Head on collision force

In the event of any collision all energies of the system need to be considered. The system would involve the 2 cars the crash dummies or any other added outside mass.

Now if the 2 vehicles were truly Identical then they both have the same amount of stored kinetic energy. That energy is then combined in the total amout of available energy for release in the system. Or better to be stored or absorbed by the effects of the collision. For this reason the Law of Conservation of Momentum apply.

Excerpt from the web:

"The Law of the Conservation of Momentum: The total momentum of an isolated system of bodies remains constant

But what does that mean? That means when you add all the momentum from each object before and after the collision, they will equal each other as long as another force does not affect the system. That is what an isolated system means, where no outside force jumps in and ruins everything. So what would be an outside force? Well, if you started with two cars and a third car smashed into them, then the system of two cars was disturbed by the third car. However, if you include the third car in the system, then it would not be an outside force anymore. It is a little complicated but understandable.

So does anyone play pool? That's a game of collisions and momentum. When two balls collide all or part of the momentum from one ball is transfered to the second ball. If all of the momentum is transferred to the second ball, then the first ball will be at rest and the second ball will move with the same speed as what the first ball originally moved at (this is called an elastic collision)."

Not sure if this really answers you question or not but I gave it a shot.

double j b --- Thanks for the explanation --

That energy is then combined in the total amount of available energy for release in the system.

Does the above statement equate to adding the 65 MPH for each car to equal the sum of 130 MPH for determining the total amount of energy available? I'm still somewhat confused. Or is the impact head-on exactly the same for two cars, each doing 65, the same as one car hitting a solid stationary object at 65?

Sorry for being dense -- but still not clear on the basic question.

You are correct. The sum of the kinetic energies of the two vehicles would be the right answer. Think of it as two vectors. Since they are both going in opposite directions the resultant force would be the sum of each other.

Another way to think of it is as if one vehicle was stopped and the other traveling at 130 mph. The crumple zones only spread the impact force over time (impulse of the collision). Two pool balls have a very short impulse time, but of great magnitude in force. Cars have a lower impulse magnitude, but the duration of the impulse is much, much longer. For a dummy (or a passenger who did not wear a restraint system (and could rightly be classified as a dummy)) the crumple zones help somewhat, but the dummy's momentum will ultimately carry it into the windshield, where that impulse will be very sharp, but of somewhat lower magnitude due to the vehicles slowing down.

The duration of the impulse is what causes bodily damage. Karate students are taught to punch quickly, but also to yank their arm back just as quick to cut the impulse duration down to a minimum and maximize body damage. The force of the blow is not that great. If you were to distribute that impulse of that same force over 60 seconds of time it would feel like a very slow and gentle push. That is what crumple zones, air bags, and seat belts do for passengers. It widens the impulse duration to allow for forces to be distributed over a longer period of time. Imagine falling off 100 feet from a building onto a concrete sidewalk. Death is almost certain. Yet people walk away from 200 mph crashes in a Formula One car all the time, which is much, much more forceful than a 100 foot fall. The energy of the impact is dissipated over time.

If the vehicles are identical, each experiences the equivalent of hitting a rock wall at 65 mph. Theoretically, neither one extends past the plane of collision and so doesn't know what is over there. None of the energy from the other car gets transferred to you - he keeps it. Obviously, in a real-life situation with crumple zones, etc, this is only approximately true.

Succinctly put. So the two impacts are essentially the same.

The flaw in the other argument is that taking one of the cars as a reference, the closing speed of the other car is indeed 130mph, but after the collision it will continue travelling at 65mph and its loss of momentum is balanced by the increase in momentum (backwards) of the previously stationary car.

Does that clarify or confuse TVP45's answer?

I should have pointed out that Kinetic Energy is not conserved in an inelastic collision. That makes it a little clearer, I hope.

That sounds about right to me.

I have to agree with the above statement that it would be like driving your vehicle into a rock at 65 miles per hour.

Of course that would be kind of like the unstoppable force hitting an impenetrable, immovable object. No one knows for sure what would happen. But it would make a helluva bang.

At any rate in the collision the vehicles sustain damage not necessarily from the combined impact speed but there own speed and mass colliding with an identical force. If I remember this corectly then much like the pool ball description if you had two pool balls of the exact same weight rolling at one another at the same rate of speed what happens when they hit?

If the speeds were combined then they would repel from one another at twice the impact speed (or shatter) but if I remember corectly they actually just stop.

This would be an experiment you could use to show your hypothesis.

That is if I have'nt just totally screwed this up. Its been 26 years since I studied physics.

Maybe STLEngineer can clear this up. He's pretty sharp on this stuff.

"If the speeds were combined then they would repel from one another at twice the impact speed (or shatter) but if I remember corectly they actually just stop."

No, they do neither. They will bounce back from each other at almost the same speed they collided. The collision will be close to perfectly elastic. A small portion of the energy will be absorbed by the balls, but the rest is used in the recoil of the two balls.

The example cited by the original post is not like hitting a rock at 65 mph. You can prove that by doing the math. It is basic physics.

