Simple (?) question: Effect of mass when placing vs dropping object on spring

[superalias]

This is my first post and I'm still a wee physics newbie, though I've learned much from these great forums in the past.

I'm looking into the degree of deformation of compressible materials (say, foam, ground, etc.) subjected to force. It's my understanding that I can use the behavior of springs as a simple model, but there's one thing puzzling me about springs right now.

Say I have a spring mounted on the floor, with spring constant k, and no net forces on the spring. Let's assume that Hooke's Law (extension/compression of spring is in direct proportion to load on spring) applies throughout any examples, and there is no plastic deformation of the spring that changes its properties.

I place a load of y Newtons atop the spring. It's my understanding that the distance by which the spring compresses – its displacement (x) – will be as follows: x = y/k.

In other words, displacement is directly proportional to the mass placed on the spring. Double the mass, and displacement doubles.

Displacement is also directly proportional to 1/k. Double k, and displacement halves.

Assuming I have that right, here's the puzzler:

Say I instead drop the mass onto the spring – or even fire it downward onto the spring as a projectile. (For that reason, I'll call it a projectile and will refer to actual velocity in examples, not to height dropped.)

The spring will momentarily compress to some unknown value of x. The easiest way to find this unknown x, I understand, is by looking at energy: when the potential energy of the compressed spring equals the kinetic energy of the impacting projectile, compression halts and x is found.

In formulae: PE compressed spring = (1/2) kx^2 = KE projectile = (1/2) mv^2

=> kx^2 = mv^2

=> x^2 = (m/k) v^2

=> x = √(m/k) v

If I've got that right, then it's clear that x is directly proportional to v. That doesn't sound surprising.

What's surprising, though, is that x is not directly proportional to m or 1/k; rather, it's proportional to √m and √(1/k).

Restating the puzzle in plain English, why do m and k have less effect on x when the load impacts the spring at some velocity, than when the load is placed atop the spring? Or as an example: Why is it that quadrupling the load's mass quadruples the compression distance when I place the load on the spring, but only doubles the compression distance when I drop the load on the spring? It sure feels unintuitive (which, of course, has nothing to do with whether it's true or not, I know…)

To add one extra piece to the puzzle: I've seen numerous experimental reports measuring depth of deformation for certain materials impacted by certain projectiles (such as steel balls fired at dirt or clay), finding depth to be directly proportional to projectile momentum. That would mean depth is proportional to v (as above) and to m (not √m, as above). That's why I expected my spring examples to yield the same results. (I guess the simple explanation is that the spring model doesn't apply to the experiments in question; does that sound right?)

I've found many interesting threads on springs and on deformation in these forums (and elsewhere), but didn't find one that directly addresses my static load vs impacting load question. My apologies if there's already a thread answering this question; I'd be happy to hear of it.

 

[bp_psy]

Say I instead drop the mass onto the spring – or even fire it downward onto the spring as a projectile. (For that reason, I'll call it a projectile and will refer to actual velocity in examples, not to height dropped.)
...
In formulae: PE compressed spring = (1/2) kx^2 = KE projectile = (1/2) mv^2

=> kx^2 = mv^2

=> x^2 = (m/k) v^2

=> x = √(m/k) v

If I've got that right, then it's clear that x is directly proportional to v. That doesn't sound surprising.

If you shoot the projectile downward you can't consider v constant and you still must include the potential.The final expression will be more complicated. The difference between the proportionality of the first and second case is due to the form of energy that you store in the spring.Kinetic energy depends on v potential on h and the final expression for x must have have consistent units.

 

[superalias]

Thank you for the quick reply. To clarify the v I speak of, I don't mean initial velocity as a projectile; I mean velocity upon impact, including any added velocity from falling. The spring, of course, has no way of "knowing" whether that v came from falling alone, or from projectile firing + falling; the cause of impact velocity should be genuinely irrelevant. In the end, it should be perfectly possible to look at KE as (1/2) mv^2.

That aside, your next point is what's of interest. I think the answer might be in there, but how to express it without bringing in h (height of fall)? (After all, we could mount the spring sideways for the projectile test, and remove any falling altogether from the example.)

Again, I'm only wondering why it is that quadrupling mass quadruples compression when we place an object of a spring and apply a given acceleration (whether from gravity, from thrusters, from anything)… yet quadrupling mass only doubles compression when we hurl the object into the spring at a given velocity (whether from falling, shooting, etc.).

If you have any more insights, I'd be grateful.