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The E&M free response was very very ridiculous, especially #2 where you had to plug in different devices at the terminal and find the current in the parallel resistor. I was completely lost.

For anyone that took it, what did you think?

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- Thread starter sadakaa
- Start date

- #1

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The E&M free response was very very ridiculous, especially #2 where you had to plug in different devices at the terminal and find the current in the parallel resistor. I was completely lost.

For anyone that took it, what did you think?

- #2

Nabeshin

Science Advisor

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- #3

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I thought the E&M FRQ's were really straightforward. #1 was ridiculously simple. #2 was just like any other circuit problem, didn't even require much analysis. I think I screwed up #3's first part though. Forgot to square root something.

Everyone was complaining about the mechanics FRQ's but I thought they weren't terribly difficult. Knowing E&M actually helped me with the first problem.

Anyway, we probably shouldn't talk about it until two days.

Everyone was complaining about the mechanics FRQ's but I thought they weren't terribly difficult. Knowing E&M actually helped me with the first problem.

Anyway, we probably shouldn't talk about it until two days.

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So, overall, challenging AP exams, but I think if someone has good preparation over the year (unfortunately I hear of all the AP Physics teachers who have no idea what they're doing, and sometimes skip over major concepts) they are not at all insurmountable. The generous curve assures that: in 2004 you only needed a 54.4% in mechanics for a 5! And E&M traditionally is curved even more than mechanics!

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"wht happens to the enrgy/voltage of a capacitor as blah blah"...this question took soo much of my time!

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Okay, 48 hours have passed, so we can now discuss the FRQs.

here is the first mechanics FRQ:

1. A skier of mass M is skiing down a frictionless hill that makes an angle θ with the horizontal, as shown in the diagram. The skier starts from rest at time t = 0 and is subject to a velocity-dependent drag force due to air resistance of the form F= -bv, where v is the velocity of the skier and b is a positive constant. Express all algebraic answers in terms of M, b, θ , and fundamental constants.

(a) On the dot below that represents the skier, draw a free-body diagram indicating and labeling all of the forces that act on the skier while the skier descends the hill.

(b) Write a differential equation that can be used to solve for the velocity of the skier as a function of time.

(c) Determine an expression for the terminal velocity V(sub T) of the skier.

(d) Solve the differential equation in part (b) to determine the velocity of the skier as a function of time, showing all your steps.

I have a couple of questions. For the free-body diagram, should the drag force point back horizontally or back up the slope? Also, what's the differential equation in part (b)? How do you solve it?

Here's what I got, but I think it's wrong.

(b) m(d^2x/dt^2)=mgsinθ-b(dx/dt)

(c) bv=mgsinθ ===> v=(mgsinθ)/b

(d) ma=mgsinθ-bv ===> mv=mgt*sinθ-bx ===> v=(mgt*sinθ -bx)/m

- #7

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The horizontal uniform rod shown above has length 0.60 m and mass 2.0 kg. The left end of the rod is attached to a vertical support by a frictionless hinge that allows the rod to swing up or down. The right end of the rod is supported by a cord that makes an angle of 30 with the rod. A spring scale of negligible mass measures the tension in the cord. A 0.50 kg block is also attached to the right end of the rod.

(a) On the diagram below, draw and label vectors to represent all the forces acting on the rod. Show each force vector originating at its point of application.

(b) Calculate the reading on the spring scale.

(c) The rotational inertia of a rod about its center is (1/12)ML^2 , where M is the mass of the rod and L is its length. Calculate the rotational inertia of the rod-block system about the hinge.

(d) If the cord that supports the rod is cut near the end of the rod, calculate the initial angular acceleration of the rod-block system about the hinge.

Here's what I got:

(a) If I made a mistake, I think I made it here. I only drew vectors for F(sub T), W(sub rod), and W(sub block). Here's my question: Does the contact point at the hinge provide a force? If so, everything else I did was wrong.

(b) ma=0=F(sub T) - W(sub rod) - W(sub block) ===> F(sub T) = W(sub rod) + W(sub block) = 25N.

