Does glass shatter when hit by projectiles with a threshold momentum or a threshold kinetic energy?

Question

Consider a sheet of glass (or any other breakable solid). Suppose you have many identically prepared copies of the same sheet, and you launch a bunch of fast-moving projectiles at the sheet from a directly perpendicular direction. You vary both the mass and the speed of the incident projectiles while holding all other factors constant (e.g. the size and shape of the projectiles, the location of the impact point on the target, etc.).$^1$ For each combination of projectile mass and speed, you record whether or not the glass fractures upon impact.

I assume that most "reasonable" target materials fracture above some speed-dependent threshold projectile mass and above some mass-dependent projectile speed. In other words, the boundary between (a) the combinations of projectile mass and speed that cause a fracture and (b) those that don't, decreases monotonically in each variable with respect to the other.

While I am less confident about this, I also suspect that for most "reasonable" materials, the shape of that boundary curve probably takes either the form $m v = p_0$ or $\frac{1}{2} m v^2 = K_0$ to a reasonably good approximation, where $p_0$ and $K_0$ are constant over a fairly wide range of masses and speeds. In other words, I suspect that there is either some (target-dependent) threshold projectile momentum that causes the glass to fracture or some threshold kinetic energy. But I don't know which. Which of these two functional forms $v_\text{threshold} \propto m^{-1}$ or $v_\text{threshold} \propto m^{-1/2}$, if either, describes the shape of the boundary curve between the projectiles that fracture the target and those that don't?

From first principles of solid mechanics, I would expect that the material would tend to fracture at some critical compressive stress known as its fracture strength. Stress is force per unit area. I assume that for most collisions that just barely induce fracture, the contact area at the moment of fracture is roughly independent of the projectile's kinetic properties, although I'm not sure of that.$^1$ So the threshold property would be the maximum force $F_\text{max}$ that the projectile exerts on the target. If we model the glass target as a simple Hookian spring with $F = -k x$ for fixed $k$, then $F_\text{max}$ equals $k\, x_\text{max}$, where $x_\text{max}$ is the target's maximum displacement at the point of impact. This model predicts that the target will fracture at some threshold displacement. For a Hookian spring impacted by a projectile with mass $m$ and velocity $v$, the maximum displacement $x_\text{max} = \sqrt{m/k}\, v$ and the maximum force is $F_\text{max} = \sqrt{km}\, v = \sqrt{2 k K}$. This would imply that there is a threshold kinetic energy $K_0 = F_\text{max}^2/(2k)$ rather than a threshold momentum. Is this correct?

Of course, we could also ask the similar question about materials other than glass, and also about the yield point when elastic deformation changes to plastic deformation in ductile materials. I assume that the answer - either threshold momentum or threshold kinetic energy - is the same for most of these variants questions, but I'm not sure.

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$^1$ Edited to add: Several commenters have pointed out that the material properties of the projectile are very important. Let's assume that the projectile is much stronger than the target material, so that we can neglect any material deformation of the projectile and idealize it as a perfectly rigid body, and only consider the deformation of the target. The contact area at the moment of fracture will certainly depend on the projectile's structural properties in general, but I don't think that the contact area will depend strongly on the projectile's kinetic properties in the regime where the projectile is much stronger than the target.

Answer 1

We are imagining impacting a target with a projectile. The projectile is like a pane of glass. It is large in a 2D geometry and thin in the third dimension. I'm going to imagine the projectile is a ball of some radius. We assume the ball is much harder than the glass so that only the glass will undergo material deformation, the projectile will not.

When the projectile hits the glass it will deflect. There will be a region on the surface of the glass where the glass deflects appreciably. My guess is that this region will be much larger than the actual physical impact area. If we treat the impact point of the projectile as a sphere with some radius $R$, I think this paragraph means that the deflection area will be only a constant or very weak function of $R$. We call the deflection area $A$. Below I am assuming the deflection area is constant regardless of how much the glass is deformed. This assumption MAY BE WRONG. For example, as the glass deflects more, the deflection region might spread out or contract in a way that the effective deflection area $A$ needs to be thought of as a function of deflection.

Materials yield or break when they exceed a certain stress. Stress is caused by pressure. In this case the area is $A$ and the force is $F$, caused by the impact of the projectile. We are assuming $A$ is constant, so we only need to find the peak force $F$.

In our simple model we treat the target/projectile system as a spring with spring constant $k$. As the target deflects it puts a force $-kx$ onto the projectile where $x$ is the deflection. Initially the system has the kinetic energy of the projectile $K = mv_0^2/2$. At peak deflection, assuming the projectile hasn't yielded or broken, the energy is purely potential, $P=kx_0^2/$. Equation these two we get $$ x_0 = \sqrt{m/k}v_0 $$ so the peak force is $$ F_0 = kx_0 = \sqrt{mk}v_0 = \sqrt{2k}\sqrt{mv_0^2/2} = \sqrt{2kK}. $$ If the threshold comes at a peak $F_{\text{max}}$ then we have $$ K_{\text{max}} = F_{\text{max}}^2/2k $$ These are the exact same expressions from the OP. I've just written in my own words the math and assumptions going in.

I think as a first pass this is a reasonable physics approach to this problem. I think for glass this will be reasonable, because the glass isn't going to deflect much before it breaks. This means the linear treatment is probably reasonable. A more stretchy material, however, could deform a lot more before it breaks. I think this is the type of thing that could break the constant $A$ assumption. The material starts behaving non-linearly, not because it is yielding, but just because where it was once flat it now has a curvature that changes the surface area of material interacting with the projectile.