60 miles + 60 miles is not force, it's only the velocity vector

I also took Vector Mechanics for Engineers in college and I got an "A’. I agree that the fundamental vector mechanics is simply 60 + 60 = 120 miles per hour. However there are other factors, that’s just the velocity vector guys! You have to take into account the weight of the vehicles. If I am driving a 350 Ford Pickup at 60 miles per hour and had the choice of running into a GEO coming at me at 60 miles/hour or hit a wall, I would choose the GEO guys! When you calculate the force on the pickup it’s going to be lower for the GEO as opposed to a stationary brick wall of a building.

I must have missed something, because the only response I can think of right now is “What?”

OK,but what issue/question are you addressing?

Re: the front end collision question at the start ofth show weekend of 23may08

Full analysis, without math

The points in question are: change in momentum and the time over which the change occurs.

Ignoring the time issue (at these speeds, it won’t be real relevant), the change in momentum for a given vehicle will be half as much for the 60-60 head on than the 120 into a SOLID wall, making this the preferable (preferable? how about neither of these…).

If the wall is NOT solid, then it depends how solid it is. Assume it is a stopped, identical car, rather than the proposed SOLID wall. After the hit at 120, the formerly stopped car will now be moving, and the fast car will be slowed. This is the same as the 60-60 collision. As the target gets heavier/more solid, the change in momentum of the fast car increases to the maximum at the case of the unmoving target, making for a harder hit.

Absolutely the Geo, from a physics standpoint.

Yes, the total velocity difference is important, but mass is also a key factor here. A train hitting a car at a crossing hardly notices the event due to the mass difference. Although the difference between the pickup and car is not nearly as great, you probably have at least a 2:1 advantage here, negating any velocity difference.

You also have to take into account “moveability” of the object. The Geo is only attached to the earth by four (or less depending how you hit it) small patches of rubber. The wall is attached to the earth along its entire base. Hitting the Geo wins here.

In addition, the Geo’s metal body should absorb more of your pickup’s energy than a brick/mortar wall. Hitting the Geo wins here.

So yes, hit the car - it’s a much better choice (at least for the physics reasons). From the spiritual and judicial liability of probably killing the Geo driver to save yourself by doing this… I guess that’s for your ethics class. :slight_smile:

  1. Your grade does not impact the basic physics in consideration.

  2. The entire chat and your reply deal with veocites as you point out. It is true that VIEWED FROM THE FRAME OF REFERENCE of one car at rest, the other car is driving 120 mph. [HUGE ASIDE: This is exactly why engineering experiments can simulate a moving aircraft in a wind tunnel. From a physics point of view, an aircraft moving through air is an unsteady problem while the air moving past the aircraft is a steady problem.] You could consider the thought experiment that Einstein discusses in Relativity. If you have a train moving and you toss bags out of the door, these bags are also moving at the speed of the train. So they are moving at (say) 60 MPG in the same direction of the train relative to the solid ground (track). If a train is coming opposite at 60 MPH, it will impact the ejected bag. Relative to the fixed ground, it appears that the second train hit the bag going 120 mph.

  3. The issue that the caller had in his head deals with the energy equation, not the vector sum of velocities. You allude to this as well but no one has followed through.

Both cars are travelling with a kinetic energy of 1/2 * mass * velocity * velocity. Assuming an ideal elastic collision (head on and both come to a dead stop), this energy is absorbed. (An inelastic collision is like pool where the balls do not absorb the energy and bounce off each other) For the cars to come to a dead stop at the point of collision, as you point out, they must have the asme mass. If one has a greater kinetic energy, the system of two cars will move in the direction of motion of the heavier car. Assume equal mass.

Each car must absorb some of the combined energy. Cars have crumple zones to do this without harming the passenger. If both cars absorb the energy of equal amounts, then each car absorbs (dissipates) 1/2 mass * velocity * velocity [joules] of energy. If the wall absorbs 1/2 * mass * velocity * velocity joules of energy then the collision is the same as the cars. If the wall does not absorb the same quantity of energy, then the car hitting the wall at 120 MPH must absorb more (or all of the energy). Thus a car hitting a wall can absorb more of the energy then two cars colliding and this energy gets dissipated by deformation of the metal. The problem that the caller was having was separating velocity from energy.

Conclusions: (Winners)

As to who is right or wrong is up to the caller and his father and wife. As long as people learn, everyone wins. If a judgment were to be rendered, the wording of the actual dispute must be considered carefully. From the impact (vector) point of view, the father is correct, the sum of the relative velocities makes the apparent velocity double. From the consequence point of view (energy considerations, potential damage,…), the caller is correct, you experience double the energy if you hit a wall at the equivalent (double) velocity. It is a matter of thinking about what question you are really asking. Everyone is right.

