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:

The letter

The letter

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/sec

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.

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,

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:

Read that out loud. Something is missing. Acceleration is measured as
**meters/sec ^{2}**. We don't have any meters or sec

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 / sec ^{2}**.

Let us look at one newton written as a fraction.

1 * kilogram * meter ____________________ = 1 newton sec^{2}

(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 sec^{2}* 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 sec^{2}* 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 sec^{2}

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.

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.

When the rocket takes off it leaves behind a column of water. The nozzle of the rocket has a area of 3.8 cm

Remember when we calculated the volume in the cylinder of the engines?

Note the word

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 cm^{3} and the nozzle area is 3.8 cm^{2}.

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:

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.

Or:

500 cm * cm * cm ________________ 3.8 cm * cmDo the division and remove the factors of one and we have: 131.6 cm or 1.316 meters.

Yes it does. Rather surprising isn't it.

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.

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.

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

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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