Kvant Physics Problem 14

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Problem

Two tubes with cross-sections $S_1$ and $S_2$ are connected to each other and plugged with pistons whose masses are $m_1$ and $m_2$ (Fig. 5). After an explosion in the space between the pistons, the pistons are ejected from the tubes. One of them leaves with a speed $v$. What is the speed of the second one if:

1. the tubes are fixed and cannot move,
2. the tubes are not fixed and their total mass is $M$?

Neglect friction between the pistons and the tube walls.

Fig. 5

Setup and Assumptions

Two pistons of masses $m_1$ and $m_2$ move inside two rigid tubes of cross-sections $S_1$ and $S_2$. The pistons are initially at rest and are driven apart by a short explosion in the common chamber between them. After the explosion, each piston is ejected along its tube with final speeds $v_1$ and $v_2$.

The gas produced by the explosion is assumed to act only for a short time and is treated as providing pressure forces on the piston faces. Friction between pistons and tube walls is neglected. Motion is purely one-dimensional along the tube axes.

In the first configuration the tubes are fixed in space. In the second configuration the two tubes form a single rigid body of total mass $M$ that is free to recoil. The goal is to determine the speed of the second piston when one piston leaves with speed $v$.

Physical Principles

The motion of each piston is governed by the impulse of the pressure force exerted by the expanding gas. The force on a piston equals the gas pressure multiplied by the piston cross-section, $F_i = p(t) S_i$, so the impulse is $J_i = \int F_i , dt = S_i \int p(t),dt$.

Newton’s second law in integral form relates impulse to momentum change, $J_i = \Delta p_i = m_i v_i$ since each piston starts from rest.

For the full system, linear momentum is conserved in the absence of external forces. In the case of movable tubes, the total momentum of pistons and tubes remains constant.

Derivation

The pressure in the common chamber acts uniformly on both pistons during the explosion interval. The impulse delivered to piston $i$ is

$$J_i = S_i \int p(t),dt.$$

Both pistons experience the same time-dependent pressure integral, so the impulses are proportional to cross-sectional areas,

$$\frac{J_1}{S_1} = \frac{J_2}{S_2} = \int p(t),dt.$$

Using $J_i = m_i v_i$, the momentum of each piston satisfies

$$\frac{m_1 v_1}{S_1} = \frac{m_2 v_2}{S_2}.$$

Solving for the second piston speed gives

$$v_2 = \frac{m_1 S_2}{m_2 S_1} v_1.$$

In the case of fixed tubes, no additional motion is introduced, so this relation fully determines both piston speeds. If one piston leaves with speed $v$, the other has speed

$$v_2 = \frac{m_1 S_2}{m_2 S_1} v.$$

When the tubes are free with total mass $M$, the system acquires recoil motion, but the pressure-driven impulse distribution between pistons remains unchanged. The tube motion does not alter the internal pressure history responsible for the piston impulses, so the same relation between $v_1$ and $v_2$ holds.

Total momentum conservation for the whole system introduces the recoil velocity $V$ of the tubes,

$$m_1 v_1 + m_2 v_2 + M V = 0,$$

but this equation determines $V$ after substituting the piston velocities and does not modify the ratio between $v_1$ and $v_2$.

Result

For both configurations, if one piston leaves with speed $v$, the speed of the other piston is

$$v_2 = \frac{m_1 S_2}{m_2 S_1} v.$$

Substituting the given speed $v$,

$$\boxed{v_2 = \frac{m_1 S_2}{m_2 S_1}, v ;\text{m/s}}.$$

This result applies identically whether the tubes are fixed or have total free mass $M$.

Sanity Checks

The expression is dimensionally consistent since $\frac{m_1 S_2}{m_2 S_1}$ is dimensionless and multiplies a speed.

If $S_1 = S_2$, the result reduces to $m_1 v_1 = m_2 v_2$, which matches equal impulse sharing between identical pressure histories.

If $m_1 \gg m_2$ with comparable cross-sections, the heavier piston moves more slowly, consistent with $v \propto 1/m$ under equal impulse delivery.

If $S_1 \gg S_2$, piston $1$ receives a larger impulse and therefore attains higher speed, consistent with $J \propto S$.

The tube mass $M$ does not appear in the final relation for piston speeds, consistent with internal-force symmetry: tube recoil affects only the motion of the supporting body and not the impulse ratio between pistons.