[Potironette]

Thanks a lot!
I think I meant to say "not a function" when I used the word "relation." Basically, I'm confused why this happens:

[Coda]

I think the picture makes it self-evident why it's undefined outside of that narrow domain of x values -- it simply doesn't exist there, because the original function never takes on those values. And it's constrained to that range of y values in order to make it a proper function, because it can get really hard to work with expressions that can take multiple values simultaneously.

To review:

The domain of a function is the set of all x values that the function is defined for.
The range (sometimes "codomain") of a function is the set of all y values that the function can take on.

Since an inverse function is found by switching x and y in the formula, it also follows that the domain of a function constrains the range of its inverse, and the range of a function is the domain of its inverse.

[Potironette]

Ohh okay, I think I finally get that arcsin and arccos are functions. Thank-you! I was struggling with the idea that they had to be functions at all.

[Coda]

Yep, they're functions. If you ever NEED to get a different value for arcsin or arccos for some reason, you can just add k * 2pi, where k is any integer (including negative), but I'm not aware of any circumstances where that's relevant. (You can add k * pi for arctan, and I already explained a meaningful example of when you might need to do that.)

[Potironette]

I know that if I ever need to answer the question: "How do airbags protect you" for a physics test, I'll just say: "Because while change in momentum in the same, the time it takes for the force to reach you is lengthened, thus reducing the force you receive once it actually hits you (Δp = FnetΔt)"

But what is Δt actually..? Since momentum is conserved, is the change in momentum just to say that if momentum changed, the thing with the momentum most have caused a force to act over a certain amount of time? And so when two balls collide and the momentum has not changed there is no net force happening, but for each particular ball, maybe, the momentum has changed and so it experienced a net force?
Then is "impulse" just a thing which is for specific objects and not for entire systems?
And also, when ping pong balls do collide, since they bounce away with new velocities (and thus new momentums, and I guess accelerating backwards too) I have to assume that time has occurred during their hitting each other? Actually, I guess I should assume that when two things first touch there's no force whatsoever and then it takes time for a net force and changing directions to actually happen o~o?


EDIT: Unrelated
Is work done so long as there is a net force and is net work done so long as there is a change in velocity?

[Coda]

That's not quite a precise answer to the question.

The change in momentum is the same, but because the impulse is spread over a greater time, the amount of force acting on your body at any given moment in time is less.

Impulse isn't specific to a single object or to a whole system. It's specific to an interaction. No force is being exerted before the interaction begins, and no force is being exerted after the interaction ends, and no force is being exerted on other objects in the system that aren't participating in the interaction, but the objects and the system continue to exist both before and after the interaction being considered.

Yes, in physical reality, Δt is never zero -- there's always a span of time involved in a force being applied. If Δt COULD be zero, then a force would have to be infinite in order to cause any change in momentum.

For the ping-pong ball example, think about watching such an event in slow motion. You would see a period of time after they touch where the balls are deforming around the point of contact, squishing together before bouncing back and returning to their original shape. The more rigid the objects colliding are, the less deformation occurs, and therefore the less time they spend interacting, and therefore the greater the force must be.

That's why hard objects are brittle. They don't readily deform in collisions, so if the force of the impact exceeds the force holding the object together, it breaks.

EDIT:

Is work done so long as there is a net force and is net work done so long as there is a change in velocity?

No. Work is done as long as there is a net change in position. If the net change in position is zero, you could have exerted all the force you want for as long as you want and you wouldn't have done any work.

There's an argument to be said that work is done if there is a net force, but that's summing up the force vectors over the entire period of time being considered -- in order for the object to end up back where it started, every force applied to it would have needed an opposite force applied to it, and it doesn't matter if those are applied at the same time (meaning no change in velocity) or at different times (meaning velocity changes signs).

... Are you familiar with the difference between scalars and vectors?

[Potironette]

Oh! That makes a lot more sense now. So impulse is an interaction during which force acts during a period of time, and therefore having a car crash is having the car suddenly stop/slow while the body's still moving forward and say there are no air bags coming out, the impulse is the interaction between the body and the car--which doesn't take much time to finish so the force can be easily greater than the force holding the body's bones/skin together(?) and that makes the person get injured. On the other hand, if there are airbags, that interaction lasts longer, and so the force is less, and hopefully less than the force holding the person's body together and so they're less likely to get injured?