In any case Tom and Ray were wrong. :slight_smile:

The vectors for a head on collision point in opposite directions: 60 - 60 = 0. This makes sense as we should expect two cars of equal mass colliding with each other, at equal velocities, to come to a stop.

Similarly, assuming an elastic collision, we expect that a vehicle colliding with an identical stationary vehicle would come to a stop and transfer its kinetic energy to the stationary vehicle, causing it to travel in the opposite direction at the same speed as the original moving vehicle. In the case of an elastic collision with a brick wall, we would expect the car to bounce off in the opposite direction with the same velocity that it hit it (think of a superball bouncing back up to the same height from which you dropped it). In other words a vehicle colliding with an immovable wall is equivalent to a stationary vehicle being hit by a moving vehicle.

Vehicle collisions are, of course, inelastic so what we actually observe, in the case of a brick wall, is that the vehicle “absorbs” the transferred kinetic energy by crumpling in heap of metal in front of the wall (which “absorbs” very little of the impact energy – in each case they are actually transforming kinetic energy into heat). Since two identical colliding vehicles cannot transfer any more energy than they already have, the cars will each “absorb” the same amount of damage as if they had both run into immovable brick walls. Ergo the caller was correct.

BTW, I’m sure I did not get an A in college physics…

As I understood it, either 1 car hits an >identical< car head-on, each going 60 for a combined speed of 120 - or one car going 60 hits an immovable wall. Keeping it to the simple, basic question that was posed, without adding crumple zones and comparing different possible wall hardnesses, the answer is very simple.

Whether it hit a wall or the other car, the caller’s car goes from 60 to 0 in the same fraction of a second. The consideration of crumple zones was outside the scope of the question and only confuses the question but it wouldn’t matter. The caller’s car would benefit from it’s own crumpling when it hit the wall. If two cars hit they would each beneift from their own crumpling. The combined benefit of two crumple zones cannot be greater than the sum of their individual benefit.

“- or one car going 60 hits an immovable wall.” This is not equivalent to two cars hitting head on.

I was O.K. in physics but not great so let’s use the KISS principle. Speed doesn’t kill but really rapid deceleration does.

Let’s slow the speed down because the examples being used we’re dead either way. Now if I was in the predicament the caller mentioned, only at a just survivable speed, being able to only choose between a solid immovable maybe brick wall and an identical vehicle approaching at the same speed in the opposite direction I probably would choose the other vehicle. Why, because nothing is identical in real life and I just might be the lucky one-there is no other reason. Now if I was going at a speed almost certain to survive I’d choose the wall for the exact opposite reason-I might be the unlucky one if I did the head on(i.e. his umbrella through my forehead). I also would rather have the insurance company deal with a nick in a wall than a collision with another vehicle.

I agree with bq with his statement that I would rather hit the GEO with a big pickup truck than a brick wall. Let’s carry it to the extreme: What’s the consequence of hitting a small bird flying at 60 miles an hour towards you? If the bird is small enough, I believe all we will have is a dirty windshield.

Looking at it from an energy point of view: The crumple zones in cars are ment to absorb energy, right? What energy are we talking about? It’s the 60 MPH that the car is traveling times the weight of the car. The total energy in the frontal collision is 2 x that of a single car. One car absorbs 1/2 of the total, the other car the other 1/2, in other words: The effect is the same as hitting a brick wall. As far as intuitively goes, instead of hitting a brick wall or another car coming at me at 60 MPH, I would rather be some place else like Cancun, Hawaii, etc. when this event happens.

In the interest of science, maybe the car guys can do a test for us. First get their cars and hit a brick wall; then get their wife’s car and try the frontal collision scenario. As my mother used to say: “Probieren get ?ber studieren”.

If “X” marks the point of impact - either the face of the wall or where the two bumpers touch, point X will be the same before, during, and after the impact. Unless something causes point X to be pushed one way or the other during the event, the energy from the impact will be absorbed at the same rate in either circumstance. It would be different if a 1-ton car hit a 2-ton car, (in which case the 2-ton car would cream the 1-ton car). Now, point X itself would be a different matter. If it had some kind of impact meter, it would register twice the impact from the two cars as it would from the car vs wall because both cars would impart their force to it.

From another view, either one car goes from 60 to 0 or two cars both go from 60 to 0 (losing a total of 120 mph). Each loses its own 60 mph worth of energy in the same amount of time. If that were not the case, do you think that each car would lose 120 mph worth of energy for a total of 240 mph worth? Whether a car goes from 60 to 0 in a split second because it hit a wall or another car is irrelevant, as long as it stops on the same dime.

