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How can I use kinematics to solve this question?

  1. Jan 4, 2004 #1
    A plane (mass=1000kg) is trying to land on a barge(mass=2300kg). The plane lands with a speed of 50m/s at one end of the barge. The barge is on a calm sea. The force of friction between the plane and the barge is 1/4 the force of gravity acting on the plane. What should be the minimum length of the barge for the plane to not fall into the sea?

    What I think:

    The only netforce acting on the plane is the force of friction. The plane undergoes a change in momentum which is transferred to the barge(which I do not see how it affects the length of the barge, we are not regarding relativistic effects). I can use the change in momentum and netforce to find the time interval. So since I have the final velocity, initial velocity, time, acceleration, how can I use kinematics to solve this question? Should I use kinematics?
  2. jcsd
  3. Jan 4, 2004 #2


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    If you already know the velocity profile, you should be able to calculate the distance the plane goes along the barge. This is the minimum length of the barge, since if the barge were any shorter, the plane would fall off the edge before it stopped.
  4. Jan 4, 2004 #3

    Doc Al

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    Staff: Mentor

    Re: momentum

    Assuming that you can ignore the drag of the water on the boat, you can easily find the final velocity of the boat + plane. The "trick" is then to calculate the distances (measured from a frame at rest in the water) it takes for the plane and barge to reach their final speeds. The plane takes a longer distance than the barge---the barge better be at least as long as that extra distance.

    To find the distances, I would use the so-called work-KE theorem, calculating it separately for each body:
    F*ΔXcm= ΔKE

    This is, of course, equivalent to finding the acceleration of each body and using kinematics:
    V2f = V2i + 2aΔX
  5. Jan 4, 2004 #4
    So basically, when the plane slows down on the barge, it accelerates the barge in the direction of its velocity. But if I was on the ship, the distance should be fixed right? So I used the equation Vf^2=Vi^2+2aÄx and got 510 m. But the books answer is 340 meters, what didn't I think of?
  6. Jan 5, 2004 #5

    Doc Al

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    Right. The plane deccelerates; the barge accelerates.
    If you mean that from the point of view of the ship, the ship doesn't move, then you are correct. But the ship accelerates, so it is not an inertial frame.
    Since I have no idea how you applied that equation, I cannot comment. What did you get for the final speed? Did you calculate the two distances like I described earlier?
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