Відео

Unfortunately those questions are not released so I can't make a video for them. Sorry I know the test was probably very different but I hope it turns out well!

@@AllenTsaoSTEMCoach Thank you! You were such a help when I was reviewing and I feel I did a good job on that exam.

Grade isn't really much of a prerequisite. The key part is you feel comfortable with algebra and are taking Precalculus concurrently (there's a few topics from precalculus you need for the later units in physics)

I am taking AP Precalc next year so I think I should fine right? I am also planning to study a bit of it over summer

Yeah that should be fine

I use One Note

I think one of the hardest things in mechanics is deciding whether to use forces, energy, or momentum in the analysis. For beginner students, I think that can be really challenging. So adding in new topics like fluids makes the mechanics portion weigh less because there will be some fluid material that you will never confuse for other mechanics topics. Even though it's more to cover, it makes the exam easier overall.

No they are not

I use One Note

I've heard the late tests are very hard. I do think the curving is more generous on them though (don't know for certain but harder tests are usually curved more generously). You did your best and just have to wait and see how it turns out.

It's not impossible. The minimum speed it needs is sqrt(gR), and the speed at that point is sqrt(4gR) which is greater than sqrt(gR). I think you mistaken the speed by dropping your g (10) at some point.

@@AllenTsaoSTEMCoach Yes, I did forget that. I was doing it mentally. Thanks for the correction. Anyway, would there be a case that normal force is not included in the FBD but the block stays in contact?

@@asherwiggin6456 if it's in contact, there's always some normal force (otherwise it wouldn't be in contact). If the contact was very very weak, then the normal force could ALMOST be 0. But if there's any contact, there's a normal force.

@@AllenTsaoSTEMCoach Thank you!

Yes that's pretty common

no they don't

good luck soldier

No they only ask for the rotational inertia of the disk. You have to set up the integral for calculating rotational inertia

Yeah that's true but they would not make you do that when asking to draw a FBD. You are only needed to have a general shape and not as detailed of a FBD as that.

I think that would be okay

Oh no that is crazy! Unfortunately they don't release any other FRQs or any of the ones from the late testing date 😢

@@AllenTsaoSTEMCoach yep, but they do make them relatively similar to each other, so if you were to, oh I don’t know, watch all the sections video frq’s youd probably have a pretty good chance at figuring out the one for the makeup

It's coming from the surface of the axle. Just because there's no friction does not mean there's no normal force. A block can slide down a frictionless ramp and have no friction but still have a normal force. Why must there be a normal force? Because we do have to assume the rod is not accelerating to the left. It's a fixed axle and it wouldn't make sense for the rod to accelerate to the left, which is what would happen if there were no other horizontal force. I agree the assumption of rotational torque being 0 is a bit of an assumption as they didn't state explicitly that it's not rotating. But it wouldn't really be solvable otherwise. I don't think it's reasonable to assume the rod can just swing left freely though.

Yes the current will start at 0 and increase to a max value. But this is a sketch of the voltage. After a long time, the voltage across an inductor is 0V. Just like for a capacitor, after a long time, the current through the capacitor is 0.

@@AllenTsaoSTEMCoach O I get it now. TYSM.

Yeah you would need the hinge force.

E field is always the derivative of voltage. In this case, we are approximating the slope by doing the change in V over the change in x. That's how we always approximate the derivative if we can't compute it exactly

Yeah anything where the coefficient of the horizontal variable is a constant and contains k is fine. As long as you explain how you calculate k from the slope for part b ii

I'm not sure that's true. The distance the spring is compressed is xc and right when it reaches t1, the block reaches xc and enters the friction region.

It's usually like 1 point off for not labeling axis

Centripetal force is not a separate force on a free body diagram. The net force (total force) towards the center of the circle is the centripetal force. But it's not an extra force.

It should be fine. And the exact angle of the axle force is not supposed to be computed so it doesn't matter the exact angle you put it.

@@AllenTsaoSTEMCoach ok thanks!

The problem is the question asked you to use the principles of work and energy. Squaring the velocity curve isn't using work and energy. Even the fact that the velocity is sinusoidal is more a function of kinematics and differential equations than work and energy.

@@AllenTsaoSTEMCoach Oh good point, forgot about that portion

That was probably the hardest portion of that question. I think most people would have done what you did and not had that inflection point and curve. I don't think it'll be a lot though. It'll be mostly on how consistent your answer was explaining the graphs shape. Maybe 1-2 points off at most.

There are always several forms of the exam. They only release 1 set every year (Form O)

yeah that would definitely work.

No it's correct. You should calculate the velocity of center of mass using momentum conservation and confirm that energy is lost, momentum is conserved, but the center of mass is moving at the same velocity.

@@AllenTsaoSTEMCoach no but if energy is lost, kinetic energy goes down, and since kinetic energy equals .5mv^2, the velocity will go down. Momentum will also go down did p=mv

Sorry, but this is a pretty common misunderstanding for students to make. If you had looked at the FRQ #1 from the 2019 exam, you'll see this identical situation play out with the inelastic collision and the velocity of the center of mass. apcentral.collegeboard.org/media/pdf/ap19-frq-physics-1.pdf Solutions here: apstudents.collegeboard.org/ap/2019-08/ap19-sg-physics-1.pdf

@@AllenTsaoSTEMCoach Okay thanks for clarifying. Now I need to confront my physics teacher who told me my reasoning was correct 🤣

Not sure how they'll score it. Technically that would be the pressure being 0 which isn't quite possible. It would be a loss of at most 1 point.

I also drew ny lines from 0 rip 😭