A couple of physics questions
Discussion
There are 2 things I’ve heard over the years, that I would like clearing up. Both of which I’ve thought, ‘“nah not sure about that” but haven’t been able to verify one way or the other.
1. 2 vehicles travelling at 60 mph crashing head-on is the equivalent, for each vehicle, of hitting a stationary immovable object at 120 mph.
2. Nigel Mansell once said something like, through a series of corners going from a left hand bend at 4g to a right hand bend at 4g is 8g in total.
Can anyone prove these statements right or wrong?
1. 2 vehicles travelling at 60 mph crashing head-on is the equivalent, for each vehicle, of hitting a stationary immovable object at 120 mph.
2. Nigel Mansell once said something like, through a series of corners going from a left hand bend at 4g to a right hand bend at 4g is 8g in total.
Can anyone prove these statements right or wrong?
Arnd said:
There are 2 things I’ve heard over the years, that I would like clearing up. Both of which I’ve thought, ‘“nah not sure about that” but haven’t been able to verify one way or the other.
1. 2 vehicles travelling at 60 mph crashing head-on is the equivalent, for each vehicle, of hitting a stationary immovable object at 120 mph.
This came up on PH quite recently. See Mythbusters1. 2 vehicles travelling at 60 mph crashing head-on is the equivalent, for each vehicle, of hitting a stationary immovable object at 120 mph.
Arnd said:
There are 2 things I’ve heard over the years, that I would like clearing up. Both of which I’ve thought, ‘“nah not sure about that” but haven’t been able to verify one way or the other.
1. 2 vehicles travelling at 60 mph crashing head-on is the equivalent, for each vehicle, of hitting a stationary immovable object at 120 mph.
2. Nigel Mansell once said something like, through a series of corners going from a left hand bend at 4g to a right hand bend at 4g is 8g in total.
Can anyone prove these statements right or wrong?
1. Assuming they both have equal deformation it is the equivalent of each vehicle having an accident at 60mph. The only thing that is relevant is the speed they are doing initially, the speed they are doing when the accident is finished, and the distance each vehicle travels between those speeds.1. 2 vehicles travelling at 60 mph crashing head-on is the equivalent, for each vehicle, of hitting a stationary immovable object at 120 mph.
2. Nigel Mansell once said something like, through a series of corners going from a left hand bend at 4g to a right hand bend at 4g is 8g in total.
Can anyone prove these statements right or wrong?
To illustrate, if a Metro is doing 60mph and has a head on with a 500 tonne train doing 60mph, it's going to end up being propelled backwards at more or less 60mph, so it will experience the equivalent of a 120mph crash while the train doesn't really experience anything
2. Our Nige, bless him, whilst being a heroically talented driver has a tendency to make some statements which are of varying accuracy... if you believed the list of injuries he claims to have suffered for example then he would have been out of action for half his career and crippled! 4G in opposite directions is just that, if you could add them together then you'd end up with zero and you wouldn't be changing direction at all!
Arnd said:
There are 2 things I’ve heard over the years, that I would like clearing up. Both of which I’ve thought, ‘“nah not sure about that” but haven’t been able to verify one way or the other.
1. 2 vehicles travelling at 60 mph crashing head-on is the equivalent, for each vehicle, of hitting a stationary immovable object at 120 mph.
2. Nigel Mansell once said something like, through a series of corners going from a left hand bend at 4g to a right hand bend at 4g is 8g in total.
Can anyone prove these statements right or wrong?
1. No - you're still doing 60 no matter whether you hit another car of a solid object.1. 2 vehicles travelling at 60 mph crashing head-on is the equivalent, for each vehicle, of hitting a stationary immovable object at 120 mph.
2. Nigel Mansell once said something like, through a series of corners going from a left hand bend at 4g to a right hand bend at 4g is 8g in total.
Can anyone prove these statements right or wrong?
Kinetic energy (in Joules) = 1/2 * Mass * velocity squared (Classical mechanics, after Kelvin, Newton, Bernoulli and Leibniz)
If you multiply velocity by 2, you quadruple (2 squared) the kinetic energy.
