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A solid (for example a steel ball) is moving with a certain constant velocity U toward a liquid in a container; I can write the equations of motion of the solid when it has a little part of it in the liquid (the forces on the solid are its weight, the buoyancy force and the viscosity) but I want to know what happens at the instant the solid collides with the surface of the liquid and why at high velocity U, the liquid acts as a solid in the collision? (the usual answer "the liquid doesn't have time to move isn't very scientific")

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The answer that the liquid doesn't have time to move is basically right, although it certainly needs some explaining. Firstly, the issue is not so much the instant of collision as what forces do you experience when moving very fast through a liquid. In common experience this happens at the surface, since most liquids are efficient at slowing things down. Also just to make the words easier let me narrow the discussion from liquid to water.

If we are willing to move slowly enough through water we can move with arbitrarily small amount of energy. The power goes like velocity squared. On the other hand it doesn't matter how slowly I'm willing to move, if I'm want to go through a mountain I'm going to need a lot of dynamite. There's are finite amount of energy needed to pass through a solid regardless of speed.

Why is that? It's not because water is "squishy". Water and many fluids are highly incompressible. Meaning you have to push very hard to change the density of water. In fact water is as incompressible as rock! If you put a piston on top of water-tight, water filled cylinder and tried to push down, it would feel like you were pushing down on concrete. It should not be so surprising that water is very incompressible - atoms don't like to be on top of each other, and in water the atoms are already quite, roughly as close as in a normal solid. (Even closer than they are in when solid ice!)

So if water is so incompressible, why is moving through water so easy? I mean there is all this water in front of you when you swim. When you are moving forward you have to displace that water. If you had to compress the water in front of you, if you had to just pushed the water in front of you into the water beyond that, swimming would be impossible. Also you would have to decompress the water behind you which is equally hard.

So you can't swim by compression - how do you swim? When you swim the water does something very kind and clever - it takes the water in front of you and redistributes it to be behind you. And it can do this without any compression and hence with almost no work([1]).

But to do this redistribution takes time - there must be communication between the water around your arms and the water around your legs for the fluid to redistribute when you swim. So if you do something fast enough ([2]) you are going to need to do it by compressing the water, since the water won't be able to communicate with itself, and arrange an incompressible movement. I.e. if you hit the water fast enough, the water in front of you has to go somewhere, but the water beyond that doesn't know it has to move, so the two are going to get smooshed into each other. That means the impact when hitting water hard are really are like hitting rock. The forces you get are from compression and the compressiblity of rock and water are comparable.

[1]I believe the work from swimming comes mainly from viscous drag or from shedding vortices, depending on what is swimming.

[2] Naively "fast enough" means fast compared to the speed of sound, but probably there is something more subtle happening in impacts.

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very good explanation, but I was wondering if there's someone who studied this problem mathematically. Consider a solid ball of radius $r$ falling into a liquid, I want to write the equation of motion of the ball; I think this is a hard mathematical problem to solve but my problem is how to translate the redistribution of the liquid and the compression of the liquid into equations. – whatever Jun 27 '13 at 8:22

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