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

The Show
141

Scenario A: Two cars each traveling 60 mph hit head on

Scenario 2: One car traveling 60 mph hits an immovable wall.

Two cars have twice the kinetic energy of a single car so there will be twice as much damage in scenario A. The good news is that the damage is divided equally between the two cars. This seems obvious from the symmetry of it. In the first scenario, the cars stop at the point the bumpers meet and in the second the single car stops where the bumper meets the immovable wall. Same distance to stop, same speed, same deceleration, same injuries. Forget about the dead simple Newtonian physics involved.

No two cars with drivers are identical and no crash is not going to be exactly head on, so one car may do better than the other and some energy will be dissipated if the cars bypass one another in an offset collision and this might lessen the damage to either car in the first scenario. No wall is immovable. The wall is going to give a bit and reduce the damage to the single car in the second scenario. In theory, there is not much difference between theory and practice. In practice there’s plenty.

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I just heard Wolfgang’s reply… I was disappointed that he didn’t talk about kinetic energy, which is key as well as total momentum.

Also, if you want to get picky, collision with the wall does NOT result in 0 momentum! The wall isn’t really immovable, just appears that way since it is anchored to an extremely massive object. So there is a tiny bit less energy dissipated in the car in the car/wall collision than in the car/car collision.

143

I believe that your Harvard Physicist’s answer was mired in theory and overlooked the reality of collison mechanics. If you plot the deceleration of the passenger compartment (and therefore the passenger), you find that it is not a smooth velocity line from 60 to zero (okay, for the purists, going in a defined direction).

It goes from zero to a maximum with a number of variations up and down as parts of the car - known in racing parlance as crumple zones, collapse to absorb energy. In a 60 mph crash, the motor is rarely lost, but does slide down and under the fire wall, creating the last cushion. The deformation of mounts and other hardware absorbs energy. Likewise, bumpers (such as they are today), fenders, wheel wells, etc., all crumple and absorb energy in this process.

As a result, to describe the change in momentum, you need a graphic representation, not just the start and end points of velocity.

When two cars collide, going precisely in opposite directions, two sets of crumple zones do their jobs, and the forces applied to passengers is reduced by these parallel event sequences. Hitting a wall - not so much. Only one set to compress, and a more abrupt deceleration for the passenger

Since the caller’s question related to the physical (not just physics) of the crash, it’s likely that the passengers would fare better if they were to collide with another vehicle (going 60) than a wall (with their car going 120). Or, for that matter, in a parked car hit by a wall going 120 mph.

I have to offer this with trepidation, since my Molecular Biophysics M.S. is from FSU, not the austere halls of Harvard. However, Paul Dirac wasn’t exactly a slouch. (though he seldom watched the Daytona 500.)

Thanks for all the great advice. And all that other stuff, too; whatever it was

144

Here’s what I e-mailed about Professor Wolfang’s statement:

Dear Car Talk,

I’m not sure if this is what you mean by “mechanics”, but Professor Wolfgang’s explanation was unnecessarily technical. A better explanation appeared on your forum. Imagine that there is a sheet of Saran wrap across the road where the two cars collide. In a perfectly symmetric collision, neither car goes through the plastic. If you take away one car and put a brick wall behind the Saran wrap, the same thing happens to the other car. That way we can omit talking about momentum and kinetic energy and all that physics stuff.

Also, there are two real-world effects that affect the situation in minor ways. First, the brick wall is not exactly immovable. It is attached to the Earth, which is about three sextillion times the mass of the car. So the collision changes the rotational velocity of the Earth by a zeptoscopic amount, reducing the car’s impact by the same amount.

A more significant effect is that the impact between two cars is not really symmetric. Even if the cars were identical, the driver’s weight would make the left side of each car slightly more massive than the right side. This would make the cars rotate slightly counterclockwise around the point of impact. Perhaps more significantly, parts of each car’s front extrude into voids in the other car. Both cars stop a little past the Saran wrap. In effect, the cars partially miss each other, and to a greater extent than they would miss the brick wall.

The effect of cars partially missing each other swamps the nigh-immeasurable effect of moving the brick wall, so there is a real advantage to running into the other car, except for the obvious problem of doubling the body count.

Dan Hoey

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Hey Wolfgang,
Conservation of momentum is not the whole story. The collision is completely
inelastic in both cases, and the final momemtum is 0 in both cases if the masses
of the two cars are the same and the wall is rigid. Two cars of equal mass
traveling at 60 MPH have half the kinetic energy of one car traveling at 120 mph,
since KE varies linearly with the mass but as the square of the speed.
This KE has to be dissipated in the crash. If the wall really doesn’t move and
doesn’t break, then all of the KE goes into deforming the single car that hits it.
When the two cars traveling at 60 MPH crash, there is half the kinetic energy and
two front-ends to dissipate it. Since the wall is absolutely fixed while the other
car deforms, I would guess that the impulse is larger when you hit the wall, but
it might be more complicated depending on exactly how the cars deform, whether
the engine breaks loose, etc.

Judah Levine

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120 mph collision has more energy and is therefore more dangerous. End of story.

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I don’t believe that Prof. Wolfgang’s statement was mired in theory. I believe it was wrong.

It’s true there’s the same amount of momentum change involved in both collisions. That means that the impulse imparted to the car is identical.

You know what else has the same impulse change? Taking your foot off the gas, and colliding with air (for a long, long time). And last time I checked, that doesn’t do a lot of damage.

