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Conservation of momemtum - skydiver letting go of a moving glider

  1. Oct 8, 2009 #1
    1. The problem statement, all variables and given/known data
    a 10m long glider with mass of 680kg (including the passengers) is gliding horizontally through the air at 30 m/s when a 60 kg skydiver drops out my releasing his grip on the glider. What is the glider's velocity after the skydiver lets go?

    2. Relevant equations
    [tex]\vec{P}[/tex]i = [tex]\vec{P}[/tex]f
    mg[tex]\vec{v}[/tex]i + md[tex]\vec{v}[/tex]i = mg[tex]\vec{v}[/tex]f + md[tex]\vec{v}[/tex]f

    3. The attempt at a solution
    for the initial momentum part of the equation, would i have to subtract the mass of the diver from the mass of the glider - since the glider already includes his mass? and in the final momentum, subtract it there as well because he is no longer attached?

    Would it kind of be like a rocket shooting off question where the rocket is loosing weight?
     
  2. jcsd
  3. Oct 8, 2009 #2

    rl.bhat

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    Homework Helper

    The glider moves horizontally, and the diver drops vertically. Hence the total initial linear momentum remains constant. Since the mass is reduced, the velocity of the glider will increases.
     
  4. Oct 8, 2009 #3
    So would it look like this after i plug everything in?

    [tex]\vec{P}[/tex]i = [tex]\vec{P}[/tex]f
    mg[tex]\vec{v}[/tex]i + md[tex]\vec{v}[/tex]i = mg[tex]\vec{v}[/tex]f + md[tex]\vec{v}[/tex]f

    (680)(30) + (60)(30) = (680-60)v +(60)(30)

    v = [tex]\frac{20400}{620}[/tex]
    = 32.9 m/s
     
  5. Oct 9, 2009 #4
    can someone check this for me?
     
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