Are springs compressed by energy, or by momentum? An object is headed towards a spring at a constant velocity, no external force will act upon it, except for any force applied by the spring itself after the collision.
Let's say the spring is of arbitrary tension and arbitrary length.
My question is whether the kinetic energy of object will make a difference in how far the spring is compressed, or whether it is only a matter of momentum?
For example, below, Object1 and Object2 have the same momentum, but Object2 has twice the kinetic energy. Would they compress the spring the same distance, or would Object2 compress it further?
Object1 has a mass of 1kg and a velocity of 1m/s.
Object2 has a mass of .25kg and a velocity of 4m/s.
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|////Spring////|   <---|Object|
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 A: As mentioned by knzhou in the comments, I don't think it's correct to really say that a spring is directly compressed by either momentum or by kinetic energy.  As dmckee's comment rightfully points out; springs are compressed by force.  The relationships between the applied force, kinetic energy, and momentum, will all depend on the situation you describe.
That said, it can easily be shown that in your example scenario, given the same momentum, we get different values for spring compression.  We know that kinetic energy can be determined by $KE = \frac 12 m v^2$ and the energy in the spring can be shown as $PE = \frac 12 k x^2$.  Due to conservation of energy, that gives us $$PE = KE$$ $$\frac 12 mv^2 = \frac 12 k x^2$$
In your example, if you solve for $x$, object 1 will compress the spring $x = \sqrt{\frac 1 k}$ and for object 2 the spring will compress $x = \sqrt{\frac 8 k}$.  You can see that increasing kinetic energy while leaving momentum the same will increase the compression of the spring.  This means that the compression is proportional to energy, not momentum; but does not necessarily say anything about the "cause" of the compression.
A: When the object collides with the spring there will be a change in both its kinetic energy and momentum. It is the kinetic energy of the object that determines the compression of the spring. If the objects have the same mass $m$ and the same velocity, they will have the same momentum and kinetic energy. 
In terms of the effect of an object on the spring, you can apply the work energy theorem, which states that the net work done on an object equals the change in its kinetic energy. In the case of the following equation applying the theorem the kinetic energy is that before the collision, and is zero at the maximum compression of the spring when the object is brought to a stop. The spring  does negative work on the object taking its kinetic energy and storing it as spring potential energy.
$$\int F(x)dx=-\frac{mv_{i}^2}{2}$$
For the spring, 
$$\int F(x)dx=\frac{kx^2}{2}=-\frac{mv_{i}^2}{2}$$
Solve for the displacement $x$.
In your example object 2, even though it has the same momentum as object 1, has twice the kinetic energy so it will compress the spring more.
Hope this helps.
A: Seems to me this question has been answered with nothing but evasiveness. 300 years ago it was understood that momentum and kinetic energy vmv where not compatible ideas. The question reveals the incompatibility... If momentum is conserved springs become free energy devices under the kinetic energy Theory. If momentum isn't conserved simple devices like levers become useless. Momentum and kinetic energy are opposing theories. The universe can only run on one of them... And if this experiment is actually done it will be proven that the "living Force" 1/2vmv was a huge scientific mistake.
