Actually, some of the MC is different for some of the different forms. IDK why…
@dfghjklasd
Is the MC different across the different forms?
@appgodxoxo Actually, I’m not sure. If they are the same, then the other forms also had those questions like the two ice skaters spinning and the corresponding double helix-shaped graph.
@AlphaDragon
No I just confirmed they’re different.
Similar but different.
They had a similar problem with the two masses and a spring, but ours had different masses, theirs was some other weird velocity thing. IDK.
What was the powder in the tube one?
What do y’all estimate the curve to be?
So far it looks like:
~63 = 5
50ish = 4
35-50= 3
@AlphaDragon @appgodxoxo I don’t remember anything like that, although I’m not very good at recalling questions after a test. I always just assumed different forms had different MC, but I really have no idea.
@dfghjklasd
At this point I’m gunning for a 4. Do you think a 55ish percent would be a 4?
@appgodxoxo I am not too sure about the powder one, but I put the one with the amplitudes at the very opening/closing (trying not to give away the problem, so I can’t say much more than that).
Yes.
For form C: in the problem with the block on the frictionless bowl, and the bowl on a frictionless table, I would say the height will be the same when the block slides down the bowl and up to the other side. The way I thought about it was conservation of energy and conservation of momentum. The system is frictionless and there will be no dissipated energy; the gravitational potential energy of the block at the beginning will be converted to gravitational and kinetic energy when it goes to is maximum height at thew other side. When the ball reaches its max height and has a velocity of 0, the bowl too must have a velocity of 0 because conservation of momentum means the center of mass of the system must have a constant velocity of 0(since the system was initially at rest), yielding mgH=msm + (1/2)m0^2 or mgH=msm, so H=h and the heights are equal.
@reqhuuya what did you get for the one about the pencil smear question parts c and d?
Nvm about the powder one. If it was an open tube at both ends, then it would probably look like a second harmonic.So the amplitudes would be 1/4 and 3/4 of the way.
@kgsoccer08 I did it earlier on, I said the smear distances would be the same.
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 from the center of the disk the pencil is pressed at. Although the pencil is pressed equally hard in both trials, 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 longer smear mark that occurs when the pencil is pushed further out (since angle times radius=length), making both trials equal.
Do you think I will be able to get a 5 if I messed up #3 completely and the last part of the spring problem?
Also, when are the scoring guidelines going to be released? Thanks!
End of July/Beginning of August
@reqhuuya Okay cool! That’s what I got too. Sorry I made you explain it again, I didn’t think to look farther back in the thread for your original explanation
Hey I’m still trying to figure out this website. Can someone go over the answers for free response question 3 in its entirety? I’m trying to prove a point to a friend that didn’t take the exam but I’m not sure what I did wrong. Thanks!
@AlphaDragon
That’s what we all got. It’s the closest of the two hills, one on top one on the bottom.
@kgsoccer08 no problem, it’s helpful to know what other people think as well
So my teacher is estimating this to be the following curve for form O-
67 percent == 5
53 percent == 4
35 percent == 3
20 percent == 2