Physics: Force of objects in motion upon impact.?

Question

I never took a physics class and was wondering which impact would cause more damage (to a vehicle and to the human body).
Scenario 1: two vehicles traveling 70mph towards each other and hit head on.
Scenario 2: one vehicle traveling 140mph hitting a completely stationary object (like a concrete column that supports an overpass).
If you could break it down mathematically, that would be fantastic! Thank you

Answer 1

This is really the kind of thing you hire engineering consultants for actually because there are so many variables to consider.

In general:
When cars crash, some of the energy of the collision is absorbed by crunching the steel, heat, noise etc.

So if two identical cars collide head-on at 70 mph, the impact would be less traumatic than 1 car hitting a stationary object that wouldn't absorb impact from the collision.

Now if a car has a head-on collision with a much more massive vehicle like a semi or tractor, it would be worse than hitting an identical vehicle or smaller vehicles.

Both of your scenarios are likely to be fatal to the driver and front seat passenger at the speeds you described even if wearing seat belts and proper air bag deployment occurs.

Cars are crash tested at only 35 mph for front impact.

Mathematically it is best described by impulse, momentum and it's roughly an inelastic type of collision.

For Scenario 1:

m1*v1 - m2*v2 = F*t

m1=mass car 1
m2=mass car 2
v1=velocity car 1
v2=velocity car 2
F=crash force
t=time in which the collision occurs

F= (m1*v1 - m2*v2)/t
for identical cars: m1=m2
v1= 70 mph = 102.7 ft/sec
v2= -70 mph = -102.7 ft/sec
m = 2000 pounds/32.2 ft/sec^2 = 62.1 slugs

So:
F= m*205.3/t
F = 62.1*205.3/t

Collisions usually last only a fraction of a second.

so we'll use 0.3 seconds

F= 4.25E4 pounds = 21.2 tons


For scenario 2:

m*v = F*t

Since the car hits a stationary object the time of collision (t) will decrease.
so we'll use 0.1 seconds
F = (62.1 * 205.3)/0.1
F= 1.27E5 pounds = 63.7 tons


Basically when the car hits a massive stationary object the time of contact is shorter than when having a head-on collision. 21 tons will crush you very much like 63 tons will. In either case the forces involved will likely kill the driver and passenger.

Air bags will increase the time (t) of collision with respect to the person, but even if it increased it to 3 full seconds, the person would still have 2 tons of force working on him or her.

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UPDATE: In regards to supastremph's comments below:

One really cannot ignore impulse and the amount of time the collision takes place in. If this was not true, airbags would not save lives.

For Example:
Scenario 3: A head-on collision between two vehicles traveling 70 mph one vehicle weighs 2,000 pounds and the other weighs 20,000 pounds and there is a giant airbag between the two vehicles:

M*v1 – m*v = (M + m) * v2

621 * 102.7 – 62.1 * (-102.7) = 682.1 * v2

v2 = 102.9 ft/sec

F*t = 682.1 * 102.9

F*t = 7.02E4

Let’s say that the giant airbag between the vehicles increases the contact time (t) of the collision to 30 seconds.

F = 2.34E3 pounds = 1.17 tons
It would be possible for the driver and passenger of both vehicles to survive this collision.

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UPDATE 2:

The previous illustration was just to demonstrate in a big way the effect time of contact can have on decreasing the blow of the impact.

I would like to offer a new version of scenario 2 that will be more equivalent to scenario 1:

Scenario 2A:

Back one vehicle so it's bumper is against the stationary object. Have an identical vehicle hit it head on at 140 mph. This way the damage caused to the car at rest will absorb some of the impact.

I have degrees in physics and engineering and create the practical models to understand what will likely happen and use ridiculous models to better illustrate the science and understand what would be necessary to do for a solution. Also, it can be amusing. (-: Nerds have to entertain themselves somehow!

Answer 2

Just to get an ideea of the thing. First of all the kinetic energy:
Scenario 1:
- one car has m*v^2/2; both have m*v^2;
Scenario 2:
- the car has m*(2v)^2/2 = 2mv^2;
So the energy of the system is two times bigger in the second scenario;
Secondly, the impact works like this: the final momentum will be zero (cars don't bounce or move around after impact, they just get stuck there), so the entire kinetical energy has to go somewhere. That's the energy that destroys the cars and kills the people inside. Now, a car can absorb a lot more energy than a wall, because it changes shape (bending a piece of metal requires energy). Just imagine a ball fill with air and a ball filled with sand. The air ball jumps back because it doesn't change shape, while the sand ball changes shape so it uses all the energy, leaving none for the lift back.

