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Conservation of momentum

  1. Nov 7, 2006 #1
    A 1000kg plane is trying to make a forced landing on the deck of a 2000 kg 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 1/4 of the plane's weight What must the minimum length of the barge be, in order that the plane can stop safely on the deck, if the plane touches down at the rear end of the deck with a velocity of 50m/s towards the front of the barge?

    My work:

    m1 = 1000kg
    m2 = 2000kg
    Ff = 1/4Fg
    V1' = V2'

    Fg = mg
    =(1000)(9.81)
    =9810N

    Ff = 1/4Fg
    = 1/4(9810)
    =2452.5N

    Fnet=Ff

    P1(plane) = mv
    = (1000)(50)
    = 50000

    P2(plane) = mv
    = (1000)(0)
    = 0

    deltaP = (Fnet)t
    50000 = 2452.5t
    t = 20.387s

    m1v1 + m2v2 = m1v1' + m2v2'
    (1000)(50) + 2000(0) = 0 + (3000)v2'
    50000 = 3000v2'
    v2' = 16.667 m/s

    a = V2-V1 / T
    = (16.667-50) / 20.387
    = -1.635m/s²

    d = v1t + 1/2at²
    = (50)(20.387) + 1/2(-1.635)(20.387)²
    = 1019.35 - 339.777
    =679.573m

    Thats what i did but the answer is 340m...wondering what I did wrong...
     
  2. jcsd
  3. Nov 7, 2006 #2

    OlderDan

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    See the annotations in the quote.
     
  4. Nov 7, 2006 #3
    i'm saying that the final velocity of the barge+plane which is 3000kg would be 16.667m/s by using the conservation of momentum equation...

    m1v1' would be 0 because the plane loses all momentum and transfers it to the barge right? if not please clarify because i'm starting to get confused...

    i dont really understand why the first part is wrong can you explain that also? thanks
     
  5. Nov 8, 2006 #4

    OlderDan

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    See the annotations in the following quote from your earlier post.

     
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