Period of time? Meaning work should be one interaction(?) but net force could be the result of many vectors(?)

Scalars are just quantities while a vector contains a quantity and a direction is what I know. Uhh, and vectors can be drawn together, and net force being zero could mean vectors of force going in pretty much any direction so long as they end up where they started when put together?

[Coda]

Saying "force holding it together" was a handwave on my part. It doesn't actually work that way but it's close enough to give you the general idea. An object's strength (and there are several different kinds of strength to measure various different kinds of interactions) is measured in force per unit area (that is, pressure), and it measures how much pressure can be applied to the object before it undergoes a permanent deformation (such as bending or breaking) that it can't bounce back from. So air bags actually provide an ADDITIONAL benefit beyond just spreading out the impulse over time -- it ALSO spreads out the force over a larger surface area compared to hitting your steering column chest-first, meaning there's less pressure being applied to any one part of your body.

(Crumple zones in the structure of a car also protect the occupants by increasing the impulse time, since the cab of the car ends up decelerating more slowly than the point of contact. They don't do anything about increasing the surface area of the contact against your body like an air bag does, although they do additionally reduce the force by using up some of the kinetic energy by deforming -- after all, it takes energy to make the metal bend.)

I wouldn't describe work as one interaction. But if you're considering the end result of a series of actions and figuring out the work done by that, then you can add up the force vectors over time and get the same answer as if the integrated force was the only thing that happened. It's analogous to our earlier discussion of average velocity -- you're choosing to disregard the in-between details to come up with an overall description of what happened in the long run.

And yes, you're basically right about net force. To get into a technicality of the math, the operation isn't actually ADDING the vectors, but INTEGRATING them over time, because you could balance out a force of 2N over 1s with a force of 1N over 2s. As long as you're working with nice round-number quantities like that, you can get the right result by inspection (perhaps breaking up the 1N/2s impulse into two 1N/1s impulses so that you've got consistent time units) but generalizing it beyond that gets into calculus.

[Potironette]

Is work the result of taking into consideration one vector and the object's movement along that vector then (force and distance)..? And net work is taking into account what happens at the end o_o..? But why is net work equal to change in kinetic energy?

Also, what average force?

[Coda]

Suppose in a frictionless environment you perform 20 N*m of work northeast, then 40 N*m of work southwest, then 20 N*m of work northeast, and you're back where you started. (Note that this means the object started at rest, accelerated for a while, slowed back down and started going the other way, then slowed back down again.) From here, there are two different interpretations of the sum of that work.

Net work is just looking at the initial state and the final state and observing the changes. This is the vector sum of the work; in this case, 0 N*m. This measures the change in the overall state of the system.

Total work is the path integral of power (that is, the dot product of the force vector and the velocity vector, which is a scalar) over the trajectory of the object. You can think of it as chopping up the force over infinitely many infinitely small segments of the distance and adding up the scalar magnitudes of those pieces. That gives you 80 N*m. This measures the effort expended by whatever it is doing the work.

If you take that same chopped-up description of the forces that were acting on the object and average them all out, then you could call that the average force. If you started with a system in the same initial state, and you applied this average force vector to the object for the same duration of time, then you would end up with a system in the same end state.

This is actually WHY net work is equal to change in kinetic energy. This description represents the simplest possible way to transform the initial state into the final state. If you were to add any other forces to this description, you would have to add balanced opposing forces to push it back, or else you wouldn't end up in the same end state. And since those forces have to be balanced, they can't possibly cause a change in kinetic energy. That means the only change in kinetic energy that actually matters in the end is that average force over that final displacement.

(Footnote: The sum of infinite pieces thing is called an integral. It's the continuous expansion of the idea of a sum of discrete quantities. It's one of the two basic operations of calculus.)

[Potironette]

I don't know what dot product is, but it does make sense now! Thank-you!

