Conservation of momemtum - skydiver letting go of a moving glider

[indietro]

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 let's go?

Homework Equations

\vec{P}_i = \vec{P}_f
m_g\vec{v}_i + m_d\vec{v}_i = m_g\vec{v}_f + m_d\vec{v}_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?

 

[rl.bhat]

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.

 

[indietro]

So would it look like this after i plug everything in?

\vec{P}_i = \vec{P}_f
m_g\vec{v}_i + m_d\vec{v}_i = m_g\vec{v}_f + m_d\vec{v}_f

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

v = \frac{20400}{620}
= 32.9 m/s

 

[indietro]

can someone check this for me?