# Spring System with Variable Tensions, functions of displacement.

1. Jun 26, 2008

### SpeeDFX

So...I wanna solve the problem where two masses on springs collide into eachother on a frictionless surface. The two masses are different, and both springs' tensions are linear functions of x.

The two mass-springs collide into eachother, each having some initial speed V1 and V2,and the springs stick together end to end. I want to find the equations of motion for each mass after the collision (1 dimensional).

I'm getting stuck on how to set up the equations, and then, figuring out what kind of equations i'm dealing with.

Right now, I have the problem set up in my head like the following. M1 (attached to to spring with K1(X1) ) comes in from the left and M2 (with K2(X2)) comes in from the right. then the springs stick together and start doing their things. I'm taking the point of view from M1.

F(onto M2) = Ktotal*Xtotal = M2*Xtotal'', where Xtotal'' is the acceleration of M2 from perspective of M1.

also, I'm calling "B" the point at which the springs stick together

F(at B)= 0 = K1(X1)*X1 = K2(X2)*X2

with these 2 eqn's, I'm stuck. Even if I use the solution to a simple harmonic oscillator, I end up having X1 and X2 inside the cosine and sine functions as well as outside. I don't know if this is OK and I don't know how to deal with it. lol. someone help me please

2. Jun 26, 2008

### SpeeDFX

woops. I meant...

F(at B)= 0 = K1(X1)*X1 - K2(X2)*X2

3. Jun 26, 2008

### tiny-tim

Hi SpeeDFX!

I'm not visualising this …

if the masses collide, where are the springs?

how can they stick together end to end?

4. Jun 27, 2008

### SpeeDFX

each mass has a spring sticking straight out of it.

I guess a better way to describe it would be the following..

2 masses with springs are attached in the followin order.

mass1_spring1_spring2_mass2

the masses have some initial velocity toward eachother

5. Jun 27, 2008

### tiny-tim

use centre-of-mass coordinates!

Hi SpeeDFX!

Ah! So two springs collide into each other on a frictionless surface, and they have masses at their other ends.

Hint: change to a coordinate system in which the centre of mass is stationary!

Momentum is conserved in all collisions, so it'll remain stationary.

(That'll make all the calculations much easier.)

Then use conservation of energy.