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Comment by cs02rm0 | Hacker News Reader

This is the same physics I was taught in school as for two cars crashing into each other at 70mph I presume? Neither has a 140mph impact, although a 140mph impact would involve more than double the energy anyway AIUI.

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No, not unless you were making relativistic corrections for the vehicles, which would be silly[1].

From the perspective of either vehicle, you do see an oncoming vehicle traveling at (approximately) 140 miles per hour.

When dealing with things traveling close to the speed of light, this doesn't work, because the relativistic correction term becomes significant. The equation they show on the blog

  (u + v)/(1+uv)

represents the velocities in fractions of speed of light. When u and v are much smaller than the speed of light, the denominator can be ignored, as it's close to 1, and we simply get (u+v).

[1] But still pedagogically sound when teaching relativity :)

For the people who (like myself) are confused by the missing c^2 in the blog, (u + v)/(1+(uv/(c^2)) is the original.

Much easier to understand that the "simplified" version, which is usually the case.

In relativity, c=1 is a common convention. It doesn't really add much to the equation because it's just a conversion factor.

Omitting conversion factors is a bit like omitting type information. If you already know what everything is, it doesn't add information, and can make the code look simpler. If you don't, not omitting them can help make sense of things when you're learning and lack context.

To get back a meaningful result for the car example, c=1 does not really help.

It is much easier to plug in u and v directly into the simpler equation that contains c^2.

Otherwise (the indirect method), to get fractions, first you do a substitution a=u/c, b=v/c, arrive at:

c * (a * b) / (1 + a * b)

Then you convert the car speeds to fractions, only to multiply back by c=299792458.

Curiously enough, this sort of thing rarely occurs in the original publications, so they are usually easier to read than obfuscated textbooks or blog examples.

No, I don't think so- two cars impacting head-on at 70mph would have a "140mph impact," insofar as that it would be essentially equivalent to one car traveling at 140mph hitting a stationary car.

Hmm, wouldn't a car of the same mass going the same speed joust counteract all the force exactly? If one of the cars had half as much mass, wouldn't that lessen the impact of the larger car and increase it for the smaller car, as the combined center of gravity afterwards would be moving inthe direction the larger car was traveling?

It's been far too long since I had a physics class...

Well, maybe we're using different terms- as the article notes, there's technically no such thing as a "such-and-such speed impact." I was speaking in terms of how much kinetic energy is liable to go towards damaging the cars involved (or into the fireball of a particle collision); in that case, if you have X energy in two cars, or 2X energy in one and 0 in the other, the sum is the same.

If we consider debris or etc flying away after the collision, then you're right that they're different- the combined velocity of the two cars is 0mph in the both-moving case and 140mph (in some direction) in the one-moving case, and the velocity of all the debris will sum to that velocity. (And the sum of velocities would be weighted appropriately if one of the cars was more massive than the other.)

On a side note, if a massive 4wd heads-on collides with Corolla, the 4wd guy will have scratches and bruises, Corolla people would not suffer, instant death.

Choosing a car, choose tons, not stars :)

It's a sad reality that just makes the road less safe for everyone and hurts the environment :(

I could have sworn that a perfect head on collision was basically the same thing as a collision at the same speed into a perfect stationary wall?

Mythbusters experiment: https://www.dailymotion.com/video/x2n9j62

No. If the wall is stationary and infinitely massive, you’d bounce off at the same velocity; if it was moving at the same speed as you you’d bounce off at three times the original velocity.

Edit: if the other object is a car, then yes, there is no difference in either case from your perspective.

If the collision is completely inelastic, then there will be no bouncing. In a car, usually the crumple zone completely absorbs and dissipates all the energy kinetic.

Yeah, this was assuming an idealized perfectly elastic collision, which is a bit difficult to have in the real world ;)

Yes, if the cars are identical and collide perfectly enough that both crumple in exact symetry. Each absorbs and so the damage is split, as opposed to hiting a non-absorbing wall. But that is basically impossible to setup.

Hitting a car is not the same as hitting a wall.

All three of these do similar damage, if the cars have equal mass:

  * 70mph car vs. 70mph car
  * 140mph car vs. parked car
  * 70mph car vs. wall.

The only real question is whether you consider "140mph impact" to mean "vs. a similar-mass car" or "vs. a wall". These are wildly different impact intensities that should not be confused.

FWIW I was thinking immovable wall rather than parked car for the 140mph impact.

The wall is complicated enough for me without thinking about multiple crumple zones, handbrakes, etc.

Kinetic energy is proportional to velocity squared. So the 140mph car has twice as much KE as 2 cars at 70mph.

It does depend how much of that energy is dissipated in the collision though.

Assuming similar vehicles that come to a stop, both cars experience a deceleration from 70 mph to zero. That is the same change of momentum per vehicle as 1 car hitting a stationary wall at 70.

I belatedly realize that almost everything I said in this thread was wrong, but it’s too late to delete it! The HN edit rules are cruel.

Depends on your reference frame. In the reference frame of the approaching car you have a 140 mph impact.

That concept still feels intuitively wrong to me.

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