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Law of conservation of momentum problem?

  1. Apr 25, 2012 #1
    1. The problem statement, all variables and given/known data

    A man of mass 65kg is running at speed of 4.9m/s, jumps in to a rowboat of 88kg that is drifting without friction in the same direction at a speed of 1.2m/s. when the man is seated in the rowboat, what is the final velocity of the boat?


    2. Relevant equations

    Momentum before = momentum after

    3. The attempt at a solution

    If
    m= mass of man = 65kg
    v= velocity of man = 4.9m/s

    m' = mass of rowboat = 88kg
    v' = velocity of rowboat = 1.2m/s

    V= required final velocity
    let
    M= mass of both rowboat and man = m+ m' = 153kg

    now
    law of conservation of momentum

    momentum before = momentum after

    i-e
    mv = M(v'+V)

    so

    V= (mv-Mv')/M

    = 0.88 m/s

    is it right method ?

    but how could the speed of rowboat is such less?
     
  2. jcsd
  3. Apr 25, 2012 #2

    tiny-tim

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    hi abrowaqas! :smile:
    yes :smile:
    noooo :redface:
     
  4. Apr 25, 2012 #3
    P = momentum, m = mass and v= velocity, and P=m*v

    So before the collision we have a total momentum of (65*4.9) + (88*1.2) = 424.1Kg/ms^-1

    After the collision, the momentum must be 424.1. This is equal to the combined mass of the man and the boat (65+88=153kg) multiplied by the final velocity of the boat. So all we need to do is 424.1/153 to give us your answer of 2.77 m/s.

    is this method right?
     
  5. Apr 25, 2012 #4

    tiny-tim

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    are you quoting someone? :confused:

    yes, that method is correct

    now rewrite your m M v V equation so that it's correct!​
     
  6. Apr 26, 2012 #5
    yes ofcourse... that answer is given by from yahoo answers.
     
  7. Apr 26, 2012 #6

    tiny-tim

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    thought so! :rolleyes:

    ok now see if you can correct your first try …
    :wink:
     
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