So on the released ap physics exam that I took, I got slightly more than half of the MC questions right because I spent the entire yr practicing formula use. The actual exam, however, really requires you to reason through the test and use your intelligence. Some of my friends getting near perfects on MC have literally no work shown for some of the tp questions, while I’m trying to derive formulas and do calculations.
For the FRQ, you should be able to connect qualitative reasoning with the quantitative calculations. The latter of which is much easier than the former for me at least. Expect to get some points taken off for FRQ, because the grading rubric is rlly specific. All I can say is review labs that you’ve done throughout the yr, and try to understand the concepts subjectively rather than objectively. I think that those that are good at SAT/ACT will do really well on this exam, if that’s helpful.
I know how to do the second FRQ. The first one looks pretty hard. Hopefully someone else will know how to do it.
So in the second FRQ, you are given the mass of the jug, the mass of the child, and the height above the ground initially. So you can first use energy conservation principles and set the GPE(Gravitational Potential energy) in the very beginning to Kinetic Energy right before the collision. The reason you can’t set the potential energies before and after equal to each other is because energy isn’t conserved in a completely inelastic collision. Then, you solve for the velocity of the child right before the collision. You should get something like this:
mgh = .5mv^2
(50 kg)(10 m/s^2)(1 m) = .5(50 kg)(v)^2
v = 4.47 m/s
Now that we know the velocity before the collision, we can use momentum conservation(Only momentum conservation works for inelastic collisions) to solve for the velocity of the jug and child together, after the collision. You should get something like this:
mv = mv
(50 kg)(4.47 m/s) = (50 kg + 4 kg)(v)
v = 4.14 m/s
Next we use energy conservation to do the last part:
.5mv^2 = mgh
.5(54 kg)(4.14)^2 = (54)(10 m/s^2)(h)
h = 0.85698 m
Therefore, the height after the collision is less than the height was before the collision. Hopefully, this helped!
@bobthebuilder13 I agree with your answer for the first one, but there’s an easier, more conceptual way of doing it.
We’re assuming no friction here, so energy will be conserved within the system completely. Therefore you can skip the entire step and just say that:
Ug initial = Ug final
Since we’re close enough to earth for it not to matter Ug = mgH
well Ug initial is just the mass of the dude (m) times g times h which in this case is one, so just mg…
Ug final is the mass of the dude AND the jug (M) times g times some H which we’re trying to solve for.
so mg = MgH, g cancles out divide M to both sides you get m/M = H. since the M is the mass of the dude and the jug while m is just the dude it will be less than one.
In general what did you guys think of the 25 or so MC questions they released? I haven’t taken the official exam yet, but I found those difficult. I think I got 18/25 or something bad like that.
mgh = mgh
(50 kg)(10 m/s^2)(1 m) = (54 kg)(10 m/s^2)(h)
h = 0.9259 m(I got 0.85698 m when solving the other way that I posted above)
The reason the height here is different from mine is because energy is dissipated into other forms of energy when the child grabs the jug. I do agree with you that there is no friction involved, but you can’t just set the energies equal to each other like that. You have to remember that this is a completely inelastic collision, and that energy will not be conserved during the collision no matter what(That’s why I used conservation of momentum for the collision part of the problem).
I might be wrong though. What do other people think?
Oh, my bad. I neglected to see part b so I thought it just wanted you to tell if it would reach 1 m or not. I didn’t know that it wanted an expression. I think you’re right in that case.
@lovesmath I have the PR book and based on the practice test I took, the frq questions are very similar (although the rest of the book is not very good). Also, PR has a lot of stuff that isn’t even going to be on the actual exam like calculations with Doppler effect.
@shizpie I personally did a lot better on the released 25 MCQs than the actual 50 MC test I took in class. However, the FRQs (esp #1) is very similar to the actual test.
Regarding FRQs, my teacher almost guaranteed that there was going to be one question about waves, natural frquency, and or resonance…
Personally, I thought the 25 MC are a just little harder than the practice test (our teacher gave it to us a month ago, and I scored 34/50 on the MC while I probably would have gotten like 16/25 on the sample)
@bobthebuilder13 YES. That’s exactly the way I did that problem( conservation of momentum since energy is not constant). The rest of my class just set energies before and after equal for the whole problem, and I was pretty sure that was wrong since an inelastic collision means mechanical energy(in this situation) is not conserved.
@lovesmath PR has way too many calculation problems, but the content review is still helpful (though 5 steps to a 5, I think, is more helpful since it is adapted to the new exam)
What are the labs we were suppose to do? I am unsure if we actually did all of them in my class.
Also, do we need to know Kirchhoff’s Laws, specifically solving problems with two batteries? Because, my class did that, but I understood from Flipping Physics and the AP Outline we don’t need to. Can anyone confirm?
@bobthebuilder13 you are right my physics teacher put that on our energy test a couple month’s back and out of 60 kids only 1 got full credit on the problem. You have to use conservation of momentum in that problem, I tried using conservation of energy and got zero credit for that in the first and third parts.
