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Man on a boat (Classical momentum problem)

  1. Feb 28, 2008 #1
    [SOLVED] Man on a boat (Classical momentum problem)

    1. The problem statement, all variables and given/known data
    This problem is from Gregory:

    A boat of mass M is at rest in still water and a man of mass m is sitting at the bow. The man stands up, walks to the stern of the boat and then sits down again. If the water offers a resistance to the motion of the boat proportional to the velocity of the boat, show that the boat will eventually come to rest at its original position. [This remarkable result is independent of the resistance constant and the details of the man's motion.]

    x1: the position of the center of the boat
    x2: the position of the man
    xcm: the position of the center of mass
    k: the resistance constant of the water
    a: the length of the boat

    2. Relevant equations
    xcm = (M x1+m x2)/(M+m)
    Initially, x1 = 0
    Initially, x2 = -a/2
    In other words, the boat starts with its center on the origin, with the man on its left side.

    3. The attempt at a solution
    I've divided this problem into two parts:
    Part 1: The man walks from position x2 = -a/2 to position x2 = x1 + a/2. As soon as the man stops walking, the position of the center of mass is x and its velocity is v, for some x and v.
    Part 2: The man stands still and waits for the boat to come to rest.

    I think I understand Part 2. It's a simple differential equation (m+M)xcm'' = -k*xcm', with initial conditions x and v. Incidentally, the proof works out if x + v(m+M)/k = (m a)/2(M+m)

    Part 1 is the issue. All we know is that the man starts on one side of the boat and eventually ends up on the other side. We don't know how quickly the man moves or whether he moves at a constant speed. It's safe to assume that while the man moves right, the boat is moving left, and the water is pushing the boat right. Also, conservation of momentum doesn't apply because the water is applying an external force. I have no idea how to determine what the boat actually does.
     
  2. jcsd
  3. Mar 2, 2008 #2
    I really, really don't want to sound impatient, because what you guys do is absolutely incredible. But, this homework is due on Tuesday, so I would really appreciate some help. Thanks!
     
  4. Mar 4, 2008 #3
    If I were to do it, I'd use conservation of momentum to find the velocity of the boat and then figure out motion resulting from the resistance offered by water. I know its not totally correct as you cannot truly apply conservation of momentum here, but the question does say that the result is independent of the resistance constant and the details of the man's motion.
     
  5. Mar 5, 2008 #4
    I guess it might be a bit late now, but what the heck.

    Let's define:

    x is position of boat (arbitrary origin)
    y is position of man relative to boat (arbitrary origin again)
    M is the mass of boat
    m is the mass of man
    g is the drag coefficient

    By simple mechanics, we have that:

    [tex]M\ddot{x} = - m\ddot{y} - g\dot{x}[/tex]

    Integrating we have:

    [tex]M\Delta\dot{x} = \m\Delta\dot{y} - g\Delta x[/tex]

    where [tex]\Delta\dot{x}[/tex] means the difference between final and initial [tex]\dot{x}[/tex], and similarly for the other terms.

    So, we have that [tex]\Delta\dot{x}=\Delta\dot{y}=0[/tex] which gives that [tex]\Delta x = 0[/tex] as required.
     
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