The expelling of fuel from a rocket (Momentum)

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Homework Help Overview

The discussion revolves around a model rocket's momentum during fuel expulsion, specifically focusing on the conservation of momentum principle. The problem involves calculating the rocket's velocity after burning a specified amount of fuel.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the application of conservation of momentum, questioning the interpretation of the fuel's speed relative to the rocket. There is discussion about the initial conditions and the relationship between the rocket's and fuel's velocities.

Discussion Status

Some participants have provided guidance on interpreting the problem, particularly regarding the relative speeds of the rocket and the expelled fuel. Multiple interpretations of the momentum equation are being explored, and there is an ongoing dialogue about the correct setup for the calculations.

Contextual Notes

Participants are working under the assumption that external forces such as gravity and air resistance can be ignored, which may influence their calculations and interpretations.

tyro008
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Homework Statement


A 4.00kg model rocket is launched, expelling 63.0g of burned fuel from its exhaust at a speed of 565m/s. What is the velocity of the rocket after the fuel has burned. Hint: Ignore the external forces of gravity and air resistance.


Homework Equations


Conservation of Momentum?


The Attempt at a Solution


I tried using the equation for the conservation of momentum by doing M1Vi=M-0.063Vf... but that did not work. I am stuck.
 
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tyro008 said:
A 4.00kg model rocket is launched, expelling 63.0g of burned fuel from its exhaust at a speed of 565m/s. What is the velocity of the rocket after the fuel has burned.

I tried using the equation for the conservation of momentum by doing M1Vi=M-0.063Vf... but that did not work. I am stuck.

Hi tyro008! :smile:

I think you've assumed that 565 m/s is the fuel's actual speed … but the question probably means the speed relative to the rocket.
 
Do you mean that this is the rocket's speed?
 
tyro008 said:
Do you mean that this is the rocket's speed?

No … the rocket starts at speed 0, and so the fuel is at speed 565.

But later the rocket is at speed v, say, and the fuel is at speed (565 - v). :smile:
 
so i did
(rocketmass X 0m/s) + (fuelmass X 565 m/s) = (rocketmass X v) + (fuelmass X (565 - v)
would this be right??
 
(this would be to find v)
 
answer - 8.89m/s...Is this right?
 
hey yeah it is! thanks so much to all you guys for helping me, i REALLY appreciate it! :]
 

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