If you want to know how “bad” a collision will be, you need to know the peak force. And that means you need to know the time that the collision takes. Shorter time = bad, high peak force. Longer time = good. That’s why cars have crumple zones. That’s why falling onto an inflatable landing pad will save your life jumping off a building.

So, let’s answer the original question properly: which is worse (i.e., which will do more damage to the car) - hitting a wall at 120 mph, or hitting another car while going 60 mph which is also going 60 mph head on?

Answer: It depends. It depends on what the car is made out of, and it depends what the wall is made out of. Is it a wall of super-strong steel 50 feet thick? If so, crash into the other car. The wall won’t give, the other car will. Is the wall a single-brick thick? Crash into the wall - the wall will probably give a lot more than the car will.

Honestly, it’s easy to see that I’m right in this by taking it to extremes. Make the wall out of Jell-O, and obviously, you’ll want to crash into the wall. (The collision might take ten minutes and half a mile to come to a stop, but hey, it’s the same amount of momentum change.) Make the car 100 feet long and make it out of NERF foam, and obviously, you’ll want to crash into the other car.

Now, let’s change the initial question again: what if the person asked “in both collisions, they come instantly to a stop.” What happens then? Easy. You die. In both crashes. An instant stop means infinite force. Thankfully, that is mired in theory, and not real.

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The total amount of energy is the same. Add it up. Two cars, at 60 mph, have 20.5(m)(60)(60) units of kinetic energy (specific units left off for brevity). A single car, hitting the wall has 0.5*(m)(120)(120) units of energy. After the collision, both sets of wrecks have zero units of kinetic energy.

Same amount.

What matters is what the car and wall are made of, and how long the collision takes. The energy has to be dissipated somehow, and the question is whether you can have it be dissipated in something else being crumpled to oblivion rather than the driver.

149

of course, if I could use computers at all, the edited post version of that would’ve been posted rather than that, in which I can do math correctly, and only the last paragraph would be there. So, uh, ignore the first two paragraphs. Pretty please.

150

If you want to get completely technical, you can determine the speed of the wall at the point of impact. The velocity is the radius of the earth at that point normal to the axis of rotation times the angular rotation of the earth in radians per second and then convert to MPH if that is what you are working in. So again, the frame of reference is very important to the problem and the conservation of momentum or energy has to be relative to something. Since our lives, the speed limit, ect. don’t apply to a space based reference frame, we use an earth fixed reference frame and assume that it is not accelerating. So you can assume away the mass and the velocity of the earth.

But, the car hitting the wall is not the same as the car hitting a car with respect to the energy of the system as many posters have adequately pointed out. Wolfgang did not explain it well enough or correctly.

151

removed per bad math admission

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use the little pencil at the bottom right to edit the original post.

153

When two 60 mph cars hit head on, the kinetic energy of both cars is converted into heat.

When a 120 mph car hits a stationary car, there is kinetic energy left over after the collision. After the collision, the 120 mph car has been slowed down to 60 mph and the static car has been accelerated to 60 mph.

The velocity of a .22 caliber long rifle bullet is about 800 mph, which happens to be the same velocity that the earth moves at around 30 degrees latitude. In years of shooting thousands of rounds out of a .22, I have never noticed a difference in results when shooting at targets due east verses targets due west. Same damage, same hold over to hit far away targets.

154

But, the caller wanted the car to hit the wall at 60 not 120. The two cases are different. He wants the qualitative answer for problems that differ.

155

Sorry Guys,
But Wait! There’s more.
Not to continue to kick a dead horse (with apologies to the winner of the Derby) the force/time history must also be considered when deciding to hit the wall or the oncoming car (or truck)
Hitting a wall at constant speed ASSUMING the wall is substantial is much worse than hitting another car. The amount of time for M.v to reach zero from th initial impact is also important in saving your bacon. An unyielding wall will cause the deceleration (yeah I know negative accleration) to occur over a very short period of time so energy dispersion per unit of time is great.
If another car is hit, the folding of metal, mashing of internal organs and lots and lots of noise indicate that the negative acceleration (OK?) is stretched out for a longer period and thus the force per unit time is less acute. Hitting a FIXED object will do more damage to the body than hitting something more ELASTIC. ( so my Associates Degree in Tomato Pruning informs me)
so , given the choice, hit the object which will spread out the dissipation of force for the same reason hitting a pile of mattresses is is better than diving head-first onto the concrete. “It’s not the crash, it’s the sudden stop” (ALfred E Neuman

156

Energy Dissapated is the key. E = m * V^2
m= mass assumed the same for each car , v = velocity of mph

Each car driving 60 mph and hitting each other, E = m * V^2 + mv^2 = 2mv^2
One Car at 120 mph hitting a wall, E = m * (2v)^2 = 4mv^2

So more energy is dissipated with things flying off faster in the brick wall scenerio

157

With all respect to Dr Wolfgang, He seems to have forgotten Newton’s Law – for every action there is an equal and opposite reaction. In this cas (2 cars, equal mass, same speed, opposite directions) one car hitting a wall has a force of the car on the wall, and an opposing force of the wall on the car. The car experiences the one opposing force. With one car (A) hiting a second car (B) there is the force of car A on B, the opposing force of car B on A, the force of car B on A and the opposing force of car A on B. Each car then experience two forces, one from the other car and one opposing force. Thus the force experienced in a head-on collision is double that of a collision with a stationary object.

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Yikes. BigRap, you might want to rethink this. The physics professor actually did get the answer to this remedial physics problem correct.

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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.Law degree | online phd degree | Criminal Justice degree