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Conservation of Linear Momentum

  1. Aug 14, 2011 #1
    1. The problem statement, all variables and given/known data
    A particle of mass 5m moving with speed v explodes and splits into two pieces with masses of 2m and 3m. The lighter piece continues to move in the original direction with speed 5v relative to the heavier piece. What is the actual speed of the lighter piece?


    2. Relevant equations
    Momentum = Mass x Velocity


    3. The attempt at a solution
    The answer is 4v, as stated in the answer sheet, but I've no idea how to get it. Appreciate any help here, thanks!
     
  2. jcsd
  3. Aug 14, 2011 #2

    PhanthomJay

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    Start by writing the conservation of linear momentum equation.
     
  4. Aug 14, 2011 #3
    I've only got: 5m x v = (2m x 5v) + (3m x -1 2/3 v)

    Does the problem lie with the term "relative speed"?
     
  5. Aug 14, 2011 #4

    PeterO

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    Certainly does!
     
  6. Aug 14, 2011 #5

    PhanthomJay

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    Yes. You should always determine speeds with respect to the ground when using this equation. If the lighter piece is moving at 5v with respect to the heavier piece, and the heavier piece is moving at a speed v2 with respect to the ground, then what is the speed of the lighter piece with respect to the ground, in terms of v and v2?
     
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