@@AllenTsaoSTEMCoach it doesn't explicitly mention that though. the only hint is it saying that it was held against the spring, and even then it could still be mistaken. i think the wording is too ambiguous to dock points.
So you think the spring is still attached to the block? Then it would have to pull the block and the block would never slide all the way to the right the hit block B. You have to assume the block isn't glued to the spring. The spring has to lose contact because the spring doesn't keep accelerating after block A reaches it's maximum speed (at the equilibrium position)
@@AllenTsaoSTEMCoach the problem is all positions given are variables/letters. "all the way to the right" is relative. the block could have just been slowing down the whole time due to the spring because we were not explicitly told the positions. it is completely possible for a attached block to reach the end without losing all its kinetic energy, depending on the length of the rest of the board/other sections(and the frictional constant), which we don't know. From what they gave use, saying the block was "held against" the spring could be interpreted as a connected block was pushed back and because of the nature of the problem with all the undefined lengths, its impossible to just assume the block is not attached. I do plan on reporting the question for ambiguity as i believe my response correctly answered one of the possible scenarios.
I think if you stated you assumed the spring was connected and answered it consistently with a properly shaped curved graph (no straight lines), then they could grant you credit. You would need to be clear that's how you assumed it though.
Question: for Q3(c), wouldn't the answer be more information is needed? since the information on how what angle the rod will be elevated is not specified, we can't deduct whether the tension is greater or less. Since the torque is calculated from the formula: Frsin(theta), if theta is smaller than 90 degrees, then the tension would definitely be smaller than before, but if it were greater than 90 degrees without any specificities, we cant determine if the tension is greater, smaller, or equal to the tension before.
The phrasing of your question makes it difficult to tell what you’re trying to ask, but I think I understand what you’re trying to say. Simply put, the video is correct, however your concern is justified. Since the question states that it is only doubling the length of the string, you can assume that the string is only becoming more perpendicular to the rod. I think the issue you’re worried about would be if the string became so long that it, in fact, started to make an angle on the other side of its perpendicular position, so much so that it ended up needing to pull with more tension. Because of the length increase specifications though, this idea can be ruled out. Hopefully this answered what you asked, and hopefully in time for what you needed it for! If not then well, sorry about that…
I looked at the videos that college board puts out, and then I just watched RUclips channels that were for reviews…I also looked at past tests, Frqs especially!
For FRQ 1 (b) (i), I drew a graph that looks like a sine graph (without a point of inflection, only concave down), as K = 1/2 * m * v^2 and argued that x is proportional to cos(t), so v is proportional to sin(t). So, shouldn't the first part of the graph look like a sine graph without a point of inflection? If not, could you please let me know how many points could be taken off? Thank you
I have no idea the points because they don't really decide those things until they start scoring. It's proportional to cos^2(t) not cos(t). I would look at those two graphs and compare how it looks for the first quarter cycle of it and see the differences.
Could there have been different answers for 2e. I put velocity vs time and then said to conduct separate tests with each varying length to see how that would affect terminal velocity. Is this right??
For question 3a on the free body diagram, I labed Force axle as F normal. Will I lose points. Additionally, the angle of the force vector is the same angle as the rod. Will I lose points, or was it supposed to be a greater angle?
For the justification of the sin^2 graph I think it made more sense to use the justification you used to get it in the first place, ie) either deriving or just stating the velocity over time graph for a spring is a sine wave, then squaring that
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.
Hello I have a question, my students are claiming they have been examined in a test which is different than sets 1 and 2 Would that be possible? Is there a version that is not published?
This might be just a technicality, but on 2B isn't it true that the drag force never becomes exactly equal to the force of gravity? Objects always approach terminal velocity, but they never reach it. This also means that the slope of the graph will never be exactly zero because there will still be some acceleration. I even worked out the math for this particular drag equation and there is definitely a horizontal asymptote at v=vmax, and the function does not cross or touch the horizontal asymptote.
You are correct that mathematically, it would never quite reach vmax. In reality, the model isn't ideal and the velocity fluctuates a lot based on the air density and other factors. And it also gets close enough that the difference is very small (it's an exponential graph). So while it's not exactly at vmax, the difference between vmax and the actual velocity is very tiny and negligible after a short time.
@@AllenTsaoSTEMCoach So how can you rigorously define a certain time t1 where it is "close enough"? For any arbitrary time you pick, won't there always be a "more correct" answer farther along the time axis? It is the equivalent to asking someone to name the exact time that an object cools to ambient temperature - it is ambiguous at best.
