@appgodxoxo The flaw in that logic is that voltage isn’t a measure of the electric potential, its a measure of the difference in electric potential that a battery or resistor creates. So the voltage across the light bulb would be by definition the change in electric potential as current flows through it. Provided the current doesn’t change, the same resistor always provides the same voltage. “Voltage per second” is actually a measure of the power of the resistor/battery and shouldn’t be relevant to the problem. Are you sure your teacher wasn’t talking about 2bi? That explanation would make more sense when applied to current, which is a measure of electrons per second.
And yeah I see what your saying in terms of error, didn’t read it carefully enough. I still think you could benefit from a little explanation of how the resistances of both would cause uncertainty, since it asks for “discussion.” But yeah, you got the gist of it.
@dfghjklasd
You are actually 100% correct. And actually, yes, that would apply to current a lot better.
For the nonohmic portion, however, the reccomended solution is to use the initial and final voltage and the initial and final current and use Ohm’s law to solve for the resistance. If the resistance across both changes, then the bulb is nonohmic. Does this sound right?
Also, do you think they will take off points for supplying what could cause this uncertainty? A common consensus in the class was that you are measuring current and voltage at different times, and because it’s running through a series of resistors, you’re not guaranteed you’re getting the current and voltage that match with each other, if that makes sense. It would be impossible to grab readings for both at the exact same time. something like that.
@reqhuuya Your explanation makes sense, I just forgot about torque when I answered the question. Interesting that they would have the same length; doesnt make a lot of sense haha but I have witnessed this year that not a lot in Physics makes sense 
@appgodxoxo I think as long as you say something that makes sense, you should get the points. I’m not the final authority though lol.
And to determine if the lightbulb is nonohmic, I would just measure the voltage with a parallel voltmeter and the current with an ammeter in series, while varying the voltage supplied by the battery. Plot a graph of the voltage (of the lightbulb) vs. the current. V = IR if and only if the lightbulb is ohmic. So if it is ohmic, the resistance, and the slope of your graph, should be constant and your graph will be a straight line. If it is nonohmic, your graph will not be a straight line.
I’m not sure what you mean by initial and final voltage and current? Do you mean you measure how the current changes as you change the voltage? If you did this and solved for R for the different recorded I’s and V’s , then you can prove whether it is ohmic. If the R’s you calculated are all the same, its ohmic. If not, then its nonohmic. Is this what you’re trying to say?
@dfghjklasd Kind of. Since the voltage is going to change as soon as it goes through the resistor, I just measured the initial and final voltage. AKA, the voltage/current BEFORE it goes through the bulb, and the voltage and current AFTER. In my experiment, i mention waiting a second and measuring the current as soon as the switch is turned on and exactly one second after. I also mention recording your voltages (i used power for some reason… IDK why. I think it made answering 2bii easier since you have a good graph of when, if at all, the Electric potential energy changes.)
And you have the initial and final current from the ammeter as well. I said use different currents from the power source, and this will vary voltage. I said to average all of your runs with the different currents, and do the analysis with that average. That way, you have ALL of the different currents and voltages and can make a nice comparison between initial voltage/current and final voltage/current.
What did the potential energy and kinetic energy graph look like for #3 about the spring compressed -D and reaching 3D?
Mine looked like the normal Potential vs Kinetic energy graph for springs(inverse parabolas) except it was very stretched out from 0 to 3D.
@AlphaDragon
IDK.
It’s NOT a simple harmonic oscillator. So you don’t have the graphs that are inverse parabolas for springs. Even though the potential energy is 1/2kx^2, there’s no oscillation.
My teacher made U a straight line, and KE a weird flipped halfways parabola going down and touching 0 at 3D.
It asks for:
i. The kinetic energy K of the block
ii. The potential energy U of the block-spring system
So it’s not oscillating. KE is reduced by friction slowly until it hits 0 at 3D.
U is not going to change: it’s just whatever it is at that point -D (compression) and it never changes.
Are there different curves for different versions? What curves are you guys predicting for Form O (the one with the circuit experimental design problem)?
For that same spring question, was the student wrong when he/she said it would stop at 6D? I put that it would stop at 12D.
Yes, it stops at 12D. I put that you have 4x the energy, so you have 4Di where Di is how far it goes originally… i’ll lose a point for sure because you are supposed to derive it with D but it’s not a big deal.
@AlphaDragon
@appgodxoxo yeah I think I made a mistake when deriving the expression b/c I ended up with something weird like position = -4D. I think the process was right though(1/2kx^2=1/2mv^2 then use kinematics). I still said in the end it would be 4x the position.
Wait, for that problem, won’t some of the spring potential energy be lost to friction?
@EveningSwan no because the spring is on a frictionless portion of the track.
Considering that the frqs have been released and we are allowed to discuss, does anyone have a google doc (just for frqs)
MODERATOR’S NOTE:
As mentioned earlier, links to google docs are not allowed; this prohibition has nothing to do with the College Board.
@AlphaDragon
My teacher did it this way but it’s the same thing basically…
1/2kx^2=W
1/2kx^2=1/2mv^2
cancel 1/2s
kx^2=W
W=FD
kx^2=F*D
k(2d)^2)=F8D
K and F also cancels out, they’re useless
(2d)^2=D
4d^2=D
d’s cancel out
left with 4D.
then, original D = 3d, so you get 12D.
i think both ways will be accepted.
he said for this problem, the college board wants to emphasize Work and Energy over kinematics, so they will prefer showing how energy becomes 4x as much.
where was the maximum vertical speed for the oscillator in part C for question 5?
@AlphaDragon
Not even my teacher could figure this one out. A lot of people said it’s at the end and the middle of the string: Nope. that’s where your nodes area.
IDK how they’re going to curve that one. lol.
@AlphaDragon you had to draw two waves (b/c second harmonic) and the two antinodes had the greatest vertical speed (since they are constantly moving up and down). In contrast, the three nodes had 0 vertical speed
I just realized I made a REALLY careless mistake on part c of the circuit question. I said to make a voltage vs. resistance graph and take the slope to find current to see if there is a direct relationship, but I don’t think resistance would be known for the lightbulb.
And on part d I really only stated a source of error because i was rushed on time.
I’m wondering if I would get any points for c and d now.