Discuss AP Physics C Free Response

conker · May 10, 2006, 9:23pm

It’s been 48 hours! I got my booklet back. For those of you who want to, let’s work out the solutions. I thought there were some interesting problems this year worth working out again.

I’ll start with Mechanics.

Mech 1.
a) Draw the forces on the block and slab.

I drew the weight and normal force on both the block and the slab. On the block, the frictional force pointed to the left; on the slab, it pointed to the right.

b) Calculate v_f.

ab = Ff/Mb = (mu)(Mb)(g)/(Mb) = mu(g) = 1.96 m/s^2
as = Ff/Ms = (mu)(Mb)(g)/(Ms) = 0.327 m/s^2

vb = v0 - abt = 4.0 - 1.96t
vs = 0 + a_st = 0.327t

vf = vb = v_s

4.0 - 1.96t = 0.327t
t = 1.75s

c) Calculate the distance the slab traveled when it reached v_f.

x = v_0t + 1/2at^2
x = 0 + 0.50.327*1.75^2
x = 0.501 m

d) Calculate the work done by friction on the slab.

Ff = mu*Mbg = 0.98 N
W = Fd = 0.98 * 0.501 = 0.491 J

Mech 2.

a) Which quantities to graph?
Graph F on the y-axis and x^2 on the x-axis. The slope represents A.

b) Calculate values for the table.
Had to label the third column as “x^2 (m^2)”

The quantities were pretty self explanatory.
0.05 –> 0.0025
0.10 –> 0.0100
0.15 –> 0.0225
0.20 –> 0.0400
0.25 –> 0.0625

c) Plot the quantities in the table. Label axes.

Graph the points, draw a line of best fit.

d) Determine A.

Pick two points on the graph and find the slope. A would be somewhere between 1600 and 1750 N/m^2. I got 1670, personally.

e) Spring is now placed horizontally on the floor. Cart of mass 0.50 kg compresses spring 0.10m. Calculate the work done by the cart in compressing the spring to 0.10m.

W = int( F dx ) = int<a href=“1670x%5E2”>0, 0.10</a> = 0.557 J

f) Calculate the speed of the cart just before it hits the spring.

Conservation of energy.
Ui + Ki = Uf + Kf
0 + .5mv^2 = 0.557 + 0
.5(0.50)v^2 = 0.557
v = 1.49 m/s

Mech 3.
a) Derive an expression for the acceleration of the center of mass of the hoop.

(Note: Imagine that @ is theta and & is alpha.)

F = Mg sin @ - Ff
a = g sin @ - Ff/M

How to determine F_f? Friction exerts a torque on the hoop.

RFf = I&
Ff = (MR^2)(a/R)/R
F_f = Ma

a = g sin @ - a
a = g sin @/2

b) Derive an expression for the speed of the center of mass of the hoop when it reaches the bottom of the ramp.

Conservation of energy, again!
Ui + Ki = Uf + Kf
Mg(L sin @) + 0 = 0 + .5Mv^2 + .5Iw^2
Mg(L sin @) = .5Mv^2 + .5(MR^2)(v/R)^2
MgL sin @ = .5Mv^2 + .5Mv^2
gL sin @ = v^2
v = sqrt(gL sin @)

c) Determine the horizontal distance from the table to where the hoop lands on the floor.

x = v*t
v = sqrt(gL sin @) <– all of the velocity is in the horizontal direction

H = v_0*t + .5at^2
t = sqrt(2H/g)

x = sqrt(gL sin @)*sqrt(2H/g)
x = sqrt(2HL sin @)

d) If the hoop is replaced by a disk, what will happen to the distance calculated in part (c).

Greater than. Because the rotational inertia of a disk is less than that of a hoop, the velocity when the disk reaches the bottom of the ramp will be greater than when the hoop reached the bottom. Therefore the disk will travel farther than the hoop.

stupak · May 10, 2006, 9:41pm

I did what you did for almost everything, except for parts 1d and 2f, which i get insanely different answers then you do… hopefully that’s isn’t THAT bad… lol

durt · May 10, 2006, 10:27pm

I forgot to draw the vertical forces in 1a… UGH! :mad: Oh well, every thing else looks good though (forgive me for not reading your post too carefully; it’s a little long :)).

mattd1688 · May 10, 2006, 10:36pm

Conservation of energy.
Ui + Ki + W(friction) = Uf + Kf
You forgot the 1/2 k x^2 where x = .11m or w/e.

