Disclaimer: The following material is being kept online for archival purposes.

Although accurate at the time of publication, it is no longer being updated. The page may contain broken links or outdated information, and parts may not function in current web browsers.

Lesson Plan #28     http://www.phy6.org/Stargaze/Lmoment.htm


              (18b)   Momentum  

   One 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,"

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

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

  • 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
. Newton actually used momentum in formulating his laws, not acceleration, but we would need a bit of calculus to follow his approach. The most important property of momentum is:
    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.

For instance, when a bullet hits a wall, where does its kinetic energy go? It turns into heat.

    [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 conservation of momentum is different--it is purely mechanical.

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?

    P = mv

--Does this depend on the direction of v?

    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.

--State the important property of momentum.

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

-- What is "an isolated system"?

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

--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?

    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.

--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 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).


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

Let v3 be the final velocity of the wreckage. The conservation of momentum gives

    P = m1v1 + m2v2 = (m1 + m2)v3

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

    v = - 5 km/s

The wreckage moves at 5 km/s in the "negative" direction in which the truck was moving.

--Is kinetic energy conserved?

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

--How much kinetic energy was lost?

    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

    --Where did the lost energy go?

      It probably went into heat.

    --(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

    --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.

    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.

                    Back to the Lesson Plan Index                     Back to the Master Index

        Guides to teachers...       A newer one           An older one             Timeline         Glossary

Author and Curator:   Dr. David P. Stern
     Mail to Dr.Stern:   stargaze("at" symbol)phy6.org .

Last updated: 10-19-04

Above is background material for archival reference only.

NASA Logo, National Aeronautics and Space Administration
NASA Official: Adam Szabo

Curators: Robert Candey, Alex Young, Tamara Kovalick

NASA Privacy, Security, Notices