Water Rocket Acceleration


Now we have the information we need to calculate the acceleration. Here are the numbers we start with:

Mass of our rocket: 543 grams Initial Thrust: 131 newtons

Take care to notice that the Initial thrust is stated. As the rocket flies, the thrust will be constantly changing. We will get to that soon. For now we will be concerned only with calculating what can be called instantaneous thrust. The word instantaneous means the thrust at any one instant in time.

Acceleration means a change in velocity. The equation for acceleration is generally written as:

F = ma

And spoken as: Force = mass times acceleration.
The letter m stands for mass, or any object that has mass.
The letter a stands for acceleration, or how fast the mass is accelerating.
This means that we start with known mass of an object. Next we apply an unknown force such that the object accelerate at a certain rate. Let us presume the acceleration is 1 cm/sec2. From that we can calculate the force that was applied.

Calculate The Acceleration

We know the force and we know the mass so we can solve for acceleration. The result is:

a = F/m

The acceleration is equal to the force divided by the mass.

What does that mean?

Back to our equation:

a = F / m

More Thought

Let us think about that a bit more. Why it is force divided by mass rather than the other way around?


Mass is the name used to describe, well, stuff. Everything that we can touch and feel. And many things that we cannot. But we need a better definition. How do we say: How much mass?

Over the years we humans have derived a method of answering that question: How much mass? Mass is the defined as the resistance to acceleration. (Deceleration is the same as acceleration, just in the other direction.) How do we do that?

We take that same equation and reformat it to: m = F / a or mass = Force divided by acceleration.

Question: But that's the same equation for Force. Are we using something to define itself?

Yes we are. Stay with me here and it will make sense.

Measuring The Mass

Imagine we are in the space station above earth and there is no gravity. In front of me is a small metal cube. It could be anything. You are to my left with a radar gun measuring the velocity of that cube. In my hand is a force meter calibrated in newtons.

After practicing a few times, I use the force meter and press on the cube with exactly one newton of force for exactly one second. Then I stop pushing and you measure the velocity of my cube and find that it is moving at 1 m/sec. The mass was accelerated for one second and the velocity was then equal to 1 m/sec.

Science has decided that the mass of that object is one kilogram.

If I put two of the cubes together and press with the same force for the same time, you will measure a velocity of 1/2 m/sec. If I take the first cube, cut it in half, then the result will be a velocity of 2 m/sec.

We had to start somewhere, so with that one experiment we demonstrated the definitions of mass, newton, acceleration, and one second.

There is more in that simple equation, F = am, than is at first apparent.

Back to the question: Why divide by m? Because if I increase the mass, the acceleration is reduced and the velocity is reduced. That means mass must be in the denominator.

Now we have the numbers for the the amount of thrust, the F part of the equation. We also have the numbers for the mass of our rocket, the m part of the equation. Plug the values into the equation. But, remember the section above: Keeping The Terms. Putting it all together we have:

a = 137.72 newtons / 543 grams

Read that out loud. Something is missing. Acceleration is measured as meters/sec2. We don't have any meters or sec2.

The Difficult Part

The term newton is a measure of force. The word newton is a short-hand term used to make discussions and equations easier. How much force is one newton? One newton is:

The amount of force needed to accelerate one kilogram at the rate of one meter per second squared.

That is the definition of a newton. Any place we have newton it can be replaced with: 1 * kilogram * meter / sec2.

Let us look at one newton written as a fraction.

	 	1 * kilogram * meter
	 	____________________   =  1 newton

(The "1" is usually presumed and not written but I want to be clear.)

Now let us take our equation: a = F/m and replace our symbols F and m with the values we have already determined. From the previous lesson our force was 132.72 newtons. Our mass was 542.65 grams. We convert the grams to kilograms to get 0.54265 kilograms. When written as a fraction we have:

		132.72 kilogram * meter 
		_________________________         = acceleration
		sec2 *   0.543 kilograms

Make sure you see the newtons and the mass. The mass is easy, the newtons is a bit more difficult.

Do you see a factor of 1 in that equation. If not, look for kilograms divided by kilograms. Lets remove the kilograms and we have:

		132.72 meter 
		_________________________         = acceleration
		sec2 *   0.543

Is this starting to look familiar? Let us do the division and clean up the denominator some.
132.72 / 0.543 = 240.73 so we have:

		240.73 meter 
		________________      = acceleration

We have distance divided by time squared.

Remember our discussion with the car. That is an acceleration. If our rocket were to accelerate that fast, after one second it would have a velocity velocity of more than 240 meters per second. At that rate the rocket would need less that one second to get to the other end of two and a half football fields. That is moving pretty fast. If a car accelerated that fast it would go from zero to more than 500 miles per hour. In one second.

