<p>Wow I just realized that I made the period from t=0 to where the above part of the wave was at a max (like I did 3/4 of a period)… darn</p>
<p>polar bears on ice</p>
<p>Oh yea! What was the answer for the Polar bear?</p>
<p>Also the pulley system questions? It was a few questions before the polar bear. I had a major brain fart around there.</p>
<p>Ah yes! Polar bear travels distance related to m/(m+M) where m is person mass and M is bear mass.</p>
<p>small mass over sum of the two masses times d</p>
<p>I gave all my answers in terms of pi on frq 1… woops. How many points do you think they would take off for that?</p>
<p>They won’t take anything off for being MORE accurate than you need to, they will probably just plug it into a calculator to see that it checks out and make grading easier</p>
<p>oh yah for mechanics number 3 they were like asking for the time it takes for that ring to travel distance L
i had the quadratic equation but i didnt solve…is that ok?
and how do you SOLVE for L???!?!?!?!?</p>
<p>I believe that works too but there was a much simpler wa of doing it. The ball starts to roll without slipping at a distance L and therefore w = V/R. Using the 2 eq u found in the previous question, u can find the time t using that.</p>
<p>I’m not going to try to solve for L here but did anybody get 3v^2/(8ug)? (u is mu)</p>
<p>Yes, that’s what I got for L, and I believe that it is the correct answer.</p>
<p>damn i got 3/2 as my coefficient somehow.</p>
<p>This is how i did 3 from what I remember
F = ma
F = m(dv/dt)
dv = (F/m)dt
Integrate both sides
V(t)=(F/m)t + C</p>
<p>V(0) = V(sub not)
so then V(t) = (F/m)t + V(sub not)
was my final answer</p>
<p>then for torque
t=Ia *a = alpha = angular accel.
and I did the same as in part a but using the rotational analogy for newtons second law</p>
<p>Also, t = Ffriction x R
becuase the firction is what is providint that torque
and then I = MR^2 (Given) and then plugged that in and for alpha i plugged in dw/dt (omega=angular vel.)
then i separated variables in order to integrate. , and bam thats what i did
also the constant of integration was simply w(sub not)</p>
<p>did anyone else do it like how I did?
and if yes lets hope we’re right :D</p>
<p>Iborpastan
Yes,
I believe the curve is dumb!
Why not just make the test a tad easier right?
I have looked at FRQ’s from 80s and 90s and they are challenging, but not RIDICULOUS like the ones from the past two or three years.</p>
<p>They should make it like before. Or maybe the test makers are running out of ideas so they make everything harder.</p>
<p>I did a and b like you. Hopefully that’s at least 4 points.</p>
<p>As for C and D
I was perplexed for five minutes because I didnt see a difference towards what they where asking.
Until I saw it.
It said EXPRESSION not Differential eq. <em>Face palm</em>…
the only thing i came up with was using the kinematics for rotation and tweaking it slightly in terms of the giving quantities.
but i don’t see what else</p>
<p>As for a and b I hope i did it right so that’s at least 5 to 6 points (:</p>
<p>For a and b, I didn’t just say F=ma or Fr=Iα; I actually used F as the frictional force causing the acceleration and then because there are no other vertical forces, I said F was μN</p>
<p>Blerb:
Yeah now that I remember I did
because Ffriction is mg times mu so Mgmu x R = torque</p>
<p>right right…I don’t know why I didn’t say so in my post but I went Fƒ=μN=μmg
I’m hoping that’s right!</p>
<p>I had no idea how to solve the second page with L though…</p>
<p>Blerb:
Yeah
I hope a and b where correct
I honestly dont see any other way… unless they got too crazy, but ughh who knows</p>
<p>yeah, that part was weird, under the time constraints i skipped it and never went back.</p>
<p>From what i’m looking at a lot of people thought this years FRQ’s were very hard
So its also good to keep in mind that around 80% of people pass this exam with 3 or more
also 25% for mech get a 5 and around 25 get a four (:
so maybe I can maintain a 4, or who knows, maybe even a 5
(I hope…)</p>