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Homework Help: Finding the maximum acceleration

  1. Dec 18, 2013 #1
    1. The problem statement, all variables and given/known data
    attachment.php?attachmentid=64903&stc=1&d=1387359170.png


    2. Relevant equations



    3. The attempt at a solution
    I don't know how to make the equations here.

    When the balls are penetrating each other, a cavity is being formed due to overlap of the negative and positive charges. The size of this cavity keeps on increasing. The trouble is I cannot figure out the instant of maximum velocity and the equations to write down. I need a few hints to begin with.

    Any help is appreciated. Thanks!
     

    Attached Files:

    Last edited: Dec 18, 2013
  2. jcsd
  3. Dec 18, 2013 #2
    The attachment is not shown. It is said to be invalid.
     
  4. Dec 18, 2013 #3
    Does it work now?
     
  5. Dec 18, 2013 #4
    If a function reaches a maximum, what happens to its derivative? What does that mean in this case?
     
  6. Dec 18, 2013 #5
    The derivative is zero but to find the maximum, I need the function. Can you please give me a few hints about finding the function? :)
     
  7. Dec 18, 2013 #6
    What is the derivative of velocity? What does that mean, physically, its being zero? When/where can that happen in this situation?
     
  8. Dec 18, 2013 #7
    Acceleration.

    It means acceleration is zero, the body is at rest or is moving with a constant velocity.
    I am not sure so I make a guess. It can happen when both the balls completely overlap. I think when the balls completely overlap, the net charge is zero so no electric force is acting, right?
     
  9. Dec 18, 2013 #8
    I agree with that. That gives you some important knowledge: you know at what separation between the centers their speed was 10 m/s. Can you proceed further?
     
  10. Dec 18, 2013 #9
    The velocity is maximum when the separation is zero, right?

    I can't think of anything about the next step. :(
     
  11. Dec 18, 2013 #10
    What is the potential energy of the system when the separation is zero?
     
  12. Dec 18, 2013 #11
    Umm...zero?
     
  13. Dec 18, 2013 #12
    Why do you think it is zero?
     
  14. Dec 18, 2013 #13
    There is no charge when separation is zero which led me to think that it is zero. Am I wrong?
     
  15. Dec 18, 2013 #14
    No charge means the force is zero, but that does not mean the potential energy is zero. We typically define potential energy in such a way that it is zero at the infinite separation. If it is defined like that, then your suggestion that the potential energy is zero at the zero separation is problematic with respect to conservation of energy. Can you see that?
     
  16. Dec 18, 2013 #15
    I am not sure so I thought of starting with the definitions.

    $$U(R)-U(0)=-\int_0^R \frac{kQq}{r^2}dr$$
    (where both Q and q are positive)

    $$U(R)-U(0)=\lim_{r\rightarrow 0}\frac{kQq}{r}-\frac{kQq}{R}$$

    I don't see how defining U(0)=0 leads to complications. :(
     
  17. Dec 18, 2013 #16
    I am not sure what that equation means. What is ##R##? Is it the size of the sphere? Then I do not see where you factor in the separation. If it is the separation, then I do not see how you account for separations less than the diameter of the spheres.

    Well, if it is zero both at the infinity and at the zero separation, then the kinetic energy must also be zero at the zero separation if it was zero at the infinity. Which, I think, quite obviously is an impossible result.
     
  18. Dec 18, 2013 #17
    Sorry, I should have been clear, I was taking a general case when a point charge q is taken from r=0 to r=R from a fixed point charge Q.

    I think I understand your point but I am still clueless about the given problem.
     
  19. Dec 18, 2013 #18
    Here is a question: what does the graph of the potential energy vs the separation look like?
     
  20. Dec 18, 2013 #19
    When the separation between the centre of balls is 2R, the potential energy is ##-kQ^2/(2R)##. For separation greater than 2R, the graph increases and tends to zero at infinity. For separation less than 2R, I am not sure how the graph would look like as they begin to penetrate each other. Would the potential energy still be ##-kQ^2/r## where r is the separation between the centre of balls and r<2R?
     
  21. Dec 18, 2013 #20
    Remember that you have already established that the kinetic energy has a maximum at the zero separation. What about the potential energy?
     
  22. Dec 18, 2013 #21
    If potential energy is still given ##-kQ^2/r##, then at zero separation, it shoots to ##-\infty##. :confused:

    But initially, the energy of system is zero, so kinetic energy at zero separation must be equal to the potential energy.
     
  23. Dec 18, 2013 #22
    What is important is that the potential energy is some constant minus the potential energy. So when one is at a maximum, the other is at a minimum. Hence, the potential energy must be at a minimum at the zero separation. It can be also seen from the fact that at the zero separation the force is zero, and the force is the derivative of the potential energy.

    Yet another way would be by saying "it is obvious that a stable equilibrium configuration of two equal shape, equal mass, opposite charge clouds is when they coincide completely".

    So, what does the total graph look like? Where is it convex, where is it concave? Where is the inflection point? What is its significance?
     
  24. Dec 18, 2013 #23
    I don't understand why you ask me about the graph when it is simply a -1/r relationship.

    http://www.wolframalpha.com/input/?i=y=-1/x
     
  25. Dec 18, 2013 #24
    The shape in the lower-right quadrant is correct only when the separation is greater than the diameter. At smaller separations, the graph is different. Which is obvious from the fact that at the zero separation it has a minimum, not a singularity.
     
  26. Dec 18, 2013 #25
    How do I find the expression for potential energy when the balls penetrate each other? :confused:
     
    Last edited: Dec 18, 2013
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