(c) I = I(sub rod) + I(sub block) = I(sub rod, center) + M(1/2L)^2 + M(sub block)L^2 = 1/3ML^2+M(sub block)L^2=.42 kg*m^2

(d) torque=I*alpha ===> alpha=torque/.42 kg*m^2

torque = L(F(sub T) - 1/2W(sub rod) - W(sub block))=(.6m)(25N - 10 N - 5 N) = 6 N*m ===> alpha = 6/.42=14.29rad/s^2

- #8

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Here's the third FRQ:

In an experiment to determine the spring constant of an elastic cord of length 0.60 m, a student hangs the cord

from a rod as represented above and then attaches a variety of weights to the cord. For each weight, the student

allows the weight to hang in equilibrium and then measures the entire length of the cord. The data are recorded

in the table below:

Weight (N) 0 10 15 20 25

Length (m) 0.60 0.97 1.24 1.37 1.64

(a) Use the data to plot a graph of weight versus length on the axes below. Sketch a best-fit straight line through the data.

(b) Use the best-fit line you sketched in part (a) to determine an experimental value for the spring constant k of the cord.

The student now attaches an object of unknown mass m to the cord and holds the object adjacent to the point at

which the top of the cord is tied to the rod, as represented above. When the object is released from rest, it falls

1.5 m before stopping and turning around. Assume that air resistance is negligible.

(c) Calculate the value of the unknown mass m of the object.

(d) i. Calculate how far down the object has fallen at the moment it attains its maximum speed.

ii. Explain why this is the point at which the object has its maximum speed.

iii. Calculate the maximum speed of the object.

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Part (b) was the easiest thing ever: 0=mg-ky ===> ky=mg, so k is the the slope of the graph. However, I can't think of how to do parts (c) and (d) right now. I haven't taken the test yet, since my school ordered the wrong test. I'm gonna take it during the late testing, and I'd like to use these as practice. Any ideas on how to do these FRQs?

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For reference, here are the original problems:

http://apcentral.collegeboard.com/apc/public/repository/ap08_physics_c_mech_frq.pdf

Drag force goes back up the slope. The force equation is really a vector equation, [itex]\vec{F} = -b \vec{v}[/itex].*v* is down the slope, so F must point exactly up the slope due to the negative sign.

(b) You're solving for velocity, so you need the equation in terms of v, not x. And solving (this is exactly like an RC or LC circuit if you've done E&M. I only remembered how to do it for the test because we recently covered solving this type of diff eq. in E&M.):

[tex]m\frac{dv}{dt} = m g \sin(\theta) - b v[/tex]

[tex]\frac{dv}{dt} = g \sin(\theta) - \frac{b}{m} v[/tex]

Seperating variables (divide both sides by RHS and multiply the dt to the right) and integrating both sides:

[tex]\int_0^{v_f} \frac{dv}{g \sin(\theta) - \frac{b}{m} v} = \int_0^{t}dt[/tex]

u-substitution:

[tex]u = g \sin(\theta) - \frac{b}{m} v[/tex]

[tex]du = -\frac{b}{m}[/tex]

Note that I put the negative sign to the right. This saves a bunch of annoying algebra later on (I forgot to do this during the test, wasted time). I'm also gonna move the constant introduced from*du* to the right.

[tex]\left[ \log\left(g \sin(\theta) - \frac{b}{m} v\right) \right]_{0}^{v_f} = \log\left(\frac{g \sin(\theta) - \frac{b}{m} v}{g \sin(\theta)}\right) = -\frac{b}{m}t [/tex]

Do the algebra yourself, I'm too lazy to finish.

http://apcentral.collegeboard.com/apc/public/repository/ap08_physics_c_mech_frq.pdf

here is the first mechanics FRQ:

I have a couple of questions. For the free-body diagram, should the drag force point back horizontally or back up the slope? Also, what's the differential equation in part (b)? How do you solve it?

Here's what I got, but I think it's wrong.

