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Conservation of Momentum- with ramp

  1. Feb 28, 2007 #1
    1. The problem statement, all variables and given/known data

    A .2kg block from rest slides down a frictionless ramp that is .5 meters long. The block lands on a flat surface and moves towards another block that is 1.4 meters away. The other block is not moving and it has a mass of .8kg. The Miu (sp?) of the flat surface is .1. Once the two blocks collide, what is the speed of the second block?

    Here is a picture of the scenario:

    http://i91.photobucket.com/albums/k285/happygolucky2442/abeapic.jpg

    [tex]v^2 = v_0^2 + 2 a \Delta x[/tex]

    2. Relevant equations

    P=MV

    MA1VA1+ MB1VB1=MA2VA2+ MB2VB2

    [tex]f_k = \mu_k N[/tex]

    [tex]\vec{F}_{net} = \Sigma \vec{F} = m \vec{a}[/tex]


    3. The attempt at a solution

    [tex]v^2=2ad[/tex]
    [tex]v^2=9.8(.5)[/tex]
    [tex]v^2=4.9[/tex]
    [tex]v=2.21m/s[/tex]

    [tex]f_k = \mu_k N[/tex]
    Ff= -.1(1.96)
    Ff= -.196

    A=f/m

    a=-.1

    [tex]v^2=2ad[/tex]
    =4.9 - .28
    =4.62
    =2.15

    MA1VA1+ MB1VB1=MA2VA2+ MB2VB2

    .2(2.15) +.8(0) = .2(0) + .8VB2

    .43=.8VB2

    VB2=.5375 m/s

    Please check and see if my answer is correct. Thanks.
     
  2. jcsd
  3. Feb 28, 2007 #2

    PhanthomJay

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    See my comments in red above.
     
    Last edited: Mar 1, 2007
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