# Average force during a collision

1. Sep 26, 2013

### CuriousParrot

This isn't actually a homework problem, I'm just wondering what I'm doing wrong:

1. The problem statement, all variables and given/known data

Say we've got a stationary brick wall, and a blob of jello with mass m is launched at the wall with speed v. The blob has length d along the direction of motion.

The blob splatters onto the wall when it hits. How much average force will the wall feel?

2. Relevant equations

The work done to stop the blob must be the same as the energy that was in the blob while it was moving, so:

F d = E

Also, from the second law, F = m Δv / Δt

3. The attempt at a solution

Solving that first equation gives:

F = E / d = 1/2 mv² / d

On the other hand, if I had used the second equation, I would get:

F = mv / Δt = mv / (d / v) = mv² / d

So, doing it this way, I calculate twice as much force. What gives? Where did I go wrong?

2. Sep 26, 2013

### Ibix

Your energy calcuation is correct.

What's different about the moving mass of jello half way through the collision compare to at the start?

3. Sep 27, 2013

### CuriousParrot

Well, half the mass has splattered onto the wall, so there's only half still colliding. I don't see how that's relevant, so I guess I'm missing the hint. I wonder if I should be getting that factor of 1/2 from an average quantity, like using Δt = d / (v/2) instead of d/v, which would make the answer consistent, but that doesn't sound right to me; I think Δt is the amount of time until the last of the jello has hit the wall, which is just d/v.

4. Sep 27, 2013

### ehild

No, that is not right. The kinetic energy of an extended non-rigid system (the blob) is the sum of the KE of the centre of mass and the internal kinetic energy of the parts. The wall acts only to the part of the blob it is in contact with, and that part exerts force on the next piece of the blob, and so on. The internal forces contribute to the change of KE, and the KE of internal motion will transfer to that of random motions: changes the temperature of the blob, of the wall and of the ambient ...
The work of the external force you have calculated is enough to stop the motion of the CM. More work is done to change the shape of the blob, warm it up, making sound and so on.

That is correct: The external force is equal to the time derivative of the momentum of the whole system. Internal forces do not change the momentum.

5. Sep 27, 2013

### Tanya Sharma

I would like to mention two things about which I am not entirely sure .

1. If FΔx = mv2/2 is applied ,then Δx is the displacement of the COM .Assuming COM lies
somewhere at the center , Δx=d/2 .So Fd/2=mv2/2 or F =mv2/d .

If FΔt = mv is applied , F=(mv)/(d/v) = mv2/d .

2. The average forces in the above cases are different .One is averaged over distance,other is
averaged over time .

ehild...Could you please check if what I have written makes sense .

6. Sep 27, 2013

### ehild

Yes, it is right, and I am not sure if my post made sense.

The work done by the wall on the blob is the force times the displacement of the point of application. Does the point of application of the contact force move at all?
Some time ago, there was a discussion about work done by a wall during an inelastic collision. And I think it was said the colliding plastic body did work on itself.

On the other hand, the time needed to stop the blob completely is not defined when we want to apply FΔt=Δp to calculate the average force.

ehild

Last edited: Sep 27, 2013
7. Sep 27, 2013

### CuriousParrot

I'm still confused.

Using the center-of-mass to wall distance, d/2, doesn't sound right to me: If the mass was distributed unevenly, wouldn't that then change the result of that calculation? Yet, shouldn't we expect the average force to stay the same? From the perspective of the wall, finding the average force means integrating over the entire collision distance.

I think I might have been wrong when I originally said the collision time was Δt = d/v. If the mass starts out at v and slows to 0, then I should use the average velocity there, v/2, hence Δt = 2d/v. On the other hand, maybe not -- if I'm picturing jello hitting a wall, it doesn't really slow to a gradual stop the way a rubber ball would. Instead, I would think the farthest-back part of the jello crashes forward at speed v until it hits the wall, while the front layers basically disintegrate on impact.

8. Sep 27, 2013

### ehild

No wonder. So am I. It is a confusing problem.
In case it is a completely soft material, the piece not touching the wall keeps its original speed and Δt=d/v. It is different otherwise.

ehild

9. Sep 27, 2013

### Ibix

Glad it's not just me...

10. Sep 27, 2013

### Andrew Mason

This is a momentum question. It is very difficult to analyse in terms of energy. You really can't say that the blob loses its kinetic energy in time d/v. If it was a smooth wall and if the blob was water, as the blob of water hit the wall it would flatten and keep spreading out along the wall after its forward momentum had been lost. If during the collision the water that has struck the wall caused very slight back-pressure on the water that had yet to hit the wall, the force applied by the wall to the water would do almost no work at all. It would simply change the water's direction. But it still changes the momentum of the water by mv.

The force during the collision is dp/dt. So the average force over the time of collision will be mv/t. If t = d/v (it would be probably be just a tad longer) the average force would be mv^2/d. But, as has been pointed out, that is the average force over time, not the average force over distance. The distance though which the stopping force acts depends on the details of the collision and could be very small.

I expect Harrylin would be interested in this question.

AM

11. Sep 29, 2013

### CuriousParrot

Thanks, this sounds really interesting. So am I on the right track to think:

1. There is no need for the time-average and distance-average forces to be the same (although sometimes they are), so there's no contradiction there.

2. The kinetic energy lost by the blob could be anywhere from zero (very "loose" blob) to 100% (very sticky blob), yet the time-average force would be the same (as long as they have the same mass, velocity, and collision time).

If I wanted to model a car crash, it would be the time-average force I care about, right? Would the distance-average force be of any use there?

12. Sep 30, 2013

### haruspex

In a car crash, you mostly care about peak forces, though it is a bit more complex when you consider the physics of the brain's anchoring. All colisions start with an elastic phase in which the force increases more-or-less linearly with impact distance. An ideal cushioning allows that to increase quickly to the max force bearable by the occupant then enters a plastic phase where the force remains constant.
If you insist on choosing one of those two averages, bear in mind that average acceleration is defined to be the time average.

13. Sep 30, 2013

### Andrew Mason

The blob does not compress to a flat pancake in time d/v. That would be the case only if there was no force on the part of the blob that has yet to hit the wall.

If the blob acted like a spring, where the force is proportional to the amount of compression of the blob, the force through the blob keeps increasing during the collision. During the collision the force is -kx where x is the compression of the blob.

Letting the mass of the blob/spring be m, its velocity before collision v, its length = the compression during collision be d (i.e. it ends up as a pancake), the "distance average" force will be KE/d = mv2/2d.

The spring constant k = mv2/d2

The time average is more difficult to determine. It will be mv/t, but the tricky part is to determine t. This is a function of the spring constant (which depends on the structure of the blob-spring).

You start with the differential equation:

$d^2x/dt^2 = -kx/m$

which has a solution x = Amaxcos(ωt) where ω = √(k/m). You can solve for t where cos(ωt) = 1 to find the time of compression:

$t = \pi/2ω = \pi\sqrt{m/4k} = \pi\sqrt{m/4(mv^2/d^2)} = \pi d/2v$

So the time average force is:

$F_\bar{t} = mv/t = mv/(\pi d/2v) = \frac{2mv^2}{\pi d}$

whereas the distance average force is:

$F_\bar{d} = \frac{mv^2}{2d} = \frac{\pi F_\bar{t}}{4}$

So if you assumed that the time average force was the same as the distance average force, you would conclude, wrongly, that the wall did more work than ΔKE.

AM

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