The Physics Hypertextbook™
© 1998-2008 by Glenn Elert -- A Work in Progress
All Rights Reserved -- Fair Use Encouraged
Start with Newton's third law of motion, toss in the impulse-momentum theorem, and see what happens.
| +F1 = | −F2 |
| +F1Δt = | −F2Δt |
| +m1Δv1 = | −m2Δv2 |
| +Δp1 = | −Δp2 |
| +(p1 − p10) = | −(p2 − p20) |
| p1 + p2 = | p10 + p20 |
| ∑p = | ∑p0 |
Momentum is conserved. The total momentum of a closed system is constant.
The symbol p may be used for momentum because it is the mirror image of the letter q. Some people use p to represent the momentum of one object and q to represent the momentum of the other object in a recoil situation. A clever way to write Newton's third law of motion is like this …
p = q
action is equal and opposite reaction
Write something.
Mass in and/or mass out.
| ∑F = | dp | = m | dv | + v | dm |
| dt | dt | dt |
There are two typical problems given to first year physics students that use this variation on Newton's second law of motion: conveyor belts and rockets.
Rockets first. The father of the academic study of rocketry and space travel is Konstantin Eduardovich Tsiolkovskiy a.k.a. Константин Эдуардович Циолковский a.k.a. Konstanty Ciołkowski (1857-1935) Russia. Tsiolkovskiy was an exceptional visionary who anticipated many features of the late Twentieth Century space programs in the Nineteenth Century.
Start with Newton's second law of motion. The net external force on the rocket is zero. Nothing is pushing the spacecraft but itself.
| ∑F = | dp | = m | dv | + v | dm | = 0 |
| dt | dt | dt |
Thus, momentum is conserved. We begin with a fancier looking version of the action-reaction or recoil problem.
| m | dv | = − v | dm |
| dt | dt |
Even though the denominators on both sides are infinitesimals, they still cancel out. We'll also drop the vector notation since this is a one-dimensional problem at it's heart.
+ m dv = − v dm
Apply the technique of separation of variables.
| + | ⌠ ⌡ |
dv = −vexhaust | ⌠ ⌡ |
dm |
| m |
Integrate over the appropriate limits.
| + (vfinal − vinitial) = −vexhaust (log mfinal − log minitial) |
Clean it up a bit.
| Δvrocket = vexhaust log | ⎛ ⎜ ⎝ |
minitial | ⎞ ⎟ ⎠ |
| mfinal |
We could quit here, but it's tradition to write the variables in a certain way.
| Δvrocket = vexhaust log | ⎛ ⎜ ⎝ |
mempty rocket + mfuel | ⎞ ⎟ ⎠ |
|
| mempty rocket | ||||
Clean it up a bit more.
| Δvrocket = vexhaust log | ⎛ ⎜ ⎝ |
1 + | mfuel | ⎞ ⎟ ⎠ |
|
| mempty rocket | |||||
This is Tsiolkovskiy's rocket equation, which was first published in 1903 -- the year the Wright brothers flew the first airplane, 23 years before Robert Goddard built the first liquid fuel rocket, 54 years before the Soviet Union placed the first artificial satellite in orbit around the earth, and 66 years before the United States sent humans to play golf on the moon. The rocket equation appeared in an essay with an odd sounding title in the 1968 English translation: "Investigation of World Spaces by Reactive Vehicles" (in russian, "Исследование мировых пространств реактивными приборами"). "World Spaces" would now be called "Outer Space" and "Reactive Vehicles" is an overly literal translation of a phrase that really means something like "Jet Propulsion". A better English translation of the title would probably be "Exploration of Outer Space by Jet Propulsion".
An interesting aside. Tsiolkovskiy saw rocket trips to outer space in his mind's eye, but missed seeing radio as the medium for communicating over large distances. (The rocket equation was written three years before the first AM radio broadcast.) When Tsiolkovskiy pondered how astronauts or cosmonauts would stay in touch with earth he thought they might use special rockets to send messages back and forth. In our current age, when people spend more time talking to disembodied voices on a cell phone than talking face to face, it's hard to imagine communicating over astronomical distances using the equivalent of rocket mail trucks or space carrier pigeons.
And now for conveyor belts …. Eh … it's kind of a let down after rockets. I think I'll end here and write a conveyor belt question for the practice problems section later.
| I | II | |
|---|---|---|
| 1st law | inertia m |
momentum p = mv |
| 2nd law | force law F = m a F = m dv / dt |
impulse-momentum theorem J = Δp ∫ Fdt = m ∫ dv |
| 3rd law | action-reaction F1 = -F2 |
conservation of momentum ∑ p = ∑p0 |

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