Kvant Physics Problem 14

Two pistons of masses $m_1$ and $m_2$ close the ends of two connected tubes whose cross sectional areas are $S_1$ and $S_2$.

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Verdicts: PASS + PASS
Solve time: 2m34s
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Problem

Two tubes with cross-sectional areas $S_1$ and $S_2$ are connected to each other and closed by pistons of masses $m_1$ and $m_2$ (Fig. 5). After an explosion in the space between the pistons, the pistons are ejected from the tubes. One of them was ejected with speed $v$. At what speed was the other ejected 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

Fig. 5

Setup and Assumptions

Two pistons of masses $m_1$ and $m_2$ close the ends of two connected tubes whose cross sectional areas are $S_1$ and $S_2$. An explosion occurs in the gas between the pistons and ejects them from the tubes. The speed of one piston after ejection is known and equal to $v$. The speed of the other piston is required.

The explosion is treated as an internal process. Friction between the pistons and the tube walls is neglected. The duration of the explosion is assumed short enough that external forces can be neglected during the process.

The pistons move only along the tube axis. Let the positive direction be toward piston 2. Then piston 1 moves with velocity $u_1<0$, while piston 2 moves with velocity $u_2>0$.

For part 1 the tubes are rigidly fixed to the ground.

For part 2 the tubes are free to move, and the combined mass of the tubes is $M$.

The quantities $S_1$, $S_2$, $m_1$, $m_2$, $M$, and the speed $v$ are given. The unknown is the speed of the second piston.

Physical Principles

The force exerted by the explosion products on each piston is equal to the gas pressure multiplied by the piston area. If the pressure in the connecting volume is $p$, then

$$F_1=pS_1,\qquad F_2=pS_2.$$

The impulse delivered to each piston is equal to the integral of the corresponding force over time:

$$J_i=\int F_i,dt.$$

Since the same pressure acts throughout the connected volume,

$$\frac{J_1}{J_2}=\frac{S_1}{S_2}.$$

The impulse of a piston equals its change of momentum. Since the pistons start from rest,

$$J_1=m_1|u_1|, \qquad J_2=m_2|u_2|.$$

Hence

$$\frac{m_1|u_1|}{m_2|u_2|} = \frac{S_1}{S_2}.$$

This relation follows directly from the pressure forces and remains valid whether the tubes are fixed or free.

For the free tube system, total momentum is conserved because external forces are neglected.

Derivation

Let

$$v_1=|u_1|, \qquad v_2=|u_2|$$

denote the speeds of the two pistons.

From the impulse relation,

$$\frac{m_1v_1}{m_2v_2} = \frac{S_1}{S_2}.$$

Rearranging gives

$$m_1v_1S_2=m_2v_2S_1.$$

Thus

$$v_2=\frac{m_1S_2}{m_2S_1},v_1.$$

This equation already determines the ratio of the piston speeds.

For the case of fixed tubes, no further condition is required. The speed of either piston immediately gives the speed of the other.

Now consider the free tubes. Let the tube assembly acquire velocity $V$ after the explosion. Conservation of total momentum gives

$$-m_1v_1+m_2v_2+MV=0.$$

The additional unknown $V$ appears because the tubes can recoil. However, the relation obtained from the pressure impulses remains unchanged:

$$m_1v_1S_2=m_2v_2S_1.$$

This equation contains only the piston speeds and is independent of $V$. Consequently the ratio of piston speeds is exactly the same as in the fixed tube case. The recoil velocity adjusts itself so that momentum is conserved, but it does not alter the relation between $v_1$ and $v_2$.

Therefore

$$v_2=\frac{m_1S_2}{m_2S_1},v_1$$

for both parts of the problem.

If the known piston is piston 1 and its speed is

$$v_1=v,$$

then

$$v_2=\frac{m_1S_2}{m_2S_1},v.$$

If instead the known piston is piston 2 and its speed is

$$v_2=v,$$

then

$$v_1=\frac{m_2S_1}{m_1S_2},v.$$

Result

For both fixed and free tubes, the piston speeds satisfy

$$m_1v_1S_2=m_2v_2S_1.$$

Hence, if the piston of mass $m_1$ is ejected with speed $v$,

$$\boxed{ v_2=\frac{m_1S_2}{m_2S_1},v }.$$

If the piston of mass $m_2$ is ejected with speed $v$,

$$\boxed{ v_1=\frac{m_2S_1}{m_1S_2},v }.$$

No numerical substitution is possible because the problem provides no numerical values for $m_1$, $m_2$, $S_1$, $S_2$, or $v$.

Sanity Checks

The factor multiplying $v$ is

$$\frac{m_1S_2}{m_2S_1},$$

which is dimensionless. The resulting speed has the correct unit of $\mathrm{m,s^{-1}}$.

If $S_1=S_2$, the pressure impulses are equal. The relation becomes

$$m_1v_1=m_2v_2,$$

which is the expected equality of impulse magnitudes delivered to the two pistons.

If $m_1=m_2$, then

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

The piston acted upon through the larger area receives the larger impulse and acquires the larger speed.

The result contains no dependence on $M$. This is consistent with the derivation because the speed ratio is determined entirely by the ratio of pressure impulses. The tube recoil affects only the common momentum balance of the whole system.

The most common algebraic error is reversing the area ratio when passing from

$$\frac{m_1v_1}{m_2v_2} = \frac{S_1}{S_2}$$

to the final expression for the unknown speed. A reversed ratio would give incorrect limiting behavior when one tube area becomes much larger than the other.