2022 AP Physics C Mechanics Free Response - Set 1
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- Опубликовано: 15 сен 2024
- Walkthrough of the 2022 AP Physics C Mechanics Set 1 Problems
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Link to the released FRQs:
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Link to my PDF Solutions:
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You make it seem so easy. Looking back I can’t believe some of the dumb mistakes I made. It’s ok, it’s a learning process I guess.
That's just experience. It's actually quite a lot to put together in 45 minutes. Honestly, it's not an easy test to do in such a short time. Making dumb mistakes is definitely a part of learning, and I'm sure you knew a lot of it.
@@AllenTsaoSTEMCoach Reading this comment made me feel better... thinking back to the e&m portion of the physics c test I took this year I can't believe that I thought the surface integral of dA was the volume of the sphere as opposed to the surface area lol, hopefully the other points I got made up for that mistake but either way thank you very much for posting solutions so early I appreciate it
Corrections:
#1: General note. The axis system they noted in the problem was right-ward positive (implicitly). I solved this problem using left as the positive direction. If you solved everything in reverse order, it's probably fine. Part (c) ii only asked for the magnitude of the force, so it didn't matter which way.
Also, you did not need to simplify sin(theta) and cos(theta) the way I did. You could have used sin(arctan(H/x)) or cos(arctan(H/x)).
part (d): I'm getting a lot of questions on the integral bounds. Technically, you can reverse the bounds, but then you'd need to use the sign convention that Ft is negative because it's pointing left. The important to note is that the integral should result in a positive value because the work was positive. If you reversed the bounds, but kept everything else the same, the integral would ultimately be negative, which is not exactly correct. So the integral stated as given in the problem is correct.
#3: part(b): Because the rope is connected directly at point P, you should place the tension at point P and direct it tangentially up and to the left. There should be a normal force from the rope down and to the right. And the axle force would need to be up and to the right (slightly). It doesn't change the torque so part (c) is still correct.
part (e): The torque from the tension force should increase. The main reason is the overall torque clockwise is now greater due to the shifted axis of rotation. Thus, the counterclockwise torque from tension (spring) would be greater to keep it in equilibrium. The torque from the normal force from the rope would increase because it was not causing a torque before, and now it is causing a torque.
For part 3b, would you agree that the forces are balanced (fnet=0) but the torques are not (Torque net is not equal to 0)? If that was the case, then if you said the force at point P acts leftward, wouldn't that mean the wheel is accelerating to the left (since there is no force directed to the right)? But we know that's not the case, because the wheel is secure to the stand.
So the axle would have to be applying a force to the top right to keep the net force 0. I should note that in the correction.
Thanks for the video! I was wondering if they grade Physics C similar to Calculus, where if you get an answer wrong but use it later you get points based on how you used it. For number 1, I was rushing since I did it last and just put F_N = mg so I'm hoping that didn't completely ruin my score.
They don't carry errors forward. So they will take into account your answer on a previous problem when factoring your score.
I don't understand his answer to question 3 part e. regarding the torque of the spring force, he stated that the point of application is closer to the axle of rotation, and therefore the total torque is less than in part B. I have two questions (bear with me):
1. Isn't the torque of the spring force applied at point P, as stated in the original problem? this point started on the top of the pulley, and at equilibrium, is right of the center of the pulley, further away from the axis of rotation than in part b. This should increase the torque, not decrease it?
2. Regardless of the answer to my last question, the point of application of the spring force should be irrelevant. more important is what happens to the net torque of all of the other forces. Since the system was at equilibrium, the spring force will have to make up for any additional clockwise torque from the other forces. if necessary, the extension of the spring x will have to increase. Since gravity now exerts a clockwise torque, and the suspended block also has an increased torque, since its torque acts farther away from the axis of rotation than in part b (regardless on if it acts on point P or if it acts where the string initially touches the pulley). To compensate for this, won't the spring force have to increase?
Any help is much appreciated!!
1. You bring up an interesting point about the rope being attached at point P. I'll have to think on that one a bit. The problem is tension for is highly simplified for pulleys in AP Physics C, so I'll have to think about whether it makes sense to talk about the force acting there instead.
I'll have to think about it some more.
@@AllenTsaoSTEMCoach Thank you for responding. I agree that tension force has been simplified, since obviously the tension force is not only acting on one location: in the real world, it is somehow distributed over some perimeter of the pulley.
