# Conservation of momentum spaceship problem

1. Nov 25, 2007

### twotoes777

1. The problem statement, all variables and given/known data
A spaceship of mass 2.00×10^6 kg is cruising at a speed of 6.00×10^6 m/s when the antimatter reactor fails, blowing the ship into three pieces. One section, having a mass of 4.60×10^5 kg, is blown straight backward with a speed of 2.10×10^6 m/s. A second piece, with mass 8.30×10^5 kg, continues forward at 1.30×10^6 m/s.

2. Relevant equations
P1+P2+P3 = P(system)
m1(v1)+m2(v2)+m3(v3) = m(total)v(total)

3. The attempt at a solution

So, ... I'm thinking this is a conservation of momentum problem... so at first I used the above equations. First off, I tried to find the mass of the third piece (since it's not given, its necessary) which is: M3 = M_t - (m_1 + m_2) = 7.1 x 10^5 kg.

Then, I substituted all of the answers into the final equation, which gives me 1.4 x 10^7 m/s. However, this is wrong. Can anyone point me to what I'm missing?

Thanks!

2. Nov 25, 2007

### Staff: Mentor

The mass looks right to me. Show exactly what you plugged in to get your answer for the speed.

3. Nov 25, 2007

### twotoes777

Clarification..

Thanks for responding!

Here we go...

m1(v1)+m2(v2)+m3(v3) = m(total)v(total)
(4.6*10^5)(2.1*10^6) + (8.3*10^5)(1.3*10^6) + (7.1*10^5)(v3) = 1.2*10^13

(1.2*10^13) - (9.66*10^11+1.079*10^12) = (7.1 *10^5)(v3)

9.955*10^12 = (7.1 *10^5)(v3)

(v3) = 1.4*10^7 :(

I have to be missing something obvious... :(

4. Nov 25, 2007

### Staff: Mentor

Careful with signs: The 4.60×10^5 kg piece moves backwards.

5. Nov 25, 2007

### twotoes777

Oh my goodness, yes, you are right.

Thanks for the help... I got it now, and that makes sense. I was wondering how it would matter if the piece went forward and not backwards.

Thanks again. :)