B Greater momentum on impact means greater force?

[dibbsy]

Then what you think is wrong.

I say it's all about the kinetic energy. Nothing to do with acceleration or the Third Law. How is this incorrect as far as physics goes??

 

[weirdoguy]

How is this incorrect as far as physics goes??

I have showed you how by explicitly wirting out Newton's second law which included acceleration. I can't do more than that. You have basic misunderstandings about what acceleration is.

 

[Ibix]

I say it's all about the kinetic energy. Nothing to do with acceleration or the Third Law. How is this incorrect as far as physics goes??

Transfer of energy is the work done, which is the force applied times the distance moved. If there's no force then there's no energy transfer.

Looking at energy transfer is another way of looking at the force, or the integral thereof. It's maybe an easier way to look at it, but the force method tells you exactly the same thing.

No time to reply to your answer to my last right now.

 

[dibbsy]

I have showed you how by explicitly wirting out Newton's second law which included acceleration. I can't do more than that. You have basic misunderstandings about what acceleration is.

You still haven't shown how what I said is wrong.

 

[dibbsy]

Transfer of energy is the work done, which is the force applied times the distance moved. If there's no force then there's no energy transfer.

Looking at energy transfer is another way of looking at the force, or the integral thereof. It's maybe an easier way to look at it, but the force method tells you exactly the same thing.

No time to reply to your answer to my last right now.

OK, so then why not conclude that the greater force of the faster car comes from the greater kinetic energy? The transfer comes from the impact. The impact involves a force. The greater the kinetic energy, the greater the force that is transferred. No mention of acceleration.

 

[malawi_glenn]

No, change in velocity can be acceleration or deceleration.

Deceleration is just a fancy word for "acceleration that decreases the speed"

Consider circular motion with constant speed. But the velocity is changing at every instant because the direction is changing. What should we call this effect? It is not changing speed...

https://www.vedantu.com/formula/deceleration-formula
It is more of an informal concept, perhaps that is why you are confused?
(in my country, we say things like the velocity limit on roads, etc. Which is wrong but it is common saying).

Acceleration is rigorously defined as (a vector) $\displaystyle \vec a = \lim_{\Delta t \, \to \, 0}\dfrac{\Delta \vec v}{\Delta t}$

Tip: try to unlearn the concept deacceleration, it does no good

 

[Herman Trivilino]

Sorry, but don't you see how absurd that sounds? If the car is going faster, the deceleration is larger. Call it 'negative acceleration' if you like, but it no more explains the larger force than my going backwards faster than you explains any damage to what is in front of me. Absurd!

The word decelerate is confusing you. Any time you speed up, slow down, or change direction you are accelerating.

To decelerate means to decrease speed. This has nothing to do with whether the acceleration is positive or negative.

 

[jbriggs444]

OK, so then why not conclude that the greater force of the faster car comes from the greater kinetic energy? The transfer comes from the impact. The impact involves a force. The greater the kinetic energy, the greater the force that is transferred. No mention of acceleration.

Force is not something that is transferred. A force is a transfer of momentum. The thing that is transferred is momentum, not force.

When you transfer momentum to something then, in the absence of other competing effects, the result is an acceleration. That is Newton's second law: $\sum F = ma$.

The transfer of kinetic energy is through a quantity known as "work". This can be numerically calculated as force multiplied by displacement in a direction parallel to the applied force: $W = F \cdot s$.

Yes, if the car is perfectly rigid, comes to rest and does not rotate as a result of the collision then all of the kinetic energy is lost by the car and will manifest as work done by the car on the wall.

Yes, the work done on the wall by the car will be energy gained by the wall. If the wall remains roughly at rest following the collision so that its bulk kinetic energy remains zero, this energy will have to be absorbed or dissipated somehow.

Yes, if the wall has no elasticity or resiliency so that the energy cannot be stored as vibrations then the energy will go into permanent deformation, breakage and thermal energy.

I believe that there is a rough proportionality -- the greater the absorbed energy is per unit mass, the more finely divided the rubble will be. More energy = more rubble or a more powdery residue.

My understanding is that one of the ways that kinetic energy was first studied was by looking at the variation in the size of impact craters in mud versus fall distance. This is a better scenario to study than cars crashing into walls. Cheaper, easier, more measurable.

 

[zdcyclops]

OK so the cars are slowed when they hit the wall. That means they decelerate. The faster the car's velocity, the greater the deceleration needed to slow it down, so the greater the force needed to slow it down. So how does that explain the greater damage done to the wall by the faster car?

