An interesting question arose as my friends and I were playing Dagorhir the other day. My one friend (Alex) said that two swords hitting each other at x speed would produce as much force as one sword hitting a stationary one at x speed, and my other friend (Bob) said that was bunk.
Now, I was called in, being a fan of physics and the most literate in the subject (I think), to settle the argument. Turns out that Alex had been quoting an experiment from Mythbusters (I hunted it down; it’s specifically the episode entitled “Mythssion Control”) to prove his point. In it, they apparently have a test where they run two clay blocks, and then two cars, into each other and then into a wall to demonstrate how the forces are the same by measuring deformation of the clay or cars. Bob cleverly pointed out, however, that the test in which the clay hit the wall was misleading because the clay, and the car, deformed while the wall did not. When the clay are run into each other, they both deform to an equal extent. To me, this is an example of elastic vs. inelastic collisions. Since our swords are more closely rigid bodies than lumps of clay under the power of our swings, I concluded the forces experienced by the swords are doubled with the collision in which both are moving.
In each of your sword scenarios, choose a frame of reference that always stays halfway between the two swords:
Scenario A: both swords are moving toward each other at speed X, and your chosen frame of reference is stationary.
Scenario B: one sword is moving at X and the other sword is stationary, and your frame of reference will be moving at 0.5*X toward the stationary sword.
it should be immediately apparent that in the latter case, the violence of the collision will be equivalent to two swords moving toward each other, each with a speed of 0.5*X. If your swords are perfectly (and linearly) elastic, then yes, scenario A - with four times the energy- should double the impact force. The reality is that if the swords hit edge-on, you’re probably going to cause some plastic deformation, i.e. you’re going to damage the blade.
The moving-clay-hits-moving-clay scenario has twice as much energy as the moving-clay-hits-stationary-wall scenario; it stands to reason that you would expect two balls’ worth of deformation in the former case, and only one ball’s worth of deformation in the latter case.
Ah, thanks. The point was that the wall didn’t deform, so all the energy was given to the clay ball, thereby leading to some confusion. So Bob is right, as I had expected.
I also prefer to take the frame of reference so that one of the balls is stationary, so we have one ball moving at x in one scenario and one moving at 2x in the other, but that’s just nitpicking. I am going to design a test where one clay ball hits one moving block of wood to show that there will be twice the deformation since the wood won’t deform.
And don’t fear us breaking our swords; they’re padded.
The clay and cars experiment was to refute Jamie’s earlier statement in the Compact Compact episode that two cars going at x and crashing into each other was equivalent to one car crashing into a wall at 2x.
And the results with the clay and cars and the immovable wall only work when you have that immovable and non-deforming wall.
Replace one of your swords with an unspecified force. Your sword (or a specified point on the sword, the impact point) moves at speed X, then it comes to a dead stop over a specified distance, let’s say 1/4 inch.
It does not matter at all if that acceleration (deceleration) is due to a sword moving at speed X in the other direction, due to a sword being held motionless by it’s position atop a cinder block, or a sword being held motionless by a Viking God. If your sword goes from moving to not moving over the same distance by a force applied at a particular point, the forces involved are identical.
However, one would expect that a stationary sword, held by a human being, would fail to stop the attacking sword over as short a distance as a sword moving the opposite direction. Thus, the impact would be less severe with a stationary sword, under those conditions.
There was a Car Talk episode where the issue came up: Which was worse? Crashing into a wall or into another car?
An MIT professor called in (after Click and Clack got it wrong) and pointed out that, assuming the two cars were the same weight, the result was the same. In both cases, the cars decelerated to 0 mph in the same amount of time. So if all cars were going the same speed, the force generated would be the same if they hit a wall or another car.
If the cars were not of equal weights, then the heavier car would decelerate less.
Depends what you mean by “stationary one”. If the stationary one deforms very little and is held rigidly in place (let’s say we embed all but the leading edge into a block of concrete), then Alex is right, the force (as experienced by any one person with a sword) should be pretty much the same. IE, it is going to hurt your hands the same amount. If the stationary sword is held in place by a human being, you are going to lose some of the force to motion in the human being and the force the sword swinger experiences would be less.
In the equivalent case of the Mythbusters car experiment.
Car head on into equal weight Car both at 50mph = Car into non-deformable concrete wall at 50mph
Car head on into equal weight car with one car stationary and one at 50mph < Car into concrete wall at 50mph (as the stationary car deforms and takes some of the impact).
