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

AI Thread Summary
The discussion revolves around calculating the velocity of a rocket when it runs out of fuel, given its maximum altitude and gravitational acceleration. Participants express confusion over the lack of sufficient information to apply kinematic equations, as there is no time or initial acceleration provided. They explore using energy methods, specifically the conservation of energy equation, but struggle with the variables involved at the point of fuel depletion. Clarifications suggest that knowing the maximum altitude allows for some calculations, yet the "short distance" aspect complicates the determination of initial conditions. Ultimately, the conversation emphasizes the need for a deeper understanding of energy conservation and the implications of the rocket's motion near the surface.
barrylwires
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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.
 
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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!
 
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