# 2 masses connected to a spring

1. Aug 3, 2014

### Karol

1. The problem statement, all variables and given/known data
Two different size masses are attached to a pre stretched string. they are released from rest.
What are their velocities when the spring returns to initial length, just before the collision

2. Relevant equations
The spring constant: $F=kx$

3. The attempt at a solution
For each mass:
$F=kx=ma\Rightarrow a=\frac{k}{m}x$
The acceleration is proportional to the displacement, i cannot integrate it because to get velocity i have to integrate acceleration with respect to time.
If i take a short interval of time Δt the acceleration is approximately constant in it:
$a=\frac{k}{m}x\cdot \delta t$
first, i don't know the time interval and secondly i will get an expression with x, what should i do with it?
$\int \frac{k}{m}x dt$

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2. Aug 3, 2014

### voko

No need to integrate anything. Use conservation laws.

3. Aug 3, 2014

### Karol

Conservation of energy yes, can i use also conservation of momentum? i guess yes but i am not sure

4. Aug 3, 2014

### vela

Staff Emeritus
You have two unknowns, so you need two equations. Conservation of energy gives you one, and conservation of momentum, the other.

5. Aug 3, 2014

### Karol

--

Yes, i know. i just wonder whether the spring force is considered an internal force, one that the masses apply on each other, since i am allowed to use conservation of momentum only when there aren't external forces, and the only forces are those that the masses apply on each other.
If i have 2 masses vertically, one is thrown upwards and the other is thrown downwards towards the first like in the drawing, but now gravitation acts on both, i assume in this case i can't use conservation of momentum, right?
I calculated the velocities using kinematics and the momentum just before the encounter is smaller than the initial momentum.

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Last edited: Aug 3, 2014
6. Aug 3, 2014

### vela

Staff Emeritus
Right, because the system consists of just the masses and spring, any force one exerts on the other is, by definition, an internal force. In your second scenario, the Earth, which is outside the system, exerts a force on the masses, so there is an external force and momentum isn't conserved.