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Conservation of mometum

  1. Sep 14, 2008 #1
    Let there be an incline which is placed on a frictionless surface. A block is placed at th top of the incline. The surface of th incline is also frictionless. The block starts falling down due to gravity. Consider only the incline and the block in the system. Is th momentum conserved in y direction.
     
  2. jcsd
  3. Sep 14, 2008 #2

    Doc Al

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    What external forces act on this system?
     
  4. Sep 14, 2008 #3
    Well, if the mass of the incline is M and that of block is m.
    The gravitational force =(M+m)g
    and the normal reaction from the ground which is equal to the weight of the to bodies. Hence the net force being zero.
     
  5. Sep 14, 2008 #4

    Doc Al

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    Sure about that? (Consider what happens to the center of mass of the system.)
     
  6. Sep 14, 2008 #5
    Thats exactly the problem, The block comes down so it will have momentum in y direction. The incline is sure not to go up. That means momentum is created in the y direction. and the there is no net force in the y direction. How is this possible?
    Yes, I think that the normal will also be equal to (M+m)g . Is it wrong?
     
  7. Sep 14, 2008 #6

    Doc Al

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    It's not.
    Yes, that's wrong.

    Example: Stand on a scale while you hold a bowling ball. The normal force (scale reading) will be (M + m)g. But what if you quickly lower the ball? Will the normal force remain constant?
     
  8. Sep 14, 2008 #7
    ok.

    Nope the reading must not be constant. This would mean that I cant take the incline and block as one system. Then how will I figure out the magnitude of th normal?
     
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