For reference, here are the original problems:
http://apcentral.collegeboard.com/apc/public/repository/ap08_physics_c_mech_frq.pdf
adartsesirhc said:
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, \vec{F} = -b \vec{v}.
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.):
m\frac{dv}{dt} = m g \sin(\theta) - b v
\frac{dv}{dt} = g \sin(\theta) - \frac{b}{m} v
Seperating variables (divide both sides by RHS and multiply the dt to the right) and integrating both sides:
\int_0^{v_f} \frac{dv}{g \sin(\theta) - \frac{b}{m} v} = \int_0^{t}dt
u-substitution:
u = g \sin(\theta) - \frac{b}{m} v
du = -\frac{b}{m}
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 going to move the constant introduced from
du to the right.
\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
Do the algebra yourself, I'm too lazy to finish.
(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.
(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.
), so yeah I found it difficult, but I think if I had studied diligently throughout the year the exams wouldn't be too difficult to score a 5 on.
