If two travelling objects collide, end on, if thier speeds are different, does the speed of one directly relate to the damage of the other?
That is: If two cars are travelling alone a road, one at 50Kph and the other at 100Kph does the first do '50Kph' of damage and recieve '100kph' from the other?
what are the factors involved?

Assuming that the two cars are the same mass (you didn't say), I'd guess that they'd each do an equally distributed total of 150kph of damage to each other. The fact that one was moving faster than the other doesn't matter. It could be 10kph + 140kph, or 0 + 150, or 75 + 75, and the damage distribution would be the same.

If they're different masses, I'm sure it gets a lot more complicated.

And by the way, I call "no SUV flamewar" for this thread right now, because I can totally see it coming.

Yeah, They would be the same mass, but for the calculation, it'd be done in newtons so the mass actually wouldn't matter; it'd just be part of the conversion.
Yes but what is the effect of the infliced damage on the other object? is it purely a 50/50 split? (given that both objects are the same construction.)

I guess that you'd always have the M1V1 + M2V2 = (M1+M2)V3 equation for an inelastic collision if that helps any. Can't remember the inelastic stuff at the moment. Rum doesn't help much either.

No, the objects don't need rum to collide.

is it purely a 50/50 split? (given that both objects are the same construction.)

Yes.

The velocities of the two objects are relative only to each other. Their velocities relative to the ground are irrelevant.

Imagine they were floating in space when they collided. Then each object's velocity would be strictly a function of the velocity of the observer. That is, if you were traveling at the same speed (and direction, of course) as one of the objects, it would appear to you to be motionless, and the other object would appear to be traveling rapidly. If, instead, you travelled at the same speed and direction as the second object, then IT would appear to be motionless and the first object would appear to be the one traveling rapidly. Yet in each of these scenarios, the only thing that was different was the velocity of the observer, which could not have any effect on the outcome of the collision itself.

tanstaafl.

Heh, another "physics problem".

what are the factors involved?

So, what's the other guy's insurance coverage like?

Law of Conservation of Momentum = M1V1 + M2V2 = M1V1' + M2V2'

Since M1 = M2 and since mass of each object will remain the same on both sides of the equasion (assuming we're not talking near-speed of light speeds), we get:

V1 +V2 = V1' + V2'

Now, since V2 is say, twice V1 we get:
V1 + 2(V2) = V1' +2(V2)'

Since the two cars are completely inelastic and stick together, V1' and V2' are the same, so we can combine to get:

V1 + 2(V2) = 3V'

Therefore:
.```` V1+2(V2)
V' = -------------
.``````` 3

If we insert real numbers (V=50)we get:

`````50 + 2(-50)
V' = --------------
````````` 3

So:
-50/3 = -16.6666666666 kph is the momentum after collision. Since both cars come to a complete stop, the car travelling slower (V1) will absorb 16.666666kpm X (mass) more energy than the faster car.

By the way, What's with the board not displaying multiple spaces? I put like 10 in a row to space the divisions, ant the spaces just went away.

Now, since V2 is say, twice V1 we get:

V1 + 2(V2) = V1' +2(V2)'

Umm... don't you mean:

V_1 + (-2 * V_1) = V_1' + V_2' (V_2' is *not* twice V_1, and V_2 is in the *opposite* direction of V_1)

-1 * V_1 = V_1' + V_2'

Since the two cars are completely inelastic and stick together, V1' and V2' are the same, so we can combine to get:

V1 + 2(V2) = 3V'

Which would make this:

-1 * V_1 = 2 * V' where V' = V_1' = V_2'

or:

V' = -0.5 * V_1

So the velocity of the new object (the two cars stuck together) is half the velocity of the slower car, heading in direction of the faster car, but I don't think that has anything to do with which of the two sustains the most damage. In that, I think that tanstaafl is correct. Imagine that instead of two cars, you have two non-identical objects that otherwise fulfill the parameters of the question. Object one is a 1-ton block of jello, object two is a one ton block of granite (suppose that the surface area where they collide is identical). Obviously, the one ton block of jello is going to sustain more damage than the one ton block of granite is, no matter which of the two objects is travelling faster. Now if you have two objects that have identical properties for dispersing the forces of impact, then they will absorb an identical amount of the forces of impact, due to that whole issue of relativity.

That's not a factor of the board, that's the way HTML works. If you want to use multiple spaces to align something, you have to use the <pre></pre> tags (or whatever the corresponding tags are in the BBS).

