Conservation of Momentum Spaceship Question

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The discussion centers on a conservation of momentum problem involving a spaceship with a mass of 10012 kg. After ejecting a 1000 kg part at a speed of 112 m/s, the speed of the remaining part is calculated using the formula for conservation of momentum. The correct calculation reveals that the remaining part, with a mass of 9012 kg, moves at a speed of -12.43 m/s, indicating it moves in the opposite direction to the ejected part. The initial miscalculation of the remaining mass was corrected from 12 kg to 9012 kg.

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Homework Statement


The mass of a spaceship is 10012 kg. The spaceship is at rest. Then one part of the ship with a mass of 1000 kg is ejected and emerges with a speed of 112 m/s. What is the speed of the other part?

Homework Equations

The Attempt at a Solution


I tried:
pi=pf
mivi=mfv2
10012 (vi)= 1000 kg * 112 m/s
solved for vi and got 11.19 m/s.
Am I correct? or did i need to assume the other part of the spaceship is 12 kg?[/B]
 
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You have to apply conservation of momentum here, where pf is the sum of the momenta of the two pieces of the ship.
 
so would it be :
pi=pf
mivi = m1v1 + m2v2
(10012 kg)(0 m/s) =m1v1 + m2v2
0= (1000 kg)(112 m/s)+ (12 kg) (v2)
v2 = -9333.3 m/s
 
Last edited:
That's more like it! Include some directions and you are golden. :)
 
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Is the original post accurate? What do you get when you subtract one thousand from ten thousand?
 
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Good point.
 
jbriggs444 said:
What do you get when you subtract one thousand from ten thousand?
oh man... i totally messed up there, thank you so much for the correction! m2 will then = 9012
(10012 kg)(0 m/s) =m1v1 + m2v2
0= (1000 kg)(112 m/s)+ (9012 kg) (v2)
v2 = -12.43 m/s
 
Last edited:

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