This one was to visually tell when it's close enough..can you see when it about hits it. Yes, mathematically it's closer as you move along. But is there a practice difference between 39.999999 m/s and 40 m/s?
@@AllenTsaoSTEMCoach So if it gets closer as you move along, would it have been correct to choose any time value after where the graph begins to level off?
No not based on the graph. You would have to argue visually why you thought a later time was closer to the terminal velocity I don't think you can visually argue it based on the graph
for 3d ii why would it not be parallel axis theorem since its asking for rotational inertia around pivot point not center of mass? I didn't know how to do it I just put the parallel axis theorem down hoping to get a point
That sin^2 question messed me up, I graphed what the velocity graph would be, not the square. Didn’t even know what sin^2 looked like off the top of my head. Wonder how many points I’ll lose
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.
@@AllenTsaoSTEMCoachok so if I did Dv/dt = bv^2-mg/m will I get sone points for that specific 2A question? Or are you talkijg about the whole Problem as a whole
So first, data has all kinds of errors in it. It's never going to be perfect. Second, you can't assume that a model may be linear forever. For example, there's a pretty linear trend for growth for kids ages 1-5 years old. It doesn't mean they will keep growing forever at that rate. This model of drag velocity may not work for very light objects. And that's fine.
can you use net force in the x and y directions equal to zero to solve for tension? i solved for the force of the pivot using net force then used that to solve for the force of the tension also using net force? I drew the force of the pivot acting in the direction of the rod, is that right?
At 31:35, wouldn't the tension be the same because according to the equation derived from the previous part, the force of tension is not dependent on the length of the string but the angle that the rod makes with the vertical. In the question it says that theta is the same so tension would be the same right?
But where did that angle come from? Part of it came from the fact that the component of the R vector perpendicular to the tension. If you change the angle of the tension, then the angle the r-vector makes with the tension will be different too. So that equation won't still be valid.
2c was so weird it looks like they didnt even finish writing the question but knowing college board i am predicting that theyre gonna have really annoying scoring guidelines where you lose points if you didnt explicitly say the words "line of best fit" or "proportional to" or something like that because it just seems a little tooo easy given the rest of the exam
@@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
18:08 i said that the nonconservative drag force would do work on the cylinder and dissipate energy, resulting in less kinetic energy for the cylinder thrown up than in the cylinder that was dropped and therefore a smaller vmax
But the height is changing. It's not asking what the speed is when they are at the same height. They're asking what the speed is when it reaches terminal velocity. It might be at a different height but it would still be the same. It would be very difficult to use energy for this comparison since it's not given it's at the same height that you're measuring speed.
@@AllenTsaoSTEMCoachdo you think saying that because it comes to rest at the top, it falls again as if it was dropped so it would have the same terminal velocity or do you think that’s not enough
for 1b(i) on the t2-t3 section wouldent it be decreasing the kenetic energy because its past the equilibrum point? spring potential increases?
It lost contact with the spring at that point and moves at constant speed.
@@AllenTsaoSTEMCoach it doesn't explicitly mention that though. the only hint is it saying that it was held against the spring, and even then it could still be mistaken. i think the wording is too ambiguous to dock points.
So you think the spring is still attached to the block? Then it would have to pull the block and the block would never slide all the way to the right the hit block B.
You have to assume the block isn't glued to the spring.
The spring has to lose contact because the spring doesn't keep accelerating after block A reaches it's maximum speed (at the equilibrium position)
@@AllenTsaoSTEMCoach the problem is all positions given are variables/letters. "all the way to the right" is relative. the block could have just been slowing down the whole time due to the spring because we were not explicitly told the positions. it is completely possible for a attached block to reach the end without losing all its kinetic energy, depending on the length of the rest of the board/other sections(and the frictional constant), which we don't know. From what they gave use, saying the block was "held against" the spring could be interpreted as a connected block was pushed back and because of the nature of the problem with all the undefined lengths, its impossible to just assume the block is not attached. I do plan on reporting the question for ambiguity as i believe my response correctly answered one of the possible scenarios.
I think if you stated you assumed the spring was connected and answered it consistently with a properly shaped curved graph (no straight lines), then they could grant you credit.
You would need to be clear that's how you assumed it though.
Question: for Q3(c), wouldn't the answer be more information is needed? since the information on how what angle the rod will be elevated is not specified, we can't deduct whether the tension is greater or less. Since the torque is calculated from the formula: Frsin(theta), if theta is smaller than 90 degrees, then the tension would definitely be smaller than before, but if it were greater than 90 degrees without any specificities, we cant determine if the tension is greater, smaller, or equal to the tension before.
