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Block Moving Down Slide

  1. May 30, 2007 #1
    A small block of mass m1 = 0.480 kg is released from rest at the top of a curved-shaped frictionless wedge of mass m2 = 3.00 kg, which sits on a frictionless horizontal surface as in Figure P9.60a. When the block leaves the wedge, its velocity is measured to be v1 = 3.00 m/s to the right, as in Figure P9.60b.

    Figure P9.60

    (a) What is the velocity of the wedge after the block reaches the horizontal surface?
    m/s to the left


    (b) What is the height h of the wedge?
    m



    You must use conservation of momentum:

    m1v1 = m2v2

    m1 - mass of block
    m2 - mass of wedge
    v1 - velocity of block
    v2 - velocity of wedge

    Plug in your values.

    (0.480 kg)(3.00 m/s) = (3.00 kg)v2

    Multiply.

    1.44 kg m/s = (3.00 kg)v2

    Divide both sides of the equation by 3.00 kg.

    v2 = 0.48 m/s to the left


    Part B

    PEi + KEi = PEf + KEf

    mgh + 0.5mvi2 = mgh + 0.5mvf2

    mgh + .5mvi^2 = mgh + .5 mvf^2

    (0.48)(9.8)(h) + .5(.48)(3^2) = 0 (as the height is now zero) + .5(.48)(.48^2)

    When i solve for this I get -.4474 which is incorrect.

    What am I doing wrong
     
  2. jcsd
  3. May 30, 2007 #2
    Wow, Am I an idiot at times.....

    I figured it out...Here is the correct way to solve the second half:

    (0.480)(9.8)h = (.5)(.48)(3^2) + (.5)(3)(.48^2)

    h = 0.53265

    So Simple lol
     
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