Car crash physics/mechanics

Collision speed = 80mph for both cars

http://www.sciforums.com/%2520%253c/Collision-physics-explanation-needed-t-104387.html

HTH.

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.

randy hickey

NIMAN wrote: »

I was just thinking about the physics of a car crash whilst reading another thread here, and have a question.

If 2 cars are in a head-on crash (similar car, same weight etc) would the person in the faster car have more chance of coming off with less injuries?

Basically Car A doing 30mph, and Car B coming the opposite direction doing 50mph. Car B obviously has more momentum and power behind it, is Driver A likely to come off worse or does it not matter?

If 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.

CiniO wrote: »

If 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.

Am I missing something here? If one car is doing 30 and the other is doing 80, then the combined speed is 80. So would the collision speed not be 80mph?

The above post would suggest that collision speeds are halved without any reason being given. If i hit a car head on, and I am doing 100 mph and the other car is doing 20mph, the inertia/collision force is 120mph, not 60mph. Lets say the other car is stationary....If i hit it at 100mph, what is the collision force/speed that the other car will experience? Not 50 as far as I can tell.

With head on collisions of objects of the same type, weight, mass etc (two identical cars), then both speeds are added together to get the end collision speed. If both cars were travelling in the same direction and the car in front was struck, then you minus the speed of the front car from the rear car to get the collision speed.

goz83 wrote: »

Am I missing something here? If one car is doing 30 and the other is doing 80, then the combined speed is 80. So would the collision speed not be 80mph?

The above post would suggest that collision speeds are halved without any reason being given. If i hit a car head on, and I am doing 100 mph and the other car is doing 20mph, the inertia/collision force is 120mph, not 60mph. Lets say the other car is stationary....If i hit it at 100mph, what is the collision force/speed that the other car will experience? Not 50 as far as I can tell.

The combined speed is indeed 80mph, but it's being shared between two cars. The effect on each car is the same as had that car hit an immovable object at 40mph.

Del2005

Anan1 wrote: »

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.

goz83 wrote: »

Collision speed = 80mph for both cars

CiniO wrote: »

If 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.

Del2005 wrote: »

The faster travelling car has more kinetic energy so should do more damage to the slower travelling car.

Why?

NIMAN

Del2005, thats what my original question was hinting at.

If one car is doing 10mph and the other 60mph, surely the 60mph car will push through the 10mph car and push it back, and go through it with more force than the slower car.

So you saying that it would be better to be in a faster moving car in a head-on?

The kinetic enery at the point of impact is the same for both cars. The only part I can't wrap my head around is why the impact is halved, just because the impact is with another car....but I guess it's got something to do with the fact that the cars are moveable objects. Still...it doesnt mean that a car hitting a stationary car at 100mph reduces the impact force to that equivalent to 50mph does it? There would have to be 0 friction for that to happen.

BrianD3

Del2005 wrote: »

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.

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.

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.htm

AvAv

If you collide, the forces are as if the cars hit immovable objects. The car cannot force the object to move, so the object forces the car to stop.

Unfortunately for the driver, if you are sitting in a car going at a high speed, your body is also moving at the high speed. When the car is stopped, the stopping force is only acting on the car, not yourself and you will continue to travel with high speed until forced to stop by seatbelt etc.

If you are travelling at a higher speed it takes more force to stop you.

If you travel and collide at 60 km/h, you will need a big enough force to slow you down from 60 to 0.

If you travel at 10, you need a force to slow you down from 10 to 0.

Which speed is better is relative to whether you're talking about your car or yourself.

NIMAN wrote: »

I was just thinking about the physics of a car crash whilst reading another thread here, and have a question.

If 2 cars are in a head-on crash (similar car, same weight etc) would the person in the faster car have more chance of coming off with less injuries?

Basically Car A doing 30mph, and Car B coming the opposite direction doing 50mph. Car B obviously has more momentum and power behind it, is Driver A likely to come off worse or does it not matter?

The momentum of each car is given by p = mass x velocity. Assuming both cars weigh the same the car travelling at 50 mph will have a higher momentum than a car travelling at 30 mph.

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.

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.

In the real world there are a lot of different factors at play - the elasticity of the collision will be determined by the types of vehicles - crumble zones, airbags etc. The actual physics at play is a tad more complicated that 30mph+50mph = 80mph collision for both drivers. But in general the heavier / faster vehicle comes out better in a straight head on collision - i.e. the one with the greatest momentum.

goz83 wrote: »

The only part I can't wrap my head around is why the impact is halved, just because the impact is with another car....

The force of the impact isn't halved, but there is twice as much crumple zone to absorb it.

Mikros wrote: »

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.

This is true.

Mikros wrote: »

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.

No, even though one ends up going backwards, both are still subject to the same deceleration.

07Lapierre

Send an email to "mythbusters"!!. They'll sort it!

tacofries

Del2005 wrote: »

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.

wrong.

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.

Anan1 wrote: »

The force of the impact isn't halved, but there is twice as much crumple zone to absorb it. :)This is true.