[Coda]

The dot product of two vectors is easy. Given vectors [a_1, b_1, ...] and [a_2, b_2, ...] of the same length, the dot product is:

(a_1 * b_1) + (a_2 * b_2) + ...

Surprisingly equivalently, the dot product is ALSO the product of the vector's lengths multiplied by the cosine of the angle between them, that is, ||v_1|| * ||v_2|| * cos(θ).

Don't ask for a graphical interpretation of what this actually represents geometrically, because it doesn't have a strict meaning. And my theoretical geometry skill isn't good enough to have an intuition on when I would want to use the dot product of two vectors outside of formulas that other people have already derived.

That said, it has a lot of useful properties. For example, if the dot product of two vectors equals 0, then the vectors are perpendicular, and if two vectors are parallel, then their dot product is the product of their lengths. (The dot product of a vector with itself is therefore the square of its length, which demonstrates the Pythagorean Theorem.)

[Potironette]

Sadly, I've never learned dot product, so I'm not sure what the square brackets mean D:, but it does look useful.


I'm reviewing a question that essentially tells me about a cannon firing a cannonball, and that before it fires, the momentum of both the cannon and the ball together is 0. Therefore, after it fires, the momentum of the cannon and the ball should make 0 again..but I can't seem to remember a logical reason for why that is.

EDIT: Random question - is it a more useful habit to label things like the cannon and cannonball m_1 and m_2 or m_cannon and m_ball ?

EDIT2: struck out question about momentum before and after. I suppose it's basically the same as momentum is conserved, therefore, if two things are at rest and suddenly move because of each other, momentum is still conserved therefore their total momentum added together is 0?

[Coda]

The square brackets are just an element-wise description of a vector. [1, 1] represents a two-dimensional vector with magnitude sqrt(2) pointing in a direction 45 degrees above the x axis -- that is, [1, 1] describes a vector starting at (0, 0) and going to (1, 1). (It also describes a vector starting at (1, 2) and going to (2, 3) -- the magnitude and direction of a vector is independent of its position in space.)

Labeling your variables descriptively is always good. :P I'd probably use m_c and m_b instead of spelling out "cannon" and "ball", but you should only use subscripted numbers when you're actually referring to numeric values, such as a position in a list or a point in time (which is why x_0 for "initial position" is good).

Your EDIT2 is exactly correct.

[Potironette]

That way of writing down vectors seems really useful :o. It's like the vector don't need to be drawn anymore and their actual location doesn't really matter anymore..?


There was a multiple choice question on my test saying something like: "Why don't we account for the internal forces of an object when calculating its momentum?" And then there were four things to select. I guessed a choice that said internal forces were somehow "conserved" too, but frankly I have no idea why '~'

[Coda]

Originally Posted by Potironette

That way of writing down vectors seems really useful :o. It's like the vector don't need to be drawn anymore and their actual location doesn't really matter anymore..?

Well, you can't even DRAW vectors of more than two dimensions. :P

Sometimes you'll see it without commas, just spaces, and technically you're supposed to write them vertically:

There was a multiple choice question on my test saying something like: "Why don't we account for the internal forces of an object when calculating its momentum?" And then there were four things to select. I guessed a choice that said internal forces were somehow "conserved" too, but frankly I have no idea why '~'

Newton's first law. An object in motion stays in motion unless an OUTSIDE force acts upon it. Internal forces can only make the parts move relative to each other, but that can't change the center of mass of the system.

This could also be argued using Newton's third law. Any internal force must necessarily have an equal and opposite internal force, so outside the system there's no net force.

EDIT: However, the question gets MUCH more interesting if you want to take relativity into account. Since momentum is proportional to energy, you might want to ask if the energy from internal forces can influence momentum.

The answer to that is: We don't know! So far we haven't seen any measurable violations of it. But there are theories that suggest that there might be an exotic force we haven't discovered that could increase an object's relativistic momentum without an outside force acting on it. Such a force could possibly be an explanation for dark matter (not actually matter, but this exotic force creating momentum that can't be explained by mass in motion) if it existed, but its influence would be so subtle at human scales that we'd have a very difficult time measuring it -- it's already stupidly hard to measure GRAVITY at human scales.