|By Callmecollege (Callmecollege) on Wednesday, May 14, 2003 - 06:03 pm: Edit|
here are the free response questions from collegeboard, could somebody please help me with question 1, all parts, to see if i did it right, thanx to anyone who contributes!
|By Callmecollege (Callmecollege) on Wednesday, May 14, 2003 - 06:11 pm: Edit|
actually, could anyone just tell me any answers that they know for a FACT are correct..all is appreciated
|By Brd (Brd) on Thursday, May 15, 2003 - 12:32 am: Edit|
Given: x(t) = 0.5 t^3 + 2 t
(a) The velocity is v(t) = dx/dt = 1.5 t^2 + 2, so at t=0, the velocity is v(0) = 2 m/s. Speed is the magnitude of velocity, so the speed is also 2 m/s
(b)(i) KE = (1/2) m v^2. We calculated v(t) in part (a). Plugging it in with m=100 kg yields KE(t) = (1/2) (100) (1.5 t^2 + 2)^2 = 50* (2.25 t^4 + 6 t^2 + 4) = (112.5 t^4 + 300 t^2 + 200) J
(b)(ii) Use Newton's second law, F = dp/dt = m dv/dt (since mass is constant here). dv/dt = 3 t, so F(t) = 100*3t = 300 t N
(b)(iii) Recall that power is the dot product of force and velocity. Force and velocity both always point in the positive x direction here, so P(t) = (300 t)*(1.5 t^2 + 2) = (450 t^3 + 600 t) W
(c) Work = Integral[F(x) dx]. But alas we have F(t). Intverting x(t) to get t(x) to do that integral is out of the question. We could compute Integral[F(t) dx/dt dt] but that's kind of a pain too. We could equivalently compute Integral[P(t) dt] since power is the work/time. Or we can remember from the work-energy theorem that the work done by non-conservative forces (which is all the work, in this case) on the box is equal to KE(2) - KE(0) = 3000 J (The second and third integrals gives the same answer if you work them out, of course)
(d) I would say greater than. Since the statement mentioned a rough surface, we must assume this net force includes a frictional force opposing the velocity. But then the force from the person on the box must be greater than the net force at all points and times, since the net force and the box velocity always point the same direction. Hence the work done by the person on the box is greater than we amount we calculated in part (c).
|By Ndhawk (Ndhawk) on Thursday, May 15, 2003 - 12:38 am: Edit|
sweet I got the first completely correct
|By Callmecollege (Callmecollege) on Thursday, May 15, 2003 - 12:50 am: Edit|
|By Eurostar (Eurostar) on Thursday, May 15, 2003 - 06:29 am: Edit|
Yes...FRQ 2. What was the checked box?
|By Callmecollege (Callmecollege) on Thursday, May 15, 2003 - 05:57 pm: Edit|
YES, could somebody please help us with the second free response question (spring problem), all parts, i think that that really murdered me and i would like to know the right way to do it
|By Brd (Brd) on Friday, May 16, 2003 - 12:00 pm: Edit|
I spend too much time on you guys.
First off, we might as well calculate the spring constant k, since it will surely come up, and the problem said to answer only in terms of given quantities and physical constants. Taking x to be positive in the "up" direction, the pan is displaced a distance -D. At equillibrium the restoring force from the spring exactly balances the force due to gravity on the pan (we assume a massless spring!): -k*(-D) - Mg = 0. Solving for k yields k = Mg/D N/m .
(a) You can certainly set up and solve the energy problem: (1/2)Mv^2 = MgH which is easy enough. But hopefully you can just immediately write down the answer without even thinking: v = Sqrt[2gH] m/s.
(b) The collision is inelastic, so kinetic energy is not conserved (some of the energy is transformed into internal energy when the clay is deformed). Momentum is conserved, however, so let's use that. The total momentum before the collision is 0 + M*Sqrt[2gH]. After the collision, the pan+clay stick together and have the same velocity, so the momentum immediately afterwards is just 2Mv. Equating and solving for v we find v = Sqrt[gH/2] m/s. Notice the total kinetic energy at this point is (1/2)2M(Sqrt[gH/2])^2 = MgH/2 which is less than the total kinetic energy MgH just before the collision.
(c) As long as you don't pull the spring so much that it's response becomes nonlinear, the undamped period for a spring-mass system is fixed by just the spring constant k and the mass. The period is T = 2pi*Sqrt[2M/k] = 2pi*Sqrt[2D/g] s after plugging in our value of k.
(d) The solutions to the equations of motion are sinusoidal, and the resulting x(t) function is zero at the equillibrium point of the spring-mass system. The velocity, being the derivative of a sinuoidal function, is thus 180 degrees out of phase with the position -- it has a maximum at the equillibrium point. But our mass is now more than it was before (since the clay stuck to the pan), so we have a new equillibrium point. Equating the force due to gravity (on 2M) with the restoring force -kx as before, we find the new equillibrium displacement is x=-2D, or a disance 2D below the empty, unstretched spring equillibrium point. You could also reason that the pan+clay will accelerate from their initial velocity, increasing velocity until they reach this position, at which point the restoring force from the spring becomes greater than the force due to gravity, and the pan+clay begins to decelerate, eventually coming to a stop and then changing direction of velocity.
