- #1
Gaderath
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Homework Statement
Two mass m1 and m2, connected by a massless spring of force constant k, are at rest in equilibrium on a smooth horizontal table. A velocity of magnitude v and direction away from m2 is imparted to m1.
a.) Find the speed of the center of mass.
b.) Determine the speed of m2 relative to the table at the instant the spring again becomes unstretched.
c.) Find the period of the oscillatory motion.
Homework Equations
Vcm = (m1*v1 + m2*v2)/(m1 + m2)
v1= v1cm +Vcm and likewise for v2
v2 = Vcm -m1v12/(m1+m2)
v12 = velocity of 1 with respect to 2 in my notation here ^^
Vcm = velocity of the center of mass in the lab frame
v1cm = velocity of mass 1 in the center of mass reference frame
The Attempt at a Solution
a.) This one I got, Vcm = m1*V/(m1+m2)
It follows from Vcm's definition above, and the conservation of momentum law for the system (m1V = m1*v1 +m2*v2) and subbing that into the numerator of the definition of Vcm.
b.) This is where I'm really stuck. After I derived v2 = Vcm - m1v12/(m1+m2), which I'm confident in, I got hung up trying to figure out the velocity of 1 with respect to 2. In the CM reference frame the total momentum of the system is 0, and so using conservation of momentum m1*v1cm = -m2*v2cm. Sorry I don't know how to do subscripts here by the way, I know that notation has got to look confusing. I've more or less been running in circles since I got this far. I thought there had got to be something I could assert from the physics and the actual motion of the system to come up w/ the velocity of 1 with respect to 2 and use "v2 = Vcm -m1v12/(m1+m2)" above with my result from a.) to get it. But as I thought about it I don't think I CAN do that, as the velocity of the 2 masses wouldn't be some constant thing based on what little info I'm given at the start.
c.) I haven't looked at this one too much, but I don't know how much trouble I'll have when I get to it after b.), so any tips/advice would be great, but I don't know how much I need it (yet anyway... :) ) I expect it to be fairly normal result from a simpler problem, but with the mass being the reduced mass, m1*m2/(m1+m2), but that's just my gut, I don't know if that's true or how to prove it yet.
So far I've taken "unstretched" to mean neither compressed or stretched, rather that completely compressed, which might actually have made the problem simpler...
Homework Statement
A rocket traveling through the atmosphere experiences a linear air resistance -kv. Find the differential equation of motion when all the other external forces are negligible. Integrate the equation and show that if the rocket starts from rest, the final speed is given by v = V*(alpha)*(1-(m/m0)^(1/(alpha))) where V is the relative speed of the exhaust fuel, alpha = absolute value( (1/k)*(dm/dt) = constant, and m0 is the initial mass of the rocket plus fuel, m being the final mass of the rocket.
Homework Equations
Not many, dp/dt = net external forces is the only one I really use (maybe I should be using others?)
The Attempt at a Solution
I started w/ dp = f*dt, where f is the external force(s). I integrated the momentum from time t to time t+dt, and got (M-absolutevalue(dM))(v+dv)+Vcm*absolutevalue(dM)-Mv
where Vcm = -V+v. I worked this down to:
f = M*dv/dt - V|dM/dt|, with the last term being the rockets thrust. The || is the absolute value again :), seemed like a better way to write it...
f here is again the net external force, which in this problem will be -kv. Knowing the mass is being lost, the sign on dM/dt is negative, so for the next step I dropped the || and added a - in front.
-V*dM/dt = M*dv/dt + kv.
I believe this is the differential equation the question asks for, however, as to integrating it... what in the world?! I really don't know much about solving differential equations, the only examples in class we've had have been separation of variables (he did a similar problem, but with no external forces where separation of variables worked really nicely). Here separation of variables does NOT work, and dividing through by M puts an M on the kv term, and dividing by v puts a v on the M terms. Theres no way *to* separate them as I see it. I also don't understand how I'm supposed to differentiate it and somehow still have a dm/dt term in the final answer (in alpha). I'm ok with there being an m and m0 in it, as I suspect it's from, the limits of the integration being m0 and m, and probably in a natural log function, so the subtraction can be written in the answer as a division like it is. But I don't know if that's the case here, as so much of this part is confusing me. The fact that he tells us in the problem that alpha = |(1/k)(dm/dt)| and that its CONSTANT seems like a huge hint to me. The rate of change of the mass is constant then, but the mass still isn't and so that doesn't help w/ the unseparability of the equation.
Heh, this assignments pretty long, and I've felt so good everytime I've finished one of the problems, lol, but it doesn't last when I get so stonewalled on these 2. I've talked to about 8 others in the class (of roughly 20 people) and they've all been just as clueless and on these same parts (or earlier)...