(b) m(d^2x/dt^2)=mgsinθ-b(dx/dt)

(c) bv=mgsinθ ===> v=(mgsinθ)/b

(d) ma=mgsinθ-bv ===> mv=mgt*sinθ-bx ===> v=(mgt*sinθ -bx)/m

Drag force goes back up the slope. The force equation is really a vector equation, [itex]\vec{F} = -b \vec{v}[/itex].

(b) You're solving for velocity, so you need the equation in terms of v, not x. And solving (this is exactly like an RC or LC circuit if you've done E&M. I only remembered how to do it for the test because we recently covered solving this type of diff eq. in E&M.):

[tex]m\frac{dv}{dt} = m g \sin(\theta) - b v[/tex]

[tex]\frac{dv}{dt} = g \sin(\theta) - \frac{b}{m} v[/tex]

Seperating variables (divide both sides by RHS and multiply the dt to the right) and integrating both sides:

[tex]\int_0^{v_f} \frac{dv}{g \sin(\theta) - \frac{b}{m} v} = \int_0^{t}dt[/tex]

u-substitution:

[tex]u = g \sin(\theta) - \frac{b}{m} v[/tex]

[tex]du = -\frac{b}{m}[/tex]

Note that I put the negative sign to the right. This saves a bunch of annoying algebra later on (I forgot to do this during the test, wasted time). I'm also gonna move the constant introduced from

[tex]\left[ \log\left(g \sin(\theta) - \frac{b}{m} v\right) \right]_{0}^{v_f} = \log\left(\frac{g \sin(\theta) - \frac{b}{m} v}{g \sin(\theta)}\right) = -\frac{b}{m}t [/tex]

Do the algebra yourself, I'm too lazy to finish.

(a) Yes. The rod must be in equilibrium. Therefore all torques and forces must balance out. The tension force has an x-component that points left. Therefore the hinge must only provide a force to the right that balances it out. It doesn't affect any of the later answers though. For part d, the tension force torque should be multiplied by sin(30 deg). In fact, there shouldn't even be a tension force, since the rope is cut.(a) On the diagram below, draw and label vectors to represent all the forces acting on the rod. Show each force vector originating at its point of application.

If I made a mistake, I think I made it here. I only drew vectors for F(sub T), W(sub rod), and W(sub block). Here's my question: Does the contact point at the hinge provide a force? If so, everything else I did was wrong.

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- #11

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Continuing, forum downtime made me miss the edit window.

#3:

Part c is definitely a conservation of energy problem. At the top, it has mgh potential energy, and at the lowest point, it has maximum spring potential energy and zero gravitational potential energy. Both points have no kinetic energy. This allows us to relate mass and height it fell.

Part d:

[tex]0 = kx^2 + mgx + \frac{1}{2} m v^2[/tex]

To maximize v based on x, solve for v, set [itex]dv/dx = 0[/itex] and solve for x. The denominator of [itex]dv/dx[/itex] is gonna be pretty long but I just ignored it and found the numerator only.

[tex]-mg - 2 k x = 0[/tex]

Then just plug x back into the equation for v.

Hope this helps. And I'm not 100% I'm right for any of this, of course.

It's pretty surprising. I think two of the hardest parts of the FRQ's (#1 solving the diffeq, #3 finding the spring compression that maximized velocity) were pretty standard problems that any textbook should have (or teacher have you do).

Now back to studying for economics tomorrow..

#3:

Part c is definitely a conservation of energy problem. At the top, it has mgh potential energy, and at the lowest point, it has maximum spring potential energy and zero gravitational potential energy. Both points have no kinetic energy. This allows us to relate mass and height it fell.

Part d:

[tex]0 = kx^2 + mgx + \frac{1}{2} m v^2[/tex]

To maximize v based on x, solve for v, set [itex]dv/dx = 0[/itex] and solve for x. The denominator of [itex]dv/dx[/itex] is gonna be pretty long but I just ignored it and found the numerator only.