No, even though one ends up going backwards, both are still subject to the same deceleration.

No not quiet. Taking the direction of travel of car1 as positive. The figures below are illustrative but hopefully you will get the idea:

Car1 travelling ---> +30 mph
Car 2 travelling <---- -50 mph

Car1car2system after collisions <

-20mph

Change in velocity of car1: -50 mph
Change in velocity of car2: -30 mph

There is a greater deceleration on car1 and the resulting forces on the driver would be higher.

tacofries wrote: »

wrong.

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.

You don't understand Newtons laws and their application to collisions.

The force each car experiences is determined by the acceleration. Acceleration is change in velocity over time. When two cars collide there is a change in velocity in a very short period of time. The change in velocity is based on the conservation of momentum - the momentum before the collision is the same as the momentum after.

Momentum is mass x velocity. The faster and heavier car will lose less velocity and therefore will experience less acceleration / force.

Think about if a train hit a car head-on both travelling at 80 km/h. The train will not experience the same force as the car because the change in velocity of the train will be almost 0. This is because the mass of the train is many many times that of a car. That is the conversation of momentum.

The "speed of the collision is 80mph" means nothing in physics.

Mikros wrote: »

Car1 travelling ---> +30 mph
Car 2 travelling <---- -50 mph

Car1car2system after collisions <
-20mph

Change in velocity of car1: -50 mph
Change in velocity of car2: -30 mph

There is a greater deceleration on car1 and the resulting forces on the driver would be higher.

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.

Anan1 wrote: »

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...

Del2005 wrote: »

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.

tacofries

Mikros wrote: »

The train will not experience the same force as the car because the change in velocity of the train will be almost 0.

Sorry, the question my post was aimed at was one about force(not the original question by the poster).

I do still stand by the car and the train experiencing the same force when they crash due to the fact that when the train hits the car, the car also hits the train. It doesn't matter if the change in velocity will be almost 0 for the train since force is equal to mass by acceleration,, and so even if the de-acceleration of the train is very small the mass of it is large enough to make the force very large. Obviously the change in the cars velocity will be very high, but only in proportion to the mass of the train.. Thus both experience the same force!

Mikros wrote: »

You don't understand Newtons laws and their application to collisions.

The force each car experiences is determined by the acceleration.

You forgot about one detail - force is determined by mass and acceleration (no solely acceleration)

Acceleration is change in velocity over time. When two cars collide there is a change in velocity in a very short period of time. The change in velocity is based on the conservation of momentum - the momentum before the collision is the same as the momentum after.

All true here.

Momentum is mass x velocity. The faster and heavier car will lose less velocity and therefore will experience less acceleration / force.

Heavier car - true - will lose less velocity.
But if they weight the same, they will both experience the same velocity change.

Think about if a train hit a car head-on both travelling at 80 km/h. The train will not experience the same force as the car because the change in velocity of the train will be almost 0.

The train and car will experience the same force.
But acceleration is force * mass (a=Fm).
So the same force will give huge acceleration to car, as his mass is small, while the same force will change train's speed only by a negligible bit.

MrDerp

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.

BrianD3

MrDerp wrote: »

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.

What if car A is doing 0 km/h and car B is doing 80 km/h - in terms of damage and injury the "car to wall" equivalent is not going to be the speed of the faster car. An 80 km/h impact into a parked car is a lot less severe (for the occupants of both the parked car and the moving car) than an 80 km/h impact into a wall.

Ronnie Beck

BrianD3 wrote: »

What if car A is doing 0 km/h and car B is doing 80 km/h - in terms of damage and injury the "car to wall" equivalent is not going to be the speed of the faster car. An 80 km/h impact into a parked car is a lot less severe (for the occupants of both the parked car and the moving car) than an 80 km/h impact into a wall.

Depends on the wall...

https://www.youtube.com/watch?v=2TbNR7L57FQ

Ronnie Beck wrote: »

Depend on the wall...

True.
Some wall lose even with small cars.

MrDerp

BrianD3 wrote: »

What if car A is doing 0 km/h and car B is doing 80 km/h - in terms of damage and injury the "car to wall" equivalent is not going to be the speed of the faster car. An 80 km/h impact into a parked car is a lot less severe (for the occupants of both the parked car and the moving car) than an 80 km/h impact into a wall.

Neither here nor there. I said nothing about damage, I was merely talking about the equivalent speed of impact. Two cars will crumple and take some impact, and a car on wheels is more moveable than a wall. Whatever the obstacle, the impact on the moving car and the other object is the same.

As, car chases on tv show us, a car hitting a bunch of boxes will scatter the boxes, a car hitting a plate glass window on it will go through the glass, and a car hitting a clothes rail will drag it down the street, with hilarious consequences. This has everything to do with what's hit and doesn't change newtons 3rd law.

A trampoline will bounce you better than a sheet, but it doesn't mean the impact is less, it's the elasticity of what your hitting that changes the results. I'd rather hit a modern parked empty car than a wall or tree for sure, just as I'd rather hit that wall in my 2007 car than a 1993 kadett