Force of collision without knowledge of the results of the collision

[InvisibleMan1]

This is not really a homework/textbook/coursework problem. It's something I ran into while studying physics, and I haven't been able to solve it. The rules seem to say that this is the correct forum for something like this, but I can't be certain. If this is misplaced, please move it to the correct forum, thanks.

Homework Statement

I'm trying to find out how to calculate the force which one object would apply to another object if the two objects collide, without knowing anything about the result of the collision. The physics tutorials I am studying were very vague in this area. Preferably, one object is at rest. The other object is moving directly at the first object, either with an acceleration or a constant velocity (I need to know how this works for both). I also need to know how this works when both objects are in motion, but I can tackle that later if it will complicate the solution too much.

Homework Equations

I'm not entirely certain what this section is for, which probably means there aren't any.

The Attempt at a Solution

I'm hoping there is a formula or something for this. I would think that the answer is related to the velocity and mass of the moving object (which appears to be momentum), but I have no way of using this in any formulas I know of... This is very frustrating; it feels like I have all the pieces to the puzzle, but one piece is missing.

A faster moving object would apply a greater force on the object it collides with. This I am fairly certain about.

I keep running into vague references about the force relating to the displacement during the collision, or some other such information... This won't work, since I would be using the force of collision to find that information, not the other way around.


Side-note: This might not be relevant, but gives some context. I am a programmer, and I plan on using my knowledge of physics in programs I create. I would think that this problem is very basic, and there would be knowledge about it everywhere... But this is either so basic that no one talks about it, or I am missing keywords. I have asked for help repeatedly in other physics communities, but I have not received a solution as of yet.

 

[PhanthomJay]

If you don't know the results of the collision, you can't calculate the impact force. You need to know the time duration of the collision, or the deformation (displacement) during impact. For example, if you dropped a brick from a certain height onto a concrete pavement, the impact force of the collision would be high (small deformation over a short collision time interval, large deceleration). If you dropped it from the same height into a pile of marshmallows, the impact force would be low (longer collision time interval and greater displacement into the marshamallows, small deceleration). In other words, you need more information to calculate the force.

 

[InvisibleMan1]

Is there a way to calculate the displacement or duration of the collision without knowing the force of impact? Would you use pre-calculated constants like when dealing with friction?

Edit: Actually, would this be considered going off-topic? I can create a new thread with the new question if required.

 

[PhanthomJay]

Is there a way to calculate the displacement or duration of the collision without knowing the force of impact? Would you use pre-calculated constants like when dealing with friction?

Edit: Actually, would this be considered going off-topic? I can create a new thread with the new question if required.

The only way I know of is through experimental testing, since the forces and deformations are largely dependent upon the properties of the materials. F =mass times (change in velocity) divided by time (F= m(delta v)/t), and although m and v at impact are known or can be calculated, F and t are unknowns without more data. Since this is not a homework problem, I suggest you post your question under the "General Physics" forum, using a different title . Maybe someone else can give a better answer.