@aao1997 Do you remember what your equation looked like? Just asking because I’m pretty sure I screwed that up pretty badly haha
Did thwy only release this or other forms as well?
g(m2-m1)/m2 is the way to derive a net force for block 2 over mass of block 2
@aao1997 That makes sense… I overcomplicated it so much
Nope. Your answer would be right if the force on m2 was (m2-m1)g, but its not. The force is m2g-T(making the force equation m2g-T = m2a). To find the acceleration you need to write the force equation for the other block: T-m1g = m1*a. The a is the same for the whole system, so you can combine both equations to cancel out the T (b/c it isn’t one of the values you’re allowed to use in your answer). You get a=g(m2-m1)/(m2+m1).
@tealpanda Only this one
@dfghjklasd is correct. Thus, when you increase the mass the acceleration decreases.
You HAVE to divide by the total mass because the acceleration refers to the system. Both m1 and m2 have the same magnitude of acceleration, just different directions.
@MileSwimmerDude for the pencil pushing on a disk, I said the smear distances would be the same. I’ll try to exaplin my thinking:
the disk slows down because of a frictional torque. This frictional torque is based on the friction force from the pencil, TIMES THE RADIUS or the distance the pencil is pressed at. Although the pencil is pressed equally hard in both trials, thus having the same frictional force, the trial where the pencil is further out will have a greater frictional torque from the pencil because of the greater radius; thus, the disc is decelerated faster and rotates through a smaller angle than in trial 1. And this compensates for the extra distance when the pencil is pushed further out, making both trials equal. There is math involved, and both exactly cancel out, making the mark the same length in both cases.
To sum up: Trial 1, disk rotating, pencil eraser pushed down, stops the rotation of the disk, leaves a mark.
Trial 2, disk rotating at same speed and pencil pushed down with the same normal force but further out from the center of the disk. The increased radius>greater torque>smaller theta & less rotation of the disk=LESS SMEAR, but also increased radius>more linear distance for the angle=MORE SMEAR. Less rotation of the disk cancels out with the increased linear distance and the smear is the SAME LENGTH for both trials.
Does it matter that i put a = -g(m2-m1)/(m2+m1)? I put that because the block moves downward, causing a negative acceleration, but the problem only asked for magnitude. For part C, the acceleration does decrease because now the denominator is m3+m2+m1.
the block is on the inside of the system though. isn’t acceleration dependent on the net external force?
@asappebble The only forces acting are in the system. Acceleration is dependent on the net force though.
For 1c, the tension is no longer constant throughout the system. Here are the new force equations:
m2: m2g - T1 = m2a
m3: T1 - T2 = m3a
m1: T2 - m1g = m1*a
Add them together to cancel out tension.
m2g - m1g = a(m1 + m2 + m3)
a = g(m2 - m1)/(m1 + m2 + m3)
The denominator is larger, thus the acceleration is less.
Or, in simpler terms, the net force on the system is still the same, but the total mass is larger, so the acceleration must be smaller.
Where are the other forms? Why is there only form O on the website?
For the lab question about circuits, it asked about uncertainty for one of the last parts, and i put that the voltmeter/ammeter could have internal resistance.
For the one asking about how to determine if the electric potential energy is different, i’m not sure that this is right or not, but i put measure the voltage of the resistors and sum up the voltage drops and they should equal zero.
None of these questions were on my test 
@AlphaDragon g is always negative while doing kinematics because there is pretty much universal agreement that up is positive and down is negative. Thus, an acceleration that pulls down must be negative. In forces, there isn’t such a set system. It depends on what’s most convenient to do calculations (see my above force equations, in one the force of gravity is negative, in the other its positive). This is why CB questions almost always ask for the magnitude of the force. To ask for a sign value, they have to set a direction to be positive, and in questions such as this that’s very hard since the force wraps around a box. I don’t think CB will take off in your case, just b/c it doesn’t really change the fact that you got the right answer, and the inclusion of “magnitude” in the problem wasn’t meant to check if you knew the concept but instead was necessary if they expected the problem to be solved. Sorry for the length, I tend to ramble.
The College Board only posts form O.
@dfghjklasd Thanks. I forgot that g was already negative to begin with. Since they ask for magnitude, I am not sure whether putting a negative sign or not matters, since it was meant to indicate direction and that was the way i set it up(m2g-T1=m2(-a)) instead of +a.
@AlphaDragon Are you referring to 2bii? If you connect a voltmeter in parallel with the light bulb, you measure the voltage across the light bulb. The voltage is the difference in electric potential. So if the voltage is anything other than 0, the electric potential changes across the bulb. The electric potential is the electric potential energy per electron. So if the electric potential changes than the electric potential energy electrons change. Sorry if this doesn’t answer your question, I’m not quite sure what it is you’re referring to.
In terms of uncertainty, you are correct that the resistance values of the ammeter and voltmeter will cause error. I’m not sure how specific they wanted you to be, but an ideal ammeter has no resistance and an ideal voltmeter has infinite resistance. So the internal resistance of the ammeter would be too large and would cause error, and the internal resistance of the voltmeter would be too small and would cause error.
My physics teacher explained 2bii a bit differently.
You’re supposed to be measuring the voltage before and after the current runs through the resistor. AKA, connect the voltmeter in parallel, and record the voltage before and after. If you want to be a lot more thorough, record exactly the voltage per second and make a voltage vs time graph. If there’s a slope in that graph, AKA the voltage changes, then yes, the electric potential changes across the bulb. That’s all there really is to it. You literally are finding if the voltage changes across the board. If the voltage = 0, that doesn’t necessarily mean that changes across the bulb. If the CHANGE IN VOLTAGE is 0, then yes, the electric potential has not changed.
For the second part, you don’t need to provide a source of error. I made this mistake too. It is just asking HOW the uncertainty will mess up your readings. AKA, a false voltage and current will produce a false resistance because of Ohm’s Law, thus messing up your ability to decide whether or not it’s nonohmic.