Obviously the total energy of the collision of two cars into each other at 50mph is greater than that of car into wall (as they have more energy to start with), but it gets split between each car.
Force may be the same, but the professor missed something very important. You’re a lot safer hitting a vehicle than a wall in any case, or at least much more likely to survive. Head-on isn’t good in any case, but cars have a lot of design features that may help you survive.
Disagree. the car you crash into has features that spread out the time of the collision, thus causing a situation where the occupant is not subject to being impacted as quickly as if you run into an immovable wall.
Consider two equal masses rushing at one another head on, one being a Fitch system, the other a car. You strike the Fitch system mass at a speed of twice your speed equivalent, and the system absorbs much of the impact energy. Without the Fitch system, your impact would be much worse, and the lessened effect of the collision is entirely due to the crushability of the object you collided with.
My conclusion is that if two cars collide head on, one without modern provisions for absorbing collision energy, the effect will be worse than if both cars have such provisions.
No. If the cars are identical the front of your car never passes the point of impact - the two cars react exactly the same way and it is as if you hit an immovable wall. The occupants of Car A will decelerate in the same time and space as if they hit the wall. The forces on the two cars will exactly cancel.
The features that spread out the time of the collision are the ones in Car A. Those same features in Car B don’t help the occupants of Car A. If they did, they would have to be less effective for the occupants of Car B since the energy all has to go somewhere.
If you put a curtain down at the point of impact so you couldn’t see whether you were hitting a wall or an identical car, what differences would you expect to see? The answer should be no differences.
Oops, missed this. My case only works if the two cars are identical. That’s the way the tests are designed. If you’re talking about cars without crumple zones or cars of different weights then the question is much more complex. In general I agree with you here but the devil is in the details.
The car you crash into is also bringing its own kinetic energy to the party; its features will be busy dealing with that car’s kinetic energy, while your car is busy dealing with its own, just as it would if it hit a wall.
You’re disagreeing with something Telemark didn’t say (and neither did anyone else). We’re talking about two identical cars running into each other; you’re talking about two non-identical cars running into each other.
Or to put it another way, the crash-protection features in the other car are exactly canceled out by the fact that the other car is coming at you at speed, instead of just sitting there immobile like a brick wall.
Interesting. I was going to post saying I would rather drive head on into another car at 25mph (with the other car travelling at 25mph in the opposite direction) than drive into a brick wall at 50mph. Intuitively, it seems like the other car’s crumple zone will help absorb the impact more than a brick wall, but actually I think you’ve convinced me that’s not the case (due to the other car’s crumple zone already being occupied by absorbing the energy from the other car).
I assume no-one disagrees that you would rather rear-end another (stationary) vehicle at 50mph than hit a brick wall at 50mph (which is clearly a different case)?
In the Mythbusters episode with the one car-wall vs. two cars test, they missed a real chance at clarifying the idea that they are the same. They only showed first one crash and then the other. If the had shown the two crashes on the screen at the same time, one above the other, it would have been incredibly obvious to the most naive doubter that they were the same.
Or they could have blanked out one half of the screen and had you guess “Did the car on the left hit a wall or another car?”
The really amazing part was the Jaimie was surprised at the result. He had thought there was some magic doubling of force going on. He had mentioned this in an early episode and the fan letters pointing out his mistake are what led to the test.
I used to work for Ford writing software to analyze the results of barrier tests (we *never *called them “crash tests”). This was the thought experiment that the engineers always used when people failed to understand that there is no difference. In either case, the car never goes past the point of impact so the deceleration is the same and therefore the damage sustained is the same.
Unfortunately the OP’s actual question is imprecise: “two swords hitting each other at x speed would produce as much force as one sword hitting a stationary one at x speed”. The force on* one sword* would be the same hitting a moving sword vs. hitting an absolutely fixed and *nondeformable *sword (not just another sword that is not moving).
Just to add, a car hitting a barrier is not at all the same as a car hitting a parked car. So the second “stationary” sword can’t just be dangling in the air.
When the head is with a similar vehicle… and its standard flat nose vehicle.
Some vehicles are more dangerous than walls, (eg snow plough attachments with angle ‘<’ front ) and some are less dangerous. (eg bus with no bull bar, because its soft all over. )
This is the Dope, though, and we love a good nitpick. Identical cars would certainly be different from a brick wall…the cars aren’t symmetrical! One car’s driver is on the north side and the other is on the south side. Suppose we designed the cars to spin to the right (or left) upon impact. They’d go different directions, and everyone lives!