Homework Help: 2005 AP Physics B

1. May 16, 2005

newo17

Does anyone have the answers with work for the 2005 AP Physics B free response section? anything would be greatly appreciated.

2. May 17, 2005

nomorevishnu

wht do u mean?

3. May 17, 2005

Pengwuino

AP tests arent released to the public for a few months, no chance your getting the answers and work.

4. May 17, 2005

Tokimasa

Actually, they are going to give us back our free response books soon (they said it would be after 48 hours, but I need to ask my teacher today). When I get mine, I could type up the problems and maybe someone here can solve them all and explain to everyone who took the AP Physics B test what you were supposed to do on all of them.

5. May 17, 2005

Chi Meson

GOT EM!

question 1:

a) a v-t graph showing a flat line at +1.5 m/s, then to zero at 9 s, then to -2.4 m/s at 19 s.

b i) -0/75 m/s/s ii) a single arrow down

c) at 4 s, a= 0, so w = mg = 686 N

question 2:

a) three vector arrows: one for weight, one for the tension in the horizontal string, one for the tension in the supporting string at an angle. Lengths and angles should be appropriate to show equilibrium.

b) T = 10 N

c) Conservation of energy gives v=SQRT(2gh). the height that the pendulum falls is L-Lcos30. Solved, v = 2.5 m/s

question 3:

a) E_net = (kq)/(a^2) (for some reason, the distance here is "a"), direction is up.

b) V_total = (3kq)/a (no direction, it's a scalar)

c i) F = (kq^2)/(a^2 + x^2) (the denominator is NOT squared)

ii) F =(2kq^2)/(a^2 + x^2) (ditto)

d) Should show appropriately angled net vector arrows such that the componant pointing to the +2q charge is obviously twice as large as the componant pointing to the 1q charge. The moving charge at point B should have a net force that is down.

I'll be back later with the rest, my class starts in 10 minutes.

6. May 17, 2005

Chi Meson

Here's more:

question 4:

a) check boxes for appropriate materials: two length measuring devices, slide holder, light intensity meter (LIM), paper OR screen, NOT the stopwatch.

b) I can't draw this right now, but look up any "two-slit" set up in your textbook

c) Intesity graph. IT should be typical for 2-slit diffraction where gap width is much less than separation.

d) here is a typical procedure:
FInd center of central max using LIM.
Find position of some nth maximum (4th for example)
measure distance between these two points, call it "x"
Measure distance from slide to screen/paper, call it "L"
Order (m) is 4, wavelength is 632 nm, space between slits is "d"
use smal angle approximation : x = mL(wavelength)/d , solve for d

question 5
a) buoyant force, simple but busywork. Trick here is they want the height of raft above surface, not the submerged height.

b) buoyant force is equal to the weight (this question is out of order, because you needed this to answer part a) F= 11,500 N

c) maximum additional mass the raft can hold is 630 kg, which would allow 8.4 people. I hope you rounded to 8 whole people.

question 6

Lame thermodynamic question (too easy, I thought)
a) isobaric expansion, use gas law but turn V into "H x A"solve for n
n= (PAH)/ RT

b) graph the data points given, find the best fit. points taken off off poor precision and poor selection of appropriate scale.

c) Use GRAPH to determine slope. POints for slope should be taken from the best fit line NOT data points.

plug n chug gets you 0.11 moles

question 7: also Lame (easy energy level / photoelectric question)
a) if you can't calculte the energy of a photon given its wavelength, ... sheesh. 10.1 eV above -13.6 eV is -3.4 eV.

b) plug n chug. momentum of a 10.1 eV photon is 5.44 x 10^-25 kg m/s
(use the "joule-second" value for h)

c)Photoelectric effect equation. THis involves a tricky mathematical process I like to call "subtraction." K_max = 5.4 eV

d) stopping potential for a 5.4 eV electron is 5.4 volts

That's it!?

That's it.

How'd ya do?

7. May 19, 2005

Chi Meson

Well, ... you're weclome! :grumpy:

8. May 19, 2005

dboy

I didn't receive my booklet back, but it looks like I did ok on that part, seeing as i kinda remember most of my answers. Did super good on the multiple choice section, shooting for a 4 or 5.

What do you think you got?

9. May 19, 2005

Fred Draco

I'm sorry... but I don't understand your answer for part (b) i)... -0/75 m/s^2? A typo perhaps?

I solved for that part using V^2 = Vo^2 + 2a(X)...

or O^2= 1.5^2 + 2a(2). X would be the distance traveled upwards between t=8 and t=10...

For 'a' I got -2.25/4, or -.5625....

Am I wrong?

10. May 20, 2005

Chi Meson

Yes, a typo, the answer is -0.75 m/s/s.

At 8 seconds the speed is 1.5 m/s. At 10 s the speed is zero.

The average acceleration in this time interval is simply delta V over delta t.

That's -1.5 m/s over 2 seconds = -0.75 m/s/s

OK now, over 200 people have been viewing... C'mon! How did you do?

Last edited: May 20, 2005
11. May 20, 2005

newo17

Thanks Chi Meson. Looks like I did alright, probably a 4. Multiple choice went well, so that probably helped my grade a bit