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Linear momentum problem

  1. May 17, 2006 #1
    From this: http://www.collegeboard.com/prod_downloads/ap/students/physics/ap06_frq_physics_c_mech.pdf
    Question 1 is what I'd like to ask about :)
    For part a, for the block it would be a simple mg down, F N up and Ff to the left...
    For the Slab, do we include the fact that it's pushing up on the block and that the block is pushing down on it, or do we just include the usual mg and FN?

    Part b, I would think linear momentum is conserved, so:
    mB v0 = mB vf - mS vf, negative mS vf because the slab would move to the left, right?
    Well, for some reason, when I solve from there, I get a negative vf. Very very fishy...

    For part c, do I assume the change in kinetic energy of the block is equal to the work done by the slab, so delta K = Ff * D and solve for D?

    d) I'm going to guess change in kinetic energy...

    I hope someone can clarify how to do this question, as it confuses the hell out of me..

    I appreciate any responses.
     
  2. jcsd
  3. May 17, 2006 #2
    don't forget Ff on the slab to the right.

    Also for the slab, you have add its mass with the mass of the block:
    (M+m)g for the equillbrium between the slab and the horizontal surface. However since the horizontal surface is frictionless, you won't have to worry with this value in your calculations. I have to get back to the lab now, but I'll help more if you need it--> send me a personal message.
     
  4. May 17, 2006 #3

    Hootenanny

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    Almost. HINT: If they are both moving with the same velocity, they must both be moving in the same direction

    Sounds good to me.

    Again sounds good to me, although I would have thought (d) would have come before (c).

    ~H
     
  5. May 17, 2006 #4

    nrqed

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    right
    The correct way to think about this is that there is m_slab g and there are *two* normal forces acting on the slab. One normal force exerted by the table (so acting up) and one normal force exerted by the block (so acting down). It turns out here that the normal force exerted by the block is equl to m_block g but this is not the case in general (it would not be true if someone was pushing or pulling on the block in a vertical direction for example).
    Also, the normal force exerted by the table happens to me (m_slab + m_block) g in this case but again this would not always be true. So I prefer to tell my students to call these forces "normal forces" and then to solve for their values as a second step (instead of trying to guess their values in terms of mg). But again, in this very simple example, the normal forces are easy to write in terms of the weights of the objects.

    And there is a kinetic friction force to the right on the slab
     
  6. May 17, 2006 #5

    nrqed

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    As Hoot said, the slab moves to *the right*!
    they want the distance travelled by the *slab* so you need the change of kinetic energy of the slab.
    This is certainly one way to do it. Another would be to use the friction force to find the acceleration of the slab. Knowing the final velocity, you could find the distance without using the work energy theorem. Might be a good idea to do it both ways as a double check.
     
  7. May 17, 2006 #6
    ok then...
     
  8. May 18, 2006 #7
    Thanks a lot guys.
    I assumed the slab travelled to the left because when I asked my teacher about this kind of question, he told me it has to travel to the left. Weird, eh?
    Just one thing though... so the slab experiences the same frictional force, but to the right?
    Anyway, THANKS A LOT!
     
    Last edited: May 18, 2006
  9. May 18, 2006 #8

    Hootenanny

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    Yes, the frictional force experience by the slab is equal in magnitude but opposite in direction to the frictional force experienced by the block. As for the slab travelling to the left are you sure your teacher said that? Or is it simply a misinterpretation? Anyway, using conservation of momentum, you need to remember that velocity is a vector and hence has a direction, therefore direction will take care of itself. In this case if by some mirical of magic the slab and the block were moving to the left, you would obtain a negative velocity. Just plug in the numbers and the direction will take care of itself. :smile: Just make sure you define your co-ordinate system before you start and stick to it :biggrin:

    ~H
     
  10. May 18, 2006 #9
    OK, just to double check to be 100% sure! :)
    Part c) The change in kinetic energy of slab = work done by slab is:
    1/2 (mass of slab + mass of block) vf^2 = W = Ff * D (I thought it would be this because the block is on the slab!)
    Or...
    1/2 (mass of slab) vf^2 = W = Ff * D ???
    Can someone explain which one is right and what the rationale is? I don't get it. :\

    THANKS.
     
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