Another way to look at it is that the speeds are relative. One could be stationary and the other traveling at 130 mph. The result is the same energy wise. Remember, energy is conserved, so you can easily do the math to prove the effect.

Imagine a newtons cradle! This is a good demonstration for the conservation of energy!

This old discussion—I first heard it in the 1972 and I assume it was old then—is really just confusion that results from placing the energy balances around each car separately versus placing the energy balance around both cars.

In essence when looking from the perspective of car A in the first scenario, the "other" car B is considered the same thing as an immovable object such as a wall.

If this were true than the mass of the immovable object could be ignored as long as it is large enough not to be moved substantially by the force of the impact—a wall is a wall is a wall.

That is not the case if car B is larger than car A so the analogy between the wall and car B breaks down.

For example, would a baseball rebounding from collision with a bat of identical mass and speed, rebound at the same speed from a stationary bat as it would from a bat that is initially moving in the opposite direction?

Of course not. The fast swinging bat would always drive the ball faster and further than a stationary or slowing swinging bat—even if the bat were moving at the same speed (in the opposite direction) as the ball.

This is actually a good opportunity for thinking about the principle of relativity, i.e., each car as well as a standing observer must see the same thing since we have only constant velocities. Thus, either car must suffer the same damage no matter the reference frame.

Tom

cant agree with that last statement with relationship to this question! Your baseball bat has follow through! Think of the fairground punch-bag! if you have a heavy baseball bat set up to hit the bag at 100km/h, the bag would get a right smacking, if you had a kids hollow plastic bat and did the same thing, do you think the bag would register such an impact? It's a vector force! The question was 2 cars of equal weight with equal velocity or a car going twice as fast!

Totally lost myself now!

Think I'll go for my siesta now....

OK, an ideal case--non-deforming, perfectly elastic ball and bat with the speed of sound in the ball and bat faster than the relative velocity of the ball and bad--think steel bar and ball bearing. I did already say that the bat and ball were the same mass.

I was attempting a simple example in which the objects colliding from different directions had a greater effect than the one object hitting a stationary object.

Kind of like the difference between physics and engineering--physicists talk about the ideal and ignore the non-ideal effects but report results to 6 significant figures and engineers take into account all of the non-ideal effects and then round off by 20%.

In response to the thread above, the two identical objects colliding head on at 65 mph only have a combined energy of half of the kinetic energy as one object hitting a stationary object at twice the speed (130 mph). Kinetic energy is one half mv squared--the faster single object would have greater energy upon impact than the two objects colliding to a standstill even when taking into account the two objects with twice the mass.

By the way, that was a great analysis at the link (Imagine a newtons cradle!) you mentioned earlier. Thanks.

Hi BuckD

You're right about the kinetic energy being doubled at 130 mph as compared to twice 65 mph. In an elastic collision (like Newton's Cradle steel balls or pool balls), kinetic energy is conserved. In an inelastic collision (like grape jello hitting a rotten tomato), kinetic energy is not conserved. And that makes all the difference. Car wrecks are mostly inelastic.

Tom

Thanks to all "VERY MUCH" for responding to this question.

Please correct me if I am misunderstanding the posts in this thread. The general idea I am getting I will try to illustrate below: (I am sure I am being overly simplistic – please forgive me)

Event 1 = Car A and Car B impact at 65 MPH head-on. --- Even though the closing speed total is 130 MPH each car will only experience the energy of a 65 MPH impact with a solid object (wall). So "Event 1" total energy is that of a 130 MPH collision but each car has essentially hit a wall at 65 MPH and only sees that energy. The other car equals, basically a wall due to the identical speed.???

Event 2 = Single car hits solid object (wall) at 65 MPH. – This energy in this event would be roughly equivalent to the energy A and B individually experience in "Event 1".

Event 3 = Single car hits solid object (wall) at 130 MPH. – This energy in this event would be totally different and would equal double the energy A and B individually experience in "Event 1". Although it would equal the total energy of event one with A and B combined.

This is how I understand what most of your responses are saying. I think I got it.

So the mechanic who said two cars each at 65 MPH hitting head-on is basically the same for each car individually hitting the wall at 65 MPH is correct.

I don't think you have achieved a consensus yet. Sorry

An observer (say that lousy fly from the previous thread) standing on the bumper of car 1 sees car two coming at him at 130 mph since the relative velocity between the cars is 130 mph.

The kinetic energy involved is 1/2mv^2 where v is the relative velocity.

The resulting damage done (regardless of elasticity) is based on the energy of the crash and therefore the damage done to car 1 would be roughly equivalent to running car 1 into a rock at 130 mph since in this case the relative velocity between the rock and the car is 130 mph. Same energy, and not enough difference in elasticity between car 2 and a rock to make any difference to the insurance company.

So, shall we start round the circle again?

Oh yes the insurance company -- a whole new issue. Maybe the wind or surf changed the insurability, Or maybe the fly on the bumper was the reason the coverage was cancelled.

