Rocket

by ktd
Tags: rocket
 P: 15 Here's another fun one: A 20,000 kg rocket has a rocket motor that generates 3.0*10^5 N of thrust. At an altitude of 5000 m the rocket's acceleration has increased to 6.0 m/s^2. What mass of fuel has it burned? I've already found the initial acceleration to be 5.2 m/s^2, then I used this formula: (Fthrust) - (Mnew)(g) = (Mnew)(a) To solve for m, I think I got the wrong algebraic process wrong - m = (Fthrust)/(g+a) because my answer isn't right!
 P: 640 Theyre asking what mass of fuel was spent..hmm I think I remeber something about this where you cannot assume F = ma, because it doesnt. F = dp/dt (change in momentum over time) F = d(m(t)*v(t))/dt = [v(t) * dm(t)/dt] + [m(t) * dv(t)/dt] assuming we can remember v = v(t) and m = m(t) F = v*dm/dt + m*a Ftotal = Fthrust-Fgravity v(t)*dm(t)/dt + m(t)*a = Fth - m(t)*g I think you do something like that. Im not completely sure but you have to remember that the mass is changing wrt time.
 P: 15 So what if I don't have a time given? (which I don't)
P: 640

Rocket

thats what im working on, you ARE given x, so you can change variables to x... but its quite tough. It looks like itll be a differential equatin, and you have initial/final data to fit into it.

Ok, I let M_tot = m + u where m is mass of the rocket, and u is mass of the fuel.
m is constant (20,000) and u is a function of time. keep this in mind.

Fth - Mg = d/dt ( M*v) = d/dt ((m+u)*v) = u' v + a (m+u)
= Fth -(m+u) * g

Fth = u' v + (a+g)*(m+u)
where u,v,a are functions of t

at t=0, v(0) =0, a(0) = ao, u(0) = uo
the inital values
Fth = (ao+g)*(m+uo)
2 unknowns...
if they meant the intial mass of the rocket + fuel was 20,000 this would be easier.
maybe we should assume that...
Ill keep working and report what I get.
 P: 15 Nevermind, I figured it out! I did use my original equation, and that answer is the "new" mass. So, by subtracting this number from the "original" mass, the answer is 1013 kg. Yay!

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