Baal did some "napkin" calculations many years ago on the MYTT forum that were pretty good.
Before that I had posted a similar calculation in "Anton's obscure question" thread. I posted a link to a view of Tacshow123 "catching" the ball and dropping it on the other side of the table. That is the video UpsideDownCarl was looking for. It is hard to find even if you know what you are looking for.
First we will make some crude estimates. With a little thought is is possible to estimate very closely. Too many think that because they can't solve a problem exactly that they shouldn't try solving the problem.
Assume the contact speed with the paddle is 10 m/s or 10mm/ms. That is the combined speed of the ball coming at the paddle and the paddle coming at the ball. The speed is measured relative to the paddle.
During the contact time the ball will slow down to 0 then speed up again in the opposite direction.
Lets assume the total contact time is 1 ms so the ball spends 0.5ms slowing down and 0.5ms speeding up.
How far will the ball go in 0.5 ms? During the time the ball is slowing down from 10mm/ms to 0mm/ms we will assume the ball decelerates at a constant rate. ( this is not right, the ball slows down slowly at first but slows down faster as the ball compresses the rubber, but we are making a crude estimate, look up Hooke's law ) Therefore the average speed during the 0.5ms of slow down time is 5mm/ms.
If the average speed is 5mm/s for 0.5ms, how far did the ball push into the paddle?
5mm/ms*0.5ms=2.5mm
Is this distance reasonable? I think we are in the ball park. I think the 2.5mm is a little too much. I think if I did some calculus and assumed the deceleration rate was not linear but was parabolic as described by Hooke's law, the deceleration distance would be shorter.
The ball will compress some so the center of mass of the ball may travel another 1-2mm closer to the paddle.
A 2mm rubber may compress 1mm. The blade may flex some but I doubt most paddles people buy will have visible or measureable flex but we know the blade flexes a little bit due the the sound.
This distance can be made shorter by decreasing the contact speed or by reducing the contact time.
A teaser for you problem solvers out there that like to strain your brain.
How can you tell how much of the ball makes contact with the top sheet?
How can you uses this to determine how far the ball's center of mass still moves after contact?
Are there any calculus wizards out there that can figure out the how the ball will decelerate if Hooke's law applies so that the deceleration rate is not constant.
How what is the peak force applied to the ball? Why do we care?
I know people will have mixed feelings about me reigniting this thread, but whatever, it's an interesting question and this is the only thread anywhere I've looked where people at least try to talk about this objectively.
For background, I'm a sensory neuroscientist, and at least as a sensory phenomenon, humans would not be able to distinguish a long and a short contact on the scales of table tennis contact times. We simply don't have the tactile equipment.
https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4893195/ for details - the thresholds discussed here are well beyond the known contact times for the ball. To our sensory system, it's indistinguishable from one instantaneous contact (if we ignore vibration afterwards)
Of course, it's a useful fiction and you can be the best player in the world while harbouring whatever myths help you play better and rationalise your technique. Knowing the physics somewhat hasn't helped me play much better! And we CAN tell some things from tactile feedback : vibration probably gives us some information about the quality or style of shot (though I think this is exaggerated and conflated with audition).
But I also wanted to better answer brokenball's question about the non-linear dynamics of the ball pressing into the rubbers. The hint to solving this is in what you already mentioned - Hooke's law. It is usually used to describe the forces involved in springs, but it is reasonable to apply it here too. But the motion of springs is very commonly sinusoidal, and for good reason, and the same applies here!
So the force (and therefore acceleration) on the ball is (negatively) proportional to the depth of penetration into the rubber, as you correctly point out. For every further mm of depth pushing into the rubber, there will be some rough corresponding increase in acceleration out of the rubber (ignoring the possibility to bottom out, let's assume reasonable rubber thickness). So some simple calculus: a=acceleration, v=velocity, s=displacement. Differentiating with respect to time, s'=v, v'=a, so s''=a, but s is proportional to negative a! Something is only proportional to its negative double derivative when we're dealing with sin waves, and that is why we see sin waves in spring motion. The portion of the sin wave we have is where acceleration is only positive. s is the same at the start of contact to the end of contact (that being the undeformed, equilibrium surface of the rubber) and (negative) max in the middle, v is reversed (assuming ideal simple energy transfer, of course it's never that simple but it's a useful assumption here) from the start to the end and zero in the middle, and acceleration starts off zero, ends off zero and is maximum in the middle. So this fits equations of the form s = -cosx, v=sinx and a = cosx over a range of -pi/2 to pi/2, where -pi/2 is start of contact, pi/2 is end of contact and 0 is the point of deepest penetration where the ball is still and experiencing maximum acceleration. (of course there will be constants and this assumes a period etc. etc.)
So it is not a parabolic acceleration dynamic but a sinusoidal one that Hooke's law describes. And the correction is reasonably easy to make. I knocked this up in desmos to show - the red line represents displacement/ball position, the blue line represents velocity and the green line represents acceleration. The black line represents the linear model you described. By working out the average speed and giving the length of contact, you were essentially integrating the first half of the black line (from -pi/2 to 0). In my graph that has an area of pi/4 (b*h/2). The equivalent part of the integration of the green line is 1 (since we've conveniently used a sinusoid with 2pi period which plays nicely with calculus rules) - the integral of the green line is actually the blue line, so take the blue line at t=0 and subtract the blue line at t=-pi/2 and you get 1). So there is a correction factor of 4/pi.
So in fact you are incorrect - the ball will penetrate further than in a linear model. And this makes sense, as there is more speed into the bat at all stages of the actual dynamics than there would be in a linear model.
So the answer would be that given your conditions of 10 m/s and 1ms, the ball penetrates not 2.5 mm but 2.5*4/pi mm = 3.18 mm in this highly idealised (but slightly more realistic than linear) scenario
Anyway, sorry to those of you who wanted this to stay buried, you're welcome to anyone who appreciates this information.
Edit : I made a mistake here and compared the black linear model to the green acceleration, where the proper comparison is with the blue velocity line in a similar way. But the same shape and same logic and same conclusions apply, so I'm not going to bother correcting it all the way.