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Conservation of linear momentum

  1. May 7, 2015 #1
    1. The problem statement, all variables and given/known data
    A 75.0 kg person runs at 3.00 m/s and jumps onto a sled of mass 10.0 kg already moving in the same direction as the person at 2.00 m/s.

    a) What is the initial momentum of the sled-person system?
    b) What will the final momentum of the sled and the person be?
    c) What is the velocity of the sled and the person after the person jumps on the sled?


    2. Relevant equations
    P = mv
    Pi = Pf (when momentum is conserved)
    m1vi + m2vi = m1vf + m2vf


    3. The attempt at a solution
    a)P = mv
    p = (75.0kg)(3.00m/s) + (10.0kg)(2.00m/s)
    p = +245 kg(m/s)

    The questions a) is little tricky, im not sure if I should get momentum of being one unit (e.g 75.0 kg + 10.0kg)(3.00m/s+2.00m/s) or what I did originally

    b) Pi = Pf (when momentum is conserved)
    Therefore pf = 245 kg(m/s)

    C) m1vi + m2vi = m1vf + m2vf
    245 kg(m/s) = (75kg +10kg)(v2f)
    V2f = 2.89 m/s
     
  2. jcsd
  3. May 7, 2015 #2
    a small error on c
     
  4. May 8, 2015 #3

    haruspex

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    No, that would be quite wrong. Nothing here is moving at 5m/s.
    How small? I get 2.882...
     
  5. May 8, 2015 #4
    Thank you, I was going to do as one unit, but I thought something was wrong, so I did the separately.
    Do you think all my answers are correct though?
     
  6. May 8, 2015 #5

    haruspex

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    Apart from the tiny numerical error in c, yes.
     
  7. May 8, 2015 #6
    Hm.. I don't recognize any error, I just rounded to 2.89 m/s from 2.882 m/s which you just wrote.
     
  8. May 8, 2015 #7

    haruspex

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    2.88 is nearer to 2.882 than 2.89 is. Unless you specifically need to round up, round to nearest.
     
  9. May 8, 2015 #8
    LoL I must've been drunken sorry
     
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