[Coda]

As for the other two questions!

And by sound "in" an object you mean it's because the atoms are in the object?

Yeah. This goes back to what I was saying about regions of different density.

From what I remember, friction happens because of temporary dipoles..and those cause a sticky effect between objects. Buut why does electrical resistance occur? From too many electrons bumping into each other o_o?

So I actually didn't know that about friction, but it makes sense. (Ever heard of cold welding? If you can get two surfaces of the same metal SO SMOOTH that you can put them together without even air getting in between them, physics has trouble telling where one object ends and the other begins, and they stick together with no adhesives or magnetism! And it's not even that hard to DO it -- junior machinists have been known to accidentally cold-weld steel nuts onto steel bolts by tightening them too far without any lubrication, and there's no way to get them apart without cutting or breaking something.)

But yes, your intuition is pretty close to right. Electrons can only go where there's room for them. Metals make room very easily. Nonmetals hold on to their electrons firmly. Since it takes energy to displace an electron, that reduces the amount of energy available to continue propagating the current.

[Potironette]

There's a whole lot more factors to sound than I'd thought :o.
By sound reflecting, do you mean that when someone knocks on a door, the door also knocks on the person's bones and then the person's bones make the reflected sound..?

I have heard of cold welding from class! But it was put in as an interesting side note and I forgot how it happened and didn't connect it with friction. But I'd though it only happened in space o_o

[Coda]

It happens more easily in space because there's no air to GET in between, but it's totally possible to squeeze out all the air on the surface with some effort (or power tools).

And no, I mean that the atoms you knock on bounce forward to hit the atoms behind them, but they also bounce back and hit the air on the same side as you. (Yes, they also knock back on your knuckles, but that doesn't make a whole lot of sound because your knuckles don't have a whole lot of surface contact with the air.)

[Potironette]

Ohh, I hadn't realized that the atoms first hit would bounce back too, woops.

Is the fact that when an object hits another object of equal mass, it can either stop, bounce back, or roll forward too (but slower) just something I need to accept and remember?

[Coda]

You already know enough about Newtonian physics to derive that result yourself, actually.

In the absence of friction, a perfectly rigid object in motion that strikes another perfectly rigid object at rest will transfer 100% of its kinetic energy. (Equivalently: in the absence of friction, two perfectly rigid objects exchange their kinetic energy when they collide. You can see this is equivalent by choosing different frames of reference for the same interaction.)

But because real objects aren't perfectly rigid (this would imply an infinite speed of sound in that material), the transfer of energy happens through a finite impulse, and because most experiments you can run don't take place in outer space, there's going to be friction. There are also additional forces acting on the system that you might not be thinking about.

Let's think of hockey pucks on a flat surface with low but nonzero friction. Why hockey pucks? Because balls roll, and spinning introduces angular momentum, which is conserved just like linear momentum is.

In order for puck #1 to make puck #2 move, it has to exert enough force to overcome the friction keeping puck #2 where it is. That force is obviously going to apply an acceleration to puck #2, and the reaction force will apply a deceleration to puck #1.

This means that the two pucks will be sliding together for some distance, remaining in contact with each other until puck #2's velocity exceeds puck #1's.

In the case where puck #1 stops and puck #2 moves away at full speed, this means that the contact time was long enough for that impulse to transfer all of the momentum. In the case where puck #1 slows down but keeps going, that means that only some of the momentum was transferred before puck #2 was moving too fast to keep them in contact.

The only remaining scenario to consider is bouncing back.

I actually don't think this is possible in a sliding-only interaction in the absence of friction if the two objects are of equal mass. Without rolling or friction, bouncing back means that puck #2 would have to have more mass than puck #1, so that the reaction force would accelerate #1 backwards faster than #2 gets accelerated forward.

The force of friction, though, shifts the balance. Static friction is greater than kinetic friction. So the reaction force has less resistance to pushing puck #1 backwards than the action force has to pushing puck #2 forwards. Over the span of the impulse, this could mean that puck #1 is accelerated enough faster than puck #2 to bounce back instead of stopping or proceeding forward.

[Potironette]

Objects being not perfectly rigid means the infinite speed of sound?
Is it something to do with how impulse requires time and if something's perfectly rigid it won't take any time?

So..impulse is force in an amount of time, so that means it takes time for momentum to be transferred and that determines if puck #1 continues in the same direction or stops.