OK... then answer me this. How come when you take a wooden bat, hold it upright, and hit it as hard as you can with another identical bat, that the bat that is stationary always is the one to snap in half?

Or the way a railgun firing a piece of plastic can demolish a brick. I doubt the brick would suffer much damage if it were fired at the same speed at a small piece of plastic. It's just a matter of inertia.

Because the bat is being held stationary -- the only method it has of dissipating force is to break. In the car example, the slower car is not being held stationary, and has the freedom to move.

that's.... 40 : 15

canuckInLA serving for the match.....

Or the way a railgun firing a piece of plastic can demolish a brick.

Uh... now you're bringing size and dissimilarity of objects into the equation. Ignoring that, though, have you ever looked at the piece of plastic after it's shattered the brick? Brick -- rigid, no method of dissipating force. Plastic -- elastic, and could deform and revert to shape. Ever dug bullets out of something, and notice how they're not all nice and pointy shaped anymore (d33zy can back me up on this one, I'm sure. )? Then there's pressure to think of, too -- objects that collide with a large surface area of impact will have less damage over a greater area than two objects that collide with a tiny surface area of impact. The smaller surface area of impact is going to create much higher pressure on one of those objects. Here's an experiment you can do to demonstrate the idea I'm talking about, since I don't think I'm doing a good job explaining it... get a fork, and hold it in one hand. Swing the hand holding the fork at your other arm, striking your arm with the flat tines of the fork. Take note that it doesn't hurt that much. Now, swinging the hand with the fork at the same rate you did last time, stab your arm with the pointy ends of the tines. Take note of how much it hurt that time, while the only thing that changed was the surface area of impact. If you want to use similar materials, think of poking someone with your finger as opposed to pushing with the flat of your palm. Same hand, same arm, same momentum, different area of impact.

The answer to physics problem #2 can only hold true in a "perfect physics world". Outside that world, you have to worry about all those silly things like minute variations in angle of impact, size of impact area on each vehicle, and the like are all going to govern the amount of damage that each vehicle sustains in the collision.

The reason I chose to use the example of the jello and granite was to show that it doesn't matter which of the two objects is moving, the jello is going to sustain the damage (although, now that I think about it some, jello may be too elastic to suffer damage in the collision, and just wobble around a bit).

I doubt the brick would suffer much damage if it were fired at the same speed at a small piece of plastic. It's just a matter of inertia.

If the plastic bit was moving free, of course not. If you were to fix the plastic bit in position, it would. No matter, it's irrelevant to the question at hand -- which of two identical bodies would suffer the most damage. With identical objects, they will have identical inertia, which is proportional to an object's mass, and is unrelated to the current velocity of the object. A stationary object has the same inertia as it would if it were moving at 30mph.

FWIW, I gave up on physics after first year university (I was passing, but not enjoying it), so if I'm bungling my explanations, speak out...

Ok. enough of the flailing of pointy things. ow note to self: don't carry out all practical explanations of things on this board
I'm not interested in surface area. it brings in whole realms of strange and mystic equations.
two identical objects; travel thru space with dissimilar speeds; impact on eachother at the same surface area.
does the speed of one directly translate into the damage of the other or is it all shared?

Picture this. A journey of sight and sound. there are two cubes of same size etc. from the viewpoint of one, the other is hurtling toward it at a treeemeeedooouuusss rate; and vice versa. Without any equations you could safely say that the damage should be shared equally.

OK... then answer me this. How come when you take a wooden bat, hold it upright, and hit it as hard as you can with another identical bat, that the bat that is stationary always is the one to snap in half?

I think you might be making an incorrect assumption there. How often have you actually done this experiment? Are you sure that's really the result?

(Assuming the bats really are identical, which you can't do because of the nature of wood as a naturally-grown substance. And assuming that you strike them so that the both contact each other at the same "spot" on the bat and the same angle, which you can't do just by hand, you'd need some kind of a machine to do the test...)

I think that both bats would be damaged equally, and whichever one had a slightly weaker wood grain would be the one to fully snap (if at all).

Also, how is each bat supported? Support forces will play a dramatic role. If the non-moving bat is secured at the base there would be shear and bending forces acting on the bat that would not exist if the moving bat hit an unsupported bat in a weightless environment. Tony also brings up good points. The problem with 'pure' physics is that things get hairy quickly once little 'real world' issues like these come into play. That's why there are engineers.

That being said, if you can prove Mr. Newton incorrect, I'd say you have a very bright future.

Best,
-Zeke

That being said, if you can prove Mr. Newton incorrect, I'd say you have a very bright future.

