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."


True and each car would absorb half the force. In the case of an immovable wall the wall would not absorb any of the force (only resist it) so the force on the car would be the same. Sort of like hitting your fist against a concrete block and hitting your fist against against each other.

The mass, momentum and energy of the system are conserved. The two systems that you describe have different total momentum and energy. Thus they are different. Momentum is described as mass (m) times (*) velocity (V) or m*V. The kinetic energy of a particle is described as 1/2*m*V*V (The cars are on flat ground so there is no potential energy). The laws of conservation of momentum and energy state that the sum of the initial momentum of the system equals the sum of the final momentum of the system and the sum of the initial energy of the SYSTEM equals the sum of the final energy of the system. The terms that the first year physics books do not give you are the dissipation terms. The terms that convert the energy from one form to another. Without the dissipation, you can't have an inelastic colision. Energy and momentum must be lost. In the case considered, all is lost at point X.

In your first case, two particles enter the infinitesimal region surrounding point X with V and -V. An instant before impact, the net momentum of the system is m*V + m*V. After impact the momentum of the system is m*0 + m*0. To conserve momentum 2(m*V) has to be dissipated. With respect to energy, 0.5*m*v*v + 0.5*m*v*v exists prior to the collision and 0 exists after indicating that m*v*v amount of energy is lost to the system.

In your second case, one particle is moving at the same V as in case 1 and the wall is not. Also note that you are assuming that the wall has large mass compared to the particle mass (i.e. it does not move). The initial momentum and energy are m*V + 0 and 0.5*m*V*V. Since V = 0 after the colision, the momentum and energy lost are m*V and 0.5*m*V*V.

Case two is clearly not the same as case 1 for conservative properties regarding the total quantities in the systems.

The initial momentum is m*V + m*(-V) = 0. The same as after the collision. Energy can be dissipated meaning it transfers to other forms such as heat and work down on crumpling cars. Momentum just transfers motion from one object to motion in another object. Momentum doesn't dissipate. In the case where the car hits a massive rock wall, the momentum lost by the car is tranfered to the rock and planet earth. The latter having such a large mass that its change in velocity is hard to detect.

Where are you with this now? The two cases (car@60-vs-car@60 & wall-vs-car@60) are the same, yes?

Yes!

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.

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 über studieren".

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

"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."

60 MPH * mass is the momentum of the car. Both energy and momentum are conserved.

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.

"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. "

Explain how momentum is conserved then? Is it the same in both cases?

Momentum is always conserved, whether the collision is elastic or inelastic. That's a Law of physics. Energy is conserved ONLY in the elastic case. In the perfectly inelastic case, all energy is converted into heat and sound. In the partially elastic case, some of the energy is converted to heat and sound, the rest is conserved as motion.

From a momentum standpoint, two identical cars going 60 mph and hitting each other = one of those cars hitting a wall at 60 mph. The caller was right. Tom and Ray were incorrect.

Momentum is a vector quantity just like velocity. Along the one dimension of the road along which both cars are traveling one of the cars (clay balls, billiard balls) have momentum mv and the other m(-v), so the total momentum is mv-mv=0. Clearly when the clay balls are both at rest after the collision the momentum is also 0. The billiard balls which recoil at the same but opposite directions also sum up to zero. So as turbonium states momentum is preserved in either case, whereas kinetic velocity is only preserved for the elastic case.

GREAT TO HEAR YOU GOT AN A.

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?)

SEE , THATS WHY YOU NEVER GOT AN A+

"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?

YES,

ALWAYS.

BUT THE NEIGHBOR IS ALWAYS CORRECT.

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=1/2m*v*v 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

This is a great way to understand the amount of energy in either of the crashes. However, it isn't clear that in the case of the head on crash, two cars will dissipate the crash energy while the energy of a single car accident with an immovable rock wall will have to absorbed by a single car. And this comes down to the consideration of the original question. The energy of a two car crash where each is going 60 mph is equal to a single car crash into an immovable rock wall at 85 mph. However, the damage to the single car will be twice the damage of either of the head-on crash cars.

In order for all cars to have equal damage, they must be going at the same speeds. As KiwiPhil mentioned, any 2 ton car traveling at 60 mph must dissipate 7194 kJ. In the case of a head on, each car brings 7194 kJ to the accident and each car absorbs 7194 kJ. To come out of the single car crash with the same amount of damage, the single car must bring and absorb 7194 kJ, meaning that it too must be going 60 mph.

If I may be so boring, allow me to present the equivalent crash energy math in another fashion without assigning values (since I went to the trouble to do it :)

Vehicle 1 & 2, each at 60 mph, impact head on
Vehicle 3 is the single vehicle hitting a wall

M1 is mass of vehicle 1
V1 is the speed of vehicle 1
...
M1 = M2 = M3 (all cars equal)
V1 = V2 (60 mph)
Solve for V3

½ * M1*V1^2 + ½ * M2*V2^2 = ½ * M3*V3^2

substitute M1 for M2 and M3 and divide both sides by ½ * M1

V1^2 + V2^2 = V3^2

substitute V1 for V2

2*(V1^2) = V3^2

V3 = V1* (square root of 2)

If V1 = 60 mph, then V3 = 85 mph

You guys make it all too complicated. Lets just put up a brick wall and have two cars hit it on opposite sides at the same time. The is they would have hit head on if the wall wasn't there. The wall is immovable whether hit by one car on one side only or hit on both sides at the same time. Now what is the difference to the vehicles if they hit one at a time or both at once? What would be the difference if the wall is removed and the two vehicles hit head on. The simple answer is that the caller was right and Tom and Ray were wrong.