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Conservation of momemtum - skydiver letting go of a moving glider

  • Thread starter indietro
  • Start date
  • #1
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Homework Statement


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?

Homework 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

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?
 

Answers and Replies

  • #2
rl.bhat
Homework Helper
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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.
 
  • #3
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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
 
  • #4
46
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can someone check this for me?
 

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