I did that in high school. No, really! We had to do an experiment on one of Newton's Laws, and the results I got
from my experiment ended up being a counter-example to the law, thereby proving it wrong. Unlike everyone else in the class, I wrote up my results without faking the data to go along with what Newton says. My mark on that experiment was rather lower than I thought it should have been.

Just think, if you'd been able to repeat your results, you might have been able to publish and move on to a life of acclaim. Alas, 98% perspiration. On the other hand, nothing ventured, nothing gained (nor, in your case, lost).

My AP physics teacher's primary job was coaching swimming. Only one person in the class (of 15) dared take the AP exam. They got a 3. (For non-US folks - AP exam is a college level exam rated 1-5. A 4 or 5 will usually get you a credit or a placement in an advanced class freshman year. 3 means squat.)

I only passed my freshman physics class the next year by making sure that the units worked right. If the answer was in the right units, it was probably not too far off.

Least favorite expression in O'Hanian (aka O'Heinous)'Physics':
"It should be intuitively obvious to the most casual of observers that: "

I didn't find the following line's statement obvious, casual observations or no. Oh well. That's the beauty of physics: it keeps right on working regardless of whether one understands it or not.

-Zeke

OK.... This is really driving me nuts. So I called my high school physics teacher. He confirmed what I thought to be true. The car travelling twice as fast will do about 4 times as much damage to the slower car as it will get. The problem above is that I was applying the wrong formulae above.

The Real issue, according to him, deals with kinetic energy, not momentum. An object of mass m moving with a speed v has a kinetic energy of ½mv². So you get ½mv² and ½mv`² Assuming speeds of 50kph and 100 kph and a mass of 1 unit (since they're the same) you get KE=1250 vs KE=5000. These units are in joules per kilogram of the mass. So if it's say 1500 kg trucks you get Car 1 issuing 1,875,000 Joules of energy into the other object. Car 2 transfers 7,500,000 Joules of energy into the other car. Since both cars come to a complete stop, the kinetic energy must be transfered into the other vehicle as damage, not movement. If you'd like, I'll scan and post the pages from my physics book.

And I'm not trying to prove Newton wrong, just a misinterpretation of his work.

The Real issue, according to him, deals with kinetic energy, not momentum.

Ah-ha! That does, indeed, sound better. Thanks for clearing that up.

Whew, glad to know I wasn't going crazy.

Since both cars come to a complete stop, the kinetic energy must be transfered into the other vehicle as damage, not movement.

Who says they both come to a complete stop at impact time? The slower car would begin moving backwards after the impact. The only way they'd come to a complete stop at impact time is if they had super-sticky tires (stickier than any formula 1 tire)...

Nothing. It was just an assumption made for the sake of simplicity. Since the crash is assumed to be inelastic, it makes the math much simpler. In the real world, the impact would NOT be completely inelastic, and in fact some of the energy would be transferred into movement, by both cars. We can't really get into impulse force dynamics without knowing the relative elasticity of the cars. So exactly how hard do we want to make this? We also need to know exactly how long it takes both cars to transfer this momentum, since Joules=kg·m²/s². We simply don't have enough information to take all this into account.

Heck, I'm just trying to avoid any actual thinking. I do it all day at work and have very little left over to add to solving this question. Heck, everyone here knows I like to snipe in and add very little information to otherwise erudite discussions. Kudos to you for actually trying to answer the question. I know the answer (at least for cars) is that anything over 20 mph is likely to result in the insurance company declaring both cars totaled, so it wouldn't really matter which sustained more damage. Besides, who has time to whack bats against eachother to see which breaks first? Those things aren't cheap! Maybe aluminum bats...

My question, is if two equally massed Volvos hit each other, does the one with the baby on board have right of way?

-Zeke

I only wish. I was doing 45 through a green light when some old man decided to turn left through the light without actually checking to see if someone was *gasp* going straight. He was doing at least 30-35. Head on collision unavoidable. When I bought that particular Cherokee (used) I paid 14,000. They estimated about 7000 of damage to the car. By the time they had finished "fixing" my car, they had spent closer to 15000. Plus rented me a car for 3 months while mine was in for repair. Then the damn thing never did run quite right again. Not to mention all the little things that I couldn't "prove" were caused by the wreck, such as the rear door hydraulic arm bolt shearing off inside the door frame. Oh well... I finally just sold it for $1000 two years later and bought a mustang. The 80-something year old man's car was totally gone. He's lucky to be alive, especially considering I had to keep my g/f from throttling him. Insurance companies suck.