Same thing with the cars. The two cars "share" the energy, so basically each of them gets (1/2)*mv^2; In the second scenario, however, there is only one car and the energy-rejecting wall, so that one car gets 2*mv^2. That's practically 4 times more energy than each car in the first case. So hitting a concrete wall is definetly worse than hitting another car. About 4 times worse :)

Mostly, in car accidents, people die when they hit a wall or a pillar or something. They have good chances to survive in a crash with another car. A 40-50 km/h crash with a wall would be enough to hurt you seriously or even kill you.

I hope I could help. Good luck.
And don't drink and drive, you could hit a bump and spill your drink :D

Answer 3

The analysis above is great, but to get to the heart of your question:

IF the two cars are identical, then the two cars come to complete stop after the collision in 1) meaning scenario 1) and scenario 2) are identical in trauma.

IF the two cars are of unequal mass, the least massive vehicle recieves the most trauma, which is more truma than hitting a stationary object. This is because of conservation of momentum. Assuming the cars become interlocked--head on collision--and one car doesn't fly off:

Mv - mv = Mv' + mv'

v' = (M-m/(M+m))v where v' is the new velocity of the wreckage and v is the initial velocity of the vehicle with mass M.

As M grows much larger than m, v approaches v', meaning the whole wreckage travels with the same velocity of vehicle M . . . meaning vehicle M won't even feel the collision, but vehicle m is forced not only to a stop, but forced in the opposite direction of which it was travelling! This is much worse than hitting a stationary object! If you are in a head on collison, would you rather be the semi, or the volkswagen, or a pedestrian?

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UPDATE: In regards to John's comments above:

Ha, ha--I assume you're my thumbs down, eh? Seriously, what collision at 70MPH between vehicles is going to take 30 seconds? You and I both know F1 = -F2 =dp1/dt = -dp2/dt, (which is independent of mechanical energy conservation). This implies the distance between you and the other person in the car is continuosly shrinking (. . . initially at 140MPH!) until the collision reaches its final state of net momentum. (And given the other vehicle is much more massive, it's momentum will have changed a lot less . . . ) 30 seconds means an awful lot of car to crumple in front of you. Airbags give you a few crucial feet, I'm not arguing about that, as I said, your reasoning is sound . . . but in this case, impractical. Maybe if you are in the back of a 50ft stretch limo you'll have that much time to work with, to the demise of your chauffeur--- but given a more reasonable 6ft distance for the hood of your car . . . if you haven't changed your state of momentum to the near final state in 60 milliseconds, it's going to be you personally hitting the oncoming car in your now-a-pancake car. From your exposition it sounds like you're an engineer? I'm a physicist--I thought you were supposed to be the practical ones. I'm just looking at this from the "There are no 500ft marshmallow-soft airbags" position. You've made your point, I just don't see the purpose of it. Are you a "giant airbag" salesman? (If so, I'll take one!)

Answer 4

just to get an ideea of the object. First of the entire kinetic vigour: difficulty a million: - one motor vehicle has m*v^2/2; each have m*v^2; difficulty 2: - the motor vehicle has m*(2v)^2/2 = 2mv^2; So the vigour of the perspective is two activities greater effective interior the 2d difficulty; Secondly, the impression works like this: the final momentum is often 0 (autos do no longer start or pass around after impression, they simply get caught there), so the entire kinetical vigour has to bypass someplace. it incredibly is the vigour that destroys the autos and kills the persons interior of. Now, a motor vehicle can soak up lots greater effective vigour than a wall, once you communicate approximately that it variations variety (bending a artwork of metallic demands vigour). in basic terms think of a ball fill with air and a ball packed with sand. The air ball jumps back once you communicate approximately that it does not distinction variety, on a similar time as the sand ball variations variety so it makes use of the entire vigour, leaving none for the bring up back. comparable ingredient with the autos. the two autos "proportion" the vigour, so virtually each of them gets (a million/2)*mv^2; interior the 2d difficulty, regardless of the indisputable fact that, there is just one motor vehicle and the vigour-rejecting wall, in order that one motor vehicle gets 2*mv^2. it is exceptionally much 4 activities greater effective vigour than each motor vehicle interior the 1st case. So hitting a concrete wall is definetly worse than hitting a greater motor vehicle. approximately 4 activities worse :) generally, in motor vehicle injuries, persons die as quickly as they hit a wall or a pillar or some ingredient. they have nicely opportunities to survive in a crash with a greater motor vehicle. A 40-50 km/h crash with a wall could be satisfactory to break you critically or perhaps kill you. I desire i could help. stable stable fortune. and don't drink and tension, you may hit a bump and spill your drink :D

Answer 5

The energy at impact would be equal in each case, I can't help with the mathematics.

Answer 6

Scenario 1, I cant explain why though...Sorry, but I know for a fact its scenario 1!

Answer 7

It would be the first one, because of the momentum of the crash, that would even out the damage.The force would be at 0.......