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Energy and momentum conservation

  1. Oct 14, 2007 #1
    1.
    A wedge mass M is at rest on horizontal frictionless table. Mass m is on wedge. There is no fricton between parrticle and wedge. Height of wedge = h and angle of incline= theta. The massm slides down wedge from rest and wedge slides left on table. How does linear momentum conservation apply?
    The question also asks Hows does the energy conservation apply in this situation. Write down an equation expressing the consequence of energy conservation.(Hint the square of a velocity is equal to the sum of the squares of its components)

    3. Can you just say that the horizontal components of the momentum must be constant before and after the mass m begins to slide? Therefore mvcos(theta) = Mu where u is the horizontal velocity of the wedge wrt the table and vcos(theta)is the horizontal component of the mass m wrt to the table?

    Does the potential energy mgh = (Mu^2)/2 + (m(vcos(theat))^2)/2

    Bit confused!
     
  2. jcsd
  3. Oct 14, 2007 #2

    Hootenanny

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    Looks ok to me.
    Why are you only considering the horizontal velocity of the mass?
     
  4. Oct 14, 2007 #3
    Should I be equating the vertical components to get

    mgh = m(vsin(theta))^2/2

    What about the kinetic energy of the wedge travelling horizontally?
     
  5. Oct 14, 2007 #4

    Hootenanny

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    You need to consider both the horizontal and vertical components of the mass and the wedge, remember that speed is a scalar quantity and direction doesn't matter.
     
  6. Oct 14, 2007 #5
    So just to clarify:

    You could say mgh = (1/2)mv^2 + (1/2)Mu^2 where v is the velocity of the mass m directed parallel to the slope?
     
  7. Oct 14, 2007 #6

    Hootenanny

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    Looks good to me :approve:
     
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