Momentum

The "Bermuda Triangle" is most of the mid to south-western Atlantic Ocean. It is underlain by truly huge and hugely extensive deposits of methane hydrate. A destabilization at depth will have positive feedback, violently frothing the ocean. As weight of volume displacement is buoyancy, ships do not float in froth. Methane is substantially lighter than air (MW = 16 vs. 29). A little methane in air (below the LEL) is a good combustion quencher for air-breathing engines. Airplanes will fall from the air.

The nice observation is satellite telemetry for methane in air via laser Raman LRS, CARS, etc.

The "Bermuda Triangle" is most of the mid to south-western Atlantic Ocean. It is underlain by truly huge and hugely extensive deposits of methane hydrate. A destabilization at depth will have positive feedback, violently frothing the ocean. As weight of volume displacement is buoyancy, ships do not float in froth. Methane is substantially lighter than air (MW = 16 vs. 29). A little methane in air (below the LEL) is a good combustion quencher for air-breathing engines. Airplanes will fall from the air.

The nice observation is satellite telemetry for methane in air via laser Raman LRS, CARS, etc.

Interesting momentum post.

Re the 'triangle' - statistically, things are no more likely to go away in that region than in any other region of similar size. it gets noticed because of its proximity to large populations, and the high volume of traffic through it. no muss, no fuss, no magic involved.

The problem with listening to that sort of program while typing on the computer, at least for me, is that my monitor ends up covered with little bits of spit. For me, it's the CoastToCoastAM show on late night radio.

And re the Nobel prize in physics, the KM of the CKM matrix, did you know that every unitary 3x3 matrix can be written uniquely in magic form with all rows and columns summing to 1? Uh, "uniquely" as far as compatibility with experimental data goes, means, up to a sign change of the imaginary unit i.

Another physics topic for you to contemplate. One that had bothered me for the last few days. What is the maximum possible muzzle velocity for an air powered gun? And how is it possible for an air rifle to shoot a projectile at 1600 feet per second?

9 Int. Science room.

CAT is sitting on a bench, LISTER on a table. RIMMER and KRYTEN stand between them.

CAT: So, what is it?
KRYTEN: I've never seen one before -- no one has -- but I'm guessing it's a white hole.
RIMMER: A _white_ hole?
KRYTEN: Every action has an equal and opposite reaction. A black hole sucks time and matter out of the universe: a white hole returns it.
LISTER: So, that thing's spewing time back into the universe? (He dons his fur-lined hat.)
KRYTEN: Precisely. That's why we're experiencing these curious time phenomena on board.
CAT: So, what is it?
KRYTEN: I've never seen one before -- no one has -- but I'm guessing it's a white hole.
RIMMER: A _white_ hole?
KRYTEN: Every action has an equal and opposite reaction. A black hole sucks time and matter out of the universe: a white hole returns it.
LISTER: (Minus the hat.) So, that thing's spewing time back into the universe? (He dons his fur-lined hat, again.)
KRYTEN: Precisely. That's why we're experiencing these curious time phenomena on board.
LISTER: What time phenomena?
KRYTEN: Like just then, when time repeated itself.
CAT: So, what is it?

We've been told all of our (physics) lives that inelastic collisions don't conserve kinetic energy, and this makes sense, because of the dissipative processes (sound generation, raising the temperature, etc).

But what I don't understand is why they conserve momentum. What allows us to make this assumption (besides the Conservation of Momentum). Certainly the momenta of the molecules in the pendulum material are increased when their temperature is increased (KE=p^2/2m). Does the effect average to zero because of the randomness of the molecular vibrations? Does it have something to do with the loss of directionality when you do p^2 (p dot p)?

And what about the sound waves? We assume that they carry energy away, but what if you treat the sound waves as a series of elastic collisions between air molecules? The momentum flows away with the sound. Or does it again average to zero because of the omni-directional nature of the generated sound waves?

Is the answer simply the v^1 vs. v^2 dependance of momentum and energy? In macroscopic situations like the ballistic pendulum, do we just throw away the momentum loss? If so, we should explain this to physics students, perhaps as a nice example of ignoring small effects that don't contribute to the final answer of a problem.

