bro i did this whole question with newton's second law instead of impulse and when my teacher said that we were supposed to use momentum, I thought I totally failed but I actually got the right equations lmao
when you divide by a fraction you can rewrite it as multiplying by the reciprocal of the fraction, so he flipped the fraction in the denominator and multiplied it with the fraction in the numerator, allowing him to cancel out F and t
Yes, you can certainly do that. Momentum is just a different way we describe Newton's Laws of Motion (F=ma), so you can always apply Newton's Laws directly. You should have gotten the same answer.
If I messed up part a ii but got the rest correct how many points would I get? ( sorry for asking so many questions if you don’t have time to respond that’s ok. Thanks for making these videos)
For part b, it says that the students starts moving with a velocity after tf so shouldn't your graph be flipped with it being constant in the beginning and from tf to 2tf its velocity is linearly increasing
No. The constant force is applied from 0 to tf, so it's accelerating during that time. From tf to 2tf, there's no more force on it, so it's no longer accelerating.
There is an equal and opposite force on the student and the disk (Newton's 3rd law of motion). From 0 to tf, the student is pushing in the disk and the disk is pushing on the student. From tf to 2tf, there is no force being applied so there's no acceleration, so the velocity is constant.
hi! I have a question about the second part of part a. why would u not use -Fh as the force that the disk exerts on the student? bc since it’s opposite but equal?
You should use the correct direction when finding velocity. But they're asking for speed, so I knew the sign wouldn't really matter in the end because speed is always positive.
@@AllenTsaoSTEMCoach Hey Allen! I put my final answer to part a ii as v_D/v_S = -M_S/M_D. should I count that as correct or incorrect because of the negative sign?
Hi can u answer me if Im wrong For part b the disk and the student go to two different ways so if students velocity is + the disk is supposed to be - correct me if Im wrong
Is doesn't seem fair that you are posting these... were these questions made public by the college board? Many students will take the online version soon, and now they have this advantage of seeing extra practice problems...
These are posted publicly on the college board website. apcentral.collegeboard.org/courses/ap-physics-1?course=ap-physics-1-algebra-based The second and third administration questions will definitely be different.
I really can't address whether it's fair that the College Board releases these questions. They have historically released the FRQs 2-3 days after the AP exam, and I wasn't sure they would this year. But it's out there now.
bro i did this whole question with newton's second law instead of impulse and when my teacher said that we were supposed to use momentum, I thought I totally failed but I actually got the right equations lmao
You can definitely use Newton's second law of motion but momentum impulse is more "traditional". But the right answer is all that matters. Nice job!
For the first one I said that Ft was equal to 0.5mv^2 because the change in kinetic energy is the work
Ft is not work. Work is force x displacement
would you mind elaborating on 'flipping that guy' at 2:57? i'm not good with deriving equations and can't seem to figure out why you did so
when you divide by a fraction you can rewrite it as multiplying by the reciprocal of the fraction, so he flipped the fraction in the denominator and multiplied it with the fraction in the numerator, allowing him to cancel out F and t
for c i, could you say that v_cm < v_1 because since the collision was inelastic, kinetic energy was lost and thus leading to a smaller velocity?
That's not an estimate though. They wanted you to estimate what vcm is, not say what the upper bound is.
Hi, for the first part- can you use f=ma, and vf = vo + at. Solve for a, and then substitute. Would that work? Or is that wrong?
Yes, you can certainly do that. Momentum is just a different way we describe Newton's Laws of Motion (F=ma), so you can always apply Newton's Laws directly. You should have gotten the same answer.
If I messed up part a ii but got the rest correct how many points would I get? ( sorry for asking so many questions if you don’t have time to respond that’s ok. Thanks for making these videos)
I would love to answer with certainty, but I'm not an expert at the scoring system. My guess is you'd lose 2 or 3 points for that?
@@AllenTsaoSTEMCoach Do you know what # question was the QQT?
For part b, it says that the students starts moving with a velocity after tf so shouldn't your graph be flipped with it being constant in the beginning and from tf to 2tf its velocity is linearly increasing
No. The constant force is applied from 0 to tf, so it's accelerating during that time. From tf to 2tf, there's no more force on it, so it's no longer accelerating.
@@AllenTsaoSTEMCoach But isn't the student exerting that force on the disk not vice versa? Which force is being applied to the student?
There is an equal and opposite force on the student and the disk (Newton's 3rd law of motion). From 0 to tf, the student is pushing in the disk and the disk is pushing on the student. From tf to 2tf, there is no force being applied so there's no acceleration, so the velocity is constant.
@@AllenTsaoSTEMCoach Thank you for the response!
I dont understand in part a they tell you to express youre answer in terms of Ms which is not included in the equation and i thought i was wrong
When they do that, they're saying you can't use anything beyond those variables. You don't have to use all of them.
hi! I have a question about the second part of part a. why would u not use -Fh as the force that the disk exerts on the student? bc since it’s opposite but equal?
You should use the correct direction when finding velocity. But they're asking for speed, so I knew the sign wouldn't really matter in the end because speed is always positive.
@@AllenTsaoSTEMCoach oop... I didn’t read the question correctly 🥲🥲🥲🥲 Thank you so much for replying!
@@AllenTsaoSTEMCoach Hey Allen! I put my final answer to part a ii as v_D/v_S = -M_S/M_D. should I count that as correct or incorrect because of the negative sign?
@@weirdrandomizer3048 it's correct. They usually won't dock for something that small.
@@AllenTsaoSTEMCoach I used F * change in time = change in momentum then let F= -F_H using Newton's third law
are you sure about 3b, because I thought that until tf, the velocity would stay the same for both
Yes I am sure. The force is constant from 0 to tf. Thus, it is under acceleration during this time (F=ma). So both are speeding up during this time
@@AllenTsaoSTEMCoach but if the student and the disk are touching together, then how do they have different accelerations
They have the same force but different masses
@@AllenTsaoSTEMCoach Oh I thought before tf they were touching so they were one person-disk system, thus having the same velocity
Hi can u answer me if Im wrong
For part b the disk and the student go to two different ways so if students velocity is + the disk is supposed to be - correct me if Im wrong
For velocity, you are correct. This is a graph of speed so it's always positive.
@@AllenTsaoSTEMCoach oh right
I didnt pay attention to that. Thanks!
if i write in part c double i that Vcm is less than Vi rather than it is nearly 0 would it count correct ?
I don't think so unfortunately. The scoring guidelines did require that you stated it was 0.
@@AllenTsaoSTEMCoach ok thank you
Would I get credit if I approximated the velocity correctly but I just said "close to initial velocity of disc" instead of putting it in terms of v0?
I would think so. I'm not a reader so I'm not sure how picky they are on it. Usually if you convey the right idea, they give you credit.
@@AllenTsaoSTEMCoach thank you. I think I might as well because it didn't say to leave it in terms of any specific variable
thank you sir
this was the worst frq question I came across in the exam, and all of them were extremely bad 😭 I just can’t stand physics-
I'm really sorry to hear that, and I'm sorry if the score didn't turn out well. It was a rough year, and it's more than OK to do poorly on a test.
Is doesn't seem fair that you are posting these... were these questions made public by the college board? Many students will take the online version soon, and now they have this advantage of seeing extra practice problems...
These are posted publicly on the college board website.
apcentral.collegeboard.org/courses/ap-physics-1?course=ap-physics-1-algebra-based
The second and third administration questions will definitely be different.
I really can't address whether it's fair that the College Board releases these questions. They have historically released the FRQs 2-3 days after the AP exam, and I wasn't sure they would this year. But it's out there now.