# Physics Rocket Question.

1. Oct 11, 2008

### MightyMan11

1. The problem statement, all variables and given/known data
A two-stage rocket is travelling at 1192.0 m/s with respect to the earth when the first stage runs out of fuel. Explosive bolts release the first stage and push it backward with a speed of 32.4 m/s relative to the second stage. The first stage is 2.00 times as massive as the second stage. What is the speed of the second stage after the separation?

2. Relevant equations
Conversation of momentum.

3. The attempt at a solution
I first set the second stage's mass as "m". The mass of the first stage then would be "2m" and the total rocket would have a mass of "3m".
Using the conservation of momentum I made an equation:
momentum of first stage+momentum of second stage= momentum of the rocket
(-32.4)*(2m) + v(m) = 1192(3m)

the m's cancel out and I solved for v:
3,640.8 m/s

however this does not match the answer in my textbook.... :/

2. Oct 11, 2008

### arildno

Note the word "relative". It is highly significant!

3. Oct 11, 2008

### MightyMan11

Wow,
I didn't catch that...
Does that mean then the first stage of the rocket is still moving up, just at a slower speed than the second stage? (1192-32.4=speed of first stage)?
And since the first stage is "slowing down", if momentum is conserved the second stage should "speed up"?

4. Oct 11, 2008

### arildno

It means that, if v1 is the fist stage's velocity, v2 the secondstage,
then we have the equation:
$$v_{1}-v_{2}=-34m/s\to{v}_{1}=v_{2}-34m/s$$
Thus, energy conservation becomes:
$$(2m)(v_{2}-34)+mv_{2}=(3m)1192$$
Solving for v2, we get:
$$3mv_{2}=m(3*1192+2*34)$$,
which you can manage on your own.
And yes, the other stage will still be moving upwards, at approximately 1181m/s

5. Oct 11, 2008

### MightyMan11

Ah.
Thank you very much for your help.