(18b)   Momentum  

   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..

Home pages:
      "From Stargazers to Starships" ....stargaze/Sintro.htm
      Lesson plan index:             ....stargaze/Lintro.htm

Goals: The student will learn

• The concepts of momentum and its conservation, using the recoil of a cannon as an example.

• That momentum is a vector, allowing its conservation to be applied to problems in 2 and 3 dimensions.

• That in the recoil of a gun, momentum is shared equally, but energy is not.

• An example shows that, similarly, in collisions kinetic energy needs not be conserved.

• How launching even a moderate payload into space may require a huge amount of fuel.

Terms: momentum, conservation of momentum, recoil.

Starting the lesson:
The 4 statements set apart below in red bold face may be put on the blackboard, to be copied by students.

  We have discussed so far mass, velocity, acceleration, force and energy, and the way Newton's laws tie them together.
  There exists one additional concept which is rather important, namely,

momentum P = mass times velocity

In an isolated system, momentum is conserved.

  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.

[In passing, that is how armor-piercing shells are designed. They can work their way through 4 inches (10 centimeters) or more of steel in a tank's armor, not by smashing the steel, which is too strong, but by melting it. Such shells usually contain a core of tungsten or uranium ("depleted uranium" from which the component used for producing nuclear power has been removed), very heavy metals whose high mass can also carry a great deal of kinetic energy.]

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.

P = mv

Yes, momentum is a vector quantity. If all motions are along the same line, we can take vector character into account by giving momenta in one direction a (+) sign and in the opposite direction a (-) sign.

In an isolated system, the sum of all momenta is conserved.

A system with no forces acting on it from the outside.

When you jump, you brace your foot against the ground, so that the Earth, too, is part of the system. When your body takes off, an opposite and equal amount of momentum has been given to the Earth, and in principle the Earth actually moves back a tiny, unmeasurable amount. When you land, your momentum is given back to the Earth, and everything is as before.

Let 1 denote the car, and let the + direction be the one in which it moved.
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).

So:

m₁ = 1500 kg    v₁ = 40 km/s
m₂ = 4500 kg     v₂ = -20 km/s

P = m₁v₁ + m₂v₂ = (m₁ + m₂)v₃

1500*40 + 4500*(-20) = 60,000 - 90,000 = -30,000 = 6000*v

v = - 5 km/s

Not likely, since each of the masses (if we consider them separately after the collision) now moves more slowly than before.

If the result is to be expressed in joules, we better convert km/hr to meters/second:

1 km/hr = (1000 meter/3600 sec) = 0.27777 m/s

Initial velocities: v₁ = 11.111 m/s   v₂ = 5.5555 m/s
Final velocity     v₃ = 1.3889 m/s

Kinetic energy =1/2 m v²

KE of the car         (1/2) 1500 (11.111)² = 92, 593
KE of the truck     (1/2) 4500 (5.5555)² = 69,444 joule

Total kinetic energy entering the collision 162,037 joule
Final KE (1/2) 6000 (1.3889)² = 5,787 joule
Loss 156,250 joule

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--Where did the lost energy go?
It probably went into heat.

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--(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?
Yes, they do

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--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.) '
It makes no difference. Suppose the colliding billiard balls have masses (m₁, m₂) and viewed from the airplane ("in the reference frame of the airplane") have velocities (v₁, v₂) where v₁ is positive and v₂ negative. The balls bounce back at velocities (v₁ ',v₂ ') and the conservation of momentum gives
m₁v₁ + m₂v₂ = m₁v₁ ' + m₂v₂ '

Viewed from the ground, each velocity is increased by the velocity v₀ of the airplane, so the conservation of momentum should give
m₁(v₁+v₀ ) + m₂(v₂+v₀) = m₁(v₁ '+v₀) + m₂(v₂ '+v₀)

This equation differs from the one preceding it only in having m₁v₀+m₂v₀ added to both sides. Since equations remain valid when equal quantities are added on both of their two sides, if one of these equations holds, the other automatically does so, too.

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Optional--Rocket Motion.

   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.

   [Calculating this "mean velocity" is actually something of a challenge. Suppose some mass m of the jet is ejected with velocity v. At first, it imparts an equal and opposite momentum mv to the rocket. However, as the rocket gains speed, one must subtract that speed from v, which first needs to overcome the forward motion. Clearly , the rocket effect gets less and less efficient!

   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

Total mass	Gives final velocity
2M	V
4M	2V
8M	3V
16M	4V
32M	5V
64M	6V

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.