The physics of my foam rocket

Introduction

I got a small foam rocket for Christmas. It works as follows, you use a pump to pressurise a small base unit. The base unit has a rod onto which you slide the foam rocket. It has a small pressure gauge which gives a rough estimate of how high the foam rocket will go. Once you’ve supplied the desired amount of pressure, you press a small release button that fires the rocket using the air you’ve compressed. At full pressure, the rocket flies to around 120 feet into the air.

Being Simon, I asked myself two questions:

1. How fast is the rocket going when it leaves the stand?
2. What is the average acceleration of the rocket before it leaves the stand?

This is an elementary physics problem, however, if we just produced the correct equations, as if by magic, it wouldn’t be much of an article so to make this more article worthy, we’re going to do the whole thing from first principles, like only Simon would.

A discussion of Work

My laptop, on which I’m writing this article, has two forces acting on it. The force of gravity pulling the laptop down to the table and the electromagnetic repulsion between the electrons on the bottom of my laptop and the table preventing it falling any further. These forces are in an exact balance so there is no overall acceleration of my laptop.

So there must be a distinction between a force that is doing an acceleration and one that isn’t. This distinction is called “Work”. Work is defined as the integral of the dot product of the force and the distance over which that force was applied.

$\begin{array}{ccc}W& =& {\int }_{s_{\text{start}}}^{s_{\text{finish}}}F\cdot ds\end{array}$

We’re going to use Newton’s theory of gravity here (the error introduced by doing so is tiny compared to the other errors in our estimation) so we can consider gravity to be a force. Since this force operates over a distance (namely, the height the rocket flies) then we must conclude work is done on the rocket throughout the flight.

To get our desired results from first principles, we need to do two things. First, calculate the work done on the rocket by gravity and second, show that the work required to stop a body in motion depends only on the mass and velocity of that body and to derive an equation that relates work to these two properties.

The work done on the rocket by Gravity

Newton postulated that gravity was a force and that the size of this force depended only on the mass of the two bodies and their separation. Put mathematically, if F is the magnitude of the force, r is the separation and M and m are the masses of the two objects involved then:

$F= \frac{GmM}{r^2}$

Where G is a fixed constant, called the gravitational constant. The distance r is the length of the line that connects the centre of masses of the bodies. In our case, we want to compare the strength of the force between the ground and about 48 metres above the ground. Using the following numbers:

$\begin{array}{rr}G& =6.67\times 10^{-11}{\mathrm{m}}^3{\mathrm{k}\mathrm{g}}^{-1}{\mathrm{s}}^{-2}\\ r_{\text{ground}}& =6378100\mathrm{m}\\ r_{\text{max height}}& =6378148\mathrm{m}\\ M_{\text{earth}}& =5.9742\times 10^{24}\mathrm{k}\mathrm{g}\end{array}$

From our definition of work in Figure 1.1 we now have enough data to calculate the work done by gravity on the rocket by the time it stops at its peak height. We deliberately omit the mass of the rocket from the integral below because we will see that we don’t need to know that value in order to solve our problem:

$\begin{array}{rlr}W& =& {\int }_{s_{\text{start}}}^{s_{\text{finish}}}F\cdot ds\\ & =& {\int }_{s_{\text{start}}}^{s_{\text{finish}}}\frac{GmM}{r^2}\cdot dr\\ & =& m{\int }_{s_{\text{start}}}^{s_{\text{finish}}}\frac{GM}{r^2}\cdot dr\\ & =& -m{\left[\frac{GM}{r}\right]}_{r_{\text{start}}}^{r_{\text{finish}}}\end{array}$

Now it’s just a question of putting in the numbers:

$\begin{array}{ccc}W& =& -m{\left[\frac{GM}{r}\right]}_{r_{\text{start}}}^{r_{\text{finish}}}\\ & =& -m{\left(\frac{GM}{r_{\text{finish}}}-\frac{GM}{r_{\text{start}}}\right)}_{}^{}\\ & =& -m\times GM{\left(\frac{1}{r_{\text{finish}}}-\frac{1}{r_{\text{start}}}\right)}_{}^{}\\ & =& -m\times (6.67\times {10}^{-11}\cdot 5.9742\times {10}^{25})\times (\frac{1}{6378148}-\frac{1}{6378140})\\ & =& -m\times (3.9847914\times {10}^{14}\times -1.1799274\times {10}^{-12})\\ & =& -m\times -470.3879\\ & =& 470.3879m\end{array}$

Finding an expression for the Work done in terms of Mass and speed.

In figure 1.1 the definition of Work is given as the integral of the dot product of the force and the distance over which that force is applied. Intuition tells us that if a force acts over a distance on a body then that body ought to accelerate and it is harder to accelerate a large body than it is to accelerate a light body. It should, therefore, be possible to find the amount of work required to accelerate a body to a speed, v, using only the mass of the body and that speed.

