Stress exerted on a wooden cube dropped on concrete floor

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I wonder if it is possible to at least estimate the stress (or the impact force) that the cube in question will experience due to the collision with the concrete floor. Assumptions:
1. The cube is dropped from height h
2. The cube is relatively small, thus we can neglect air resistance

[tex]\frac{\Delta p}{\Delta t} = F[/tex]
[tex]F = \frac{m \Delta v}{\Delta t}[/tex]
[tex]\Delta v = -\sqrt{2gh}[/tex]
[tex]F = \frac{-m\sqrt{2gh}}{\Delta t}[/tex]

Now, I do not have an idea how to find the distance of the collision. It seems that the only way would be to measure the deformation of concrete under the cube (I assume the cube does not bounce upwards).
Is there a way to estimate how long the collision will last?
 
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The concrete won't deform much but the wood might if dropped on a corner. Otherwise I think the only way to do it is to measure it.

We used to fit accelerometers to computers and drop them to test packaging.
 
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jim mcnamara,
Actually I already used impulse in the equations I wrote.
The only thing that is problematic is the time of the collision / the deformation of the ground.
I guess there is really no way to calculate it, it needs to be measured.
 
You could estimate a lower limit of the force...

For example if the corner of a 1kg cube was compressed say 5mm by a drop onto concrete from 3m you could use 5mm as an estimate for the stopping distance. That gives you a deceleration of >5880m/s2 (about 600g) and a force of 5880N.

Perhaps worth a look...

https://www.endevco.com/news/archivednews/2009/2009_02/tp321.pdf
 
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It is possible to model all this by treating the ground and the cube as elastic materials, and using solid deformation mechanics to calculate the deformation histories of the cube and the ground. This will also give you the time history of the contact force, the contact pressure distribution, and the time of contact. This kind of calculation is typically carried out using finite element method (FEM) software. If the ground is rigid, then the contact time is equal to 2 times the side of the cube divided by the speed of sound in the cube material.
 
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