# What is the glider's velocity just after the skydiver lets go?

1. Jun 18, 2008

### habibclan

1. The problem statement, all variables and given/known data

A 10-m-long glider with a mass of 680 kg (including passengers) is gliding horizontally through the air at 30 m/s when a 60 kg skydriver drops out by releasing his grip on the glider. What is the glider's velocity just after the skydiver lets go?

2. Relevant equations
Pi= Pf
m1v1= m2v2

3. The attempt at a solution

m1v1=m2v2
(680)(30)= (680-60)v2
v2= 32.9 m/s

Glider's velocity just after the skydiver lets go is 32.9 m/s.

This is how I solved the question. however, for the concept of conservation of momentum, it has to be an isolated system right. So when one skydiver lets go, do we ignore the force of gravity on the skydiver? This question seems too easy this way but I can't seem to relate it to the concept of the isolated system. Thanks in advance!

Last edited by a moderator: Jan 7, 2014
2. Jun 18, 2008

### dirk_mec1

right.

The force due the gravity doesn't play a role when finding the velocity (in this case at least).

3. Jun 18, 2008

### habibclan

Thanks a lot! You're like my physics saviour for today :D. I really appreciate it!

4. Jun 18, 2008

### alphysicist

Hi habibclan,

I don't believe your equation is correct. The left side has the momentum of everything, but the right side does not have the horizontal momentum of the skydiver who let go. When the horizontal momentum is conserved, it means that the total momentum of the entire system is unchanged--and here the system is the (glider + skydiver).

(Your equation would represent the situation if the skydiver, instead of just letting go, pushed off backwards just enough so that his horizontal velocity was zero. Then his final horizontal momentum would be zero and would not appear on the right side of the equation.)

5. Jun 18, 2008

### habibclan

So then how to do I take into consideration the momentum of the skydiver who let go? Because the skydiver is in freefall.

6. Jun 18, 2008

### alphysicist

Remember that you're only looking at the momentum in the horizontal direction. The force of gravity won't affect that.

If you want to write down the equation taking into account the skydiver's momentum, you just add the term to what you had before:

m1v1=m2v2+m3 v3
(680)(30)= (680-60)v2 + 60 v3

where v2 is the horizontal velocity of the glider and v3 is the horizontal velocity of the skydiver, both velocities being right at the instant the skydiver has let go.

Do you see how that helps? When the problem says the skydiver just lets go and falls, what information is that telling you?

7. Jun 18, 2008

### habibclan

The velocity of the skydiver, v3, is 30 m/s, as the skydiver retains that horizontal velocity in freefall?

Last edited by a moderator: Jun 18, 2008
8. Jun 18, 2008

### alphysicist

That sounds right to me.

I think this is more of a conceptual question than anything. Think of a collision. Momentum is conserved, yet the individual velocities change. What makes them change? With the way this question is worded, does it make sense that we do not have to worry about that for this problem?

9. Jun 18, 2008

### habibclan

There is no impulsive force, so velocities don't necessarily change. right? :D

10. Jun 18, 2008

### alphysicist

That's the idea. One of the great strengths of the conservation of momentum equation is that it allows us to find the effects of internal forces (internal to the system) without knowing what those forces are (which means they don't appear in the equation).

But here, like you said, when the skydiver just lets go, there are no internal horizontal forces involved in the release process, and so not only is horizontal momentum conserved, but also the individual horizontal velocities themselves are unchanged. (There's nothing to change them.)

11. Jun 18, 2008

### habibclan

Thanks a lot alphysicist! I love how you take words right out of my textbook and relate them to these problems. It makes so much more sense to me now =). I feel so guilty though for posting so many questions..it's just that I have my physics exam next Monday and I want to prepare as much as possible so I can ace it =).