# Conservation of Linear Momentum of spaceship

1. Oct 1, 2006

### User Name

With the engines off, a spaceship is coasting at a velocity of +230 m/s through outer space. The ship carries rockets taht are mounted in firing tubes, the back ends of which are closed. It fires a rocket straight ahead at an enemy vessel. The mass of the rocket is 1300 kg, and the mass of the spaceship (not including the rocket) is 4.0 x 10^6 kg. The firing of the rocket brings the spaceship to a halt. What is the velocity of the rocket?

I'm pretty sure the equation I have to use is: mAvA + mBvB = (mA + mB)V'

I'm just unsure on which side to set equal to 0, because it's difficult for me to picture this in my mind. I have a hunch that the left side is equal to 0, because the rocket was in motion initially?

So if I did that, it would be 0 = (1300)(v'A) + (4.0 x 10^6)(230)

And then solve for V'A.

Also, I'm stuck on this problem:

Batman (91 kg) jumps straight down froma bridge into a boat (mass = 510 kg) in which a criminal is fleeing. The velocity of the boat is initially +11 m/s. What is the velocity of the boat after Batman jumps into it?

I tried to use linear momentum equations but none of them get me the right answer of 9.3 m/s ~_~' Any ideas on what I am supposed to do?

Last edited: Oct 1, 2006
2. Oct 1, 2006

### Noein

Is the total momentum of the system (spaceship + rocket) equal to zero before the rocket is fired? How about after the rocket is fired?

3. Oct 3, 2006

### speckledot

the formula you have to use is:
mAvA + mBvB = mA'vA' + mB'vB'

batman has no vA so vA is 0.
[(510m/s)(11)]=(91+510)(vB')

4. Oct 3, 2006

### radou

(mass of the spaceship + mass of the rocket)*velocity of the spaceship = mass of the rocket*velocity of the rocket.

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