However, that question is irrelevant to my main point, which is that the radial distance to the center of rotation plays no role here. comparing torques based on radial distance is only relevant if the perpendicular force remains constant, which is not the case here. the spring must obviously extend further than in part B, increasing its torque. there is no way to calculate the torque based on radial acceleration alone.
the only way to calculate the spring torque is by the fact that the system is in equilibrium. therefore, since the clockwise torque has increased relative to part b, than the counterclockwise spring torque must also have increased, in order to maintain a net zero torque in static equilibrium. we know that the clockwise torque increases, since the torque of gravity is now nonzero, and also, the suspended mass has the same linear force, but is applied farther away from the axis of rotation.
in another person's walkthrough (the channel is called Physics with the Nanoscale Bison), I actually saw him assume that force acts upon point p, but he then gave the same line of reasoning as you (but obviously achieving the opposite answer). He said that since the point P is farther than it was in part B, the torque of the spring has increased. I agree with this answer, but not his explanation. he also didn't explain that radial distance is irrelevant to answering the question.
now we have three different explanations for the same problem; yours, his, and mine. I would love to be shown why I am wrong. And thank you for the great content!!
Okay I think I do agree. What I was initially confused on the comparison was that in part b, the block was cut off. So I think in my head, I discounted the block being there. So I thought the spring was just in equilibrium with the mass of the disk causing a torque.
However, in rereading, I was indeed supposed to include that mass of the block. So I agree that the torque from the spring force would increase by your reasoning.
The force on a pulley is tougher to analyze because normally, it's not a tension force that acts on a pulley. It's the friction between the rope and pulley that causes a torque. So this attachment makes it harder to consider because yes, the tension pulls there but usually there's friction that's causing the torque so it becomes much more difficult to think about in this case.
In any case, you are correct and I should probably make a correction that the tension applies at point P as well
I completely forgot that Tangent was a trig function and I ended up completely messing up the first question. Thank you for this video!
Oof that's frustrating. I thought it was weird to ask that question, but I think they wanted it to help you get setup for the rest of the problem. Sometimes, it's just good to say "use the theta I would have found in part (b)" in all your answers if you leave theta in there.
same I had a feeling the answer was a simple trig functoin but bc of pressure and time I didnt think it could be so simple
@@AllenTsaoSTEMCoach Thanks for replying back! I’m going to college and the physics and math classes I’ve taken have been the only ones that I really enjoyed. Although I’m doing a Finance Major, I’ve been considering doing a Physics Minor just because I enjoy the subject a lot.
@@impanthering That’s the one thing that I really dislike about the whole standardized testing process. The time crunches are insane to work around and it puts a lot of pressure on the mind.
I think you made question 1 a lot more difficult than it had to be. I think you could've just said that theta = arctan(H/x) and then used that for the rest of the problem instead of having to use square roots and longer trig expressions.
You can but then I'd have a lot of sin(arctan..) or cos(arctan...) statements which I personally don't like to see. It's fine if you left it that way though.
been looking for this thank you!
allen ur the goat
Question about the last part of the last question: it says the system comes to equilibrium again. Since the CW torques both increase, shouldn’t the CCW torque from the tension/spring increase as well? The spring would stretch farther creating a larger force even if r decreases. I assumed that it was talking about the point when it has come to equilibrium again.
Yes, you're right. I noted that in the corrections comment.
Man I’m so mad I knew how to derive the normal force, the net force, and the integral expression, I just didn’t know how to do the derive the angle question and I panicked and barely wrote anything. I had lang that morning so I was brain dead taking physics too. Prolly got a 4 on mech and a 5 on E&m
Oh no that's a rough schedule. You do know the physics so be proud of that!
same I had lang right before and my brain was just fried I couldnt think straight
AP test tmmr, wish me luck!
In question 1 part e, Ftcos(theta) is smaller as you get closer to the edge of the table. The angle will increase, which means the cosine of the angle will decrease to zero when the block is at the edge of the table (which you explained in the video). This would mean the string does less work as it moves from L to L/2, right? L is farther away from the edge of the table than L/2.
But the two intervals, it travels the same distance. From L/2 to 0 is the same distance as L to L/2
Oh I got it now. I didn’t read the question carefully enough before. Classic mistake 🤦🏼♀️
Oh I am guilty of the same!
I think you got the interval of the integral flipped because in the question the block moves from L to 0
Oops you're definitely right. Good catch!
Actually on second thought, I think my integral will give a positive value for work, which is what I wanted in the end. If I swap the bounds, I think the result of the integral will be negative, which would be the wrong sign.
It's probably a double error because FT should technically be negative because it's pointing left. So missing that negative + missing the swapped bounds = right answer! Two wrongs make a right.
@@AllenTsaoSTEMCoach Appreciate it!
so disappointed in myself I made some of the dumbest mistakes ever and I also ran out of time so I couldn't give full explanations for the last question, but I probably got points here and there
I wish I could retake the exam I feel like I studied so hard for nothing
It's frustrating to not do well on a test. But I don't think it was a waste. You probably learned a lot of physics over the year, and that will set you up for future success. You know you can do well at it. You don't have anything to prove.
