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1D Collision / Charges / Coulomb's Law

  1. Jan 29, 2015 #1
    1. The problem statement, all variables and given/known data

    Two frictionless pucks are placed on a level surface with an initial distance of 20.0 m. Puck 1 has a mass of 0.80 kg and a charge of + 3 E-4 C while puck 2 has a mass of 0.4 kg and a charge of +3 E-4 C. 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?

    Untitled-6.jpg

    2. Relevant equations

    KE 1 + KE 2 + EE = EE' + KE'
    Pt=Pt'

    3. The attempt at a solution

    First I used conservation of momentum to calculate v for each mass at the min separation.
    Both pucks have the same velocity.
    Then just some algebra to isolate v'.

    m1v1+m2(-v2)=(m1+m2)v'

    v'=(0.80)(12)+(0.40)(-8)/0.80+0.40

    v'=5.33 m/s [East]

    This part I can use conservation of energy (kinetic and electric potential). I don't think the energy here would be negligible and doing calculations I see big differences in the answer.

    KE 1 + KE 2 + EE = EE' + KE'

    ½ m1v1² + ½ m2 (-v2²) + kq²/r = kq²/r' + ½ (m1 + m2)(v')²

    ½ (0.80)(12)² + ½ (0.40)(-8)² + (9.0 E+9)(+3.0 E-4)²/20 = (9.0 E+9)(+3.0 E-4)²/r' + ½ (0.80 + 0.40)(5.33)²

    r' = 7.34 m

    I've found so many attempts at this and there is so many answers.. Many I know are wrong and this one I think is right.

    Please can you review this and see what you think?

    Thank you
    Julian
     
    Last edited: Jan 29, 2015
  2. jcsd
  3. Jan 29, 2015 #2

    DEvens

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    I have a little difficulty following what you are doing. I see a bunch of equations with little explanation of what they are. And then suddenly at the end out pops a distance.

    What is the condition you used to determine the minimum separation? Can you put in a few words to explain what each of those equations is? I can kind of guess what you are trying to do, but it's pretty foggy.
     
  4. Jan 29, 2015 #3
    I might be doing the calculations wrong. I get 7.34 or 8.63... Wolfram gave me 8.63 and I keep checking the arrhythmic. I think it might be my scientific calculator has issues rounding.

    I tried to be more specific if you look back I did an edit.
     
  5. Jan 29, 2015 #4

    BvU

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    DE's post made you add comments, which now makes his (her ?) post look strange, but never mind.

    Doesn't look right to me. Did you forget the second term on the right ?

    A few comments from my part:

    Do you realize the 5.33 m/s is the center of mass speed ? Working the whole thing in the c.o.m. is a little easier to follow.

    The exercise mentions a surface, not a rail. But I agree that a minimum distance is achieved when the puck move head-on.
     
  6. Jan 29, 2015 #5

    BvU

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    And sure enough, the diffference (7.34 or 8.63) is the 17.0667 J kinetic energy of the c.o.m :)
     
  7. Jan 29, 2015 #6
    1/2(0.80)(12)^2 + 1/2(0.40)(-8)^2 + (9.0e9)(+3.0e-4)^2/(20) = (9.0e9)(+3.0e-4)^2/r + 1/2(0.80 + 0.40)(5.33)^2

    This gives me 8.63 m which now I think this is right...

    But I'm curious as to which second term I may have forgot... I don't exactly have a math mind. Also I don't know what you mean about c.o.m. I'm new, sorry.
     
  8. Jan 29, 2015 #7
    The centre of mass of the system of particles has a velocity given by [(m1)(v1)+(m2)(v2)]/(m1+m2). Here v1 and v2 are the respective velocities and not just speeds. You start seeing the two pucks in the centre of mass frame. Minimum separation will be attained when the velocity of both the pucks becomes the same as the velocity of the centre of mass,.i.e., in the centre of mass reference frame their kinetic energy becomes zero. Equate the initial energy K.E.+P.E. in the centre of mass frame of reference to the final energy in the same frame(when there will be only potential energy and no kinetic energy) and you'll get the answer. Your approach also seems right. I don't know why you're not getting the answer.
     
