Impulse Time Elapsed in relation to Hardness and Momentum I am trying to calculate the time elapsed of an impulse during a collision of two objects.
I will give three examples below in hopes that they make clear what I am stuck on.

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*Two object of the same mass of 1kg each, and same hardness of Shore 60A.
If object1 hits object2 head on, what is the time elapsed of the collision?


*Object1 again with mass of 1kg and hardness of Shore 60A hits a wall and bounces back, what is the time elapsed of the collision? Is it the same as in case 1? Or different?


*Two objects, object1 with mass of 1kg, and object2 with mass of 500kg. Both have the same hardness of Shore 60A. Object1 hits object2 head on, what is the elapsed time of the collision?
Thank you in advance to anyone that can help me figure out what the relationship is between hardness and momentum.
 A: Let say the objects are two uniform spheres. The contact physics is governed by Hertzian contact theory (page 3, sphere-sphere contact).
The deflection $d$ is a function of the force $F$ $$ \frac{d}{R} = \sqrt[3]{\tfrac{9}{16}} \left( \frac{F}{E\, R^2} \right)^\tfrac{2}{3} $$
The problem is that the stiffness amount is not constant, but depends on the depression the contact makes, as it spreads out more and more.
Overall you can equate the incoming kinetic energy to the elastic energy stored in the contact.
$$ KE = R \int  \frac{d}{R}\,{\rm F} = \sqrt[3]{ \left( \frac{243 F^5} {2000 R E^2} \right)} $$
From the above, you solve for the peak force $F$
$$ F = \left( \frac{2000 E^2 R}{243} KE^2 \right)^\frac{1}{5} $$
The above tells us nothing about the time it takes. For that you need the total impulse $J$ exchanged in the impact and $J = \int F \,{\rm d}t$. Specifically, we linearize the contact dynamics to find the shape of $F(t)$ is close to the upper half of a cosine curve (with peak at time=0) in which case
$ J \approx \frac{2}{\pi} F \Delta t $ or
$$ \Delta t  \approx \frac{\pi\,J}{2 \,F} $$
