If two cars of the same weight (mass), size, shape, etc. hit head on at 50 mph, does that equate to a relative force (for passengers) as if the car was traveling at 100mph? I can understand the relative approach speed to impact as 100mph, but would think the passengers would only feel the force of 50 mph ? as if the car hit a solid concrete wall at 50mph instead of another car.
What do you think?
If a moving car hits a movable object (parked car, sign post, etc.) the movable object will absorb some of the momentum providing a slower de-acceleration giving the passengers a “softer landing” so to speak. A head-on collision between two vehicles of equal momentum is effectively like hitting an immovable object. This will create a more rapid de-acceleration on the passengers. The 50% fatality rate for seat belted passengers is a 34 MPH delta V (instantaneous change in speed). Delta V can be going from 34 mph to zero or from 100 mph to 66 mph. If you are going to be in a accident try not to hit the on-coming car.
Twotone
UK accident statistics in this report: http://tinyurl.com/y9u2e7l
You really can’t generalize as to the force on one car or another except in laboratory conditions . Depending upon the engineering, what the passengers “feel” depends as much upon the energy absorption of the vehicle they are in. Would you rather be in a well designed, high strength steel constructed lighter car, or a good old heavy weight from years ago that would perhaps impale every passenger on impact, let alone two cars of the same weight and size but different years. If you’re comparing two identical bricks, then yes, some conclusion can be made. Otherwise, even identical cars loaded differently with one slightly lower than the other will present different results.
If you hit the oncoming car perfectly straight the forces inside the cabin are much greater than if you hit a flat fixed wall. The exact effect on impact is a “multiplier” of the force and so the question of how much greater is “a whole lot” without having a computer simulation program to do the math.
The physics laws relate to “kinetic” energy. The fixed wall has no kinetic energy at the time of impact. All the kinetic energy is in the car that is moving into the wall at the moment of impact. In the case of a head on crash both vehicles are carrying kinetic energy as they move toward each other. At the moment of impact you have much more energy to disperse as the cars collide. Therefore more energy to be dissapated by crush zones, and more energy transmitted to the cabin and the passengers in the cabin.
Usually the impact “head on” with another vehicle is not square and the twisting and spinning that results reduces the cabin impact somewhat. These twisting and spinning motions are whip like forces that compromise the ability of the passengers to survive the impact. Hence the advent of side impact air bags.
Neither impact, head on or fixed object, at 50mph would be pretty.
[i] If you hit the oncoming car perfectly straight the forces inside the cabin are much greater than if you hit a flat fixed wall. …
The physics laws relate to “kinetic” energy. The fixed wall has no kinetic energy at the time of impact. All the kinetic energy is in the car that is moving into the wall at the moment of impact. In the case of a head on crash both vehicles are carrying kinetic energy as they move toward each other. At the moment of impact you have much more energy to disperse as the cars collide. [/i]
Almost right, but not quite. In your example the fixed wall does no absorb any energy, it does not move. In the two car example, you have each car absorbing half the total energy, so in both cases the cars all absorb the exact same force.
Now a small amount of difference will result because the two cars are not mirror images so some parts may come loose and cross the line of impact or the cars may not stay exactly straight and will end up with parts of the car crossing the line of impact so you would be very slightly safer, on average, with the impact into another car. Of course striking a parked car would be even better since that car would absorb part of the total force, but would not provide any force to the event.
You describe the US DOT test - almost. They use a deformable barrier to simulate the second car. The offset frontal crash test conducted by IIHS is similar in that it also uses a deformable barrier, but tests a slightly different collision. As the other posters said, a solid wall is not an adequate simulation. You need the deformable one.
I heard this from a friend after He was in a sideways crash,
your weight x your speed, is the force you feel.
in this one, it would be your weight x 100 MPH, as apposed to weight x 50 MPH.
because you were indeed traveling at 100 MPH relative speed.
In a collision of two objects, there are two things to be considered: momentum and energy. Momentum is always conserved; energy is only conserved in an infinitely elastic rebound. (I.e. when an object hits the immovable wall, it can either “bounce back” at the same velocity it started with, it can just “splat” and not bounce at all, or it can do some combination. For all intents and purposes, cars just “splat.”)
Since the car crash is not elastic, energy must be dissipated. This energy is dissipated by bending metal (less bad) and causing greivous bodily harm (more bad).
Now consider the “immovable wall.” Since the wall is anchored in the ground, you’re effectively running into a planet. The momentum (car+planet) is the same as before the accident, but since the mass of the Earth is SO MUCH larger than that of the car, the correspoding induced velocity in the Earth can be disregarded, and assumed the Earth/wall combo does not move. Thus, the car “just stops.”
In considering two identical vehicles, moving at identical speed, the two cars collide, and ultimately “just stop.” In this sense, it is the same as the “wall”; the car has to dump all of its energy into the collision. The “crumple zone” is doubled between the two cars, but the “closing velocity” is twice as fast, so that the energy is dumped in (pretty sure) same time frame.
Seems like a wash to me.
Two identical cars traveling at 50 have the same impact as hitting a concrete wall at 50, not 100.
Paint a white line at the point of impact. That’s where the front bumpers stop in either case. The rate of deceleration is the same.
Now if one vehicle weighs twice as much as the other, it’s a completely different story…The heavy vehicle is not stopped by the impact but continues forward at 25 mph. It’s occupants have only suffered a 25 mph deceleration…The poor folks in the lighter car suddenly find themselves coming to a dead stop and then accelerating to 25 mph in the opposite direction. It’s like they hit a concrete wall at 75…
in this one, it would be your weight x 100 MPH, as apposed to weight x 50 MPH.
because you were indeed traveling at 100 MPH relative speed.
No. Each car and the occupants absorb half the energy. The immovable wall absorbs none.
then what velocity do you measure?
if they were headed at each other at 100 MPH relative speed and the impact was split in half between the 2, that would be equal to 50 MPH against a wall that doesn’t move.
…I think I answered My own question. never mind. Your right.
That’s exactly what I had in mind with this question. If the plane (your white line) is the point of exact impact and not crossed by either car then I would consider replacing the other car with the infinitely stiff ‘concrete’ barrier an equivalent substitute - therefor my thought that a 50mph crash with another equal car is about the same as one car hitting the immovable wall. Now approach speed is different and I think people mix the two often.