Don’t worry, I don’t need an answer fast, and I swear this is a totally hypothetical situation, not a “could my ‘friend’ be proven to be at fault” type of deal. It actually came up in a puzzle I heard recently (on Car Talk, if you must know).
The scenario is this: two cars of approximately equal size and weight (in the puzzle, they were identical, but whatever) are driving toward each other at the same speed. One car swerves partially into the oncoming lane, and the two cars hit each other on the left headlight. So, the impact is not across the entire front of the cars; it’s the left 1/4 (let’s say) of one car hitting the left 1/4 of the other.
Now, from what I understand about physics (which is just this side of diddly squat), I would imagine that the cars would maintain their forward momentum, but the point of impact would become a kind of axis, and the cars would spin around it to the right. So if I were in one of the cars heading west, the impact would be on the south side of the front of my car, and the rear of my car would swing north (and then west, and then south, and so on, depending on how fast I was spinning). And I imagine that the speed of the spin would be increased with the speed of the cars at impact, and decreased by the friction of the tires moving sideways.
I never got an answer, because this wasn’t relevant to the solution to the puzzle, but it wasn’t ruled out, either. So basically, I’m wondering, would it happen as I envision? If not, why not? If so, are there other conditions that would need to be present? For example, would it not happen above/below a certain speed? What would happen if one of the cars was stationary at impact? What if one car was significantly bigger? What if Toonces was driving?
It also seems to me that the spin would decrease if the point of impact moved toward the center of the cars, with zero spin happening in a perfectly aligned (right headlights to left headlights) impact, and increase as the impact moved out toward the edge (although I imagine that at a certain point, the cars would basically just scrape past each other).
I’ll leave it to others to fill in more of the details, but you’ve got the basic idea right. The cars have very little linear momentum (they’re travelling at the same speed in the opposite directions, so their linear momenta cancel out), but they each have a fair bit of angular momentum as measured relative to a point in the middle of the road. The frictional force of the tires being pushed sideways after the collision would then exert a torque that would slow the rotation.
I agree with this but would tweak the wording a bit on a technicality:
Heart of Dorkness is also correct that no spin would happen if the cars were perfectly aligned.
I used to write software to analyze barrier tests at Ford (we never called them crash tests). Another tangential point is that many people think that a hard going 30 MPH having a head-on collision with another car at the same speed is the same as one car hitting a fixed barrier at 60 MPH, since the closing speed of two cars is 60 MPH. This is incorrect. It’s the same as one car hitting the barrier at 30 MPH.
The type of imact you are describing is called an “offset frontal” collision. This is one of the specific scenarios for which manufacturers test their vehicle’s crashworthiness. here’s a video of a crash test; the car doesn’t spin around over and over (crash speed wasn’t that high), but you can see that the ass end gets kicked out to the vehicle’s right, pretty much as you describe. With higher speeds, a greater offset, and a slippery road, it’s not hard to envision complete revs on a vertical axis.
Whuh? You can count me among the many people who (incorrectly) think that.
So, you’re saying that if I, driving at 30 mph, hit a stationary car head-on, that the impact would be the same as if the other car was also moving toward me at 30 mph?
Oops, missed the edit. Anyway, thanks JoeFrickinFriday, that’s *exactly *how I imagined! Though interestingly, it looks like the tire friction doesn’t play a huge role in slowing the spin, since the car tends to bounce off the ground.
On this last one the car is spinning on rebound. It’s center of gravity is moving backwards. I would have expect the car to be spun on impact, not on rebound. This demonstrates that the car’s body is designed to deform on impact to absorb some of the energy of the impact.
That’s not what I said I didn’t say hitting a stationary car, I said hitting a fixed barrier. A fixed barrier will not absorb any of the crashing car’s kinetic energy (except for a relatively small amount of heat). A stationary car will deform and probably move if it is hit by a car going 30, absorbing a significant amount of kinetic energy, so that the moving car is deformed less.
Do this thought experiment. Imagine a car going 30 MPH heading towards a curtain. It hits the curtain, and crashes. The curtain doesn’t budge, at all. You can’t tell whether the curtain is hiding a massive concrete wall, or an identical car that came the other direction at the same speed.
in laymen’s terms…
if you hit the other car at 0 mph then you and the other car would continue plowing forward. a barrier completely stops your car, which would happen if the other car was going at your car at the same speed.
