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Crumple zone in cars and ways to measure its effectiveness

  • Thread starter komender
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Homework Statement


I am trying to prove the efficiency of crumple zone in a presentation, but I think that I'm am doing something wrong. Basically I'm a grade 12 physics student, and I am suppose to present crumple zone. I understand that we need to impulse and momentum in here, and I tried setting up a problem that proves the effectiveness of crumple zone.

In one case, I show a car of mass 4000 kg that has a rigid front and travels in uniform motion of 30 m/s and it hits the wall. the front only crumples by 10 cm (this is an assumption that I'm making, based on nothing really) and I compare it to the same car, driving at the same velocity but with crumple zone at the front, making the crumple 30 cm.

The question is, am I making a good assumption? or am I suppose to account some other factors as well?

Homework Equations


Δd=vΔt-(1/2)aΔt^2.
FΔt = Δp.


The Attempt at a Solution


For the first case I am using the formula Δd=vΔt-(1/2)aΔt^2, where i substitute "a" with (v2-v1)/Δt, which results in 6.67*x10^-3 seconds to decelerate from 30 to 0.next I am using the equation FΔt = Δp to calculate the force that is exerted on the passengers, which results in 120000/6.67x10^-3 = 1.8x10^7 N.

I am doing the same calculation for the second case:
0.3=-(1/2)(-30)Δt = 0.02 seconds
120000/0.02 = 6x10^6N.

which proves that crumple zone (in this case) is reducing the force on the passengers by 66%.
But I am not sure if that is the force that the people are experiencing or the force that the car absorbs. Any suggestions.
 
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Answers and Replies

  • #2
haruspex
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You got the right answer, but you have the wrong forces on the passengers.
You correctly calculated the deceleration of the vehicle. Since the passengers are restrained to a constant position within the vehicle, they get the same deceleration. What force does that mean the passengers are subjected to (F=ma)?
 
  • #3
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Do you mean that I should use the deceleration that I found and multiply it by the car's mass to get the force that acts on the people? So what is the significance of the force that I calculated using the momentum and impulse equation? is it just the force that the car experiences? and if so, wouldn't that mean that the people inside the car would feel the same force?
 
  • #4
haruspex
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Do you mean that I should use the deceleration that I found and multiply it by the car's mass to get the force that acts on the people?
No. Forget the car for the moment and concentrate on the people. You know their deceleration rate. What force would produce that deceleration?
 
  • #5
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Wouldn't that simply be the reaction force of the wall?
 
  • #6
haruspex
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Wouldn't that simply be the reaction force of the wall?
Forget the car, forget the wall. A person weighing, say, 80kg is decelerated at 4500 ms-2. What force is required to do that?
 
  • #7
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Oh you mean the actual number... I see what you mean. I thought you asked what is the source of this force. So its not the force that comes from the crash that is important, but the deceleration. But what about the energy that is generated from the collision? Wouldn't that have a factor as well? or will it result the same thing?
 
  • #8
CWatters
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What kills people is their internal organs impacting the inside of the chest. The force with which they do that depends on how fast the body decelerate. I would look at another equation..

V2 = U2 +2as

from here..
http://en.wikipedia.org/wiki/Equations_of_motion#SUVAT_equations

That can be rearranged to show how the deceleration "a" relates to the stopping distance "s". The stopping distance is the length of the crumple zone.
 
  • #9
haruspex
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Oh you mean the actual number... I see what you mean. I thought you asked what is the source of this force. So its not the force that comes from the crash that is important, but the deceleration. But what about the energy that is generated from the collision? Wouldn't that have a factor as well? or will it result the same thing?
At the same speed with the same masses, the total energy must be the same.
As I wrote, if what you care about is the percentage reduction in force on the passengers, your calculation was fine. But the two values you had for those forces were much too large.
 

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