Acceleration and deceleration are the same thing. There is no time variable in the equation. Acceleration is the velocity. Reverse the equation m x a = F

 

[weirdoguy]

Acceleration is the velocity

What?

 

[PeroK]

What I want to know is why there is no damage in this scenario?

 

[Mark44]

Acceleration is fundamentally about changes in velocity.

And by "fundamentally" I assume you mean by definition.

Acceleration is the velocity.

Not even close. Acceleration is the change in velocity.

 

[FinBurger]

Dibbsy: Your lack of linguistic rigor suggests to me that you are not ready for physics. You use words and equations without understanding or definition. You will never get anywhere doing that.

As to your question, you ask about "damage". What is damage? How do I measure "damage". Until you figure that out, trying to answer the question, "Which car does more damage?" is impossible.

 

[hutchphd]

Even shouting at a blind man will not make him see.
The title is nonsense:
Trying to understand force and momentum. If F=ma, then force is about acceleration, not velocity. But what about greater impact with greater momentum and no acceleration?
What is an "impact..... with no acceleration"? Try to understand that this is an impossibility.

 

[berkeman]

Thread closed for Moderation...

 

[jrmichler]

TL;DR Summary: Trying to understand force and momentum. If F=ma, then force is about acceleration, not velocity. But what about greater impact with greater momentum and no acceleration?

Sorry for this beginner's question, but...if F=ma, then force is all about acceleration. But if vehicle A moving at constant velocity V hits a wall, and vehicle B moving at constant velocity greater than V hits the wall, then B hits the wall with greater momentum than A and does greater damage etc., so it must have hit the wall with greater force than A. So nothing about acceleration in this scenario, but force is greater. What am I missing here? Thank you in advance.

There are a number of things going on in this question. So we need to simplify things. First of all, we simplify by considering two kinds of impacts - elastic and plastic. In elastic impacts, things bounce. Billiard balls are an example. In plastic impacts, things crush without bouncing. A vehicle crashing into a hard wall is a plastic impact, it crushes without bouncing. Modern vehicles are specifically designed to crush with constant force in order to minimize crash forces on the occupants.

More simplifications: Two identical vehicles, both hitting the wall head on, and a rigid wall. The two vehicles are at different speeds. The rigid wall is a physics wall, it does not move, nor is it damaged. Assume that the first car is traveling 30 MPH (44 ft/sec), and crushes 18 inches. These numbers are pulled out of thin air to illustrate the point, and are not intended to represent reality.

Since the rate of acceleration is constant during the crash, the velocity decreases at a constant rate, so the average velocity during the crash is 44/2 = 22 ft/sec. Then the duration of the crash is the distance divided by the average velocity = 1.5 ft / 22 ft/sec = 0.068 seconds. Note that this impact, like all impacts, has a finite duration.

The acceleration (see earlier discussion of deceleration vs acceleration) is the change in velocity divided by the duration of the impact = (44 ft/sec - 0 ft/sec) / 0.068 seconds = 645 $ft/{sec^2}$.

The second vehicle, identical to the first vehicle, is traveling faster, 45 MPH (66 ft/sec). Since it is identical to the first vehicle, and since vehicles are designed to crush at constant force, then it will crash at the same acceleration of 645 $ft/{sec^2}$. The crash duration is the starting velocity divided by the acceleration, or $(66 ft/sec) / (645 ft/{sec^2}) = 0.102 sec$. Since the acceleration force is constant during the crash, the velocity decreases at a constant rate, and the average velocity is 66/2 = 33 ft/sec. The crush distance is the average velocity times the crash duration = 33 ft/sec X 0.102 seconds = 3.38 ft = 40.5 inches.

Note that both crashes have the same acceleration during the crash, but the second crash does a lot more damage. Since the acceleration during each crash is the same, and the mass of each vehicle is the same, then the force is the same because $F = ma$. The damage in the second crash is larger because the force happens over a longer time. In this example, the acceleration of 645 $ft/{sec^2}$ is equal to an acceleration of 20 G's (645 $ft/{sec^2} / 32.2 ft/{sec^2} = 20 G's$. The force on a 3000 lb vehicle would thus be 3000 X 20 = 60,000 lbs.

You may need to review the basic physics of acceleration, velocity, and position in order to properly understand all of this. It would be time will spent.

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