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Sir Isaac Newton

  1. May 21, 2007 #1
    From the Sir Isaac Newton contest:

    A 1.0*10^3 kg plane is tryin to make a forced landing on the deck of a 2.0*10^3 kg barge at rest on the surface of a clam sea. THe only frictional force to consider is between the plane's wheels and the deck, this braking force is constant and is equal to one quarter of the plane's weight. What must the minimum length of the barge be for the plane to stop safely on deck, if the plane touches down just at the rear end of the deck with a velocity of 5.0*10^1m/s towards the front of the barge?

    Sorry, I don't quite understand this question. When the plane touches down, the barge will move as well, due to the frictional force. Then, how should I understand the motion of the plane? : should the plane be moving ahead at 51m/s, with deceleration determined by the braking force.... or do we also have to take into consideration of the barge's motion as well?

    Please, does someone have a solution for the problem? Thank you!
     
    Last edited: May 21, 2007
  2. jcsd
  3. May 21, 2007 #2

    Dick

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    The motion of the barge is important because when the plane has stopped on the deck of the barge the barge and plane are moving forward. So the final velocity of the plane is not zero. Figure out the final velocity of plane+barge using conservation of momentum then figure out the length of deck you need to stop.
     
  4. May 21, 2007 #3

    If I want to determine the motion of the plane, with respect to the Earth, is its motion = 50m/s - decel due to friction force? or from here, do we still ahve to take into consideration the motion of the barge?
     
  5. May 21, 2007 #4
    What you need to determine is the velocity of the plane relatively to the barge... The velocities in respect to the water can be determined by approaching the problem like an elastic collision...
     
    Last edited: May 21, 2007
  6. May 21, 2007 #5

    Dick

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    INELASTIC collision.
     
  7. May 21, 2007 #6
    ? How would you determine the velocities if it's inelastic? There wouldn't be enough information; one equation and two variables.
     
  8. May 21, 2007 #7

    Dick

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    Inelastic collisions still conserve momentum.
     
  9. May 21, 2007 #8
    Yes, hence our first equation... we are missing the other one.
     
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