# Could this question be calculated without knowing the change in length?

1. Jan 7, 2012

### +polarity-

In order to measure the energy of projectiles, a test facility consisting of a horizontal insulated tube, closed at one end and with a frictionless piston of mass 1.5 kg at the other is used. The tube has a cross sectional area of 0.05 m2 and contains 1.15 kg of air at 20 °C and 100 kPa. During a test the piston is struck by a bullet of mass 0.15 kg travelling at 800 m/s, which remains embedded in the piston. It rapidly moves and compresses the air bringing the piston and bullet to rest.
For air R = 0.287 kJ kg–1 K–1 ; Cp = 1.005 kJ kg–1 K–1 .

Assuming the bullet is embedded in the piston, and momentum is conserved, and reversible adiabatic compression of the air in the tube, calculate the temperature of air at the end of compression when the piston has stopped.

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P1 = 100 kPa; T1 = 293K and so using PV=mRT, I calculated V1 to be 0.967 (3dp) and so the length of the chamber would be the V/A => 19.34m and if I was given the length it was compressed to (or if someone could kindly point out how to calculate it) I could work out T2 using T2 = T1 (V1/V2)^k-1 as it's adiabatic.

Any help would be really appreciated!

2. Jan 8, 2012

### 256bits

The solution to this problem is not a plug into one formula and find the answer. It involves kinematics and thermodynamics and tests how well you understand the concepts.

Problem states that momentum is conserved, so that is your first clue for a calculation.
What next might be important, well maybe the kinetic energy of the piston imbedded with the bullet at impact.
Then the problem states that the piston stops, so you would have kinetic energy go from some initial value at impact to zero.
Well, now there must be some force acting on the piston to slow it down, and we all know that F=ma and W=Fd, so that looks a little promising, but ...

Aha - all the kinetic energy must go into compressing the gas, and since it is an s adiabatic process then the work on the gas can be calculated from a formula,

And since PV^gamma=C, the temperature can be found also.

( Hope my post is not too lame or missing a step, as I haven't had my morning coffee yet )

3. Jan 8, 2012

### +polarity-

http://img208.imageshack.us/img208/2708/workingzl.jpg [Broken]

Unfortunately that isn't correct but that is most likely just me doing it wrong. Thanks so much for the help 256bits, you explained it perfectly and is much more logical in comparison to my attempt :)

Last edited by a moderator: May 5, 2017
4. Jan 8, 2012

### 256bits

Your calculations look finebut I guess it is the incorrect method.

Try
Cv = R - Cp

dU +dW = Q = 0 ==> dU = dW

$\Delta$U = -P$\Delta$V

$\Delta$U = m Cp $\Delta$T

And the change in internal energy would equal the change in kinetic enrgy

solve for $\Delta$T

5. Jan 11, 2012

### rude man

I would't try that one!
I'll try to get involved here later if the right answer is still not forthcoming.