Kinematics, work-energy, conservation, more, less?

Join the discussion
Registration is free. Ask a follow-up in this thread, or start your own.
4 replies · 2K views
barrylwires
Messages
4
Reaction score
0

Homework Statement


Rocket launched runs out of fuel after accelerating "for a short distance," coasts to a maximum altitude, then falls back. Known: rocket mass, max. altitude, g. Find velocity when rocket runs out of fuel.


Homework Equations


There doesn't seem to be enough information to use kinematic eqs: no time for anything, no initial acceleration; only maximum height. We know the rocket force must be at least mg, but it's clearly more, since it coasts upwards.

The Attempt at a Solution


Seems to need energy methods, but I can't see the way. E=K+U, but E=U (at top) doesn't equal E at bottom, where K and U are both zero. And what can we know about K+U at yfuelanyway? Do I need work, force? Do I need Wnonconservative=∆K+∆U? Does momentum enter this? I know that force changes from rocket to gravity at yfuel, but I'm not sure what to do with that. I could use a shove in the right direction.

If I try K1+U1=K2+U2 at y1=no fuel and y2=max. altitude, I just wind up with v2=2g(y2-y1), the kinematic eq., and I have two unknowns. I need another expression with one of these variables in it. Momentum? Kinematics? Thanks.
 
Physics news on Phys.org
barrylwires said:

Homework Statement


Rocket launched runs out of fuel after accelerating "for a short distance," coasts to a maximum altitude, then falls back. Known: rocket mass, max. altitude, g. Find velocity when rocket runs out of fuel.

Homework Equations


There doesn't seem to be enough information to use kinematic eqs: no time for anything, no initial acceleration; only maximum height. We know the rocket force must be at least mg, but it's clearly more, since it coasts upwards.

The Attempt at a Solution


Seems to need energy methods, but I can't see the way. E=K+U, but E=U (at top) doesn't equal E at bottom, where K and U are both zero. And what can we know about K+U at yfuelanyway? Do I need work, force? Do I need Wnonconservative=∆K+∆U? Does momentum enter this? I know that force changes from rocket to gravity at yfuel, but I'm not sure what to do with that. I could use a shove in the right direction.

If I try K1+U1=K2+U2 at y1=no fuel and y2=max. altitude, I just wind up with v2=2g(y2-y1), the kinematic eq., and I have two unknowns. I need another expression with one of these variables in it. Momentum? Kinematics? Thanks.

Welcome to PF.

If you know the max altitude you know a lot don't you?

Vi2 = 2*g*x

since you know the max altitude is where vf = 0

The only thing off is the "very short distance".
 
Hello and thank you. Yes. y2 = ymax does tell me a lot, but I don't see what resources I have to determine either a) the distance y1 is above the ground = y0 or b) what time the rocket ran out of fuel. I do see that v2 and v0= 0 and that v12 = -2g(y2-y1). Are you saying there is a way to determine y1?
 
barrylwires said:
Hello and thank you. Yes. y2 = ymax does tell me a lot, but I don't see what resources I have to determine either a) the distance y1 is above the ground = y0 or b) what time the rocket ran out of fuel. I do see that v2 and v0= 0 and that v12 = -2g(y2-y1). Are you saying there is a way to determine y1?

The problem statement says short distance.

If you have Δy << y, and v2 = 2*g*y

then I'd say (y/(y-Δy))1/2 ≈ 1
 
Great. Got it.
When the problem said "Assume it was not far from the surface [of the moon] at that time [when it ran out of fuel]," I thought the point might have been to use mgMOON, rather than GmRmM/r2, but I'm happy to know that there isn't a way to find an actual value from the information given.
Many thanks!