In all seriousness are any of the events 1-2 or 3 in post 15 remotely accurate? The way I read your comment is "event 1" in post 15 is closely related to your response. ??? Am I not understanding well??

Steve,

I disagree. The elasticity makes all the difference.

In an elastic collision, each car would remain "as good as new" as in a billiard ball collision. The kinetic energy would be conserved and the 130 mph idea applies.

In an inelastic collision, a fair amount of the kinetic energy goes into crumpling the car(s) and thus is not conserved. Total energy is conserved, but not kinetic energy.

Tom

Tom, I agree with you, and what I was trying to say is that amount of damage is related to elasticity, but that is not relevant because in the original statement of the problem elasticity is a given.

"One person was saying that the closing speed for two is 130 MPH and the impact would be similar to hitting the stationary wall at 130 MPH. The other argued that there is absolutely no difference. He said that the impact for two would still be the same as a 65 MPH impact for the car vs wall impact."

You are completely correct, if we were talking about billard balls there would be no damage at all, only energy disapated in sound and motion (CLICK!!!!).

My point is that the elasticity of a car, is close enough to that of a wall (although not the same), that the resulting damage done to the cars running them into each other would be very similar to the damage done running a car into a wall at 130 mph.

The key issue is that in either case the impact energy is based on relative velocity, and the relative velocity is 130 mph.

Hello Steve,

but that is not relevant because in the original statement of the problem elasticity is a given.

I missed that.

Tom

I'm sorry,

I see where you are coming from now. In the head on collision half the total impact energy (1/2mv^2 with a relative v of 130 mph) is shared in the destruction of each (assumed equal mass and elasticity) car, so that is the basis for the conclusion that it is equivalent to running one of the cars into a wall at 65 mph? Because only half the impact energy goes to destroying each car.

Again I am sorry for re muddying the water...

Ok so basically wher I lost it was, I understood the kinetic portion but not the inelastic portion. That is if I am understanding correctly.

Let me take a stab.

Because of the inelasticity of the vehicles then the total kenetic energy involved in the colision imparts into each vehicle. So therefore the total collision speed will equate to the 130MPH wall hit.

Like I said in my post it's been a long time.

I started to get real confused for a bit there.

I am much less confused but I wonder if Fyzz is around, he always has such a unique way of discribing things sometimes.

In simple form,

Car A and Car B are travelling at 65 mph each and collide head on, totalling 130 mph.

But energy is distributed to each vehicle based on mass. Since vehicles are identical, each feels the effects of hitting a theoretical immovable object at 65 mph.

Next Scenario, Car A has a mass of X and car B has a mass of 2X, both travelling at identical speeds and collide head on.

The energy should effectively exchange so that car A feels 2X of the energy from the collision while car B only feels X energy.

Right, wrong, maybe???

Or is it that each vehicle feels 1.5X of the energy from the collision, so that effects can me determined by:

1.5X energy/mass of vehicle

???

In which case, the bigger one still wins.

Forget cars for a minute and consider trolleys with springs attached to the front like we used at school for proving out the laws. Start imagning two identical mass trolleys, each with springs on the fornt of identical stiffness. The impact will result in the trolleys bouncing apart with the same exit velocities, the springs having deflected by the same amount. Now try doubling the mass of one trolley. The springs still have the same rate so which will deflect more??? What will be the exit velocities?

Go back to the trolleys with the same mass and double the spring stiffness of one of the (put two springs on if you like). Now what happens?

Etc..

OK, 2 trolleys moving toward each other down the same set of tracks. One weighs 1000 lbs. and the other 2000 lbs., both have big honkin springs on the front and both are moving at 65 mph.

for the sake of units:

65 mph = 1144 inches/second

1000 lbs. = 1000/386.4 (gravity in inches /sec) = 2.58799 and 2000 lbs. = 5.17598

we can calculate each force independently and add them together or just tally the two and run with newton's 3rd law.

so let's say it takes 1 second to stop each trolley this means a deceleration of 1144 inches/sec^2 x 2 totaling 2288 in./sec^2

2288 x a combined mass of 7.76397 yields 17,763.96336 lbs. of force (or in other terms BANG!)

dividing this force by 2 gives us 8,881.98168 lbs of force on each trolley.

dividing the force by the mass of each trolley yields it's acceleration away from the point of impact/spring collapse.

the 1000 lb. trolley = 8881.98168/2.58799 = 3,432 in/sec^2 for 1 second

the 2000 lb. trolley 8881.98168/5.17598 = 1,716 in/sec^2 for 1 second

Conclusion: each car will feel roughly what the opposing car would feel hitting a "brick wall". The forces exchange with each other. The accelerations are simply representative of the actual effects/energy.

This also means that hitting a truck head on while driving a economy car is worse than hitting a brick wall.

Why assume each trolley stops? By my reckoning there is an exit velocity that is positive in the direction of the heavier trolley, it doesn't stop, the 1000lb trolley bounces off. I did suggest starting with trolleys of the same mass and springs of the same stiffness, then they would stop, both of them (by my reckoning).