Polish Physics Olympiad: Proving Adiabatic Reversible Process

[marcnn]

Homework Statement

(56th Polish Olympiad in Physics, 2007) We have a tube of mass $M$, consisting of two segments of diameters $d_1, d_2$. The pistons (see the picture http://www.kgof.edu.pl/archiwum/56/of56-2-1-R.pdf) have mass $m_1, m_2$.

At the start the air inside had pressure $p_0$ equaling the pressure outside the pipe. The tube itself and the right piston weren't moving. The left piston was moving to the right with velocity $v_p$.
The force air acts on an element of a piston or a pipe doesn't depend on the element's velocity.

The process is adiabatic and reversible and the pistons are hermetic. We neglect the friction.

Homework Equations

It is suggested that if $a_1, a_2, a_3$ are the accelerations of the left piston, right piston and the tube respectively, $S_i = \frac {\pi d_i^2}4, i = 1,2$, moreover $\Delta S = S_1 - S_2$ and $p$ is the difference of pressures, then
$$m_1 a_1 = -pS_1 ~~~~(1)$$
$$m_2 a_2 = pS_2 ~~~~(2)$$
$$M a_3 = p \Delta S ~~~~(3)$$

Is my attempt of proving this correct?

The Attempt at a Solution

The formulas (1) and (2) are obvious and come from the formula $F = pS$. It's only left to prove the formula (3).
Let $p_1$ be the momentum of the left piston, $p_2$ - of the right piston and the tube. The momentum of the whole system is constant, so $p_1 +p_2 = 0$. Hence the difference $d(P_1 + P_2) = dp_1 + dp_2 = 0$. If it happens over the same, very short time, we have
$$ 0 = \frac {dp_1}{dt} + \frac {dp_2}{dt} = \frac {m_1 dv_1}{dt} + \frac {m_2 dv_2 + M dv_3}{dt} = m_1 a_1 + m_2 a_2 + M a_3 $$

Hence
$$Ma_3 = -m_1a_1 - m_2 a_2 = p(S_1 - S_2) = p \Delta S$$

Is it correct?

 

[TSny]

I believe that's correct. Note that it agrees with the notion that the net force on $M$ is due to the difference in pressure $p$ acting over the regions of $M$ shown below.

 

[marcnn]

Thanks!