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Work - energy theorem and conservation of momentum question

  1. Apr 19, 2010 #1
    1. The problem statement, all variables and given/known data

    There is a hemispherical track of radius R = 4m. A block B is lying at rest at the bottom of the track. A block A is pushed along the horizontal portion of the track with speed u. Collision between A and B is perfectly elastic. Find u such that B reaches P (the end of hemispherical portion of track). (See figure for clarity)

    Given: mass of A = 2 kg
    mass of B = 4 kg

    track is frictionless



    3. The attempt at a solution


    velocity of B after collision should be [tex]\sqrt{}[/tex]2gR


    let velocity of A after collision be v

    u = v + 2[tex]\sqrt{}[/tex]2gR

    [tex]\sqrt{}[/tex]2gR - v = u

    ------------------------------------------ adding
    3[tex]\sqrt{}[/tex]2gR = 2u

    u = 12 m/s (approx)



    by conserving momentum and condition for elastic collision, u = 12 m/s (approx)

    is the solution correct? the answer is supposed to be 10 m/s.
     

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    Last edited: Apr 19, 2010
  2. jcsd
  3. Apr 19, 2010 #2

    Doc Al

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    Staff: Mentor

    Where's your solution?
     
  4. Apr 19, 2010 #3

    Doc Al

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    Staff: Mentor

    This looks good.

    Where did you get this?
     
  5. Apr 20, 2010 #4
    Thanks doc, for showing interest in this question.


    for elastic collision, coefficient of restitution is 1.

    so,
    velocity of separation = (-1) velocity of approach

    [tex]\sqrt{}[/tex]2gR - v = -( 0 - u)
    = u
     
  6. Apr 20, 2010 #5

    Doc Al

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    Staff: Mentor

    OK, I see what you did. Makes sense.

    The problem is that you found the pre-collision speed of A at the bottom of the track, but they want the speed while it's still on the horizontal portion. Redo it with that in mind and you'll get the right answer.
     
  7. Apr 20, 2010 #6
    Thanks, i got it. I got to avoid such mistakes, really.....
     
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