The expelling of fuel from a rocket (Momentum)

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SUMMARY

The discussion revolves around calculating the final velocity of a model rocket after expelling fuel, utilizing the principle of conservation of momentum. A 4.00 kg rocket expels 63.0 g of fuel at a speed of 565 m/s. The correct approach involves setting up the momentum equation as (rocket mass × initial velocity) + (fuel mass × fuel speed) = (rocket mass × final velocity) + (fuel mass × (fuel speed - final velocity)). The final velocity of the rocket after the fuel has burned is determined to be 8.89 m/s.

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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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