Greater momentum on impact means greater force?

[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.

This thread will stay closed.