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Linear Momentum of system of two blocks

  1. Apr 9, 2016 #1
    Untitled 2.jpg 1. The problem statement, all variables and given/known data

    The problem is shown in the photo above. I would like to discuss part B.
    2. Relevant equations

    conservation of linear momentum
    3. The attempt at a solution


    the solutions say that we should have the following approach:
    Mb*vb = [Mb + Ms] * vFinal.

    Plug in the numbers and get vF = 0.57 m/s

    However, I do not understand this. I thought that momentum is a vector and that while the block is moving forward, the slab is moving backward, so I would need to have the following:

    Mb*vb = Mb*vF + (Ms* -vF)

    notice the negative vF for the final momentum term of the slab. I put a negative because the slab would be moving backwards with a negative velocity.

    Why is my approach not correct? I got a different vF than 0.57.

    Thanks
     
  2. jcsd
  3. Apr 9, 2016 #2
    i think the first part is important -when you draw the FBD for the two bodies your confusion will get settled- remember when two surfaces in contact move relative to each other the frictional forces on the bodies also act equally but in opposite direction. if one is pulling behind the other will be pushing ahead. let us draw the part a, that is the forces.
     
  4. Apr 9, 2016 #3

    PeroK

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    Why do you think the slab would move backwards?
     
  5. Apr 9, 2016 #4
    I did the first part allright. For both of those objects, I will have a normal and gravitational force. the block will have a leftwards frictional force and the slab will have a rightwards frictional force.

    The block is going to be going forward until the frictional force stops it.
    The slab will thus have to go backwards to conserve momentum
     
  6. Apr 9, 2016 #5

    PeroK

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    The slab moves forward because there is nothing pushing the slab to the left. All the momentum is in the forward direction.
     
  7. Apr 9, 2016 #6
    The block has ƒ = 0.5 × 10 × 0.2 = 1 N, so it has a = -2 m/s2
    And the slab has same ƒ, so it has a = ⅓ m/s2

    Then v1 block = v2 slab

    Sorry if I only up to here
     
  8. Apr 9, 2016 #7
    How is that possible? I know that when I stand on a piece of cardboard and I move forward, the board will move backwards to conserve momentum [but I know that there is friction force here]. Are you saying that there is no friction between the block and slab that will cause the slab to move in reverse? [this is false because there is friction]

    They say the speed of the slab and block is the same, so the magnitudes of their velocities are the same, but one should have a plus sign, but the other should have a negative one.

    Please advise.
     
  9. Apr 9, 2016 #8

    haruspex

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    I assume you are thinking of standing still on the cardboard then walking forwards. That is not the same as set up here. The right analogy would be to get up speed on a (frictional) floor then jump onto the cardboard and try to stop on it. Which way will the cardboard move now?
     
  10. Apr 10, 2016 #9
    I and the cardboard will move forward. But how do we know if this analogy is valid, especially with all those frictional forces we have drawn in the free body diagram?

    What makes my approach wrong?
     
  11. Apr 10, 2016 #10
    First and foremost
    The frictional force acting on the small block is in the backward direction (I assume the right direction to be the forward direction and the left direction accordingly)
    The frictional force (which equals mu times N and is purely kinetic accelerates the slab
    And the same frictional force which acts on the small block decelerates it to a certain speed vf
    And the final velocity of the slab also becomes vf
     
  12. Apr 10, 2016 #11
    You're lucky
    The problem doesn't involve the length of the slab (over which the block moves)
    Otherwise it would have been too cumbersome
     
  13. Apr 10, 2016 #12
    No!!!
    Why would the slab have a backward velocity?
    When you run over a floor and gain speed
    And then jump on a cardboard, the cardboard moves along with you
    Why?
    It's because of the points which are in contact (here, they're your legs)
    Considering the cardboard offers friction
    Your feet are moving forward w.r.t to the cardboard
    However, consider the motion from your frame
    When you move forward
    Don't you momentarily see the cardboard moving behind you in the other direction? (However, this isn't the final speed of the cardboard, just the negative of the initial speed you possessed
    And that too for a small instant
    And now since w.r.t your feet, the cardboard tries to move backward, what will friction do?
    It will oppose relative motion at the "point of contact" and push the cardboard forward!:)


    UchihaClan13
     
  14. Apr 10, 2016 #13
    Since there's no external force acting on the slab-block system (friction is an internal non conservative force)
    Gravity is an external force here
    But its component in the horizontal direction is m×g×cos 90=0
    So using the principle of conservation of momentum
    The answer you got is correct
    Nothing wrong!!




    UchihaClan13
     
  15. Apr 10, 2016 #14

    haruspex

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    Friction acts to oppose relative motion (actual or potential) of the two surfaces in contact. Note that exact wording. So it acts on each of the two surfaces in opposition to its motion relative to the other surface.
    The initial state is that the block is moving and the slab is not. Which two contacting surfaces are in relative motion? Which way does friction act on each to oppose that relative motion?
     
  16. Apr 12, 2016 #15
    Let's see. The slab and the block are in contact and are in relative motion.

    Ahh... I think I see now.

    My force diagram dot for the slab shows the friction force going rightwards, so the friction force must accelerate the block. What an interesting question this is.

    Thanks a lot for the help!
     
  17. Apr 13, 2016 #16
    You are welcome!
     
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