@@AllenTsaoSTEMCoach Thank you for the encouragement! I also feel kinda better seeing other comments know that I’m not the only one that struggled lol
how nice, I didn't get this form ;( I got form H and I have no idea what it stands for... and I completely bombed that one T^T sigh...
☹️ Keep in mind the curve is a bit generous on AP Physics C. You may have done better than you realized.
@@AllenTsaoSTEMCoach ty! i just looked at your other video and saw that I got that set of FRQs instead :o i actually got some correct so i was super happy =DD thanks for the videos!
Shouldn't the bounds on the integral for 2d be from L to 0 rather than 0 to L?
I noted it in my comments. I was doing everything left as positive direction, so that's why my integral is a little confusing.
If you wanted to use rightward as positive (following the x-coordinate system listed in the problem), then you have to place a negative sign in front of the tension force because it's technically going left.
In the end, either way would be the same because reversing the bounds just adds a negative sign to the integral.
FBD on Q3.
1) Shouldn't the tension actually one be at attachment point? (I know you corrected this in a comment. But keep reading.)
2) If rope is drapped over, even if it is massless, there should be a net normal force between rope and disk pulling down and to the right (exact direction hard to predict since the magnitude of the force that actually operates all throughout contact between rope and pulley probably changes between attachment point and the point where rope is facing down).
3) Doesn't the above mean that the axle force could be tilted left or right from being vertical depending on whether that tension (see comment 1) and normal force (see comment 2) are stronger
My intuition tells me the axle force may still be vertical because the horizontal components of this normal and tension may just cancel out. I think this because if one replaces the spring with a weight but then allows the whole system (including the pulley itself) to free fall, I suspect no rotation would take place. And I think if that axle was to fail at the exact moment the rope is cut, the pulley woulld fall straight down.
Thoughts?
Thanks for the great questions.
I did note in my corrections that the tension should be at the attachment point.
And while technically, there should be a normal force, we have never had to draw that on the FBD of a pulley before (or at least I'm not aware of having to do that before). We usually ignore that. But the net normal force that the disk exerts on the rope should be up and to the left. To see this easily, consider what happens when the rope is placed evenly around the top of the disk. The normal force on the rope from the disk would be directly upwards. In this instance, you don't have a portion of the rope on the right side, so the net normal force from the disk onto the rope is up and to the left. Thus, the rope's force on the disk would be down and to the right.
And yes, the axle force may or may not have a horizontal component based on that.
I feel like this question was a lot more complicated in nuance. Because *usually* in a pulley, it's a frictional force (not a tension force) that's acting on the disk. We usually assume that's static friction (rope isn't slipping) and that is equal to the tension force. So we make a lot of simplifying assumptions with pulleys that aren't as appropriate for this problem.
I'll also add that for that "normal" force caused by drapped rope may be at minimum (zero) at 12 and 9 o clock points along pulley circumference with a maximum somewhere between them (Hard to say where but maybe right between them). It also increases as you go clockwise from 12 o clock to attachment point but with vectors facing down and to the left. But all vectors between 12 o'clock and that connection point have their horizontal components canceled by those between 12 o clock and a point in bilateral symmetry with the connection point along an axis running from 12 to 6 o clock. So that'd leave a downward vector only for that section. Past that point of symmery all vectors would be down and right such that overall net vector would just be down and right.
Your way of thinking about it in terms of comparing it to a symmetric dropping was indeed a simpler way of thinking about it. But I like to justify it thinking of vector addition.
I read your reply and yes in this case it is different since rope is attached to the pulley. It is indeed not the usual setup where you roll without slippage.
I think the axle force is most definitely partially fighting that normal force. And I think that possibly balances out the horizontal component of tension such that all you have left is a force down on the pulley that is lineareraly cancelled along with weight by the axle force but causes a net torque since it acts on edge of pulley unlike pivot and weight that have no torque since they act at pivot. This is what js making it spin - by pulling down along edge of pulley (as you originally solved it but called tension).
If on your lab you build an Atwood machine in L shape setup and the pulley axle failed, the rope would most definely push the pulley onto the corner of the table. The axle force on a pulley in that setup is diagonal to fight both the weight and this normal force that definitely shouldn't be "ignored" in a diagram of forces acting on a pulley. And yes friction would usually be the thing spinning that wheel. But this is a little different. In this case, it's actually that normal force between rope and pulley that's doing it. I think. The downward part of it anyhow.
I suspect if the axle failed here (removed instantly). The pulley would fall straight down.
