# Spring Projectile problem: center of mass

1. Feb 22, 2005

### 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 lenght l (relaxed).

m1
|
| _ > spring, at lenght 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

2. Feb 22, 2005

### 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.

3. Feb 22, 2005

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)

4. Feb 23, 2005