Let’s examine our definition of Work again:

$\begin{array}{ccc}W& =& {\int }_{s_{\text{start}}}^{s_{\text{finish}}}F\cdot ds\end{array}$

Newton’s second law is that a force accelerating a body is defined as the rate of change of momentum of that body:

$\begin{array}{ccc}F& =& \frac{\text{d}mv}{\text{d}t}\end{array}$

In almost all cases the mass in a system is kept constant. So Figure 1.6 becomes:

$\begin{array}{ccc}F& =& m\frac{\text{d}v}{\text{d}t}\end{array}$

Next, we substitute Figure 1.7 into our definition of Work and do some rearrangement:

$\begin{array}{ccc}W& =& {\int }_{s_{\text{start}}}^{s_{\text{finish}}}F\cdot ds\\ & =& {\int }_{s_{\text{start}}}^{s_{\text{finish}}}m\cdot \frac{\text{d}v}{\text{d}t}\cdot ds\\ & =& {\int }_{v_{\text{start}}}^{v_{\text{finish}}}m\cdot \frac{\text{d}s}{\text{d}t}\cdot dv\\ & =& {\int }_{v_{\text{start}}}^{v_{\text{finish}}}m\cdot v\cdot dv\end{array}$

This step might be a little confusing so I’ll go through it. Fig 1.7 shows dv by dt multiplied by ds. What Figure 1.9 shows is that we can write this as (dv times ds) by dt. We can then take dv outside to give (ds by dt) multiplied by dv. ds by dt which is clearly the rate of change of distance which is v. So we can write v multiplied by dv in the integral. It looks like a bit of witch craft but we can get away with it..

For our problem, we want to consider the work done in accelerating a body from speed 0 to some speed V_finish. So we solve Figure 1.8 with v_start=0:

$\begin{array}{ccc}W& =& {\int }_{0_{}}^{v_{\text{finish}}}m\cdot v\cdot dv\\ & =& {\left[\frac{m{v}^2}{2}\right]}_{0_{}}^{v_{\text{finish}}}\\ & =& \frac{m{v_{\mathrm{finish}}}^2}{2}-\frac{m\cdot 0^2}{2}\\ & =& \frac{m{v_{\mathrm{finish}}}^2}{2}\end{array}$

It’s clear that Figure 1.9 is an equation relating Work require to accelerate an object to speed V_final is dependant on only that speed and the mass of the body - exactly the point we wanted to derive.

Putting it all together

My first goal for this article was to determine, from first principles, how fast the foam rocket leaves the stand. With Figure 1.9 and 1.5 we are now in a position where we can determine this.

Figure 1.5 tells us the amount of work done on the rocket by gravity over the 48 metre height that the rocket climbs to. The secret to cracking this problem is realising at the top of the 48 metre climb, the rocket momentarily stops. Gravity is working on the rocket right from the moment it leaves the stand so all the energy to climb all 48 metres must come from the initial speed the rocket carries.

To see this, imagine for a moment we could make time run backwards. We’d see the rocket start at 48 metres above the ground and accelerate to the ground. The speed of the rocket just before it leaves the stand would be equal to the amount of work done by gravity over the 48 metres. This allows us to determine the speed when it left the stand by equating Fig 1.9 and 1.5:

From figures 1.5 and 1.9:

$\begin{array}{ccc}W& =& 470.3879\cdot m\\ W& =& \frac{m{v}^2}{2}\end{array}$

So:

$\begin{array}{ccc}\frac{m{v}^2}{2}& =& 470.3879\cdot m\\ \frac{v^2}{2}& =& 470.3879\\ v^2& =& 940.776\\ v& =& 30.672{\text{ms}}^{-1}\end{array}$

30.672 metres a second translates to 110.419 kilometres per hour or 68.611 miles per hour. Quite fast, I think you’ll agree.

The next thing to tackle is the acceleration of the rocket as it leaves its stand. The acceleration we calculate here is an average since the way the acceleration actually happens as a function of the height up the stand is probably very complicated. For our argument, we assume that the rocket accelerates at a constant rate.

Once again, we bring out our trusty definition of work:

$\begin{array}{ccc}W& =& {\int }_{s_{\text{start}}}^{s_{\text{finish}}}F\cdot ds\end{array}$

Combine this with our definition of a force in Figure 1.6 then we get:

$\begin{array}{ccc}W& =& {\int }_{s_{\text{start}}}^{s_{\text{finish}}}F\cdot ds\\ & =& {\int }_{s_{\text{start}}}^{s_{\text{finish}}}\frac{\text{d}mv}{\text{d}t}\cdot ds\end{array}$

Since the mass of our rocket is constant and we’re assuming that the acceleration, a=dv by dt, is also a constant then the integral above simplifies to:

If we define the bottom of the stand to be zero, then S_start=0, then Figure 1.12 becomes mas_finish. The work done that we’ve derived here must be equal to work done in accelerating the rocket from rest to its take-off velocity. so equating Figure 1.9 with this result we obtain a formula for the average acceleration experienced by the rocket:

$$\begin{array}{lcr}ma{s}_{\text{finish}}& =& \frac{m{v}^2}{2}\\ a& =& \frac{v^2}{2s_{\text{finish}}}\end{array}$$

The stand is roughly 0.2 metres long so all that is left to do is to put in the numbers and out comes the answer:

$$\begin{array}{lcr}a& =& \frac{v^2}{2s_{\text{finish}}}\\ & =& \frac{(30.672{)}^2}{2\times 0.2}\\ & =& 2351.928{\text{ms}}^{-2}\end{array}$$

That might sound like a huge number for a little foam rocket but it’s actually quite reasonable when you consider it does 0-60mph in 0.2 metres. Put another way, if you were accelerated that much you’d momentarily weigh 240 times what you do normally. Quite remarkable!

The analysis in this article is approximate. It neglects all frictional forces and doesn’t make exact measurements of the height attained by the rocket or the length of the stand.. Also, the jet of air might continue to accelerate the rocket once it leaves the stand. This said, it’s a reasonably good estimate of the speed and acceleration of the rocket and it derives the mechanics from first principles so I’m quite proud of the article. One day, I might do a follow up with a more exact analysis! If you’ve got this far, then thanks for reading and have a nice day.

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Copyright Simon Johnson 2004 - Remastered in to MathML 2025