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Block on Wedge

  1. Aug 25, 2009 #1
    1. The problem statement, all variables and given/known data

    Attached in the files, scan from Resnick and Halliday. Also the answers (notice that I use a different meaning for [tex]u[/tex], but it shouldn't matter.

    2. Relevant equations

    Conservation of momentum, and conservation of energy.

    3. The attempt at a solution

    Initial Energy of the system is [tex]mgh[/tex], initial momentum is 0.
    I'll call the velocity of the wedge [tex]u[/tex] and the velocity of the block [tex]v[/tex].
    The equations:
    [tex]2mgh = mv^2 + Mu^2[/tex] - all potential energy goes to the velocities of both objects.
    [tex]Mu = -mv \cos \alpha[/tex] - conservation of momentum - only the horizontal component of [tex]v[/tex] is taken into consideration.

    I end up having this as an answer:
    [tex]u = \sqrt{\frac{2gh}{M^2 + Mm \cos ^2 \alpha}}m \cos \alpha[/tex]
    Which is, not correct according to the answer.
    What did I do wrong?
     

    Attached Files:

    Last edited: Aug 25, 2009
  2. jcsd
  3. Aug 25, 2009 #2

    kuruman

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    I can't see the pictures pending approval, so I guessed at what the question is asking. I assumed that you are looking for the speed of the wedge and I got the same answer as you. What is the answer in the back of the book?
     
  4. Aug 25, 2009 #3
    I'll type it in:
    A block of mass m rests on a wedge of mass M which, in turn, rests on a horizontal table, as shown in Fig. All surfaces are frictionless. If the system starts at rest with point P of the block a distance h above the table, find the velocity of the wedge the instant point P touches the table.

    Answer is:
    [tex](\frac{u \cos \alpha}{\sqrt{1-u \cos ^2 \alpha}})\sqrt{2gh}[/tex]

    [tex]u = \frac{m}{m+M}[/tex]

    It has some similarities to my answer, so I guess the general idea I had in mind was correct. My guess is my conservation equations are not accurate (because I don't feel sure about them).
     
  5. Aug 25, 2009 #4

    kuruman

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    The problem with your expression is that the x component of the block's velocity is vblock*cosα relative to the wedge, not relative to the table. Relative to the table it is vblock*cosα - Vwedge.
     
  6. Aug 25, 2009 #5
    so I need to replace it only in the momenton equation? What about the energy equation?
     
  7. Aug 25, 2009 #6

    Doc Al

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    You'll need to modify both equations.
     
  8. Aug 25, 2009 #7
    Meaning in the energy equation the vertical component of the block's velocity stays the same while the horizontal is relative?
    I'll try.
     
  9. Aug 25, 2009 #8
    So here is yet another attempt.
    I'm out of ideas.
    Where did I go wrong?
     

    Attached Files:

  10. Aug 26, 2009 #9

    Doc Al

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    I haven't had a chance to look at your attempt in detail (I will a bit later), but at first glance:
    (1) Your energy equation is OK.
    (2) Correct your momentum equation: The velocity of the wedge is -u (u is positive).
     
  11. Aug 26, 2009 #10

    kuruman

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    It works. You should correct your momentum equation as Doc Al has suggested. From the momentum equation you should get

    [tex]v cos\alpha = \frac{M+m}{m}u[/tex]

    In order not to confuse your u with the book's u, let

    [tex]\frac{M+m}{m} = \frac{1}{\beta}[/tex] so that

    [tex]v cos\alpha = \frac{u}{\beta}[/tex]

    Put that in your energy equation and it should come out.
     
  12. Aug 28, 2009 #11
    Ok, I got it. Thanks!
     
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