  9. Jan 29, 2015 #8
    Is 8.63 right? I think so... I think it's my scientific calculator and me working together poorly because when I do it with wolfram I get that answer. Maybe the calculator doesn't round well enough for some decimal place or maybe I missed a parenthesis Maybe I should stop being lazy and calculate things separately. :P

    I'm putting together what you mean, I'm really uncomfortable with this charge stuff which scares me because I know its easy just I'm so uninterested with these types of problems.

    Thanks for your response.
     
  10. Jan 29, 2015 #9
    Haha... yes 8.63 should be the right answer...
     
  11. Jan 29, 2015 #10

    BvU

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    Well, it's definitely not the calculator.
    You can claim to be uninterested, but you sure did a good job composing a decent post and did all the work to obtain an answer. Even bothered to ensure it's the right answer. So why not harvest the reward and feel a little satisfied ? I know it's not going to get you the Nobel prize, but it is a nice exercise :)
     
  12. Jan 29, 2015 #11

    BvU

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    Since we already have the answer, I can dump your calculations here .
    Alas, it can't serve as a demo on how to do it: the right way to do it is to set up
    one concise formula using symbols only (so that you can check what cancels
    and check the dimension).

    Code (Text):

    Given:
    m_1        0.8      kg
    v_1       12        m/s
    q_1        3.00E-04 C

    m_2        0.4      kg
    v_2       -8        m/s
    q_2        3.00E-04 C

    kc         9.00E+09 Nm^2/C^2

    r_0        2.00E+01 m

    Calculated:
    v_cm       5.33     m/s     =(m_1*v_1+m_2*v_2)/(m_1+m_2)

    p_1        9.60     kg m/s  = m_1*v_1

    p_2       -3.20     kg m/s  = m_2*v_2

    p_tot      6.40     kg m/s  = p_1+p_2

    Ek_1      57.60     J       = v_1*p_1/2
    Ek_2      12.80     J       = v_2*p_2/2
    Ek_tot    70.40     J       = Ek_1+Ek_2

    E_el_0    40.50     J       = kc*q_1*q_2/r_0

    E_tot_0  110.90     J       = Ek_tot+E_el_0

    E_tot_f  110.9      J       = E_tot_0

    E_k_f     17.07     J       = v_cm*p_tot/2

    E_el_f    93.83     J       = E_tot_f-E_k_f

    r_f        8.63     m       =(kc*q_1*q_2)/E_el_f

    Forgot E_k_f ?
                7.30     m       =(kc*q_1*q_2)/E_tot_f
     
    (shows how dinosaurs can use this horrid excel to do things. Think I'll try Wolfie too, some time..)


    very useful, this markup stuff -- if you know how to use it :)
    Ah,wait, the preview isn't as nice as the final post.
     
  13. Jan 29, 2015 #12
    I find my casio scientific will give me 4.3 if an answer is 4.39 with a high power. I wonder how that affects things when you have long functions that very between regular and large powers. I know it wouldn't be that much and it goes down to the purpose of sig digits.

    I'm only interested so far as I know it's going to relevant to understand basic things like this in university. And its really going to matter on exams I don't want to mess something stupid like that up because you can lose a lot of the final grade. bI love to learn about theories in QM and relativity and everything about space is so cool,
     
  14. Jan 29, 2015 #13
    Ahhh so its excel... I always wondered where people did that. I thought there was some secret program or script I didn't know about. The more you know.
     
  15. Jan 29, 2015 #14
    I like the way wolfram can get you any algebra problem really fast but you have to pay to get step by step solutions. I'm debating whether it would help or "ruin" the learning process.
     
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