Edit: Assume both cars are identical
In: 0
I’m not a physics wizard but …
If I’m not wrong, the crash force is mass * velocity, in the first example, you would have (X+Y)*v where X and Y is the cars masses and v Is the velocity in the hit point, I’d say it’s 50mph expressed in m/s.
In the second scenario, it’d be just X*v
EDIT I’m using velocity when it’s acceleration BUT…in a crash, the velocity goes to 0 so I asume v = a in this case
So, if both cars have the same mass, there’s no difference, BUT if the cars doesn’t have the same mass, it could be better or worse depending on who you are crashing against.
Imagine a focus (1300kg) against another focus at 50km/h, (1300+1300)*833ms. It’s 2.166T of force
A focus (1300kg) against a smart (1000kg) at 50km/h, (1300+1000)*833. 1.9T of force
A focus (1300kg) against a hammer (3000kg) at 50km/h, (1300+3000)*833, 2.5T.
So imagine, hitting your focus against a wall at 100 km/h would be 1300*1666, 2.16T
It just depends on the other car
Ps. Please check my maths
[deleted]
Depends on other factors. What lane are you in? What car are you driving? The instant of the collision is the same, but the aftermath isn’t.
The 50v50 collision leaves you both stationary. The odds of any more damage happening after the accident are pretty slim. Your cars aren’t flying off the road at 50mph.
In the 100v0 scenario, both cars are now flying off at 50mph and probably no longer in a position to steer or stay on the road.
It’s basically the same.
The momentum is the same, the crash forces will be the same.
The only difference is using a different frame of reference, so you’ll end up with some terms cancelling out in the force and momentum equations.
**Momentum:**
1.) p=mv –> mv initial = mv final
2.) m-car1 * v-car1 + m-car2 * v-car2 = m-final * v-final
3.) m-car1 * v-car1 + m-car2 * v-car2 = (2 * m-carN) * v-final (assuming the car masses combine)
let’s say the masses are the same and just divide by 1 car’s mass (this will work with different masses, though the change in speed will be smaller for the more massive vehicle – both scenarios will be identical.)
4.) v-car1 + v-car2 = 2 * v-final
So here we have the difference, you can do 100 + 0 or 50 + (-50). Either way, the cars will be going half the final differential speed relative to their initial speed.
You can also shift the frame of reference:
to car1: 0+(-100) vs 0 + 50 + (-50) [going car ground car for the 2nd case here]
to car2: 100 + 0 vs 50 + (-50) + 0
Note they all add up to the difference in speeds. So car 1 is going from 100 to 50, or 50 to 0 (-50mph), and car 2 is going from 0 to -50, or 50 to 0 (-50mph)
**Energy:**
Then you move energy imparted – which is based on the change in velocity. Both cars see a delta of 50mph in each scenario, so will see equal damage.
**Force:**
To get the peak forces imparted on the passengers, you need to know about the crumple zones. Car crash rails are designed to [crush with a constant force](https://www.youtube.com/watch?v=TM5dyY8zfxs), but I think you see an increase in force as you get further into the vehicle. (Imagine crushing a pyramid, you will get into more material the further you go)
To calculate forces, I would look at the energy imparted, then see how much distance that gets you into the crumple zone, then back out how much force that would impart.
**Real world:**
There is one difference here, you have the “ridedown” – in one scenario, post impact, the cars are at 0mph relative to the ground, in the other they are both going 50mph. After a crash like this, the cars won’t have the wheels and brakes all attached/in good shape. (Even if they were, I don’t think either driver would be able to apply sensible control) At 50mph they’d be able to roll along for about a mile or steer into something and hit it.
So if I had to choose a scenario, I would rather be in the +50 -50mph crash.
Here’s a more detailed breakdown of the math: https://www.youtube.com/watch?v=JyCVl95UDxw
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