The phrasing of your question makes it difficult to tell what you’re trying to ask, but I think I understand what you’re trying to say. Simply put, the video is correct, however your concern is justified. Since the question states that it is only doubling the length of the string, you can assume that the string is only becoming more perpendicular to the rod. I think the issue you’re worried about would be if the string became so long that it, in fact, started to make an angle on the other side of its perpendicular position, so much so that it ended up needing to pull with more tension. Because of the length increase specifications though, this idea can be ruled out.
Hopefully this answered what you asked, and hopefully in time for what you needed it for! If not then well, sorry about that…
I surprisingly got most of these right despite not even being in an AP physics class…I’m so lucky
Could u tell me some resources u may have used to self study the course? Thank you
I looked at the videos that college board puts out, and then I just watched RUclips channels that were for reviews…I also looked at past tests, Frqs especially!
For FRQ 1 (b) (i), I drew a graph that looks like a sine graph (without a point of inflection, only concave down), as K = 1/2 * m * v^2 and argued that x is proportional to cos(t), so v is proportional to sin(t). So, shouldn't the first part of the graph look like a sine graph without a point of inflection? If not, could you please let me know how many points could be taken off? Thank you
I have no idea the points because they don't really decide those things until they start scoring.
It's proportional to cos^2(t) not cos(t). I would look at those two graphs and compare how it looks for the first quarter cycle of it and see the differences.
Could there have been different answers for 2e. I put velocity vs time and then said to conduct separate tests with each varying length to see how that would affect terminal velocity. Is this right??
For question 3a on the free body diagram, I labed Force axle as F normal. Will I lose points. Additionally, the angle of the force vector is the same angle as the rod. Will I lose points, or was it supposed to be a greater angle?
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!
For the justification of the sin^2 graph I think it made more sense to use the justification you used to get it in the first place, ie) either deriving or just stating the velocity over time graph for a spring is a sine wave, then squaring that
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
16:14 is the drag force not equal to bv^2 or is it bv? Edit: Just saw you corrected it lol
Do you think I would lose my point on question 3 part (d)(i) for not putting units? I had the correct number, I just didn’t put kg for some reason…
Depends on how much it's worth. I don't know how often they actually dock points for units.
Was wondering the same thing lol 😂
Hello
I have a question, my students are claiming they have been examined in a test which is different than sets 1 and 2
Would that be possible?
Is there a version that is not published?
Yes there are several forms of the exam..the released sets are the most common ones but the other ones will not be released
And there is scaling for the exam right?
And the scaling is different from set to set, true?
Correct if am wrong please
Yes that is true. The scaling is different for each form.
This might be just a technicality, but on 2B isn't it true that the drag force never becomes exactly equal to the force of gravity? Objects always approach terminal velocity, but they never reach it. This also means that the slope of the graph will never be exactly zero because there will still be some acceleration. I even worked out the math for this particular drag equation and there is definitely a horizontal asymptote at v=vmax, and the function does not cross or touch the horizontal asymptote.
You are correct that mathematically, it would never quite reach vmax.
In reality, the model isn't ideal and the velocity fluctuates a lot based on the air density and other factors. And it also gets close enough that the difference is very small (it's an exponential graph).
So while it's not exactly at vmax, the difference between vmax and the actual velocity is very tiny and negligible after a short time.
@@AllenTsaoSTEMCoach So how can you rigorously define a certain time t1 where it is "close enough"? For any arbitrary time you pick, won't there always be a "more correct" answer farther along the time axis? It is the equivalent to asking someone to name the exact time that an object cools to ambient temperature - it is ambiguous at best.
This one was to visually tell when it's close enough..can you see when it about hits it.
Yes, mathematically it's closer as you move along. But is there a practice difference between 39.999999 m/s and 40 m/s?
@@AllenTsaoSTEMCoach So if it gets closer as you move along, would it have been correct to choose any time value after where the graph begins to level off?
No not based on the graph. You would have to argue visually why you thought a later time was closer to the terminal velocity
I don't think you can visually argue it based on the graph
For question 2e, will I lose points for plotting Vmax^2 instead of Vmax on the vertical axis?
That is an equally good answer.
@@AllenTsaoSTEMCoach thats good to know thanks!
