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Basic Momentum Questons

  1. Apr 6, 2009 #1
    1. The problem statement, all variables and given/known data

    a) KE=(1/2)(m)(v-u)2
    Conversion: u = (31km/hr)(1hr/3600s)(1000m/km) = 8.61m/s
    v = (64km/hr)(1hr/3600s)(1000m/km) = 17.78m/s
    KE = (1/2)(1850kg)(17.782-8.612) = The Answer for a)

    b)Momentum= m(√(v2+u2)), v=14.17i u=11.39j
    = 1850(√(17.782+8.612)) = Answer for b)

    c)tan θ = (8.61/17.78) = Degrees South of East

    2. Relevant equations

    p=mv

    Momentum= m(√(v2+u2))


    3. The attempt at a solution

    What I can't figure out is why they are taking the square root of the two squared numbers, can anyone help explain this? The question is asking for the magnitude of the momentum, but I thought that was p=mv.
     
    Last edited: Apr 6, 2009
  2. jcsd
  3. Apr 6, 2009 #2

    rl.bhat

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    Homework Helper

    When two momentum are perpendicular to each other, the resultant momentum is calculated by the above method.
     
  4. Apr 6, 2009 #3
    Wow. One reply can really make a world of difference. Thank You, it makes sense now.
     
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