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A boy that jumps off a skate, v of the skate=?

  1. Jun 9, 2012 #1
    A boy of 30kg is riding a skateboard of 4.9kg and both are going on with a velocity of 2.2m/s. In a moment the boy jumps forward, pushing the skate to the other direction. Immediatelly after the jump, the horizontal velocity of the boy in relation to the skate is 3m/s. Whats the velocity of the skate in this moment?

    What I've tried to do:

    Momentum before=Momentum after
    (mass of the system skate+boy)(initial velocity)=(mass of boy)(velocity of boy)+(mass of skate)(velocity of skate)

    I think what im probably missthinking is the velocity of the boy, because in the question, he says "velocity of the boy in relation to skate", so i suppose, it would be: (mass of the boy)(3 + velocity of the skate)?

    It sounds wrong, i dont know why XD

    Can somebody give me a light here?
     
  2. jcsd
  3. Jun 9, 2012 #2
    Hi frank! Welcome to PF!!

    The question has given you the velocity of the boy relative to the skate. That means,

    [tex]V_B - V_S = V_{BS} = 3[/tex]

    So yes, you got it right!! :smile:
     
  4. Jun 9, 2012 #3
    So can you check if im doing it correctly?

    look:

    (34.9)x(2.2)=30(3+Vs)+4.9Vs
    76.78=90+30Vs+4.9Vs
    -13.22=34.9Vs
    Vs=-0.3787m/s

    EDIT: thanks for helping so quickly :)
     
  5. Jun 9, 2012 #4
    Yep, your method looks correct! :approve:
     
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