Not So Fast

(Pun intended)
Alas, our rocket does not accelerate to anywhere near that fast. Why not? Because we run out of rocket fuel in much less than one second. Still, when that plug popped out of the nozzle, it took off like, well, like a rocket.

Our rocket probably did accelerate to close to 50 mph. That is pretty fast. If you were in the way it could really injure you. That little bit of pumping we did put enough energy into our rocket to make it jump up to some 50 mph and leap up into the air.

Pretty neat.

Bonus Question

How far did the rocket go before it ran out of fuel? We can make a quick approximation.

When the rocket takes off it leaves behind a column of water. The nozzle of the rocket has a area of 3.8 cm2. So its is leaving a column of water behind. Said differently, it leaves a cylinder of water behind with a cross sectional area of 3.8 cm2. The column falls apart very quickly, but we can use our imagination and see it in our mind's eye.

Remember when we calculated the volume in the cylinder of the engines?
Note the word cylinder. It does describe the shape of both that part of the engine and our plume of water.

We multiplied the area of the piston by the length of the stroke. The area of piston corresponds to the area of the nozzle. The stroke of the piston in the engine corresponds to the length of the water column left by the rocket. In the engine we had the stroke and multiplied by the area of the top of the piston to get the volume.

In this problem, we have the area (of the nozzle) and we already have the volume of the water. We want to calculate the length of that cylinder, the length of the column of water. So divide the area into the volume of water. The volume was 0.5 liters or 500 cm3 and the nozzle area is 3.8 cm2.

Remember all the discussion about terms. Write down the equation, do the division, and see if you get a length in centimeters. Hint: we will be dividing cubic centimeters by squared centimeters. Said another way the division problem will contain these terms:

( cm * cm * cm ) / (cm * cm )

That is cm3 divided by cm2. Look at what is above and below the vinculum. (Note, in this format the vinculum is replaced by the forward slash.) What factors of one can you find and remove. What will be left over from that division?

Hint: That column of water is not very long. The rocket is surprisingly low when it runs out of fuel and coasts the rest of the way up. And that is an indication of how fast it is going when the fuel is gone.


Written as a fraction we have: volume of water / cross sectional area of the column of water.
		500 cm * cm * cm
		3.8 cm * cm 
Do the division and remove the factors of one and we have: 131.6 cm or 1.316 meters.


Does that mean the rocket ran out of fuel by the time it was about as high as our heads?

Yes it does. Rather surprising isn't it.


The next phase is to calculate the velocity. That will be another lesson. And it will be more complex. You might ask why?

Because as the water squirts out the back, the rocket has less mass. That means it should accelerate faster. But, at the same time, the air expands to take up the space the water vacated. That means that the pressure is reduced and there is less force on the water. Simultaneously, the air expands very quickly causing the temperature to go down. That further reduces the pressure.

The result is that the mass is being reduced, but the force (the air pressure) Thai produces the thrust is also being reduced.

Which has the greatest effect?

The change in mass, or the change in pressure?

Think about it. From these few lessons you have sufficient information to answer the question. But it is a difficult question to answer. You will not answer it for certain until you put some numbers together. And they have to be put together just right. We will start putting those facts together and come up with our answer in the next lesson.


If you do want to try that problem, here is a hint. Use the two extremes. First, use the mass and pressure at liftoff. That is the point where the mass is the greatest and the pressure is the greatest. We just did that and calculated the acceleration. Said more correctly, we calculated the instantaneous acceleration. The acceleration at the moment the plug popped out.

Second: For the second extreme, consider the point just before the end of powered flight. This is the time where the mass and the pressure are both almost the least values. Think about what happens when there is a small amount of water left, say 10 cc of water. Just enough water left so that all the air doesn't go out with a quick whoosh. That is called the water out event and it changes all the thrust calculations. We will consider that on another day.

When only 10 cc of the water remains, how must volume must the air occupy? Remember, making a gas change shape, size, and pressure is fairly easy.

From the mass of the air and the volume of the air you can determine the pressure of the air just before the water out event. This will probably be the difficult part.

Now recalculate the gross mass of the rocket. Subtract out all that water that has been ejected out the nozzle. You have a new gross mass and a new pressure. Put those values back into our Force equation and determine the thrust just before the water out event. Then determine the new acceleration. Just like we did here but with the new numbers.

Compare the two acceleration numbers and see what you have discovered.

With the calculated acceleration at both ends of the powered flight, you can probably determine which has the greater effect, lower mass or lower pressure.

Dec 2014
Bryan Kelly
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