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Linear momentum

  1. Apr 14, 2009 #1
    1. The problem statement, all variables and given/known data
    a simple pendulum consist of 1.50Kg mass connect to a cord without any mass or friction. initially the pendulum is vertically positioned when a 2.00Kg mass collides with it, causeing the pendulum to displace vertically upward 1.25m. after the collision, the 2.00Kg mass travels along the frictionless horizontal surface until it meets a 30.0˚ incline with coefficient of kinetic friction of .400. if the mass travels a maximum distance of 1.125m up the incline, determine the initial velocity that the 2.00Kg mass strikes the pendulum with.


    2. Relevant equations
    Pο = P
    mVο+ MVο = MV +mv


    3. The attempt at a solution
     
  2. jcsd
  3. Apr 14, 2009 #2

    Kurdt

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    Where is your attempt?
     
  4. Apr 14, 2009 #3
    I don't know where to start?
     
  5. Apr 14, 2009 #4

    Hootenanny

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    I think that conservation of energy might me a more appropriate concept to use here.

    HINT: How much energy does the 2.00kg mass transfer to the 1.50kg mass?
     
  6. Apr 18, 2009 #5
    so here is what i tried, but I don't know if I'm right i just followed an example in the book that is somehow similar to this problem.
    m1=2.00kg
    m2= 1.50kg
    height to which the block and pendulum swings hf=1.25
    (m1+m2)Vf= m1Vο + 0
    Vο = (m1+m2/m1)Vf
    Vο= (m1+m2/m1) √ghf
    Vο = (2.00kg + 1.50kg / 2.00kg)√2(9.80m/s^2)(1.25m)
    Vο = 8.66M/s?


    http://s44.photobucket.com/albums/f6/kingLA/?action=view&current=04132009360-1.jpg
     
    Last edited: Apr 18, 2009
  7. Apr 18, 2009 #6
    PE=mgh= (2.00kg)(9.80m/s^2)(1.25m) =24.5J
     
  8. Apr 19, 2009 #7

    Hootenanny

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    Does the 2kg mass travelling 1.25m upwards?
     
  9. Apr 19, 2009 #8
    well, what happens is that the 2.00Kg hits the 1.50Kg mass which cause it to swing and then keeps moving straight forward till it reaches 30 degree incline and travels a distance of 1.125m
    I posted a link that has the diagram up above if you like to look at it.
     
  10. Apr 20, 2009 #9

    Hootenanny

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    So if the 1.50kg mass swings upwards, why have you used 2.00kg in your calculation of the potential energy?
     
  11. Nov 13, 2011 #10
    I was wondering if anyone ever figured this one out completely? I think that if you get the initial velocity of the 2.00kg mass at the base of the incline then that can be used as the final velocity after it strikes the 1.50 mass, and so can be used in this equation
    MVinitial + mVinitial = MVfinal + mVfinal
    Is this correct thinking?
     
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