Momentum: Person drops out of a Hanglider

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SUMMARY

The discussion centers on a physics problem involving momentum and the effects of a skydiver dropping from a hang glider. The initial momentum of the glider is calculated as 19247.4 kg*m/s. After the skydiver, who has a mass of 57.0 kg, releases their grip, the glider's new velocity is determined to be 31.60 m/s. The key insight is that the act of letting go does not impart any acceleration to the glider, as the change in momentum occurs instantaneously at the moment of release.

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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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