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Car crash physics/mechanics
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09-12-2012 02:28PM#1
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The energy will be equally shared between the two cars, the effect on each car will be the same as if it had hit an immovable object at 40mph.Collision speed = 80mph for both carsIf both cars are the same weight and model, then they both will be affected like each on of them crashed at 40MPH into a wall.
Wrong, the energies involved is the square of the speed.
The car going at 50 will have a K.E of massx1250 and the car at 30 will have a K.E of massx450, non SI units used so only a rough estimation. The faster travelling car has more kinetic energy so should do more damage to the slower travelling car.0 -
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I think momentum and change in velocity rather than KE is what the OP is asking about. The KE does indicate the potential for damage but it will be dissipated by both cars as both have the exact same crumple zones.Wrong, the energies involved is the square of the speed.
The car going at 50 will have a K.E of massx1250 and the car at 30 will have a K.E of massx450, non SI units used so only a rough estimation. The faster travelling car has more kinetic energy so should do more damage to the slower travelling car.
A car colliding with an identical stationary car at 80 km/h results in the same damage as two identical cars colliding with each other doing 40 km/h
If KE before the collision was the full story then if a moving car hits a stationary truck you'd be better off in the car. This clearly isn't the case.
Going back to the OPs query i think for the occupants in identical vehciles a 50 km/h + 30 km/h collision is the same as an 80 km/h + 0 km/h collision or a 40 km/h + 40 km/h collison.
The human body doesn't distinguish between being decelerated from 80 km/h to 0 km/h in 1 second by a seatbelt or being accelerated from 0 km/h to 80 km/h in 1 second by a seatbelt. Therefore it doesn't matter whether you are in the car doing 80 km/h or the car doing 0 km/h.
I may be wrong of course and some of the posts at the below link would suggest so but others would agree with me.
http://www.abc.net.au/science/k2/stn/q&a/notes/060518-6.htm0 -
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The force of the impact isn't halved, but there is twice as much crumple zone to absorb it.The only part I can't wrap my head around is why the impact is halved, just because the impact is with another car....
This is true.In the event of a head on collision the total momentum is conserved. Because the momentum of the faster car is greater, the final velocity of the combined system (2 cars together) will be in the direction of the faster car after the collision. In other words the slower car will be pushed backwards.
No, even though one ends up going backwards, both are still subject to the same deceleration.That means the driver of the faster car experiences less acceleration (same direction of velocity) than the driver of the slower car does (direction of velocity is reversed). The person in the slower car is likely to come off worse, all things being equal.0 -
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wrong.
Wrong, the energies involved is the square of the speed.
The car going at 50 will have a K.E of massx1250 and the car at 30 will have a K.E of massx450, non SI units used so only a rough estimation. The faster travelling car has more kinetic energy so should do more damage to the slower travelling car.
both cars exert the same force on each other no matter what the difference in speed is. This can be seen by newtons third law: if a body 'a' exerts a force on a body 'b' ,then body 'b' exerts an equal but opposite force on body 'a'.
thus each car experiences the exact same amount of force in the crash.
Collision speed is 80mph.0 -
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Where are you getting those figures from? If one car is doing +30mph, and the other is doing -50 then the wreckage will be doing -10, ie a reduction in velocity for both cars of 40.
Your right of course. For some reason I was thinking one car was heavier than the other.
Something about reading the question comes to mind...0 -
Wrong, the energies involved is the square of the speed.
The car going at 50 will have a K.E of massx1250 and the car at 30 will have a K.E of massx450, non SI units used so only a rough estimation. The faster travelling car has more kinetic energy so should do more damage to the slower travelling car.
In your calculation you assumed, that during the accident car going 50 would stop to 0 and car going 30 would stop to 0.
That's wrong, as this would break the rule of conservation of linear momentum.
In fact what is going to happen, car going 50 will slow down to 10, and car going 30, will be accelerated backwards to 10 (so both cars will travel 10 in the same direction).
Obviously this speed will be lost very quick due to friction between cars and the ground (wheels might not operate anymore, brakes might get engaged, etc).
But generally assumption that cars will stop still at the place of accident is only valid when they hit each other in exactly straight line at the same speed, and they will be both the same kind of cars with the same weight.
If speed or weight differs, then after the accidents car will still move.
So according to rule of conservation of momentum (assuming that those 2 cars are isloate system without any other forces working on them) we have somethign like that (assuming cars weight the same).
p=m*v (momentum = mass * velocity)
p(total) = p1(car 1 momentum) + p2 (car 2 momentum)
assume m=1 (tonne)
p1 = 1*50 = 50
p2 = 1*-30 = -30 (minus because other car is going in opposite direction)
p(total) = p1 + p2 = 20
So the total momentum of both cars is 20, and it must remain the same after the accident (due to rule of conservation of momentum)
After the accident:
p(total after) = 20
P(total after) = (m1+m2) * v
v= P(total after) / (m1+m2)
v = 20 / 2 = 10
So the speed after the accident will be 10MPH.
And as you said with kinetic energry - both cars will suffer change of speed of 40mph, so their kinetic energy loss will be equal, therefore damage should be equal.0 -
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07Lapierre wrote: »Send an email to "mythbusters"!!. They'll sort it!
They already did
http://ca.autoblog.com/2010/11/21/road-myths-physics-of-a-head-on-collision/
The nuts and bolts of the results are:
1. If two cars collide head on, both doing 50km/h, the result is the same as hitting a wall at 50km/h
2. It is not the same as a 100km/h car to wall collision, that was the myth busted
Forget the maths and equations for a minute:
- If two objects hit at the same speed, it's the same for each as if they hit a wall at that speed
- Car 1 effectively hits a wall at 50km/h; Car 2 effectively hits a wall at 50km/h
Why?
- Newton's 3rd law is often simplified to every action has an equal an opposite reaction
- Roughly speaking, the energy taken into the collision is exerted back.
- It seems intuitive that the relative velocity is 100km/h and so double the impact, but there is 4 actions (2 actions and 2 reactions) dissipating the energy in 4 ways:
1. Car A exerts an action, actionA, onto Car B
2. actionA has an equal and opposite reaction, reactionA, onto Car A
3. Car B exerts an action, actionB, onto Car A
4. actionB has an equal and opposite reaction, reactionB onto Car B
- actionA = reactionA
- actionB = reactionB
Chucking the physics and maths aside, think of the reaction as the impact. If you hit a wall (immovable object), the reaction is equal to the action, and the reaction crushes your car.
A car travelling at the same speed is a brick wall. Since both cars pop with the same energy (assuming equal weight here), they each have the same reaction as if they hit the wall at 50km/h.
Now, where it gets interesting is if the cars are at different speeds. Lets say Car A is at 50km/h and Car B is at 100km/h.
- action A = reaction A
- action B = reaction B
- BUT the speed is not equal
Effective car to wall impact speed? 100km/h.
- Car B hits a 'wall' at 100km/h [action B = reaction B]
- Car A has a 'wall' coming at it at 100km/h. It hits at 50km/h, but the wall hits back at 100km/h.
You see:
- The action brought by car A in hitting car B has an equal and opposite reaction on car A, but car B is hitting car A too, which in turn has an equal and opposite reaction from car A to car B.
- The action/reaction pair on A must be the same as the action/reaction pair on B
The 'effective speed' (car to wall equivalent) of the crash is the speed of the faster car.
In short, the momentum of the car does not change just because there's something coming towards it.0 -
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