How does the repeated expansion of a balloon on top of a car affect the force of air expelled from it?

Answer 1

As the balloon is repeatedly stressed, by being blown up, its material loses its ability to return to its original shape. (The elastic limit has been exceeded and permanent deformation has resulted)
As a result of the restorative property having been lost, the wall of the balloon exerts less pressure on its contents.
Less pressure by the material equals less pressure on the gas.
Less pressure on the gas equals less exit velocity.
Less exit velocity equals less force.

Answer 2

An expanded balloon is in equilibrium when the pressure inside it equals the pressure outside it PLUS the force of the stretched rubber wanting to shrink back to its uninflated size. In math talk; F(in) = F(atm) + F(elastic); where F(in) = p(in)A = force inside the balloon due to pressure (your hot breath) acting over the area (A) of the balloon's innards. F(atm) = p(atm)A; where the force due to atmospheric pressure (p(atm)) acts on the same area (A) but outside the balloon.

F(in) = Ma = p(in)A(nozzle) when the balloon is let go with the nozzle open; where M = mass of the balloon plus the mass of the included air and a = the acceleration of the balloon/air as you let the balloon go with its nozzle open to an area A(nozzle)....sprrrrrrrrrrrt. That is to say F(in) is the driving force that propels a balloon through the air when let go. And it is affected by both the atmosphere and the stretched balloon.

F(elastic) is proportional to the elasticity of the rubber and to the volume (size) or surface area the balloon is stretched. It's something like a spring being stretched, the force there is F(spring) = k del x; where k is the spring coefficient and del x is the distance stretched.

From F(in) = F(atm) + F(elastic), we see that if we blow up the balloon so that F(elastic) = constant, the same value each time, then F(in) will remain the same each blow because F(atm) is also a constant at a constant altitude. Thus, if the elasticity (e) remains fixed, we need to blow the balloon to the same volume each time for F(in) = constant and, consequently for the force of the expelled air to remain the same each time.

But what if e <> constant; what if e gets smaller with each expansion?

As you stretch and deflate a balloon, F(elastic) weakens due to material fatigue. All materials weaken when stretched or bent repeatedly. That weakening causes elasticity (the ability to regain shape after some stress) to diminish; in math talk e(n) < e0, where e(n) is the elasticity coefficient after n expansions. Because e is getting smaller, to keep F(elastic) constant, we need to increase the volume (V) of the balloon to compensate for the loss of e(n) with the repeated expansions.

Bottom line...repeatedly expanding the balloon weakens the elasticity, which means the elastic force will diminish if you blow the balloon up to the same volume each time. Thus, from F(in) = F(atm) + F(elastic), F(in)n < F(in)0 if V = constant = V0 the initial blow volume of the balloon. This follows because F(atm) = constant and has no effect on the inequality.

However, if V(n) > V0, where V(n) is the volume you blow the balloon up to after n expansions and V(n) is sufficient to offset the loss in e(n), then F(elastic) = constant and, consequently, F(in) are constant. In which case, the rate of air expulsion will remain the same no matter how many times the balloon is expanded.

To sum up, in the case of constant volume V(n) = constant, the force of air expelled will diminish as the rubber weakens through fatique. In the case of V(n) > V0 to offset e(n) < e0, the force of expelled air will remain fixed over each expansion.

One note of caution, as V(n) gets ever bigger and e(n) gets ever smaller, eventually the rubber will tear...then pop. This is yet another effect of material fatigue...you can stretch things just so far before they rupture.