How Does a Force Affect Spring Length Between Two Equal Mass Blocks?

[Jorgen1224]

Homework Statement

[/B]
There are two blocks that have equal mass m, they are connected by a spring (with a coefficient of elasticity k) that has length l0.

Instantly on the left block a force F begins to act. What will be the min. and max. length of this spring during this movement.

There is no friction

Homework Equations

Newton's second law
F=kx
Eventually kx²/2
And that's all i figured out

The Attempt at a Solution

I know that answer is l0-F/k and l0 (confirmed answer), but i don't know how it actually happened. (I actually know where does l0 in the elongation comes from :P)

These are 3 ideas on how to solve it.

(N = Tension)
1st. From the 2nd law (let's call these m1 and m2 to make it easier, for me at least)
m1⋅a=F-N and m2⋅a = N ---> a=N/m2

m1⋅N/m2 = F - N
m1⋅N= F⋅m2-m2⋅N ---> N=F⋅m2/(m1+m2)

Now because N = F·m2/(m1 + m2)
Then shortened spring will be x = N/k = F·m2 / (k·(m1 + m2))
Therefore length L = l0 - x = l0 - F·m2 / (k·(m1 + m2)) so it's actually l0 - Fm / k⋅2m therefore it's l0 - F/km

2nd (It's actually not mine idea)
F= kx²/2
F1=ma
kx=m2a2
N=m2a2
m1a1-N = F, a=N/m
F=2N
x = N/k = 2F/k
l=l0 - x = L0 - 2F/k

3rd
I thought something about centre of mass, but it's not a very highly developed idea, i don't even know how to describe it.
Forgive me for any language mistakes

 

[haruspex]

m1⋅a=F-N and m2⋅a = N

The masses will not have the same acceleration.
But they have the same mass, so you can make the algebra easier by just using m instead of m1, m2.

 

[Merlin3189]

The Attempt at a Solution

[/B]These are 3 ideas on how to solve it.

(N = Tension)
1st. From the 2nd law (let's call these m1 and m2 to make it easier, for me at least)
m1⋅a=F-N and m2⋅a = N ---> a=N/m2 I don't know what 'a' is, but if it is the acceleration of m1, it won't be the acceleration of m2.
They move at different rates, as implied by the question, when it asks about the variation in spring length.

m1⋅N/m2 = F - N
m1⋅N= F⋅m2-m2⋅N ---> N=F⋅m2/(m1+m2)

Now because N = F·m2/(m1 + m2)
Then shortened spring will be x = N/k = F·m2 / (k·(m1 + m2))
Therefore length L = l0 - x = l0 - F·m2 / (k·(m1 + m2)) so it's actually l0 - Fm / k⋅2m therefore it's l0 - F/km

2nd (It's actually not mine idea)
F= kx²/2 Why?
F1=ma F1? Isn't that just F? ma? Is that m1.a1 ? But, F = m1.a1 ignores the spring force.
kx=m2a2
N=m2a2
m1a1-N = F, a=N/m Which a is this?
F=2N Why?
x = N/k = 2F/k
l=l0 - x = L0 - 2F/k