How much the gas has increased the speed of the plug?

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The discussion revolves around calculating the speed increase of a plug shot from a test tube due to gas expansion. The initial speed of the plug is given as 12.134 m/s, derived from kinematic equations. Participants debate whether to consider friction and the effects of temperature and humidity on the gas pressure inside the tube. There is uncertainty about the coefficient of kinetic friction, which is not provided in the problem. The goal is to confirm that the final speed aligns with the calculated value of 12.134 m/s.
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Homework Statement


The cannon is made from V=30ml test tube completely closed with a plug (diameter d=1,5cm; length l=3cm; mass m=5g). Before closing the test tube, V'=1ml of water is poured into it. Room temperature is 20oC , pressure p=105 Pa , relative humidity is σ=20%. When the temperature in the test tube becomes t'=40oC the plug is shot. I need to find how much the escaping gas has increased the speed of the plug after it had been shot.

I know that the initial speed with which the plug starts to move is v=12.134m/s.

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Homework Equations


Ideal gas laws.

The Attempt at a Solution


Vplug=πr2=5.3 cm3
Vfree space=30ml-5.3ml=24.7ml[/B]

I know that:
t/Vfree space=t'/V

What now?
 
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Is the plug completely embedded in the tube such that the pressure in the tube acts on the plug for a distance l=3cm?

How do you know v=12.134m/s? Is this the initial velocity while the plug is inside the tube or is this the velocity exiting the tube after traveling 3cm?

Are we to assume there is no friction acting on the plug inside the tube?
 
The plug is completely embedded in the tube and the pressure does act on the plug for a distance l=3cm.

v=12.134m/s is the speed which I've calculated using laws of kinematics. I know that h=17cm, α=45o, s=5m. This is the speed with which the plug starts moving when it exits the test tube. There is no doubt that this is the correct speed.

It is said that the plug's and the test tube's friction force is proportional to their contact area. I don't really know how to use this information.
 
Can you show your work on that calculation? I think you have left out some important information here.

Also you'll have to assume some coefficient of kinetic friction.
 
Here's the calculation. As I've said, there's no doubt that this is the correct answer.
20150407_105518.jpg


I don't know the coefficient of kinetic friction, it is not given in the problem.
 
I seem to remember this part of the problem from a previous thread.

Do you think you could you use conservation of energy for this problem?
 
I suppose so, but first I have to know what temperature and humidity has to do with all of this.
 
Well do you agree there is a gas trapped inside the tube that has a certain pressure and temperature?

And just to be clear, what we are trying to show is that the final answer will come out to be the same as the calculated value of v=12.134 m/s.
 
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