# Conservation of Momentum Spaceship Question

## 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
jbriggs444
Homework Helper
Is the original post accurate? What do you get when you subtract one thousand from ten thousand?

qwertyqwert321
Good point.

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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