Help...please..ASAP?

Question

If a can begins empty, sitting at rest in a sea of pressurized air and the valve is opened, then what would happen to the can?
Wil the can move to the left, right, or not move once the valve is opened and air is let inside

Make the following assumptions: Pressure outside can = P ; mass of can = m; Area of can ends(each) = A; area of valve hole = s.

Newton's second law :

m*a = P*s*e^(-bt) - C*v(t)

a = d^2x / dt

v = dx / dt

m*d^2x / dt = P*s*e^(-bt) - C*dx / dt

m*d^2x / dt + C*dx / dt = P*s*e^(-bt) >>> Differential equation

the solution for this differential eq is:
y’ = Ps/(mb-c) { e^ (- c/m * t) - e^ (- bt)}

^^ according to this solution, which is velocity, the can will move to the right, but how can i prove that? we need to find something in common between all these relationships that prove it moves to the right.........

the valve is located on the right and the can is on its side...there must be something existent in the velocity equation that shows us and proves to us that the can moves to the right...

the actual question is on:
http://answers.yahoo.com/question/index;_ylt=AuXbSuiFQmy84jsLSUidAuPty6IX?qid=20070601130745AAK5cl8

and the way it was solved is on:
http://answers.yahoo.com/question/index;_ylt=Aq5ZO0qSlr2b85jBBkEdNjPty6IX?qid=20070603120520AATkArz

y’ = Ps/(mb-c) { e^ (- c/m * t) - e^ (- bt)}

How can i show that (mb-c) and { e^ (- c/m * t) - e^ (- bt)} are both positive?
granted, that we say positive=the right direction

Answer 1

With your equation, you can say that P is the pressure from outside on the left side area equal to the valve hole, which is directed to the right. Therefore, for the equations purpose it is a vector.

You do not, as you expressed, want to prove that mb-c and e^(-ct/m)-e^(-bt) are both positive. You simply want to prove that they are OPPOSITE, so that the overall equation is positive. This is relatively simple:

If e^... is to be positive, the first term must be greater than the second (since e^anything is always positive). Therefore, you need to prove that c/m is less than b (and therefore greater when multiplied by -t). Going from there:

c/m < b
c < mb (we know that m, c, and b are all positive)
0 < mb-c

So mb-c is positive also, and since the numerator and denominator are positive, the overall equation is positive.

However, if e^... is to be negative, the first term must be less than the second, and you need to prove that c/m is greater than b (and therefore less when multiplied by -t). Going from THERE:

c/m > b
c > mb
0 > mb-c

Therefore, the numerator and denominator are both negative, so the overall equation is still POSITIVE.

Since you know that the force acting on the can (from pressure) is directed to the right, you can prove that the velocity of the can is directed in the same direction.

Hope that helped. Have a good day.

Answer 2

Without seeeing those equations (I am poor at that part), I can tell you that the can moves to the right with the given conditions. When the valve in the can is opened, the outside air rushes in and acts as a rocket jet, propelling the can forward.

If the can was pressurised and the valve was opened, the can will move left (opposite direction). This is Newton's Third law in action.