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Relative velocity of blocks

  1. Apr 29, 2013 #1
    1. The problem statement, all variables and given/known data
    (see attachment)


    2. Relevant equations



    3. The attempt at a solution
    Consider the system at time t (see attachment, I am representing the blocks as point masses)
    The component of acceleration along the string is ##a_0\cos \theta##. This is the acceleration of the other two blocks at the sides. The component of acceleration along the horizontal direction is ##a_0 \sin \theta \cos \theta##.
    [tex]v\frac{dv}{dx}=a_0 \sin \theta \cos \theta[/tex]
    As ##x=l\sin \theta##, ##dx=l\cos \theta d\theta##
    [tex]vdv=a_0 \sin \theta \cos^2 \theta d\theta[/tex]
    The problem is, what should be the limits for ##\theta##? And is my working even correct?

    Any help is appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Apr 29, 2013 #2

    mfb

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    I would consider this in the frame of block B. You get a very common physical setup, which can be solved without differential equations.

    The limits for θ are given by the initial layout and the final one (collision).
     
  4. Apr 29, 2013 #3
    So there will be pseudo force acting on the other two blocks? But then how would I calculate the acceleration in the horizontal direction?

    Are the limits 0 to pi/2?
     
  5. Apr 29, 2013 #4

    mfb

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    Once you see the analogy, it is easy ;). You can even treat it as real force...

    Sure.
     
  6. Apr 30, 2013 #5
    Sorry, I am still lost on this one. :frown:

    But that doesn't give me the right answer. I had the following expression:
    [tex]\int_{0}^v vdv=\int_{0}^{\pi/2} a_0 \ell \sin \theta \cos^2 \theta d\theta[/tex]
    [tex]\frac{v^2}{2}=\frac{a_0 \ell}{3}[/tex]
    This is incorrect as per the answer key. :confused:
     
  7. Apr 30, 2013 #6

    mfb

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    Both blocks will feel a fictious force of m*a0 "downwards" acting on them.
    Just consider that as real force, and you get a
    pendulum in a gravitational field.
     
  8. Apr 30, 2013 #7
    How is that a pendulum in a "gravitational field" when the motion takes horizontally? :confused:

    And why does my DE gives the wrong answer? :(
     
  9. Apr 30, 2013 #8

    mfb

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    The motion is the same as if the setup would be in a gravitational field (with gravity acting "sidewards"). It is not in a relevant real gravitational field, of course.

    I don't know. What is v (in which frame)? Why did you introduce x?

    The question asks to find the relative velocity of A and C, while you just calculated the velocity of A (or C). That factor of 3 should not appear, however.
     
  10. Apr 30, 2013 #9
    Ah yes, I get your point. It turns out to be same as if the motion takes place in gravity.
    [tex]ma_0\ell=\frac{1}{2}mv^2 \Rightarrow v=\sqrt{2a_0 \ell}[/tex]
    Hence the relative velocity before striking is ##2\sqrt{2a_0 \ell}##

    Thank you mfb! :smile:

    v is in inertial frame.

    Acceleration is vdv/dx so I had to introduce x and express it in terms of ##\theta##.
     
  11. May 1, 2013 #10

    mfb

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    That is complicated. In particular, you don't get the relative velocity with that, and "v" as integration limit does not make sense (it is a vector).

    Anyway, treating it like a pendulum is way easier.
     
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