Rebound acceleration of a falling object really independent of mass?

Question

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?

Answer 1

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.

Answer 2

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.