# Energy and impact -- Are these scenarios too different?

1. Oct 15, 2015

### thepatient

Let's say that there is a sort of large concrete wall, that applies an average force of 12,900 N for .04 seconds while you're initially at rest. After the impact, your momentum is now 516 N*s. Your mass being of 50 kg, means your speed is 10.32 m/s. So, in terms of kinetic energy, you've gained .5(50 kg)(10.32 m/s)^2 = 2663 J.

In terms of the damage that this would cause, is this equivalent to having a gravitational potential energy of 2663 J, and falling onto the concrete wall?

2. Oct 15, 2015

### A.T.

For damage peak force is more relevant than average force.

3. Oct 15, 2015

### CWatters

I calculated that this means the you are accelerated over a distance of about 20cm.

What sort of stopping distance would you have when hitting a concrete wall? Ok so your body would deform and it wouldn't (all) stop instantly but could it average 20cm? I suspect not. I think it would stop in less distance meaning peak deceleration and force is higher.

4. Oct 15, 2015

### thepatient

That's true, concrete won't deform enough, which means the force will be higher due to that shorter stopping distance. Stating a material makes this a bit difficult.

I guess the real question is, given a collision with an object that accelerates you to a certain amount of kinetic energy KE, is the collision equivalent to dropping with an amount of potential energy PE (equaling the KE of earlier) onto the same object itself?

5. Oct 15, 2015

### A.T.

No.

6. Oct 15, 2015

### thepatient

Why not? Assuming that the acceleration distance from the collision is equivalent to the compression distance of the landing medium.

7. Oct 15, 2015

### A.T.

Why should it be?

8. Oct 15, 2015

### thepatient

Conservation of energy is the reason why I highly suspect it can be.

9. Oct 15, 2015

### Staff: Mentor

How you acquired your original kinetic energy is irrelevant (looking strictly at it from an energy standpoint). I can shoot you out of a cannon or drop you off of a mountain and the impacts will be the same as long as your velocity and KE and all is the same for both collisions. That being said, since you will be traveling in different directions in each case, gravity will affect each collision a little differently. If you fall onto a surface, gravity is continuing to accelerate you downward during and after the collision, which is not the case if you impact a wall while moving horizontally.

10. Oct 15, 2015

### A.T.

And how does conservation of energy imply equivalent damage for these two cases?

11. Oct 15, 2015

### thepatient

Drakkith, thanks, but I think I answered my own question, but feel free to refute. This problem is more of an engineering problem. I'm only looking for an approximation (and it isn't really applied to anything. Just something I became curious about, which is why I am only looking into average impact force.) I had to review my old Mechanics of Materials textbook. There is this thing called elastic strain energy that can be used with the concept of impact loading. During impact loading, the kinetic energy of the object that is about to impact will be equal to the maximum elastic strain energy on the object being impacted. U = .5 mv2 For a rod with uniform cross section, the strain energy is derived as: U = P2L/(2AE), where P is the axial force applied, L is the length, A is the cross-sectional area, and E is the Young's Modulus.

So the answer is yes and no. The question comes in when you take into account the material and it's Young's Modulus. The material is now restricted to motion at it's fixed point. So if you're considering concrete as the object that impacts you with an average for Favg, the same material can't be used for the falling case, as the material would require a much larger amount of force to compress the material to the point that the strain energy equates to the kinetic energy. In other words, it hurts much more to fall onto a rigid object than to be impacted by it given the same impact energy.

So in order to equate the two scenarios as best as possible, I would need to change the material onto which the object falls on. It would require a lower Young's modulus, but enough so that the average force and compression energy for the fall is equivalent to the average force and kinetic energy in the case of the collision. The impact areas should also be the same, because damage occurs not by force alone, but by force per unit area. Also, the fact that KE = Um will will take into account that Um is a function of the maximum force, not an average force. Because in the elastic range, the force is proportional to displacement, it is fair to say that Pavg = Pmax/2. I know that Pmax in this case is different to the maximum force on a collision, but since there is no real way to determine the max forces of an impact without instrumentation and experimentation, an average is suffice for me. I know that in reality, the force of a collision over time spikes up, then shoots back down in a short period of time.

12. Oct 15, 2015

### Staff: Mentor

I'm not quite sure what impact energy means, but if the velocity is the same in both situations, then the collisions are identical. In other words, you hitting a brick wall at 60 mph is identical to a brick wall hitting you at 60 mph.

13. Oct 16, 2015

### A.T.

If the relative speeds before the impact are the same, the collisions are identical (ignoring gravity). But the OP equates the speed after the wall impact with the speed before the ground impact.

Exactly. And that peak force can determine whether a bone breaks etc. If you want to relate damage to energy, you have to look at the KE dissipated in the collision. Instead of comparing KEafter in one case with KEbefore in the other case, you would find the difference KEbefore - KEafter in both cases, and compare those differences.

Last edited: Oct 16, 2015
14. Oct 16, 2015

### Staff: Mentor

Oh. Well in that case ignore everything I said.

15. Oct 16, 2015

### thepatient

I guess I should have been more precise when I said "impact energy." What I meant was, if we consider scenario A being the collision and B the fall scenario, the kinetic energy gained by the object after the collision in A equaling the kinetic energy lost by the falling object in scenario B.

But like I mentioned earlier, the material matters in the second case, because the compression distance would be much smaller in B than the push distance in A. Meaning that the force acting on the falling object would be much larger on average. So the material would have to be changed to something that, with the same kinetic energy before the falling impact, will provide the same average force to slow down the falling object (with contact surface area remaining the same as in scenario A.) If I'm not mistaking.

16. Oct 17, 2015

### A.T.

Why would this comparison be relevant for damage? It says nothing about the energy dissipated doing damage in either collision.

17. Oct 17, 2015

### thepatient

I mentioned earlier that for scenario B, the reaction force on the falling object can be related by equation U = P2L/(2AE). Where U is the maximum compression energy of the stationary object on the ground (which equates to kinetic energy of the falling object before the collision), P is the maximum reaction force. L is the length of the material, A the cross sectional area, and E is young's modulus. This assumes a uniform cross-sectional area. Also that E remains constant, within the elastic portion of the stress/strain curve.

Now for A, I think you brought something interesting up. You're right, the energy gained for a collision of a big object on a stationary small object isn't the same as the deformation energy. The deformation energy is more closely related to the difference in initial energy of the system, and final energy of the system. (Since energy isn't conserved in collisions that are not perfectly elastic.)