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Homework Help: Impulse of a falling ball under gravity onto a solid surface

  1. Mar 12, 2010 #1
    1. The problem statement, all variables and given/known data
    A 0.5-kg rubber ball is dropped from rest a height of 19.6 m above the surface of the Earth. It strikes the sidewalk below
    and rebounds up to a maximum height of 4.9 m. What is the magnitude of the impulse due to the collision with the sidewalk?

    2. Relevant equations
    J=[tex]\Delta[/tex]P


    3. The attempt at a solution
    mgh=.5mv2
    v=[tex]\sqrt{}2gh[/tex]

    Just before collision:
    v=[tex]\sqrt{}2*9.8*19.6[/tex]
    v=19.6 m/s

    Just after collision:
    v=[tex]\sqrt{}2*9.8*4.9[/tex]
    v=9.8 m/s

    J=m(vafter-vbefore)=.5(9.8-19.6)=4.9 kg m/s

    The answer is 14.7 N.s, so I'm not sure what I'm missing.
     
  2. jcsd
  3. Mar 12, 2010 #2
    Re: Impulse

    Well velocity is a vector - is this how you would subtract them?
     
  4. Mar 12, 2010 #3
    Re: Impulse

    J=.5[9.8-(-19.6)]=14.7 kg m/s

    Wow that was a stupid mistake, thanks!
     
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