Calculating Force of Car hitting wall

[poole_3]

I am trying to calculate the force of a 2300 lb car hitting a wall at 40 mph. I know that momentum is mass x velocity and that force is mass x acceleration, but I'm not sure how they both relate to find the force on the brick wall.

Any ideas?

 

[marcusl]

To calculate the force, you need the acceleration. You can estimate this by assuming the acceleration (actually deceleration) is uniform and occurs in a time tau. It is complicated by safety features (crumple zone) in the auto that absorb and dissipate energy.

The following thread contais some details:
https://www.physicsforums.com/showthread.php?t=365317"

 

[Jonnyb42]

To calculate the force it would apply you'll have to know how long the car is impacting the wall for. You can estimate it if you don't have any other information, or calculate it somehow knowing certain properties of the materials of the car if you know that much. I, not having any other information, could estimate the impact to last let's say .2 seconds. Assuming the car impacts the wall to zero speed, the impulse will be equal to the moment it has.
p = mv
$$\Delta$$p = F$$\Delta$$t
and since final momentum is zero,
mv = F$$\Delta$$t
or,
F = $$\stackrel{mv}{/Delta t}$$
2300 lb's $$\approx$$ 1043 kg (mass of car)
40 mph $$\approx$$ 18 m/s (mps) (velocity of car)
We can just estimate $$\Delta$$t to be .2 seconds duration, and you get F to be,

F = $$\stackrel{(1043 kg)(18 mps)}{(.2 s)}$$ = 93870 N = 21103 lb's

(I converted the weight of the car to the mass of the car in Kg and all other units to SI units, then converted final force back to lb's)

 

[poole_3]

My problem is actually two cars traveling in opposite directions at 40 mph. On car veered into the other lane and hit the side of the other car at a 20 degree angle. I need to calculate the force of that impact. Do the velocities of both vehicles need to be taken into account?

Also, is .2 seconds a reasonable assumption for the time of this impact?

Thanks

 

[DannoXYZ]

The velocities and mass of the two cars needs to be taken into account. Also the area of the cars that were impacted will determine the impact time. Some areas of the car are quite stiff, like the firewall and suspension areas. While direct frontal & rear impacts have the softest impacts. So each car may have a different impact/crush time depending upon what parts were hit.

What is interesting in this puzzle is that two cars of equal mass hitting head-on will have the same impact as each one hitting a brick wall independently.

 

[poole_3]

Hmmm, this is getting complicated. I just need to calculate the force of a car traveling at 40 mph hitting another car traveling at 40 mph at a 20 degree angle. Are there any conservative assumptions we can make? Conservative in this case would be less of a force since we're trying to prove that this collision could do the damage that it did.

 

[DannoXYZ]

The other factors you need to account for are passenger weights and fuel-tank weights. The kerb-weight listed in the manufacturer's specifications are typically with the tank 50% full. The actual cars in the collision needs to be weighed as a full or empty tank can vary the weight by +/-100 lbs, leading to differing KE figures.

Without extensive FEA data on the two cars in question, there's no accurate way to model a collision. Simplest method is to carry out the actual collision with the demo cars loaded identically to the actual crashed ones, and videotape it for analysis.

 

[FoxCommander]

Ok so you have two cars one both going forty miles an hour. One hits the other head on at a twenty degree angle. So you must say momentum is conserved. So these two cars, assuming that they stick together will travel at a NE direction assuming that the car that was heading straight was heading due east and the other car due NW. Then you will see the change in velocites, You have the time so you will have the acceleration. Now all you have to do is multiply this times the mass

One problem i can forsee is that the cars will change directions aswell. I don't know if this affects my method i have stated above. I believe it does but I am too tired to get my calculator and i just did this in my head ha ha.


FoxCommander