I thought one axis was already labeled x (m) and so I graphed F(N)/x vs x. I didnt get my sheet back so I dont know the exact answers I put.

3 d = I put that and then said how its like whatever / (1+B) where B = I/MR^2.

stupak · May 10, 2006, 10:44pm

wait its 2/5 MR^2 if its a disk, not B+1.

and ouch durt =(.

i’ve yet to get my sheet back so im a little worried

ccmbr002 · May 10, 2006, 10:50pm

a lot of ppl seem to have gotten perfect scores/near perfect scores on this exam’s FRQ’s…(unless we all made the same mistakes? lol) I’m feeling a harsh curve coming for it… :/

stupak · May 10, 2006, 10:52pm

its gonna be up to the mc this year-

and i feel sorry for next year’s bunch- for every easy ap test, they usually rebound with a hard version the year after

odan · May 10, 2006, 10:56pm

I didn’t answer any of the free response questions but I wrote a really good essay and drew some sick pictures.

ccmbr002 · May 10, 2006, 10:58pm

yah omg i bet they have FRQ’s from hell haha

conker · May 10, 2006, 11:11pm

Hmm…I don’t think you can use the expression .5kx^2 because the spring is nonlinear (and you don’t know k anyway). However, you do know the total elastic potential energy from your work in part e), which is what I substituted into the Conservation of Energy equation. Hooke’s Law and the potential energy derived from Hooke’s Law only apply to linear springs. Please correct me if I’m wrong.

Stupak, what values did you get for 1d) and 2f), or how did you do them differently?

By the way, for those who did not receive their booklets yet, I think the 2006 FRQ questions may be posted on the AP site. Take a look. :)

Earth-dragon · May 10, 2006, 11:26pm

There’s a second force going down on the slab, the normal force of the block?

I got 2.154 m/s for the 1b part? (I forgot my work exactly…)

gregofj · May 10, 2006, 11:28pm

wait,you can’t use conservation of momentum for 1b?

i thought inelastic collission was how you find velocity final!

Earth-dragon · May 10, 2006, 11:30pm

no, because you have outside forces working on it, so momentum isn’t conserved…right?

Also, I just want to say, except for question 1 (where my v_f differs), everythingelse I agree with.

alwaysfaithful · May 10, 2006, 11:33pm

i used conservation of energy for 1b, damn it

ccmbr002 · May 11, 2006, 12:21am

Earth-dragon: yah same everything but 1b i have the same answers as conker. My answer is wrong though I know. Damn…I could’ve made a perfect score on an AP free response X_X o well i guess i have to allow for one or two dumb mistakes…gufah but yah…i really feel bad for people who missed problems on the FR b/c their scores are going to be shot…then again i guess CC isn’t exactly a representative population of the kids who took the AP exam lol

Sogui · May 11, 2006, 12:53am

I made a silly error in 1b, and while I got the correct acceleration for the block, and the correct force acting on the slab, I divided the force acting on the slab by the mass of the slab AND the block, instead of just the slab… this gave me an incorrect Vf for part 1-b.

Asssuming that ALL other work was correct except for this one mistake, how many points could I expect from this section???

conker · May 11, 2006, 5:39am

Sorry, I didn’t complete my work for 1b). I calculated the time and not the velocity. So here’s the rest of my work.

v_f = 0 + 0.327*1.75 = 0.572 m/s

For the people whose answers didn’t match mine for 1b), does this resolve it?

Earth-dragon · May 11, 2006, 7:43am

as = Ff/Ms = (mu)(Mb)(g)/(M_s) = 0.327 m/s^2

<- why are you using (Mb)? It should be Mb + M_s???

conker · May 11, 2006, 10:17am

Hmm…isn’t the force of friction just mu*Fnormal? The Fnormal on the block is mu(Mb)(g), and the frictional force on the slab is the same as the force on the block (Newton’s Third Law). Am I missing something here?

I think you may be thinking of the friction between the slab and the surface it is sliding on, in which case the frictional force would be mu(Mb + Ms)g. However, the problem has stated that there’s no friction between the slab and the surface.

ccmbr002 · May 11, 2006, 3:57pm

I’m pretty sure that conker’s answer is right: .572m/s was the velocity. I looked over all my problems again and reworked that one after i realized i did it wrong. I also asked my physics teacher to check my work. We both got .572…

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