Thanks lectric. I guess we have a winner.

So if it's say 1500 kg trucks you get Car 1 issuing 1,875,000 Joules of energy into the other object. Car 2 transfers 7,500,000 Joules of energy into the other car. Since both cars come to a complete stop, the kinetic energy must be transfered into the other vehicle as damage, not movement.

No. 1,875,000 plus 7,500,000 joules (9,375,000 altogether) are converted from kinetic energy to "damage" (sound and friction). But there's no rule that says that all of car 1's energy goes into car 2 (rather than being partially dissipated in car 1 itself), or vice versa. In the zero-momentum frame, the situation is completely symmetrical. Each car will sustain about 4,687,500 joules of damage.

If you'd like, I'll scan and post the pages from my physics book.

If it genuinely concurs that all car 1's energy will leap out into car 2 causing car 1 no ill-effects, and vice versa, then just scan the title page. It'll let us know which authors to avoid

Peter

The car travelling twice as fast will do about 4 times as much damage to the slower car as it will get.

By this argument, you could hit a stationary car going as fast as you like and only the stationary car would be damaged. "It should be obvious to even the most casual observer" that this is complete bollocks.

Peter

See? This is why I gave up on physics in University.

I so wanted to say something like this, but after a fortnight of lots of my common sense being refuted I thought I should hang back and wait...

I definitely agree that the force on A from B has to equal the force on B from A. So the damage really all comes down to construction, angle/position of collision, speed difference, and what happens to the cars after the impact.

OK, sorry I'm so late into this thread. I would let it die, but I think I can shed some light on the subject (I am a mechanical/strength engineer, or at least I used to be).

The problem with this discussion is the word "damage" and some confusion about the nature of "inelastic" collisions. These things are related. "Inelastic" means that the cars don't bounce off of one another. It means that all of the energy of the impact is *absorbed* by the bodies rather than remaining kinetic energy.

In an elastic situation (as if the cars were billiard balls), the two cars bounce off of one another and sustain *no* damage. Their speeds change is all. Like billiard balls. I can't remember the relative speeds used in the example, but in a perfectly elastic situation, the kinetic energy is conserved and the slower vehicle speeds up and the faster one slows down.

The inelastic situation is different. When we speak of inelastic collisions, we are entering into the realm of "deformable body mechanics". That means the bodies in question "deform" or go beyond the elastic region of the material and "bend" or "break". As they do so, they absorb the energy of the collision. [Technical folks will remember that the total energy absorbed is the area of the hysteresis in the stress/strain curve.] This is why cars are designed to "crumple". By designing a car to crumple, the energy of an impact is dissipated by the crumpling, if you will. If you're clever about how you design the vehicle to crumple, you can lengthen the duration of the energy transfer quite significantly so that the forces are significantly smaller.

So, you can see the relationship between the inelasticity and "damage". Damage is what happens when the kinetic energy of the impact is absorbed by the bodies and the materials deform or fracture.

You can see that "damage" is a function of design. *If* the vehicles are of identical design (which the won't be, because the front of one car is hitting the rear of another) and the impact is totally symmetrical, then they should sustain nearly identical damage. This is never the case, obviously.

The speed that the vehicles are travelling relative to the road is irrelevant. As mentioned above, it is their speed relative to each other that matters.

The high school physics teacher is correct that the faster car has more kinetic energy, but what matters is the relative speed and the energy that must be absorbed to bring the cars to the same speed. All things being equal (which they never are), the cars will share the impact energy and sustain the same damage.

To answer the question about the baseball bats: it is because the systems are not of identical design, as mentioned above. Swing two bats at each other and see what happens.

Hope this helps,

Jim

Ah bugger it all. This is why I fix computers for a living. This all just makes no sense. I mean, When I was playing baseball, if I ran full throttle and creamed the catcher, He sustained far more "damage" than me. More than a couple were knocked out, and I barely felt it. As was mentioned in other threads, I was slight of build as a child, so if it was mere mass, I'da always lost, especially since most catchers were simply power hitters and therefore a little on the beefy side.

Basically, I DO understand what you're saying, it just doesn't make sense to me. So I give up.

I'm guessing you "aimed" at those poor beefy catchers? You didn't lead with your face, did you? You might've hit them low, or with your feet? Did you guys just bump bellies?

:-)

Grumble..... I dunno, I just kept running, ignoring them. 90% of the time they dropped the ball and I was safe.