URGENT. Simple satellite / rocket seperation problem... is my thinking wrong?

[Tschew]

Rocket mass M-kM and satellite kM (k < 0) are seperated by an explosion which releases energy Q and lets rocket M-kM come to a stop relative to the observer and satellite kM continue at velocity v. Initial velocity u.

One has to show that v² = 2Q / kM(1-k) the solution of which is rather trivial, when using the following relationship:

(1) 1/2 (M-kM) u² - Q = 0

and then

(2) 1/2 kM u² + Q = 1/2 kM v²

replacing u² from (1)

However, first of all, if all energy is used up to decelerate the rocket, where does the extra kinetic energy of the satellite come from? Shouldn't half the energy go to the rocket and half to the satellite? (Of course, all this considering an instantaneous transfer of energy)

Also, since momentum should be conserved then shouldn't this be true?

M u = kM v [since M-kM comes to a halt]

Now, replacing everything with numbers:

M=5 k=0.2 kM=1 u = 3

1/2*(4)*9 = Q = 18

1/2*1*9 + 18 = 1/2*1*v²

9+36=v²

45 = v²

v = 6.7

--> 5 * 3 = 1 * 6.7 ? NOPE

Also: v² = 2Q / kM(1-k) = 36 / 0.8 = 45 which is certainly the same as otherwise, but still momentum is not conserved.. although it should be right?

Wouldn't it be much more useful to say:

M*u = kM*v

v = u/k ?

Please clarify the issue as I seem to have serious problems understanding it even though it is a simple linear momentum / energy problem! grr!

Thanks.

[Doc Al]

One has to show that v² = 2Q / kM(1-k) the solution of which is rather trivial, when using the following relationship:

(1) 1/2 (M-kM) u² - Q = 0

and then

(2) 1/2 kM u² + Q = 1/2 kM v²

replacing u² from (1)
I'm not sure where you get those equations, but they are incorrect. Note that if you add them, they imply that the KE of the system did not change! What happened to Q?
Also, since momentum should be conserved then shouldn't this be true?

M u = kM v [since M-kM comes to a halt]
Absolutely. Start with this equation for momentum conservation, then add the fact that the KE of the entire system is increased by Q. (Make the assumption that all the energy of the explosion goes into the mechanical energy of the system.)