@sscharter for #7 when your using (PAT) dont you have to use the total mass times the distance squared which is now 2m (r/2)^2. or is it since that is your new axis you dont count that mass
last questions if someone could please do 30 and 35, i appreciate it!
Hi guys! why are the FRQ’s different for the 2014 exam in the two links. Aren’t they the same exams?
http://mralsterscience.■■■■■■■■■■/uploads/3/8/0/0/3800138/ap_physicsc_m_practice_exam_2014.pdf
http://media.collegeboard.com/digitalServices/pdf/ap/ap14_frq_Physics_C-M.pdf
Comparatively, I think that the 2015 Princeton Review practice MC practice mech tests are much harder than the official College Board’s - what do you all think?
@RexArcadia hey could you also post the 2013 ap physics c multiple choice exam and 2014 ap physics c electricity and mechanism multiple choice please?
Hi,
@Sajidur4 , I was initially confused for that also. They are not the same exam; the 2014 full test is a practice test which was provided by the college board for students to practice/prepare. It was never actually administered. The full test for 2014 hasn’t been released; the collegeboard has only published the free responses.
@Mathinduction
For the 2008 AP physics C mechanics 2008 FRQ Mech 2 part A), why is the Hinge force at a 45 degree angle inward and not directly upward?
Anyone feeling confident about the exam tomorrow?
I bet the MC’s are going to be different from the ones we’ve practiced considering that we’re now allowed the equation sheet and our calculators.
Could someone please explain #5 from this link: http://media.collegeboard.com/digitalServices/pdf/ap/ap-physics-c-course-description.pdf#page=46 ? The answer is A. Thank you!
Also #9, a few pages later? The ring has a greater rotational inertia than the disk, and so I thought that the disk would move farther.
rockinman… I believe you don’t count that mass.
@sscharter because your new axis is through that point?
sleepdeprived4 (haha, luv ur username… That’s been me all this year!)
For 5) you do the forcedisplacement in each component… The work will end up being a scalar so don’t worry about actual vector addition. You would do 312-4*5= 16 j
For 9, i just thought of it in real world cases… Since the ring has a larger diameter, or a larger circumference, it will go the farther distance.
Hi all,
For the question 2, part c on the 2014 practice, I set work equal to the change in kinetic energy by the work energy theorem and got the negative answer of the answer key. Should the final answer for F0 be positive or negative? I always thought that for conservative forces, work either equals the change in kinetic energy (positive change) or the negative change in potential energy (thus F=-dU/dx). Am I right in including the negative sign (and the answer key just missed it), or am I doing something wrong?
@sleepdeprived4
For number 5, an IMPORTANT concept you must understand is that work is positive in direction of motion and negative opposite direction of motion. So it does positive work in the x direction and negative work in the y direction 6: (12N x 3m) - (4N x 5m)=16 work or joules
For number 9, they have different rotational inertia but the same angular momentum. L=mvr which is the same for both: same mass, velocity, and radius but L also equals Iw and since the rotational inertia of the ring is greater the ratational velocity is smaller; thus the rotational kinetic energy is smaller(1/2 Iw^2). a smaller rotational kinetic energy means a smaller potential energy or mgh.
write this on your formula sheet: L=mvr=Iw
By the way, the AP Pass website for calculating AP scores is actually really unreliable because they’re using the 2009 and 2004 curves, but using the 2011 changed way of calculating. We will be graded the 2011 changed way, but on harsher curves.
Before 2011, incorrect answers caused a deduction in points, like the SAT.
Also any help with 1984MC numbers 3, 22, and 29.
Thanks in advance.
Following up on my question above, I’ve been thinking about it more, and it seems like the force should be positive (in the same direction as the work). However, as the work-energy theorem indicates a negative force, do you just have to flip the signs in orbit problems involving work like these? So that normally W=change in kinetic energy or negative change in potential energy, except in this kind of problem where W=negative change in kinetic energy or positive change in potential energy. I just want to confirm this in order to be clear on the signs before tomorrow.
Thanks!
@Woandering #3: Period for a pendulum is T=2pi(root(l/g)). Mass is not in the formula at all, therefore period does not depend on mass of the hanging bob.
22: C. Draw a free body diagram of your rod. Weight down, tension left-up, therefore you need a vector to the right to have a net force of zero.
29: B. Redraw the three wires as point masses at the individual Centers of mass. Call the bottom wire y=0. Center of mass of the y coordinate, then, is (0.5m+0.5m)/(3m)=1/3. Point that is closest to 1/3 of the way up the wire is B.
@punctiliouseye Thank you! If the rotational velocity of the ring is smaller than the disk, though, why would the ring move farther than the disk?
@Kyuutoryuu Thanks! Still confused about 22, though. It’s not asking for a force to set the net force to 0. It’s asking what the vector is for the force that is applied by the string.
If I misunderstood your explanation, please explain a bit more, and if not, please help again, if you can. Thanks!