So what's the answer? Maybe it's really simple and I just forgot. I suppose I could go to my copy of HRW and find out, but I'm also sort of bringing up a point about the dogmatic way I was taught - lots of instructors repeating "in inelastic collisions, momentum is conserved, but not kinetic energy" and not offering much in the way of justification. The students certainly deserve more than that, even if it won't serve them right away. Someday they may randomly think about an inelastic collision and remember "oh, that's why".

Answering DG:

No (new) assumption is needed, because we have already assumed Newton's Third Law. That ensures that the *internal* forces between bullet and pendulum are equal and opposite. Whatever impulse B applies to P is equally (and oppositely) applied by P to B. Hence the change in momentum of the bullet is equal and opposite to the change in momentum of the pendulum. Notice that the vector nature of momentum is crucial here.

Those internal forces can also do work. In any non-elastic process, that work is non-zero. [In those cases the work can be positive, for explosive collisions, or negative. In the special, and rare, cases of elastic scattering, there is a spring-like behavior where force A on B does equal amounts of positive and negative work. See diagram I can't draw.] That work changes the kinetic energy of each object, but because the displacements are opposite along with the forces being opposite, the sign of the work on P is the same as the sign for the work on B.

This is tied up in work being a scalar, but also in the fact that delta-t is the same sign for the two impulses (while the forces are opposite), but delta-x is opposite in sign for the two works (while the forces are opposite). So the internal impulses cancel but the internal works add.

This is most easily seen in an "explosion" where you put two spring-loaded carts next to each other, at rest, and release the spring. Net momentum afterward is still zero, but total KE is positive.

And to answer the questions in the second paragraph, Yes and Yes. Thermal energy (unlike the mechanical energy of the original bullet) consists of incoherent rather than coherent motion of the molecules in the object. The momenta have a stochastic distribution of vector directions, summing to zero, but the kinetic energies still add. You can think of thermal energy as a kind of rest energy, since the total momentum is zero yet the object still has energy.

All that matters with sound is that it eventually ends up in your ears or dissipated in the air or walls of the room. As long as it doesn't all get back into the object, it is gone, and thermodynamics won't let it go back.

PS - Nothing dogmatic in my class. I demonstrate this effect with the spring carts while doing (fake) work and impulse estimates as an example. However, whether my students retain it, only time will tell.

PPS - The same thing goes on in clever angular momentum demonstrations with a bicycle wheel. You quickly realize there is a lot of work being done by the internal forces as soon as you are the one doing the demo.

Thanks CCPhysicist, that was actually very helpful. Sometimes it's all about putting the stuff you already know together in the right way.

Energy as a scalar is key. It's kind of like the question "If E&M forces are so much stronger than gravity, why does gravity dominate over long distances?" Gravity always adds, like energy. But the charges can cancel each other out if randomly distributed between + and -, like the momenta of thermal vibrations (except restricted to 2 degrees of freedom).

Physics is fun, huh?

But ... the balistic pendulum swings upward, comes to a halt, and then swings back down again. If it comes to a halt at the top of its swing, where has all the momentum gone?

Huh? What about that?

Yes, I understand that it drags the building and then the entire earth forward (by just a tiny bit), but you can see that it's a difficulty when handwaving "conservation of momentum allows us to determine the velocity".

We're not interested in the momentum at the top of the swing. All we need momentum conservation for is to relate the velocity just before the impact to the velocity just after the impact. After the collision, it's just energy we're working with.

Impulse - change in momentum - is force through a time, and essentially no time (relative to v/g) goes by during the impact. Therefore the momentum of the system doesn't change during the impact. That fact allows us to formally justify our procedure.

Evasive answer, Matt. We don't have to make it look like a magic, arbitrary choice. He was interested in what happened to the momentum, so tell him.

The answer to #10 is that the momentum has changed due to the impulse supplied by gravity. The net force vector (gravity minus the tension in the pendulum), integrated over time, changes the momentum after the collision. [It has no effect during the collision for the reason Matt articulated.] We *could* use impulse to calculate *when* it gets to the top (although it is a messy vector integral), but we didn't measure the time - we measured the height! Figuring out how high it got by using momentum (or F=ma) is hard, but it is easy to use energy conservation to answer the question about how high it went. So we use energy for that part of the problem.

The choice is a practical one: do each part of each problem the easiest way.

Any problem that can be done with F=ma and the differential equations for motion can also be solved with energy and/or momentum, not to mention Hamilton's and Lagrange's methods or even relativistic field theory.