No, hitting a stationary brick wall (that stays stationary during and after the collision) at 30 is the same as hitting an oncoming car head-on with both going 30. Hitting a brick wall at a given speed is not the same as hitting a parked car at the same speed, because the car will be moving after the collision. If you’re going 60 and hit a parked car, that is in fact roughly equivalent to a head-on at 30.
Oh, sure; that totally makes sense: hitting an oncoming car when you’re both going 30 mph is like driving into a building at 30 mph, because the other car’s forward motion counteracts its absorption of your forward motion, and so it reacts as a wall would, more or less.
I’ve just never heard impacts of different objects compared that way. I’ve always heard it as “hitting an oncoming car when you’re both going 30 mph is like hitting a stationary car at 60 mph”. Which it is, roughly. Right?
Since the friction of the tires never disappears, but the amount of spin decreases as the speed of the cars, if the cars are going slow enough, the spin force won’t be able to overcome the friction of the tires (which is of course what you think would happen - imagine two cars bumping at the slowest speed you could imagine. No spin, right?).
Just pointing out that this is a physics-problem-set-land perfect, idealized brick wall, that doesn’t move a fraction of a millimeter when a car crashes into at 30 mph.
In the real world, I’d rather hit a tree at 30 mph than have a head-on collision at 30mph each, because the tree will move at least a little bit.
Let’s assume a frictionless scenario, such as if the parked car is in neutral without the parking brake on. To conserve momentum in a frictionless scenario the two cars would meet then continue to move at 30 MPH in the same direction (assuming an inelastic collision where they kind of glom together, unlike a billiard ball collision). The speed of the moving car was reduced by 30 MPH, which is what happens when hitting a barrier at 30.
However, damage is generally considered to be directly proportional to kinetic energy. The kinetic energy of the system (where each car has mass m) would be (leaving out units for the sake of expediency)
.5(m**60[sup]2[/sup])=1800m*
before the collision and
.5(m**30[sup]2[/sup])=450m*
after the collision. That means that the two cars had to absorb 1350m of energy, or 675m each. A car going 30 hitting a wall only has to absorb 450m. So hitting a parked car at 60 will cause more damage to the cars.
Although the tree won’t change position, it will certainly flex more than a concrete wall, absorbing some of the impact. How much depends on the tree.
Math error, there: You forgot that after the collision, both cars are moving, so you need two cars worth of mass. That brings the kinetic energy to 900 after the collision, for a change of 900, or 450 absorbed by each car, just like the single car hitting a wall.
Yes, but a 2 foot diameter tree that you hit at bumper height still doesn’t flex very much. And the impact was centered on a 2’ wide part of the car, rather than across the full width, so it ended up penetrating deeper into the engine compartment than if I hit a wall. Good thing I was in a '79 Impala with a very big engine compartment.
All told, hitting either a wall or a tree isn’t much fun.
You know, I just realized I never did explain what the puzzle actually was. I’m kind of surprised no one asked.
If anyone’s curious, it started with the scenario I described above, and went on to say that although these cars were identical, and hit each other in a symmetrical fashion, driver A was very badly injured, while driver B suffered only minor scrapes and bruises. The question was, what accounted for the difference in their injuries?
My hypothesis was that there was a third car following a short distance behind driver A. As I now know would indeed happen, the cars spun on impact so that the drivers wound up sitting perpendicular to their respective lanes, and driver A ended up getting T-boned by the third car. It seemed like getting rear-ended by the third car wouldn’t have caused that much more injury, so it was significant to me that the cars spun.
Of course, this was *not *the solution. Not even close. Their answer was:
Driver B’s car was retrofitted as a rural mail delivery vehicle, and so the steering wheel, and the driver, were on the right side of the car.
I didn’t say hitting a stationary car, I said hitting a fixed barrier. A fixed barrier will not absorb any of the crashing car’s kinetic energy (except for a relatively small amount of heat). A stationary car will deform and probably move if it is hit by a car going 30, absorbing a significant amount of kinetic energy, so that the moving car is deformed less.
