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Help: Two frictionless pucks are placed on a level surface

  1. Aug 19, 2014 #1
    1. The problem statement, all variables and given/known data

    "Two frictionless pucks are placed on a level surface at an initial distance of 20.0 m. The mass of puck 1 is 0.80 kg and it has a charge of +3.0 X 10^-4 C., while puck 2 is 0.40 kg with a charge of +3.0 X 10^-4 C. the initial velocity of puck 1 is 12.0 m/s (E) and the initial velocity of puck 2 is 8.0 m/s (W). Find the minimum separation of the two pucks."


    2. Relevant equations

    Pti= Ptf
    Etki= Etkf+ Ee

    3. The attempt at a solution

    (I dropped the units just so reads easier)

    m1v1+m2(-v2)=(m1+m2)vf
    vf=(0.80)(12)+(0.40)(-8)/0.80+0.40
    vf=5.33 m/s

    Ek1+Ek2=Ee+Ekf
    1/2m1v1^2 + 1/2m2(-v2^2) = kq^2/r + 1/2(m1+m2)(v')^2
    1/2(0.80)(12)^2 + 1/2(0.40)(-8.0)^2 = 1/2(0.80+0.40)(5.33)^2+(9.0X10^9)(+3.0X10^-4)^2/r
    Isolate for r
    r= 1/2(0.40)(-8.0)^2= 15.2 m= 15m

    Can someone advise if this is correct? I am unsure about this answer as it isn't that different from the original distance. In a similar example (as per homework manual), it is taking the initial electric potential energy as negligible. Is this right?
     
  2. jcsd
  3. Aug 19, 2014 #2

    TSny

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    Hello, Jaime.

    Is there a reason that you didn't make use of the initial distance of 20 m in your energy calculation?
     
  4. Aug 21, 2014 #3
    I think that I am having trouble understanding how the conservation of energy is applied to the moving charges.
    From what I am now understanding is that the conservation of energy equation, when applied to these problems should be...Etki + Eei= Etkf+ Eef. Is this right?
    In this case, I am to interpret that because they are at a set distance apart (not very far distance apart), there is electric potential energy initially so I need to factor in the initial distance with the Ee equation. So...

    Etki + Eei= Etkf+ Eef
    1/2m1v1^2 + 1/2m2(-v2^2) + kq1q2/ri = 1/2Vf^2(m1+m2) + kq1q2/rf
    1/2(0.80)(12)^2 + 1/2(0.40)(-8.0)^2 + ((9.0X10^9)(+3.0X10^-4)^2))/20= 1/2(0.80+0.40)(5.33)^2 (9.0X10^9)(+3.0X10^-4)^2/r

    r= 810/93.86= 8.63m= 8.6m

    Can you also give some advice on how you approach these problems? Very frustrating as I don't feel like I am "getting this" quickly. I'm having trouble sorting out how to approach a problem like this and when to use cons. of energy eq. or cons. of momentum eq. Thank you!
     
  5. Aug 21, 2014 #4

    BiGyElLoWhAt

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    Similar type of problem, but using something that you might know a little better:

    I have a spring with spring constant k, and I put a block of mass m ontop (the spring is attached to the ground at one end and oriented vertically) I stretch the spring by pulling the block upwards 1 meter. What's the closest the block gets to the ground?

    If you can solve this you should be able to solve the original problem. =]
     
  6. Aug 21, 2014 #5

    BiGyElLoWhAt

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    I guess i should give the spring a length =/

    5m haha
     
  7. Aug 21, 2014 #6

    haruspex

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    You have to think about whether there's any work lost or gained (internal friction, external forces, impacts) and whether momentum, angular momentum could change (external forces and torques).
    For momentum, you may be able to find a particular direction in which it will be conserved, because the only external force is orthogonal to that. Similarly, you may be able to find a reference point about which angular momentum is conserved, because the only external forces act through that point.
    Whichever conservation laws pass this test can be used. Which are the most useful for a particular question is another matter, but most questions are set such that you need to use all conservation laws that apply.
     
  8. Aug 23, 2014 #7
    Thanks everyone!
     
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