Projectile coming to a full stop at impact versus breaking through and continuing on In the following scenario neglect any deformation forces within materials and assume the collision is inelastic. There are no instances of rebounding of the projectile at impact. There is no flying debris or shrapnel. Assume all projectiles used are of identical mass. 
Essentially, what I am after here is an understanding of both momentum and kinetic energy as it pertains to harder and harder collisions, even after it has exceeded the breaking point of the target object.
First I will ask about momentum and then about kinetic energy. 
Imagine a projectile being shot or thrown at some larger, stationary object that is immovable at very low impact collisions. Say that this object can sustain a 50 lbs force before it breaks and yields to whatever is in contact with it. 
The first projectile hits the object and it comes to a full stop, in the process it sets up a force of 30 lbs. The second projectile hits the objects, comes to a full stop, and sets up a force of 40 lbs. The third projectile hits the object, breaks the object, and is stopped in the process. 
Since the object can only "hit back" at the projectile, per Newton's Third Law, with as much force as it can withstand before breaking, does this mean that there is a maximum possible change in momentum for this type of collision? For instance: say the first projectile goes from 10 m/s to full stop. the second projectile goes from 20 m/s to full stop, etc. all in same amount of time. Once the largest force is created, where the object breaks, the maximum amount of momentum that acts to decelerate the projectile is reached. I am not sure if this is true, I am just trying to figure it out and also relate it to kinetic energy. 
Now, regarding kinetic energy:
If the maximum force between projectile and object is set up, it will always be the force experienced by the projectile and object no matter how much faster the projectile is going beyond the "breaking point" of the object. Since it will also always be traveling the same distance during the breaking of the object, does this imply that no matter how hard the impact is the same amount of kinetic energy is lost by the projectile? That seems absurd but I feel lead to believe that because the same force is being applied over the same distance in every case. But if this were true, wouldn't that also mean at extremely high kinetic energies the velocity would have to slow down extremely little compared to previous projectiles traveling more slowly? If it slows down less and less then the momentum can't be changing by the same amount at each impact that exceeds the "breakthrough point".
If force, distance, and time are all the same at a certain point then shouldn't the change in momentum and kinetic energy also be? Unless I am not reasoning properly (which I suspect is the case). 
I feel like this has something to do with work-energy theorem and impulse of momentum but I can't figure it out. 
Additionally, I also have noticed a few threads on this forum and elsewhere about the physics of karate chopping boards and it is very close to what I am trying to understand. 
Some people are saying that when you fail to break the board it is because a large impulse occurs where your hand is decelerated to zero velocity in short time, so there's a large force. They then go onto to say that breaking the board means your hand continues through after the break and so it slows down some over a longer period of time. If this were the case, wouldn't that mean breaking the boards takes less force than having your hand stopped by them? That doesn't seem to make much sense. 
The board should only break if you supply enough force to break it, and no less. If you fail to break the board, not enough force was set up at the collision. But how is that worse on your hand?  Also, if you successfully break the board your hand and arm would arguably be traveling faster to begin with, and when a larger force is set up (enough to break the board) the deceleration would still seem to be larger on your hand than if you didn't break them (by creating less force, when your hand goes from a slower speed to zero). 
Can someone make sense of this?
 A: Force is the change in momentum over time. This can't be instantaneous. The object must deform as it resists the entry of the projectile. Initially the force will increase with the deformation nonlinearly. At some point the material will yield and the force will decrease. Experimentally we can measure these forces though the results aren't a great depiction of what is going on internally due to limitations in the sensors and techniques. There will be some variation of how this curve looks with distance at different speeds as fractures are very complex phenomena.
If we simplify things and assume the force depends on distance only (i.e. by time only as it relates to distance) there will be a fixed maximal change in energy that is determined by integrating the force distance curve over the thickess of the material (work is force times distance).
A distinction should be made on what "breaking" means. Penetration occurs when an object has enough energy to travel through thre object. A projectile may have enough energy to "break" it by exceeding the yield point of the material but not penetrate it; remaining embedded inside it. A material is also modelled as having an elastic limit below which deformations are linear and above which they are nonlinear. There is also a point, whose name escapes me atm, below which deformations are reversible and above which there is creep (deformations have an irreversible component). Both of these are well below the yield point. Any impact that causes irreversible deformation will also alter the material's force over distance curve and the yield point; typically reducing the critical energy needed to penetrate the material.
A fixed change in energy for projectiles above the penetration threshold will not mean a fixed change in momentum. Faster projectiles will need a smaller change in mv in order to have the same change in 1/2 mv^2.