(e) This question is a gimme. Again, the period for SHM in a spring-mass system (this time with mass just M, not 2M) is T=2pi*Sqrt[M/k]. You don't even have to plug in for k -- this is obviously less than the result T = 2 pi Sqrt[2M/k)] from the situation in part (c).
|By Callmecollege (Callmecollege) on Friday, May 16, 2003 - 10:09 pm: Edit|
|By Callmecollege (Callmecollege) on Saturday, May 17, 2003 - 12:23 am: Edit|
|By Eurostar (Eurostar) on Saturday, May 17, 2003 - 01:28 pm: Edit|
Thank you so much Brd! So if I got all of the questions in FRQ 1 and FRQ 2 (except for part d)...that would be enough for a 5 right there, right? Considering I did equally on the MC...
|By Brd (Brd) on Saturday, May 17, 2003 - 01:57 pm: Edit|
Gosh I am such a pushover. Although, I haven't done problems like this in quite a long while. It's good for me to realize how much I tend to overthink them. Anyways,
Glancing at the problem, it looks like we'll need to calculate the moment of inertia. Since we are assuming a massless bucket, cup and arm, we can just use the formula for point masses: I = Sum[m_i*r_i^2]. In our case, we I = 10(12)^2 + M(2)^2 = 4M +1440.
(a)(i) The data look vaguely proportional to Sqrt[M] (or maybe Log[M]). I imagine this is a gimme question as long as you draw something reasonable.
(a)(ii) By hand, I would probably never bother with anything more complicated than linear interpolation. Drawing a straight line between the first two points, you can eyeball a value of about x ~= 31 m for M = 250 kg.
(b)(i) Easy kinematics. 0 = 15 + 0*t + (1/2)*g*t^2. Solving for t yields t ~= 1.75 s.
(b)(ii) What they want is actually pretty simple. Measuring the angle A from the horizontal down to the small arm, we find the (gravitational) potential energy (with respect to the ground) of the counterweight is Mg(3-2Sin[A]). For the projectile, it is 10*g(3+12Sin[A]). Adding these, and collecting the Sin terms, we get U = 3g(M+10) - 2g*Sin[A](M-60) J.
(b)(iii) Well, let's neglect possible losses to internal energy and assume that the final kinetic energy is equal to the change in potential energy. Then KE =2g*Sin[A](M-60) J. Now, let's further assume that all this kinetic energy goes into launching the projectile. How can we find the linear velocity of the projectile when it leaves the cup? Here we need to use the moment of inertia: KE = (1/2)Iw^2. But the angular frequency w = v/r, so there is how we can find v. Putting all this together (I'll let you work out the details, since HTML sucks for math notation), you get v(M) = r*Sqrt[2KE/I] = 12*Sqrt[g(M-60)/(M+360)] m/s. Note since we are interested at how fast the end with the cup is moving, we use r=12 m.
(c)(i) x=v*t, so just plug in the value of t from (a)(i) and the v from (b)(iii) and get x(M) = 1.75*12*Sqrt[g(M-60)/(M+360)] = 21*Sqrt[g(M-60)/(M+360)] m.
(c)(ii) Using a value M=300 kg in this result, we get x ~= 39.65 m. This is slightly larger than the observed value of 37 m. This makes sense -- in calculating the moment of inertia tensor, we neglected the masses of the bucket, arm, and cup. In reality, it takes energy to move these parts of the system, which would leave less energy for launching the projectile, resulting in a smaller distance thrown than our simple model predicts.
|By Brd (Brd) on Saturday, May 17, 2003 - 02:06 pm: Edit|
No problem, Eurostar. I'll be TAing physics classes soon enough; it's good for me to practice explaining problems like these.
|By Eurostar (Eurostar) on Saturday, May 17, 2003 - 05:03 pm: Edit|
Wow...so you're a college student, Brd?
|By Brd (Brd) on Saturday, May 17, 2003 - 06:21 pm: Edit|
A few times over. I start graduate school in physics in the Fall.
|By Miscanon (Miscanon) on Thursday, May 29, 2003 - 05:01 pm: Edit|
I was just reading this...I think I did something completely different for FR # 3 parts b ii. onwards
|By Miscanon (Miscanon) on Thursday, May 29, 2003 - 05:03 pm: Edit|
brd...if you have the time can you attempt the E & M problems?
|By Miscanon (Miscanon) on Thursday, May 29, 2003 - 05:08 pm: Edit|
okay...I see what I did differently...I forgot to include rotational quantities for part b iii. onwards...now...since c is dependent upon b will they penalize me twice over...or use the ansewr that I put in (b) as the "correct answer"...so I lose points only once? Morever...air resistance is also an answer right brd?
|By Miscanon (Miscanon) on Friday, May 30, 2003 - 04:01 pm: Edit|
|By Miscanon (Miscanon) on Friday, May 30, 2003 - 11:50 pm: Edit|
please....is anyone reading this?
|By Brd (Brd) on Saturday, May 31, 2003 - 12:33 am: Edit|
Miscanon: If I were grading the test, I would say that neglecting air resistance, neglecting losses to internal energy and ignoring the mass of the arm/bucket are all assumptions that would contribute to the model's overestimating the actual measured experimental result. Of course, I'm not grading the thing, but that's what I'd say if I were.
I don't know the answer to your other question about how the later part will be scored.
|By Miscanon (Miscanon) on Saturday, May 31, 2003 - 08:58 am: Edit|
Alright, thank you Brd for your response. I appreciate it.
Report an offensive message on this page E-mail this page to a friend
|Posting is currently disabled in this topic. Contact your discussion moderator for more information.|
|Administrator's Control Panel -- Board Moderators Only|