On D you needed to explain the effects of the source of error.
And you’re right, you wouldn’t know the resistance. Only the current and the volts.
I just said you know the initial and final voltages/currents so just calculate R using ohm’s law. if there’s any significant difference it’s nonohmic. You can’t make a graph unless you state “record the voltage every time you change current.” you could have said average the voltages and average the currents and run off that to reduce error…
@AlphaDragon
Thanks @appgodxoxo On the FRQ, i meant to make a voltage vs current graph and take the slope to find resistance but I’m still not sure whether i would get any points for what i wrote.
@appgodxoxo If you don’t change the voltage yourself, then neither the voltage or current will change after they go through the resistor. It seems you have a fundamental misunderstanding of both. Voltage doesn’t travel through the circuit. Each resistor or battery in the circuit has a voltage value, this is a measure of the difference in electric potential (which does travel through the circuit) that this resistor or battery causes. You can think of current as traveling through the circuit, but a resistor won’t take any current away. Current only divides when it has multiple paths to follow.
Nonohmic does not mean that after a current flows through a resistor, it decreases. It means that the resistor does not have a set resistance. Its resistance depends on the amount of current that flows through it.
Also, you seem to be under the impression that you can alter the current of a circuit to change the voltage. The problem says you have an adjustable power source, so you directly alter voltage, which will alter the current. Current, however, can not be adjusted on its own, you must either alter voltage or resistance.
@AlphaDragon Technically yes, you have no way to determine the resistance of the bulb, but your method is a conceptually correct way to determine if the lightbulb is ohmic. I’m sure you’ll get the majority of the points.
thanks @dfghjklasd I was just worried that points would only be given for saying make a voltage vs current graph.
For 3c, the graph will have the typical simple harmonic oscillator parabolic shape until x = 0. Both U and KE will have half parabolas, U decreasing to 0 and KE increasing to the original U value. U will stay at 0 for the rest of the graph while K decreases in a parabolic shape till it becomes 0 at 3D.
Above explanation is for 3a, sorry about that.
For 3c, 12D is correct. @appgodxoxo, you need to use both energy and kinematics. Energy to find the speed at x = 0, and then kinematics to find how this speed translates into distance. @AlphaDragon, I also originally got that the distance changes by a factor of -4. You used the equation vf^2 = vi^2 + 2a(deltax), probably, because that’s what I used. Its the only equation without time, which you don’t have any info on. If you solve for delta x and set vf to 0, you get that delta x = -(vi^2)/2a. The negative, however, is actually the coefficient -1, which is a constant like 2 and a. You don’t take it into account with proportion. Again, you have the conceptually correct solution and I doubt that you’ll lose any points.
Oops, @appgodxoxo, your explanation is also correct. I didn’t even think about doing it that way. Although I think saying that force and the spring constant are “useless,” instead of constant, is a bit misleading lol.
@dfghjklasd
How many points will they take off for the 4D answer? A lot of people put that. Is it even inherently wrong?..
yes, because you want it in terms of D. but a lot of people thought D meant the displacement of the object, not the distance compressed. IDK. we’ll have to see.
Also your explanation for 3A is spot on. I missed that the KE and U have to increase first 
You FOR SURE got a 5 on this test. I am really hoping for a 4. I screwed up the first question, it looks like I kind of messed up on the second, I got the third mostly right, the fourth correct, and the last mostly correct.
@appgodxoxo I think only one at most, as long as you show the work that shows it changes by a factor of 4. Forgetting to multiply it by 3 is a pretty minor error, especially considering the time constrictions and stress that go with ap tests.
Just remember the huge curves on AP tests. Looks like you got around 60 - 70% right, which is definitely 4-5 range. And thanks for the vote of confidence! I actually had a different form than the one they released, and I think it was probably a little easier.
Do you mind explaining the last question, part c?
And how would you do the graphs and free body diagrams for question 4?
Sure. For 5c, there are two loops on the string, with nodes at each end and in the middle. The antinodes will have the greatest average vertical speed, so you mark at the 1/4 and 3/4 lengths. Think of it this way. Every individual point on the string travels up and down in the same amount of time. The only thing that differs is the distance they have to travel in that time. The antinodes have the greatest distance to travel in that set amount of time, so they have to have the greatest speed.
For 4a, both just have an mg arrow straight down. Its the only force acting on objects in free fall. B is moving to the right, but it does not have a force acting on it nor is it accelerating in that direction. It just still has its initial horizontal velocity.
For 4b, sphere A’s graph would be a straight line at 0, and sphere B’s would be a straight line at vo. There are no forces in the horizontal direction so neither’s horizontal velocity changes.
@dfghjklasd
How many points do you think those two will be worth? I guess not much because they only have one specific answer.
@dfghjklasd Hi, I have a question on the last two parts of the circuit question. So I designed my circuit with one ammeter in front of the light-bulb and one after it (both in series) and a voltmeter in parallel with the lightbulb. I said that I would vary the current in the circuit by adding more resistors, since the total current that travels through the circuit depends on the circuit’s overall resistance ( I did not even pay attention to the “adjustable power source”). I assumed that the voltage across the lightbulb would remain constant. So with the same constant voltage across the lightbulb and varying currents due to varying numbers of resistors within the circuit (I noted that all resistors would be located after the lightbulb), I said that I would graph the current values on the x-axis and the the calculated resistance values on the y-axis (using the equation V= IR to calculate the resistance). Due to the the R= V/I relationship, I said that if the graph showed an inverse relationship between the two graphed values, then the lightbulb would be nonohmic. Does my reasoning make sense?? How many points do you think they would take off?
@16apkb Yes, adding more resistors would alter the current, so that’s correct. The problem is that the voltage of the bulb won’t be constant if you alter the current. You do have a measurement of how the voltage changes for the different currents from the voltmeter in parallel. Did you write that you used that to record the voltage?
The other flaw is that you cannot then use V = IR to calculate resistance. V = IR is true only if the lightbulb is ohmic. If you used this equation to find the resistance, then your graph would show the lightbulb is ohmic because you have already assumed it is. A better solution would have been to plot a V vs I graph and then look at the slope. If its constant, it ohmic. I think I explained this in more detail in an earlier post.
I really have no idea what the scoring will look like, but you’ll definitely get some points for your setup and measurements and mentioning how a graph should show the relationship between the values.
I really don’t think this was a well-written question because it didn’t have a very good explanation of what it means to be “ohmic” or “nonohmic.” Even just saying “an ohmic resistor follows Ohm’s law” would have gone a long way. Unless you had a prior knowledge of what ohmic means, it would be pretty hard to answer correctly.
@appgodxoxo I’d assume that as well, but your guess is as good as mine.
@dfghjklasd Welp it’s looking like I got around a 50% on the FRQ. Which is the same as I did on our practice test. How did you find the multiple choice?
Anyone else think the Form O multiple choice was pretty easy?
Isn’t the multiple choice the same across all of them? @AlphaDragon
I thought the multiple choice was pretty easy, although I can’t remember which form it was. E, maybe?