# Conservation of Momentum Spaceship Question

• qwertyqwert321
In summary, the mass of a spaceship is 10012 kg at rest. When a part with a mass of 1000 kg is ejected with a speed of 112 m/s, the speed of the other part is approximately -12.43 m/s. This is found by applying the conservation of momentum equation, where the initial momentum is equal to the sum of the momenta of the two pieces of the ship.

## 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?

## 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]

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

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That's more like it! Include some directions and you are golden. :)

qwertyqwert321
Is the original post accurate? What do you get when you subtract one thousand from ten thousand?

qwertyqwert321
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

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