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From what i understand, you can calculate the impact force of a falling object hitting a surface using the impulse-momentum theorem.

So for example, if you dropped a $10\text{ kg}$ ball from a height of 2m onto a hard surface, the average force acting on the ball at impact would be

$$F_{avg}(t_2-t_1) = m(v_f-v_i) \\\implies F_{avg}=m\frac{\sqrt{2gh}}{t_2-t_1}$$

It is $F_{avg}$ that causes the ball to rebound (i.e. move in the opposite direction)

What I'm having a tough time grasping is that it seems like the rebound acceleration of the object due to this force is independent of mass.

$$F_{avg}=m\frac{\sqrt{2gh}}{t_2-t_1}=ma,\\ \text{so}~~ a=\frac{\sqrt{2gh}}{t_2-t_1}$$

Is that right? Am i missing something? Just like falling, all objects rebound at the same acceleration?

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To understand this intuitively just think that a heavier object will hit harder on the ground so it will feel more force upwards given to it by the ground. Thus it's a bit like gravity that even though heavier objects have more weight, they fall by same acceleration as their mass is larger. So in the end the collision time and the difference of $v_{impact}$ and and $v_{rebound}$ is what determines the acceleration.

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  • $\begingroup$ In theory, how do you explain the scenario of two objects of different masses (M1 and M2) where M2>M1 AND both objects are infinitely hard and the surface is also infinitely hard? So what would be acceleration1 vs acceleration2? $\endgroup$ – PhysicsMakesTheWorldGoRound Jun 13 '17 at 18:27
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Looping back to your assumptions

Is that right?

You've taken a set of assumptions, and derived a result. Your derivation is correct, so under your assumption, your result holds.

Am i missing something? Just like falling, all objects rebound at the same acceleration?

Yes, you are missing something.

As I said, you took some assumptions, and you now believe your result holds true regardless of those assumptions. That is incorrect.

Let me be clearer. The set of assumption I am talking about is:

  • The rebound occurs between a time t1 and t2
  • The difference $t2-t1$ is a constant

You imposed a defined speed change, over a defined time interval. Yes of course the average acceleration will always be the same, because that average acceleration has to be $Acc_{Avg}=\frac{\Delta V}{\Delta t}$. That is just your assumption you're getting back, nothing else.

Reality check

In reality, different objects can have differing rebound times. Two elastic ball of young modules $E$, with masses $m$ and $M$, will have very different rebound times. I expect the heavier ball to deform more to absorb the energy of the impact, thus having a longer time in contact, thus a lesser average acceleration.

To impulse, or not to impulse

Note this is no longer an impulse collision. However you dropped that idea yourself when you introduced a rebound time.

Impulse is instantaneous, therefore the concept of "average acceleration" is doubly wrong: - there can be no average - there can be no acceleration under impulse response.

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  • $\begingroup$ If impulse is instantaneous, then what is t1 vs t2? Shouldnt t1=t2 for an instantaneous scenario? I understand that in reality the value of t2-t1 will differ based upon materials etc, but theoretically, how do you explain the scenario of two objects of different masses (M1 and M2) where M2>M1 AND both objects are infinitely hard and the surface is also infinitely hard? So what would be acceleration1 vs acceleration2? $\endgroup$ – PhysicsMakesTheWorldGoRound Jun 13 '17 at 18:23
  • $\begingroup$ Well you introduced it haha! Two ways to look at it. If it's a pure impulse, T2=T1 then it's a maths problem (a limit) and acceleration does not exist, you only built a mental construct. If it's approximately an impulse then T2 is so close to T1 that the rebound force becomes enormous. That means that no other force has the time to do useful work. So you disregard the rest and perform calculation in that very small time frame... Whose length may vary case by case, remember. If you ask "smthg infinite smthg" you're not talking physics, but maths. Both approach above bring answers, you choose! $\endgroup$ – MrBrushy Jun 13 '17 at 20:34

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