[tex]-mg - 2 k x = 0[/tex]

Then just plug x back into the equation for v.

Hope this helps. And I'm not 100% I'm right for any of this, of course.

It's pretty surprising. I think two of the hardest parts of the FRQ's (#1 solving the diffeq, #3 finding the spring compression that maximized velocity) were pretty standard problems that any textbook should have (or teacher have you do).

Now back to studying for economics tomorrow..

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- #12

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The second problem is very unlike the typical rotation problems you usually see on the AP exam. However it's not atypical in the sense of being harder, it's atypical in the sense of being easier. The rotation problems are usually either collision problems or dynamics problems. Until the last part this was a statics problem. In the class that I taught, we did a lab just like that problem... and boy do I hope that my students remembered it!

Most of the third problem is a breeze provided that you can follow instructions, make plots and find best fit lines. I make my students use a ruler just because even if you think your line is straight and thin enough, the reader might disagree. Play it safe. Especially since you have to use your best fit line to find the slope.

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(e) On the axes below, sketch a graph of the acceleration a of the skier as a function of time t, and indicate the initial value of a. Take downhill as positive.

I just made a piecewise graph: a straight line with negative slope from the initial a to some point on the t-axis, and then a straight line at a=0 from then on.

As for the initial value of a, I used mgsinθ, since the initial velocity is zero, making the drag force zero.

- #14

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1. b) dv/dt=gsinθ-(b/m)v

c) v(sub T)=Mgsinθ/b

d) v=(mgsinθ-me^(-bt/m))/b

2. b) F(sub T) = 50N

c) I = .42 kg*m^2

d) alpha = -21.4 rad/s^2

3. b) k = slope

- #15

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(e) On the axes below, sketch a graph of the acceleration a of the skier as a function of time t, and indicate the initial value of a. Take downhill as positive.

I just made a piecewise graph: a straight line with negative slope from the initial a to some point on the t-axis, and then a straight line at a=0 from then on.

As for the initial value of a, I used mgsinθ, since the initial velocity is zero, making the drag force zero.

You screwed the plot up, but I bet you'll get full credit. Your plot reflects your wrong solution, and they grade based on consistency and not correctness (that is to ensure that they don't penalize repeatedly for the same mistake).

The acceleration is an exponential decay. You could have reasoned it through even if you did the math wrong before because as you reach terminal velocity the acceleration goes to zero, also the equation never changes so the curve needs to be everywhere continuous. Initially since it starts from rest, there is no drag force, and the initial value should be simply gsin. If you drew a curve with everywhere positive concavity that was continuous everywhere, started at gsin and asymptotically reaches 0 at infinity you would have had it.

Limiting behavior is the key to do all of those sketch a plot or identify a plot without breaking a sweat.

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- #17

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Yeah I screwed up my graph on the test. Drew a linearly downsloping line and made it zero after it hit the axis. Terminal velocity is never actually achieved, acceleration actually never hits zero, only in the limit as time goes to infinity.

They want acceleration as a function of time, not as a function of velocity.

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Ohhh... I see what you mean. Thanks!

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for number one, how would you determine the charge induced on the inner and outer surfaces?

is it simply a matter of surface area?

im not allowed to post the url, but the questions are available at apcentral.collegeboard.com/apc/public/repository/ap08_physics_c_em_frq.pdf

- #20

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for number one, how would you determine the charge induced on the inner and outer surfaces?

is it simply a matter of surface area?

im not allowed to post the url, but the questions are available at apcentral.collegeboard.com/apc/public/repository/ap08_physics_c_em_frq.pdf

The positive inner sphere polarizes the outer sphere, pulling negative charges to the inner surface, and leaving positive charges on the outer surface.

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from origin to a

=zero

from a to b

=zero

from b to c

=zero

from c to d

decreasing 1/r^2

this is the only thing i can conclude based on the field within a charged sphere being equal to zero

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- #23

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how were the FR questions in the late physics C mechanics exam?

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they were easier than expected. and i got that answer too! what about the other questions?

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