I was pretty much guessing that an object of equal mass could bounce back because I wasn't certain what it could do '~'.
So..with lots of friction or greater mass for puck #2, then because puck #2 with friction, or puck #2 with greater mass will push puck #1 the same, puck #1 will accelerate backwards?

[Coda]

I mentioned a couple posts ago that the speed of sound is how fast forces can propagate through an object. A perfectly rigid object -- one that wouldn't deform at all when you push on it -- would therefore have to have an infinite speed of sound, in order to keep the far side of the object exactly the same distance from the near side of the object at all times.

It is indeed related to impulse taking time.

Sounds like you've got a pretty good handle on this!

[Potironette]

Can synthetic division not be used for fractional rational roots?

(I was doing a problem to solve for all real roots of p(x) = 3x³ - 5x² - 8x - 2 and I got -1/3 as a real root using rational roots theorem. Trying to find the rest of the roots by using synthetic division with -1/3 gave me a wrong answer, but long dividing the original 3x³ - 5x² - 8x - 2 by 3x + 1 worked fine.)

[Coda]

Erf. Geez. I haven't done synthetic division in like fifteen years, maybe longer. This is going to take research. ^^()

[Potironette]

If you don't want to research it, I can also ask a math teacher tomorrow xD

[Coda]

Okay, so brushing up on the technique...

-1/3	|	3	-5	-8	-2
.	|	_	-1	_2	_2
.	.	3	-6	-6	0

3x² - 6x - 6? Seems right to me. You must have made a mistake in your arithmetic somewhere.

EDIT: Equivalently, x² - 2x - 2, because we're finding roots; 0 * anything = 0 so we can multiply the whole thing by 1/3 and get the same results.

EDIT 2: Which... um... I usually go for the quadratic formula here:

(-(-2) +/- sqrt((-2)² - 4*1*-2)) / (2*1)
(2 +/- sqrt(4 + 8)) / 2
(2 +/- 2sqrt(3)) / 2
1 +/- sqrt(3)

The remaining two roots are irrational -- (x - 1 - sqrt(3)) and (x - 1 + sqrt(3)). You CAN, theoretically, do synthetic division on one of those to get the other, but you'd be juggling those silly radicals through the whole thing and it's probably not worth the effort.

[Potironette]

By multiplying the whole thing by 1/3, do you mean multiplying the (x + 1/3) by 3 and multiplying (3x² - 6x - 6) by 1/3 ?

I'm not sure what I did wrong, but maybe I messed up the quadratic formula..? I tried to use the quadratic formula on (3x² - 6x - 6) and ended up with x = 3 +/- sqrt(15) instead of 2 +/- sqrt(3) for

Ohhh I figured out where I went wrong. And I guess that applies to the bit about multiplying the whole thing too. I completely forgot that quadratic formulas needed to have the equation be simplified. Woops o_o.

Thanks for the explanations!

EDIT: Uh, quadratic equations don't need equations to be simplified right? I think I just messed that part up (forgot to multiply the bottom of the fraction by 3 as part of 2a --> 2(3))

[Coda]

No, I mean multiply both sides of 0 = 3x² - 6x - 6 by 1/3. Don't screw with the factor you've already pulled out; that one's already locked down.

The equation doesn't have to be simplified to use the quadratic formula. If you don't simplify it, then the /2a term will simplify it for you in the end. I was simplifying it to see if I could find the remaining factors by inspection.

[Potironette]

Ohh I see. Normally I'd simplify because of habit. I didn't because I forgot. Thankfully it seems okay to forget.

Oh! I see, so because in trying to find zeros for that, multiplying and dividing wouldn't affect anything!
If I were trying to multiply the factors together to make the original equation, is it okay to screw with the factor..?

[Coda]

If you're multiplying the factors together to get back the original equation, then you can always be off by a constant multiple. That is, (x² - x) and (2x² - 2x) have the same roots (x and (x-1)) but they differ by multiplying the entire expression by a constant.

If it matters, and you don't know, write k(x² - x). If it matters and you need to find out, then you need at least one non-zero point on the curve and then you can plug that into the expression and solve for k.

[Potironette]

Ohh, I see!

I have another question, does it matter if I write a factor as (x + 1/3) or (3x + 1)? Is it more convenient or more proper to write it one way or the other?