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Conservation of momentum: throwing bricks

  1. Feb 15, 2015 #1
    1. The problem statement, all variables and given/known data
    A 59.8kg person holding two 0.902kg bricks stands on a 2.14kg skateboard. Initially, the skateboard and the person are at rest.

    Calculate the final speed of the person and the skateboard relative to the ground if the person throws the bricks one at a time. Assume that each brick is thrown with a speed of 16.7m/s relative to the person.
    2. Relevant equations
    m1v1+m2v2 = m1v1'+m2v2'

    3. The attempt at a solution
    0 = m1v1' + m2v2' + m3v3'
    0 = (59.8+2.14)(v1') + (.902)(16.7)+(.902)(16.7)
    v1' = 0.486 m/s

    Why is this incorrect? The initial momentum is zero because they are at rest, then if I consider the system as the skateboard, bricks, and the person, the final momentum has to equal zero too. So the momentum of the two bricks, each moving at 16.7 m/s, and the momentum of the skateboard and the person should equal zero.
     
  2. jcsd
  3. Feb 15, 2015 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    One obvious point- v1' will have to be negative. The solution to the equation will have to be negative. You have dropped a negative sign when you solved the equation.

    Much more important- you have misread the problem. You have done this as if both bricks were thrown at the same time, when the person is stationary so that "speed relative to the person" is the same as "speed relative to the ground". But the problem says that the bricks are thrown one at a time and both at 26.7 m/s relative to the person. When the first brick is thrown that is the same as "relative to ground" but not when the second brick is thrown.
     
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