get an answer of about 8.6 meters for the minimum distance.
     
    Last edited: Jan 18, 2012
  9. Jan 18, 2012 #8
    40.5 n?
     
  10. Jan 18, 2012 #9

    SammyS

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    Well 40 Newtons produces a fairly large acceleration on objects with masses of 0.4 kg and 0.8 kg.; 100m/s2 and 50 m/s2 respectively.

    However, I get a force of more like 10 Newtons at 8.6 meters separation. Even that gives pretty big accelerations on the pucks.
     
  11. Jan 18, 2012 #10
    How did you get to 8.6 m seperation?


    this equation?

    EE+EΔK=EE+EK

    Did you also calculate the mass and momentum as one singular object i.e (M1+M2)V2 for the right side of the equation ?


    I really want to understand this question so I appreciate any help you can offer a lot!
     
  12. Jan 18, 2012 #11

    SammyS

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    Actually, I did it in the center of mass system, but that doesn't change the answer.

    (M1+M2)V2 is not momentum. Velocity should not be squared. (M1+M2)V is the momentum of the system when the pucks are at minimum separation. Conservation of momentum gives that V=(20/3) m/s @ min. separation. Remember, v2 is negative, initially.

    The Kinetic Energy of the system is (1/2)(M1+M2)V2 at minimum separation.

    What is the K.E. at 20 m separation?

    What is the Potential Energy at 20 m separation?

    Conservation of energy then gives the P.E. at minimum separation. Solve this for r .
     
  13. Jan 18, 2012 #12
    Should it look like this?

    KE (@20 m) + EE(@ 20 m) = KE(@ distance of closest approach) + EE (solve for r)

    and KE= (M1+M2)V2 ( on both sides of the equation)?
     
  14. Jan 18, 2012 #13
    so I get 10.67 m/s as velocity at minimum seperation M1v1+M2V2=(M1+M2)V

    v=10.67 m/s

    EK1+EK2+EE=EE+(M1+M2)(10.67)

    1/2m1v12+1/2m2v22 + Kq2/20= kq2/r+1/2(M1+M2)V ( I didn't square V here)

    R= 7.75 m

    What am I doing wrong?
     
  15. Jan 18, 2012 #14
    OK I found out how to get 8.6 I was just confusing myself, now how are you getting 10 N?

    I used

    (9x10^9)(3x10^-4)2/8.6

    EE= 94.18 N
     
    Last edited: Jan 18, 2012
  16. Jan 18, 2012 #15

    SammyS

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    The velocity at minimum separation (which is also the velocity of the center of mass) is (16/3) m/s. I had the wrong value in my previous post.
    Remember, v1 and v2 have opposite signs.​

    What is momentum, (M1+M2)(10.67), doing in the following equation?
    EK1+EK2+EE=EE+(M1+M2)(10.67)​
    Take that out & you may get the right answer.
     
  17. Jan 18, 2012 #16
    Did you see this?
     
  18. Jan 18, 2012 #17

    lightgrav

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    Harley, the initial (smallest) Force is q kQ/r^2 , with r starting at 20m => 2.025 N.
    ... IF it would get to 8.6m distance, then (9E9)(3E-4)^2/(8.6)^2 = 10.95 N.

    Your post #13 is pretty much right, except that you did not square V(closest)
    ... which doesn't matter because (using c.o.mass frame) V(closest) is zero.
    use velocities relative to c.o.m. and you'll get 19m as closest approach.
    (WTF?)
    the big mass starts off 1.333 m/s faster than it is at closest approach,
    the small mass starts off -2.666 m/s different than it is at closest approach.
    2 N causes the big mass to accelerate a=F/m = 2.5 m/s/s ,
    so the closest approach occurs in about dv/a=dt= .53 seconds.
    the big one's _avg_ speed relative to c.o.m (10.67 m/s) was 0.666 m/s
    so it went about 0.333 m relative to the c.o.m. in that time.
    small mass moves twice as far (0.666 m) relative to the center-of-mass
    before they're both going 10.666 m/s.

    the center-of-mass moves 5.6 meters before the closest approach, so a LOT of Energy gets transfered!
     
  19. Jan 18, 2012 #18

    SammyS

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    You need to square the 8.6 .

    That will then give you 10.95 N .
     
  20. Jan 19, 2012 #19
    I don't think R is squared in electric potential energy , only in electric force and field?

    So are you guys calculating electric force? Its unneeded for the answer ( I just need distance of closest approach) Im just curious

    thanks everyone for the help!
     
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