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[AMF]

Science rocks.

[Xrati]

It's not the fall that will kill you! It's the sudden STOP at the end.

[El_Phil]

Indeed. I have no problem with heights, but falls terrify me.

[Renegade 13]

Will said:

With all the science people on these boards, nobody's called bull**** on this yet? The weight of the vehicle does NOT increase the safety. If anything, extra weight makes car more dangerous.

You should recall from Newtonian physics, that inertia is the tendancy of mass to remain at it's current velocity (or speed and heading, if you prefer). When the vehicle is moving, the inertia causes it to keep moving at its current speed until outside forces slow it down. In a wreck, this outside force is applied during a fraction of a second. So, the equation for kinetic energy, E=(0.5)*m*(v^2), says that a vehicle that is twice as heavy will have to withstand twice as much energy on impact. With older cars, the extra weight does not add to the structure of the vehicle overall, and simply adds on to the stress the frame needs to deal with to protect the occupants of the vehicle.

Ok, lets use your equation, E=1/2(m)(v^2). My vehicle weighs in at, oh lets say 2000 kg. Yours weighs 1000 kg (all that plastic ). We're both travelling 100 km/h. My kinetic energy = .5(2000)(100^2) = 10,000,000 or 1.0 x 10^7 (Joules, I think). Your kinetic energy is .5(1000)(100^2) = 5,000,000 J or 5.0 x 10^6 J.

So we're both in a head-on crash. For my forward kinetic energy to be dropped down to 0 J (and therefore 0 km/h) I would have to encounter a force of -10,000,000 J in the direction that I'm travelling. However, your vehicle is only capable of supplying a counter-force of -5,000,000 J. Therefore, after my vehicle impacts yours, I should have 10,000,000 J - 5,000,000 J = 5,000,000 J of kinetic energy remaining in the motion of my vehicle. Reverse the above kinetic energy equation, and you get 5,000,000/.5/2000kg = 5000. Take the square root of 5000 to get the velocity of 70.7 km/h remaining. So, based upon what little I know of physics, I would decellerate about 30 km/h in a very short amount of time, while you, if you were lucky in your lighter vehicle, would decellerate a minimum of the 100 km/h you were formerly travelling. A situation such as this is why a semi-truck (or any other large truck for that matter) is more likely to survive in the event of a crash with a smaller vehicle. They sure don't have that much plastic on them either!

Now, since I'm not the best when it comes to physics, can someone please point out all the flaws in my thinking?

[Kamog]

I agree with Renegade. In a head-on crash, you'll be safer in the heavier vehicle. Because of the greater kinetic energy and momentum, the heavier vehicle will be slowed down less, and therefore you'll experience less deceleration.

I'm not sure which car would be safer if the vehicles individually crashed into a solid wall, but we're discussing a collision with another vehicle, which is what he said in his original post.

[Will]

Renegade, almost right. First, to get Joules out of E=(0.5)*m*(v^2), you need mass in kg (you got that), and v in m/s (not km/hr). 100km/hr == about 27.8 m/s.

So, in the hypothetical full head-on collision (which, by the way, would almost never happen, nearly all are offset collisions), the two cars would have:

E=(0.5)*2000*(27.8^2) = 772840 J
E=(0.5)*1000*(27.8^2) = 386420 J

And right at the point of impact, things get fuzzy. In an ideal collision, the larger vehicle would have 386420 J of energy remaining in its direction, and so the COMBINED speed would be v=sqrt(386420/1500)=16 m/s = 57.6 km/hr

However, this would not be an ideal collision. The cars bounce and deform, there's friction from the road, most likely there will be a conversion of some forward momentum to torque, making the vehicles spin, etc. This is where the more intangible stuff begins to take effect, like how well materials will crumple up around the frame.

What would likely occur in the full head on collision would be the lighter car would begin to turn away from the larger, beginning from the rear. So, the rear of the lighter car will arc around, possibly until the direction the lighter car is facing is the same as that of the larger, but probably more at a perpendicular angle. Basically, the lighter car is designed to get the most of that impact energy dispersed on something other than the passengers. The bigger vehicle will not turn as much, probably won't crumple as much, and a lot of the decelleration gets applied to the squishy passengers.

Regardless, in this particular situation, at that high speed, none of the passengers is going to feel "fine" afterwards. So, what it comes down to is, do you want to have a lower chance of injury, or do you want to point at the other guy's car and say "I win!" (likely while wearing a neck brace)?

Oh, and the semi... have you ever seen just how much open space is in and around those cabs? This is the one case where having a lot of size (as in volume) and mass would pay off. Because semis are higher off the ground, some of the impact energy goes into trying to LIFT the semi, and so gets dispersed. Then, there's a good sized space between the front of the truck and the cab, spreading out the impact energy over time. Plus, the mass is so much bigger than any conventional vehicle because of the cargo, the reduce in speed is not going to be very high. But does this mean that you're going to start driving a semitruck around? What happens when everybody does this?

*/really now, end thread hijack*

[Renegade 13]

Arrrggg, I knew I messed up somewhere! m/s instead of km/h...I knew that too! It's just been a while since I took physics

I think we both agree on the physics part of it, just not on how much decelleration will be applied to the passengers. Since I don't think either of us are physicists (we aren't crazy enough!), I doubt either of us is qualified to say exactly what will happen, especially as real life isn't nearly as antiseptic as some equation. Oh well, agree to disagree Often a very productive thing to do!