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Inelastic ball problem

  1. Oct 11, 2011 #1
    1. The problem statement, all variables and given/known data
    Now the racquet ball is moving straight toward the wall at a velocity of vi = 12.4 m/s. The ball makes an inelastic collision with the solid wall and leaves the wall in the opposite direction at vf = -8.2 m/s. The ball exerts the same average force on the ball as before. Mass of ball =.247 kg



    2. Relevant equations
    What is the magnitude of the change in momentum of the racquet ball?


    3. The attempt at a solution
    so I took
    m*v(initial)=m*v(final) and got 3.0628 and -2.0254 respectively. so I subtracted the two and got -5.0882. I dont feel like this is correct.
     
  2. jcsd
  3. Oct 12, 2011 #2

    kuruman

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    Gold Member

    What is incorrect is to say m*v(initial)=m*v(final) because that is simply not true. It is correct to say that the change in momentum Δ(mv)=m*v(final)-m*v(initial), so if the final momentum is negative and the initial momentum is positive, you end up essentially adding two negative numbers and that's that.
     
  4. Oct 12, 2011 #3

    PeterO

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    Many strange "formulae" and calculations, but probably almost the correct answer.

    Generally change in a quantity means Final - Initial [which you actually ended up doing!!].

    Depending which direction you decide to call positive will mean a final answer of -5.etc, like you got in an unusual way, or +5.etc.

    The question only asked for the magnitude of the change, so the question of a +ve or -ve answer is irrelevant.
     
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