Spring Projectile problem: center of mass

[Zenshin]

Hello everyone. I'm burning my head at this one, maybe someone could offer some insight? Here's the deal: imagine two masses, m1 and m2, united by a massless spring, of length l (relaxed).

m1
|
| _ > spring, at length l (relaxed)
|
___m2____ ground

The system has this config at rest. Then, it'f given a velocity v , upward, to the mass m1, and the question is: what is the position of the masses in any given time? I've figured out that I've to break the problem in two parts: find out the center of mass motion and then determine the position of the masses relative to it. However, how can I do that? I've already found out the center of mass motion, but I can't find the position of the masses relative to it. I assume it's a oscillatory motion, but I can't find it's amplitude. The only thing I have is the initial velocity of the mass m1, nothing else. Please, any advice would be HIGHLY useful hehehe

Thanks in advance!

 

[arildno]

All right:
Let us set up the individual laws of motion here:
$$-m_{1}g-k(x_{1}(t)-x_{2}(t)-l)=m_{1}a_{1}$$
$$-m_{2}g+k(x_{1}(t)-x_{2}(t)-l)=m_{2}a_{2}$$
where $$x_{1},x_{2}$$ are the respective positions of the masses, with associated accelerations $$a_{1},a_{2}$$
This is a second-order linear system with constant coefficients which is readily solvable.

 

[klmdad]

since the motion is vertical, we use y instead of x (y positive upward). We divide this soution into two parts. 0.5mv1^2 + mgy1 = 0.5mv2^2 + mgy2
0 + mgh = 0.5mv2^2 + 0
v2 = squ root2gh)

 

[arildno]

since the motion is vertical, we use y instead of x (y positive upward). We divide this soution into two parts. 0.5mv1^2 + mgy1 = 0.5mv2^2 + mgy2
0 + mgh = 0.5mv2^2 + 0
v2 = squ root2gh)

Whatever are you talking about?
Where, for example, have you included the potential energy of the system contained in the spring?

Zenshin:
While I believe that the equations given are the ones you were asked to find, nevertheless it is important to realize that these equations are only valid under the assumption NO NORMAL FORCE ACTING ON THE SYSTEM FROM THE GROUND. If you want to include the possibility of a non-zero normal force as well, you've got a much trickier situation.