# Momentum and isolated system

1. Nov 18, 2005

hi,I already solved the problem but my question is partly related to it

A person with mass m=80kg is climbing up the helicopter ladder that is
hovering in the air.Person climbs with speed 0.5m/s relative to the ladder.
With what speed is helicopter descending?

If we think of helicopter and a person together as isolated system then any forces the two exert on each other are considered internal and as such can't change the momentum of the system.

Since person has momentum p = m*v=40 helicopter must also have same momentum in opposite direction

40=mass_helicopter * speed
speed = 1.6cm/s

Although I solved it, I'm not totally shure I understand what is going on. If we consider both the helicopter and a person as one object with some net momentum, then net momentum will stay the same as long as there are no external forces acting on an object.

I solved it as if net momentum was always zero and I considered mass_person * 0.5m/s as a change of momentum (previously both the person and a helicopter each had momentum=0), so helicopter also had to have the same change in momentum.
But couldn't net momentum actually be the momentum person climbing up the ladder has? In that case for net momentum to stay the same helicopter's momentum should stay zero.

Also confusing is the fact that if this was to be isolated system,then once person has momentum p=40 it should remain 40 even if person doesn't try to push itself up anymore. But in reality as soon as it does that its momentum will be zero due to gravity. So why call this isolated system ?

And finally,as you can see from above I have a hard time figuring out what is isolated system and what not. Two cars going with constant speeds and crashing with each other could be considered one, if it were not for friction with the road.

thank you

2. Nov 18, 2005

### Staff: Mentor

I assume that they want you to assume that the net force on the system (helicopter + person) is zero: that the upward force of the air on the helicopter equals the the downward force of gravity. So the total momentum remains the same.

I further assume that they want you to assume that initially the person was not moving, then started climbing the ladder. Note that the speed of the person was given with respect to the ladder, not the ground. So your method is not exactly right (though the answer is practically the same).

3. Nov 18, 2005

I missed that.Then it is

v1 = v2+0.5

p1=m1*v1=40+80*v2

p1=p2=40+80*v2=m2*v2

Can you comment a little on my question about what is isolated system in this and similar cases(since gravity does in a way affect this person )?

4. Nov 18, 2005

### Staff: Mentor

Right.

I would say that total momentum = 0, so:
m1*v1 + m2*v2 = 0

(Note that v2 will be negative, since the helicopter moves down.)

Sure. Momentum is conserved if the net force on the system is zero. If a system is isolated from its environment, that means there is no outside interaction or force on it. So if a system is isolated, then its momentum is conserved (since the net force on it is zero). But a system doesn't need to be isolated in order for momentum conservation to hold. Let's consider your two examples:

(1) Helicopter + person: Since both air and gravity act on this system, I would not consider it isolated. Nontheless, since the net force is presumed to be zero, momentum is still conserved.

(2) Two cars crashing: As you point out, if there's no friction then momentum is conserved since there's no (net) external force acting on the cars. Without friction, you could think of the cars as being isolated from the environment (as far as horizontal forces go, at least).

Sometimes collisions take place so quickly, that even if there are external forces acting, they don't have enough time to change the momentum much. In that case you can apply conservation of momentum to the collision. (Momentum just before the collision will equal the momentum just after the collision.)