<p>This is a thread to post consolidated FRQ answers for all AP Exams:</p>
<p>Calculus BC: 2008</a> Free Response Solutions
Comment: "btw on #6, d is the answer to c since c wasnt put up"
<thank you,="" chickenboi8008.="">
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<p>Physics C:
[Mech]
Mech1.
1. a) There are 3 forces: normal force n, gravitational force Mg, and drag force bv.
b) Mgsin(theta)-bv=M(dv/dt). v is the unknown.
c) At terminal velocity, acceleration is 0. Thus, v=Mgsin(theta)/b
d) This is a separable differential equation. Separate variables, do indefinite integration, use the initial velocity to solve for the constant of integration. The answer I got is v=(Mgsin(theta)/b)(1-exp(-bt/M))
e) Differentiate the above equation, and sketch the graph. I got a=gsin(theta)exp(-bt/M). Thus, at t=0, a=gsin(theta).</p>
<p>2.
a) There are 4 forces: tension T, mg for mass, mg for rod, hinge force FH.
b) The hinge force is entirely horizontal. Thus, Tsin30 balances out the two masses, and T= 49N.
c) Moment of inertia of system about hinge = """ of rod about hinge + """ of block about hinge
Use parallel axis theorem and the definition of moment of inertia, I got I(total)=((1/3)M+mass of block)L^2=0.42kgm^2
d) Torque = Ia a= Torque/I=(mass of rod<em>g</em>L/2+mass of block<em>g</em>L)/I=21rads/s^2</p>
<p>Mech 3.
a) Standard graphing.
b) slope=25N/m
c) Use conservation of energy. Initially, there is GPE. In the final state, there is only SPE. Thus, solving the equation for m, I got 0.69kg.
d) Think about it this way: the mass accelerates all the way down until the cord stretches sufficiently enough such that the weight of the mass is balanced completely by the cord spring force, i.e. acceleration is zero. This is when maximum speed is attained.
Let extension of cord at max speed be x. Then, kx=mg, and x=0.27m. Thus, it dropped 0.87m.
To solve for the max speed, use conservation of energy. Initially, there is GPE. In the final state, there is maximum KE and SPE. v=3.80m/s
<thank you,="" samuelc="">
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<p>[E&M]
E&M1.
a) i) Inner surface is -Q, outer surface is +Q.
b) i) E=0 because in conductor charges are on the surface.
ii) use spherical gaussian surface to get E=kQ/r^2
iii) E=0 because in conductor charges are on the surface.
iv) E=kQ/r^2 because the outer shell has 0 net charge.
c) The graph is basically the graph of kQ/r^2 with two chunks erased.
d) First, find potential difference between 10r and infinity through definite integration. Then, dU=qdV=-dKE. Solve this equation for v to get -Qq/20m(pi)(e0)(r)</p>
<p>E&M2.
a) i) 500V
ii) at t=0, it is as if the inductor has R=infinity. Thus, 900V.
iii) At t=0, it is as if the capacitor has R=0. Thus, 4500/11V.
b) I got a graph with a straight line, an exponential decay graph for C with y-int above the straight line, and a log-like graph for L with asymptote greater than the straight line.</p>
<p>E&M3.
a) i) up using right hand rule
ii) use Biot-Savart and algebra/trig to get B=4(u0)(I)/(25R^2)
b) Bnet=2B because the two Bs have same magnitude and direction.
c) Bnet<em>s^2
d) the dot product expression has coswt, so E = Bnet</em>s^2<em>sinwt</em>w
<again, thank="" you="" samuelc="">
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