Bro, I am sorry to point out but your solution to the last question is incorrect. The torque increases. You assumed that the force does not change, but since the torque clockwise by the mass of the wheel and the block increased, the torque counterclockwise which is by the spring should also increase so equilibrium is achieved. So it should be increase for all three of the forces.
I’m kinda mad cuz it was crunch time for me when I got to 3e, and I didn’t notice the problem said that the block was hung again.
yeah I somehow thought it should have been removed too. It was a weird comparison to me, but I get it now. But no sweat, that's partly why the curve is pretty generous on the exam. You don't have to be perfect (or even close to it)
Hello: With respect to FRQ1, could you please clarify why Work Done is dependent only on Tension Force (FT) when in fact, there is Kinetic friction acting against FTCosTheta. Isn't Work Done dependent on Net Force and not just FT?
The question is asking about the work done by the string. it's not asking for the total work done.
Thanks you so much for clarifying! How silly of me not to have read carefully enough. Thanks again.@@AllenTsaoSTEMCoach
For part c ii of question 1, I set the horizontal net force as friction minus tension instead of tension minus friction. So basically, I set the rightward direction to be positive and the leftward direction to be negative. Do you think this will be ok?
It shouldn't matter too much. The question was for the magnitude of the net force, so they didn't really care which way it was, but the net force is technically to the left.
@@AllenTsaoSTEMCoach the diagram actually defines right as positive, so friction minus tension should be acceptable
But technically magnitude is a positive quantity. So strictly speaking, it should be tension - friction. Friction - tension would be a negative value. But I don't think the graders will be that picky about it.
I tried to derive theta from net force equation (Tcostheta) and COMPLETELY messed up problem 1 with how much time trying that took LOL, how did I not see that
Seems like I also screwed up 3.....at least I seem to have aced 2?
In question 2 part A, wouldn't the correct answer be less than? Although the two impulses are equal in magnitude, wouldn't the impulse on cart 1 from cart 2 be negative. The question asks about the impulse on cart 1 from cart 2 vs the magnitude of the impulse on cart 2 from cart 1. It doesn't ask about the magnitude of the impulse on cart 1 from cart 2 vs the magnitude of the impulse on cart 2 from cart 1.
Why is it negative? That depends on which way you say is positive.
You make a good point that it doesn't say magnitude for the first part. I'm 99% sure they meant magnitude. But as long as you were clear, they may give you full credit
Yeah, I agree, it’s probably just a little unintentional wrinkle. It’s just that you showed the momentum of cart 1 to decrease after impact later in the question in the momentum graph, which I believe defines the impulse on cart 1 as negative. They probably wouldn’t look at that graph and then go back to part A to see if it lined up with that though. Thanks for the vid though, was nice to see the questions worked out!
You're welcome!
While that's true, I think Ft would have to be negative too because it points left.
@@AllenTsaoSTEMCoach I accidently deleted my first comment about the limits....if you used a neg Ft wouldn't that make all the previous equations backwards as well? The net force equations were written as left positive and right negative, weren't they?
For Q1 part (c) you established you're own positive directions even though on the image it displays the positive directions being right and up, does writing your own positive directions take authority over a given one?
Nevermind, ive just read the corrections i see that you pointed it out
Isn’t integral otherway around?
Read the pinned comment.
why is the answer not just tan theta = H/x because its asking for as a function of x
Which part are you asking about?
Are you going to the E&M problem sets?
Yep I just finished set 2. It's processing now.
Do you know when form I for international students will be released?
I don't think they will release those forms
Would it have worked to plug in tan-1(H/x) anywhere instead of theta
Yes
what if for frq 1 when they asked to derive the expressions as a function of x you substituted "theta = arctan(H/x)" for the theta's (I thought that's why they made us do that in part b) would u get full credit or would it have to be simplified as u did it?
You can definitely do that. You don't have to simplify it the way I did.
In 3a shouldn’t the spring force has a formula of -kx why ignored the negative sign
Negative sign is there to tell you direction. It's letting you know the direction is opposite of the displacement.
If you draw the FBD correctly, the direction is already handled by which way is positive and which way is negative.
I wish I’m as smart as you are 😔
If you're taking AP Physics C in high school, then you know a lot more physics than I did at your age. I couldn't draw a free body diagram until I studied physics in college.
Don't mistaken my greater experience in physics as me being "smarter".
@@AllenTsaoSTEMCoach
I don't know if you are a teacher or not. But thank you for encouraging students with this comment. You are a great man.
This message is to all students who took AP courses in high school and completed the exam. No matter what score you got, you need to know that you are smart and motivated, and you should be proud of yourself.
@@Canda-fh4xc I hope others see both mine and your comment.