I like what my boss says better (although unprintable as he says it, so I’ll say it in Russian): ??? ???

It’s all about conservation of momentum.

If the collision is perfectly elastic, such as when two billiard balls collide moving at 1 m/s (meters per second) each, they each leave the point of impact at 1 m/s in the oppposite direction. That’s the same impact as one ball hitting a solid wall at 1 m/s and rebounding at 1 m/s.

In the other extreme where the collision is perfectly inelastic, such as when two balls of clay hit each other and stick, there is no difference between two balls of clay moving at 1 m/s each, colliding, and coming to rest, and one ball hitting a solid wall at 1 m/s and coming to rest.

Since cars are somewhere between elastic and inelastic, the collision where each car is going 60 mph, is exactly the same as one car hitting a brick wall at 60 mph.


I think you should have studied harder,you are capable of an A+

if a tree falls in the woods and no one is around to here it(what is the velocity of said tree?)


Assuming an ideal elastic collision (head on and both come to a dead stop), this energy is absorbed. (An inelastic collision is like pool where the balls do not absorb the energy and bounce off each other)

You’ve got it backwards. Pool balls colliding would be a nearly elastic collision – all energy that goes into deformation is returned and the balls end up none the worse for the wear. Two cars colliding would be inelastic – there is much deformation and it’s permanent.

“if a tree falls in the woods and no one is around to here it(what is the velocity of said tree?)”

…and if a husband speaks in the woods where his wife can’t her him, is he still wrong?

richa60 is correct. If two identical cars collide (each going 60), it’s the same as one car going 60 into a very solid wall. The point of collision (where the front bumper contacts the other car or the wall) will be essentially where the car comes to rest (less any bounce back due to elasticity in the car structure). The crumple zone is irrelevant (except that the driver may survive with it). Whether this is equivalent to one car doing 120 hitting one doing 0 is an interesting question – the total scalar momentum is the same as 60x60 (0 + m2v = mv + mv), but the total energy is twice as much (0 + .5m4vv vs. .5mvv + .5mvv). I suspect that there would be twice as much deformation shared equally between the two cars, equal to one car hitting a wall at 1.414*60.

Speed Bump,
Above, you wrote “kinetic energy of 1/2 * mass * velocity * velocity”, then you wrote "you experience double the energy if you hit a wall at the equivalent (double) velocity. This is not quite correct. You experience the same impact at 60mph hitting a wall or an equal car head on. We agree on that. The total energy absorbed (or converted to heat) in the head-on is double that of one car hitting the wall.
However, one car hitting the wall at 120mph has 4 times the energey of one car hitting a wall at 60mph and still double the total energy of both cars together in the 60mph head-on. When the speed doubles, the enrgy squares. The impact to your body is 4 times greater at 120 than at 60 in either case. Staggering!

Having read all of the replies I will add to them in the hope of clarification. Speed Bump has the clearest description so far.

As I understand the problem a caller asked:
?If two identical cars travelling towards each other at 60 mph crashed head on, would it be the same as one of the cars crashing into an immoveable wall at 120 mph??

There are two parts to the problem, speed (velocity) and energy.

First speed:
From the point of view of the driver of car A, the total speed of both crashes is the same. Car B is perceived as approaching at 120 mph, as is the wall.
But if you consider the deceleration you get a different answer. Assuming that when the two cars meet they stop at that exact point, car A would decelerate 60 mph, but when car A hits the wall it would decelerate 120 mph, a big difference.

Now Energy:
It is the energy in the crash that does the damage which is why we need to think about energy instead of just speed. The equation to find out the kinetic (moving) energy of an object is: Energy = 1/2 Mass (or weight) * Velocity * Velocity. Because velocity is squared, if you double your velocity you get four times the energy.

(We will assume that the cars weigh 2 tonne each for simplicity. I will use metric units, sorry Liberia, Myanmar and the United States.)

Car A at 60 mph will have a kinetic energy of 7194 kJ. So will car B. So the total energy in the head-on crash is 14389 kJ. But as there are two identical cars, they absorb half of the energy each (7194 kJ).

BUT, Car A travelling at 120 mph will have 28778 kJ of energy when it hits the wall. Way more than the total energy of the head on crash. To have the same energy as the head-on crash Car A should travel at 85 mph.

So if you have to be in one of these crashes, it should be the head-on crash.

Boring maths for those that care:
The SI (metric) units to use for the equation E=1/2mvv are:
E (Energy) in J (Joules)
m (mass) in kg (kilograms)
v (velocity) in m/s (metres per second)
60 mph = 26.8224 m/s
2 tonne = 20000 kg
85 mph is 60 mph * Square root of 2