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Momentum question

  1. Nov 18, 2007 #1
    1. The problem statement, all variables and given/known data
    A 1000kg plane is trying to make a forced landing on the deck of a 2000kg barge at rest on the surface of a calm sea. The only frictional force to consider is between the plane's wheels and the deck, and this braking force is constant and equal to a quarter of the the plane's weight. What mus the minimum length of the barge be, in order that the plane can stop safely on deck, if the plane touches down just at the rear end of the deck with a velocity of 50m/s towards the front of the barge?


    2. Relevant equations
    F=ma
    m1v1 +m2v2 = m1v1' + m2v2'


    3. The attempt at a solution

    first i found the friction force
    Ff=1/2mg
    =1/2(1000kg)(9.8)
    Ff=2450N

    then i found the acceleration of the plane
    F=ma
    -2450/1000=a
    a=-2.45m/s^2

    then i used kinematics to find the displacement
    V2^2 = V1^2 + 2ad
    0 = 50^2 + 2(-2.45)d
    d=510m

    i dont know if this is the right answer, but since this is a question from a momentum assignment, is there a step involving momentum that i missed?
     
  2. jcsd
  3. Nov 18, 2007 #2

    Doc Al

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    Staff: Mentor

    Yep. What's the final speed of the plane + barge?

    Also realize that while the plane moves forward, the barge moves backward.
     
  4. Nov 18, 2007 #3
    shouldnt the planes final velocity be 0?

    mpvp + mbvb = mpvp' + mbvb'
    1000(50) + 0 = 0 + 2000vb'
    vb'=25m/s
    the equation says that the barge is moving forward
     
  5. Nov 18, 2007 #4

    Doc Al

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    Staff: Mentor

    The final velocity of both barge and plane had better be the same. Redo your momentum equation.
     
  6. Nov 18, 2007 #5
    exactly, v1' = v2' because th exlane and the barge will move at the same velocity after the landing, as one object persay.. (for simplicity) so the equation is m1v1 = m1v' + m2v'

    m2v2 = 0 as the barge is at rest
     
  7. Nov 18, 2007 #6
    ok so its like this?
    mpvp + mbvb = mv(barge and plane)
    1000(50) + 0 = (1000+2000)v
    v= 16.67m/s
     
    Last edited: Nov 18, 2007
  8. Nov 18, 2007 #7

    Doc Al

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    Staff: Mentor

    Exactly.
     
  9. Nov 18, 2007 #8
    ok so using that velocity from above i subbed all the info into this equation
    v2^2 = v1^2 +2ad
    16.67^2 = 50^2 +2(-2.45)d
    d=453m
     
  10. Nov 18, 2007 #9

    Doc Al

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    Staff: Mentor

    Careful. You found the distance the plane moves with respect to the water. But as I said earlier, don't neglect the movement of the barge.
     
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