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Momentum and Energy- Major help needed

  1. Oct 25, 2006 #1
    A small block of mass m1 = 0.300 kg is released from rest at the top of a curved wedge of mass m2 = 2.00 kg, which sits on a frictionless horizontal surface as in Figure P6.59a. When the block leaves the wedge, its velocity is measured to be 4.00 m/s to the right, as in Figure P6.59b.


    (a) What is the velocity of the wedge after the block reaches the horizontal surface?
    I found the velocity to be -.6 m/s. I did 0=m2v2 + M(block)V(block)

    (b) What is the height h of the wedge?
    I dont know how to get the height of the wedge, please help

    I know the energy equation will be

    mgh= .5mv^2

    Please help
  2. jcsd
  3. Oct 25, 2006 #2
    please can anyone help?
  4. Oct 25, 2006 #3


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    If you find the velocity of the block and the wedge, you should be able to apply concervation of energy to determine the height of the wedge. This means one point you have to apply the formula to is when M1 is at rest at the top of the wedge. use mgh = (whatever you decide) to determine the height of M1 relative to the ground. This is the same as the height of the wedge.
  5. Oct 26, 2006 #4
    I still dont understand. Can you show me how to set up the equation

    Would it be

    mgh= .5mv^2 Now which mass and which velocity should I use to find h here
  6. Oct 26, 2006 #5


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    Homework Helper

    As AVD said, you have to use conservation of energy on the system consisting of the block and the wedge. So, write down the sum of kinetic and potential energy for these two momens and set them to equal one another; a) the block rests on the wedge, b) the block is on the horizontal surface.
  7. Oct 26, 2006 #6
    ^ I am really sorry, I tried setting it up but I am getting it wrong.
  8. Oct 26, 2006 #7
    would it be?

    KE + PE = KE + PE

    and go from there?
  9. Oct 26, 2006 #8
    I tried that

    .5(.3)(4^2) + .3(9.8)h= .5(32)(.6^2) + 2(9.8)h

    is that right

    It didnt work for some reason, I got .112 is height, or am I doing something wrong
  10. Oct 26, 2006 #9
    am I showing in notes from that this may be an inelastic collision (m1v1i + m2v2i = (m1 + m2)Vf)... sorry Im not help to you :(
  11. Oct 26, 2006 #10


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    PE(block) = KE(block) + KE(wedge), where PE(block) is for the moment the block is resting on the wedge, and KE(block) and KE(wedge) are for the moment when the block is on the horizontal surface, and the wedge has the velocity you calculated in part 1.
  12. Oct 26, 2006 #11
    ^ Its ok, I hope someone else helps
  13. Oct 26, 2006 #12
    Thanks so much radou, but why dont we use the PE of the wedge?
  14. Oct 26, 2006 #13


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    The wedge has no potential energy, since it is on the horizontal surface all the time.
  15. Oct 26, 2006 #14
    ^ oh ok thanks a lot
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