Momentum: Person drops out of a Hanglider

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The discussion revolves around calculating the glider's velocity immediately after a skydiver drops out. The initial momentum of the glider is calculated to be 19247.4 kg*m/s. The user attempts to find the new velocity by adjusting the mass of the glider but questions the validity of their approach. Key hints suggest considering whether the act of letting go imparts any acceleration to the glider and the timing of this acceleration. The user expresses confusion about how the release could affect the glider's motion.
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Homework Statement


A 10.1 m long glider with a mass of 666.0 kg (including the passengers) is gliding horizontally through the air at 28.9 m/s when a 57.0 kg skydiver drops out by releasing his grip on the glider. What is the glider's velocity just after the skydiver let's go?


Homework Equations


P= mv
F=dp/dt
dp=mdv


The Attempt at a Solution


P= mv
p=(666)(28.9)
= 19247.4 kg*m/s

P= mv
p/(666-57) = v2
31.60 m/s =v2

what did i do wrong?
 
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huynhtn2 said:
What is the glider's velocity just after the skydiver let's go?

Hint question: Does the act of letting go of the glider impart any acceleration to the glider? If so (and here's the key), how long has that acceleration been acting on the glider? In other words, how long is "just after?" ;)
 
mugaliens said:
Hint question: Does the act of letting go of the glider impart any acceleration to the glider? If so (and here's the key), how long has that acceleration been acting on the glider? In other words, how long is "just after?" ;)

I don't understand how letting go could accelerate the glider.
 
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