Please release Set 2 whenever you can!!
for 3d ii why would it not be parallel axis theorem since its asking for rotational inertia around pivot point not center of mass? I didn't know how to do it I just put the parallel axis theorem down hoping to get a point
No cuz rod isnt uniform
@@ihatekitcat7931 so? the rotational point isn't at center of mass regardless, so you would still need to apply parallel axis theorem
@@ihatekitcat7931 nvm I just realized why lol
That sin^2 question messed me up, I graphed what the velocity graph would be, not the square. Didn’t even know what sin^2 looked like off the top of my head. Wonder how many points I’ll lose
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.
For question 1b wouldn't the kinetic energy drop right at t3 since there is an inelastic collision?
True but you only needed to plot to t3. I think it's fine if you did or didn't add that.
For frq 3 b how was the tension at cos theta? The tension looked like it was already acting completely perpendicular
It’s from the angle relative to the pivot and the normal line of the rod so it’s 90- theta or cos
For 2A, I did bv-mg instead of mg-bv,, will I lose all points even if I did the dv/dt correct? Just messed that up
You won't lose all points. Errors don't carry forward
@@AllenTsaoSTEMCoachok so if I did Dv/dt = bv^2-mg/m will I get sone points for that specific 2A question? Or are you talkijg about the whole
Problem as a whole
for the line of best fit, how can an object that has a mass of zero have a non zero terminal velocity? Like why is there a y intercept that isn't zero
So first, data has all kinds of errors in it. It's never going to be perfect.
Second, you can't assume that a model may be linear forever. For example, there's a pretty linear trend for growth for kids ages 1-5 years old.
It doesn't mean they will keep growing forever at that rate.
This model of drag velocity may not work for very light objects. And that's fine.
can you use net force in the x and y directions equal to zero to solve for tension? i solved for the force of the pivot using net force then used that to solve for the force of the tension also using net force? I drew the force of the pivot acting in the direction of the rod, is that right?
Unfortunately you don't know the direction of the pivot force. It can be in any direction, so that's why that method wouldn't work.
very helpful, thank you so much ❤
Why isn’t the diff eq mg-bv^2?? It says drag force is bv^2
I fix that later. It was a mistake.
You wrote V instead of V^2 on the drag problem
Yeah I fix that much later on in the video.
question can u take our an x from part a i in question 1?
At 31:35, wouldn't the tension be the same because according to the equation derived from the previous part, the force of tension is not dependent on the length of the string but the angle that the rod makes with the vertical. In the question it says that theta is the same so tension would be the same right?
But where did that angle come from? Part of it came from the fact that the component of the R vector perpendicular to the tension. If you change the angle of the tension, then the angle the r-vector makes with the tension will be different too. So that equation won't still be valid.
@@AllenTsaoSTEMCoach ohhhh I see now. thanks!
The angle theta is the same. Theta is not the angle that the string makes with the rod, but rather the angle of the rod to the wall
2c was so weird it looks like they didnt even finish writing the question but knowing college board i am predicting that theyre gonna have really annoying scoring guidelines where you lose points if you didnt explicitly say the words "line of best fit" or "proportional to" or something like that because it just seems a little tooo easy given the rest of the exam
I never know how to do the graphs specifically concavity
HOLY SHIT IT WAS VMAX AND LENGTH GOD IS GOOD
I thought v max and length was too obvious to where it would be wrong 😭
yea same I did like linear density
25:50 why cant the Faxle be straight upwards?
He explains that forces need to cancel, so there needs to both be a horizontal and vertical component, not just a vertical
19:05 I made the line go through the origin what do yall think
i am thinking about this more and i do not like my answer
bv^2 bro
Anyone else here to see the answers for the redo exam. Whole state of Maryland got switched to second time because our elections were held on the 14th
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
Uh I got 20 as the b value of the slope I’m so screwed gg
18:08 i said that the nonconservative drag force would do work on the cylinder and dissipate energy, resulting in less kinetic energy for the cylinder thrown up than in the cylinder that was dropped and therefore a smaller vmax
That’s also what I said
Because you lose energy when you go up and you lose even more energy coming down
But the height is changing. It's not asking what the speed is when they are at the same height. They're asking what the speed is when it reaches terminal velocity.
It might be at a different height but it would still be the same. It would be very difficult to use energy for this comparison since it's not given it's at the same height that you're measuring speed.
You’re right
@@AllenTsaoSTEMCoachdo you think saying that because it comes to rest at the top, it falls again as if it was dropped so it would have the same terminal velocity or do you think that’s not enough
Im cooked gg
goat
I'm so screwed I literally calculated the center of mass instead of rotational inertia and I completely misread parts of question 2🥲🥲I am going to cry