(18b) MomentumOne useful consequence of Newton's 3rd law is the conservation of momentum, as is shown by analyzing the recoil of a cannon. The example shows that while the gun and shell get equal amounts of momentum, by far most of the energy is given to the shell.
Here mechanical energy is not conserved--some ends up as heat. However, one can also visualize perfectly elastic collisions--like the gravity-assist maneuver in section #34, when a spacecraft encounters a moving planet--and there the total kinetic energy is conserved, although some may be passed from the planet to the spacecraft, or vice versa..
Part of a high school course on astronomy, Newtonian mechanics and spaceflight
by David P. Stern
This lesson plan supplements:|
Section #18b "Momentum,"
Note to the teacher|
Momentum is a fundamental concept in Newtonian mechanics, but in the original version of "Stargazers" it was left out, in the interest of brevity. The concepts of rocket motion (#25) and of planetary gravity-assist maneuvers (#34), which are usually presented as applications of the conservation of momentum, were covered there in different ways, using the concepts of centers of gravity and of frames of reference. Teachers who choose to include this lesson may also relate those sections to the conservation of momentum.
Goals: The student will learn
Terms: momentum, conservation of momentum, recoil.
Starting the lesson:
We have discussed so far mass, velocity, acceleration, force and energy, and the way Newton's laws tie them together.
Now you already know that energy is conserved, but there exists a big difference: energy can change into other forms, say turn into heat. Therefore
mechanical energy, (potential + kinetic), is not always conserved--some of it may change into other forms.
For instance, when a bullet hits a wall, where does its kinetic energy go? It turns into heat.
The total momentum going (say) into a collision always equals the total momentum coming out of it--there is nothing else momentum can convert to. It is therefore something we can always rely on in a calculation. The momentum given by a rocket to its gas jet is always equal to the momentum which it itself receives, regardless of the details of the process.
The way momentum will be introduced here is through an actual example.
Here go into the lesson, the calculation of the recoil of a cannon.
Guiding questions and additional tidbits:
-- What is the momentum P of a mass m moving with velocity v?
--Does this depend on the direction of v?
--State the important property of momentum.
-- What is "an isolated system"?
--When you jump across a ditch, your body clearly has a momentum P = mv during the jump. It did not have that momentum earlier and does not have it afterwards. How can you then say that P is conserved?
--A 1500 kg car going at 40 km/hr smashes head-on into a 4500 kg truck going in the opposite direction at 20 km/hr. The cars end up locked together. In what direction does the wreckage move (initially), and how fast?
Let 2 denote the truck, moving in the - direction.
For the equations of the conservation of momentum, the units are not important, as long as the same ones are used before and after the collision (i.e. as long as we compare quantities measured in the same units).
m2 = 4500 kg v2 = -20 km/s
v = - 5 km/s
--Is kinetic energy conserved?
--How much kinetic energy was lost?
1 km/hr = (1000 meter/3600 sec) = 0.27777 m/s
Final velocity v3 = 1.3889 m/s
Kinetic energy =1/2 m v2
KE of the truck (1/2) 4500 (5.5555)2 = 69,444 joule
Total kinetic energy entering the collision 162,037 joule
--Where did the lost energy go?
--(Optional) "Humongous Airlines" publicized the smooth ride of its new "steadijet" airliner by installing a billiards table in its first class cabin. While the plane is flying at a steady velocity v0, do collisions of two billiard balls in it conserve momentum?
--Which velocities do we have to use in such a calculation--velocities relative to the airplane or to the ground? (For simplicity, assume the balls collide head-on and move along the direction in which the airplane flies, so that all motions are along the same line.) '
Suppose we have a rocket of total mass 2M, of which M is payload and M is fuel. As the fuel is burned, it is ejected with an average velocity V relative to what the rocket had at the start.
To avoid this (the full derivation requires calculus) let us just define the mean velocity V by conservation of momentum, without trying to puzzle out its relation to the jet speed v. At burn-out, we say, the jet has imparted momentum MV to the payload, and has itself carried an equal and opposite momentum backwards.]
The payload now has gained velocity V.
But we need more! so we build a rocket of mass 4M, of which 2M is fuel, while the payload, also of mass 2M, is the smaller rocket described above, serving as second stage. When the fuel of the big rocket is finished, we reach a velocity V, then the second stage is ignited, adding another V to the velocity, for a total of 2V.
Still faster! Now the rocket has mass 8M of which 4M is fuel of the first stage, while 4M is the two-stage rocket of the preceding design. The first stage gives velocity V, to which the other two add 2V, for a total of 3V.
By now you can see the trend. If the mass of the final payload is M, then
Each time the velocity increases by one notch, the mass doubles.
One cannot avoid this sort of thing by giving up staging--say, in the rocket of mass 8M, by firing all the 7M of fuel in one blast. That is because (as already noted), as the payload (+ remaining fuel) gain speed, less and less momentum is transferred, the jet first having to overcome the forward motion, Indeed, the correct derivation (which uses calculus) gives the equivalent of a huge number of little stages, fired one after the other. The same exponential result is still obtained.
This is one of the great problems of spaceflight, especially with the first stages which rise from the ground: even a small payload requires a huge rocket. Perhaps some day space explorers will be able to shave off some fuel weight by using air-breathing rockets ("scramjets") but only for the lowest 1/4 to 1/3 of the orbital velocity. Launching from a high-flying airplane--like Burt Rutan's "SpaceshipOne", or the "Pegasus rocket--also helps cut air resistance, another factor. But no other shortcuts are in sight. Once in orbit, of course, more efficient but more gradual ways of generating thrust can be enlisted, like ion propulsion.
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Author and Curator: Dr. David P. Stern
Mail to Dr.Stern: stargaze("at" symbol)phy6.org .
Last updated: 10-19-04