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A Momentum Problem

  1. Nov 6, 2003 #1
    A peron on a boat with a total mass of M are stationary in the river, the man throws his shoes mass m at a velocity v successive, find the expression for the boat's final velocity.

    Here is what I did.

    First let u be the velocit of the boat
    After the first shoes:
    (M + m)u = mv
    u = mv(1/(M +m))

    After the second shoes:
    Change in momentum of the boat = Change in momentum of the shoe

    So, Mu'- (M + m)u = mv, but (M + m)u = mv

    Therefore Mu' = 2mv and u' = 2mv/M

    u + u' = mv(2/M + 1/(M + m)) is the final velocity of the boat.

    But the correct answer is mv(1/(M + m) + 1/(M + 2m)), can anyone please tell me where I've got it wrong?
  2. jcsd
  3. Nov 7, 2003 #2
    The 2nd shoe ist not at v relative to the water, since it's thrown from a moving boat.
  4. Nov 7, 2003 #3
    So this is the correct expression?

    Mu'- (M + m) = m(u + v)? Because I still can't find the correct expression...
  5. Nov 7, 2003 #4
    Nor can I reproduce that answer.
    There's also something unclear in the problem: It says 'total mass M'. Does that include the shoes, or not?

    Let u be the boat velocity after 1st throw.
    Let u' be the boat velocity after 2nd throw.

    If we assume that M includes the shoes, then

    I. (M-m)u = mv
    II. (M-2m)u' - (M-m)u = m(v-u)

    This leads to
    u' = mv(1/(M - m) + 1/(M - 2m)).

    Now if M doesn't include the shoes, this will sure mean the transformation
    M -> M+2m, yielding
    u' = mv(1/(M + m) + 1/M).

    I can't see how to get closer to the 'correct' answer.
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