speed and relative damage in car crash


by Avila
Tags: crash, damage, relative, speed
Avila
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#1
Mar2-12, 05:03 AM
P: 3
Hi,

This week I was in an accident - one track road head on round a bend.

My impression was that the other car was faster and yet their car was more damaged.
I know that make and design is a huge issue in this but trying to recall my physics and failing.

In the not real, frictonless piston tradition -

If both cars were identical models and one was more damaged does that mean they or the other one were going faster?

I expect the claim to go through as 50/50 but wondering if there is a chance of one side blaming the other - ie them blaming me based on more damage to them.

So any thoughts on my hypothetical version?
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Drakkith
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#2
Mar2-12, 05:05 AM
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If both cars are identical then the damage is also identical as long as they are hit in the same spots. My guess is that your own vehicle was larger, heavier, and a little sturdier than theirs.
Avila
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#3
Mar2-12, 07:46 AM
P: 3
My car is smaller, but apparently tougher!

But in the hypothetical case, if the damage is not equal then it follows that the force must be different, and if mass the same then velocity would be the variable. My question is which would be going faster? Is it as obvious as the faster one causes more damage to the other, or could that force affect it itself??

tiny-tim
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#4
Mar2-12, 09:15 AM
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speed and relative damage in car crash


Newton's third law

the force of your car on their car is the same as the force of their car on your car.
nsaspook
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#5
Mar2-12, 11:12 AM
P: 492
Once you get above a slow speed crash the forces exerted on the car frame and occupants are huge and get to the point where there can be limited additional damage at impact points. http://www.youtube.com/watch?v=6dI5ewOmHPQ
Avila
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#6
Mar2-12, 12:25 PM
P: 3
Quote Quote by tiny-tim View Post
Newton's third law

the force of your car on their car is the same as the force of their car on your car.
Ok, so as said above, but I didn't grasp why - the damage on equal cars would be equal regardless of a difference in speed.

Thanks for that.

I am someone who likes to scratch an intellectual itch when it comes even when not in my field - so thanks folks.


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