# Balloon homework question

## Homework Statement

When the air rush out of a balloon, it shoots out. What physics principle is responsible for this behavior?

## The Attempt at a Solution

Is it due to Newton's 2 nd law or 3 rd law??

We can compare this to a rocket propulsion. We can explain this using Newton's second law (conservation of momentum when Fnet = 0). However, it is true only if net force = 0. But we know that the balloon also will shoot up, if the balloon's open end is pointing down. How do I explain this because now force of gravity is acting on the system.

## Answers and Replies

Huh? I don't think you need take that approach at all, because a blown-up balloon is a very high pressure confined area, by 'letting' it go, you're allowing the pressure to 'equilibrate?' (is that a word? :P), basically the air in the balloon is bieng crushed together, and when you let it go it's free to move apart to an area of low pressure (where it wants to be)

minger
Science Advisor

It is a combination of the two. If you draw a free-body diagram at the nozzle, you have two components on each side. There are pressure forces on each side, high inside the balloon, and ambient outside. Multiplied by the area causes a net force.

You also have mass flow, which causes thrust. This can all be lumped into conservation of momentum. It is a textbook problem when discussing that topic in fluid mechanics.

So force on air molecules is some thing like, (Pin - Patm) * A where A is the area of the balloon's nozzle. Since this force is caused by pressure difference, and balloon itself did not exert this force on the molecules, can we say that Newton's 3 rd law is not at work here?

I still think, only newton's 2 nd law is the one causing the balloon to fly backward. I vaguely remember doing a problem of rocket propulsion where we considered the force of gravity. We had a differential equation and we found the velocity of the rocket after some time has elapsed.