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Reflected particle stream

  1. Sep 26, 2016 #1
    1. The problem statement, all variables and given/known data
    A one-dimensional stream of particles of mass ##m## with density ##\lambda## particles per unit length, moving with speed ##v##, reflects back from a surface, leaving with a different speed ##v'##, as shown. Find the force on the surface

    2. Relevant equations


    3. The attempt at a solution
    The incoming momentum of the particle stream is ##\lambda m v^2## and the outgoing momentum is ##\lambda m v'^2##, thus the change in momentum of the particle stream by the mirror is ##\lambda m (v^2 + v'^2)##. Thus the force on the mirror is also ##\lambda m (v^2 + v'^2)##.

    Is this correct?
     
  2. jcsd
  3. Sep 26, 2016 #2

    mfb

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    Nearly.
    How did you get this?

    Is the particle density still the same for the outgoing stream?
     
  4. Sep 27, 2016 #3
    Well, ##\lambda## is particles per unit length. If we multiply by m we get the mass of the stream per unit length, which is ##\lambda m## If we multiply this by v we get the momentum per unit length, and if we multiply by v again we get the momentum per second, which is exactly the incoming change in momentum. So ##\lambda m v^2## is the incoming change in momentum. Now it seems that the outgoing change in momentum would be ##\lambda m v'^2##, but you are saying that the particle density may not be the same for the outgoing stream, but I don't see how the velocity of the particles changes the number of particles per unit length... Wouldn't they all be the same length from each other, except going slower?
     
  5. Sep 27, 2016 #4

    PeroK

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    Have you ever watched a motor race? Like the formula 1. What happens when the cars get to the fastest part of the track? And the slowest?
     
  6. Sep 27, 2016 #5

    mfb

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    What is the conserved quantity in the stream? It is not the density.
     
  7. Sep 27, 2016 #6
    Is the momentum conserved?
     
  8. Sep 28, 2016 #7

    PeroK

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    Given you are calculating the force caused by the change in momentum, that can hardly be conserved.

    You didn't answer my question about the racing cars. It's the time interval not the spatial interval that remains constant. Think about it.
     
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