2 cars means double the kinetic energy.
1 car going twice as fast means 4 times the kinetic energy.
If you multiply velocity by 2, you quadruple (2 squared) the kinetic energy.
2 cars means double the kinetic energy.
1 car going twice as fast means 4 times the kinetic energy.
Arnd said:
There are 2 things I’ve heard over the years, that I would like clearing up. Both of which I’ve thought, ‘“nah not sure about that” but haven’t been able to verify one way or the other.
1. 2 vehicles travelling at 60 mph crashing head-on is the equivalent, for each vehicle, of hitting a stationary immovable object at 120 mph.
2. Nigel Mansell once said something like, through a series of corners going from a left hand bend at 4g to a right hand bend at 4g is 8g in total.
Can anyone prove these statements right or wrong?
On 1), no. Identical cars hitting each other at speed x is the same as one hitting an immovable object at x.1. 2 vehicles travelling at 60 mph crashing head-on is the equivalent, for each vehicle, of hitting a stationary immovable object at 120 mph.
2. Nigel Mansell once said something like, through a series of corners going from a left hand bend at 4g to a right hand bend at 4g is 8g in total.
Can anyone prove these statements right or wrong?
2, it’s not at all clear what that means. I’d definitely say that switching repeatedly from one way to the other is going to feel a lot more than a force being placed in one direction than removed. If you are pushing one way against a 4g acceleration then yes, the sum of you pushing and the acceleration pushing in the same direction may feel for a moment like an 8g force.
Kent Border Kenny said:
2, it’s not at all clear what that means. I’d definitely say that switching repeatedly from one way to the other is going to feel a lot more than a force being placed in one direction than removed. If you are pushing one way against a 4g acceleration then yes, the sum of you pushing and the acceleration pushing in the same direction may feel for a moment like an 8g force.
I have no idea how the reversal of cornering accelerations feel in an F1 car, but in aerobatics the physiological effects of quickly switching from positive G to negative G can be quite unpleasant. Short periods of high positive G can be ameliorated by bracing your torso and maybe having a bit of a yell (less embarrassing if you're on your own), but negative G your only option is largely to turn your head and tuck your chin into your chest (you're generally looking down the wing at the sighting device anyway), but swapping rapidly between the two can mess you up - a 4g pull rapidly followed by a 4g push would be more likely to see you graying out than, say, an 8g pull. Obviously it depends in your G acclimatisation and body type.1) I think a 60mph crash into an oncoming car doing 60mph is equivalent to a 60mph crash into an immovable object. The kinetic energy that needs dissipating is double in the head on crash, but its split equally between the cars. Hitting an immovable object would cause the same deceleration, and therefore force as hitting an oncoming car. The kinetic energy that would need dissipating would be half that of the head on crash, but all of it would have to be dissipated by the one car.
2) I don't think you can add opposite accelerations like that. Its 4g followed by -4g. They aren't happening at the same time. You go through 0g in between too.
2) I don't think you can add opposite accelerations like that. Its 4g followed by -4g. They aren't happening at the same time. You go through 0g in between too.
Edited by Gulf7 on Sunday 12th July 14:04
Gulf7 said:
2) I don't think you can add opposite accelerations like that. Its 4g followed by -4g. They aren't happening at the same time. You go through 0g in between too.
No you don’t. Acceleration is a vector. The magnitude of 4G is the force. The sign is the direction. It is possible to change the direction of force instantaneously or continually without altering its magnitude. Edited by Gulf7 on Sunday 12th July 07:11
At one point the body is experiencing a force equivalent to 4G on the left as it is being pressed by the side of the cockpit, which is accelerating to the right, then it (almost) instantaneously experiences a force of 4G to the right and the car turns (accelerates) left. It doesn’t drop to zero.
As said above is the change in direction of force that I presume Nigel was referring to.
Edited by V8LM on Sunday 12th July 08:15
Very interesting,
The first quote was from an uncle of mine who was a police officer and he was referring to crashes he had seen, but the Myth Busters clip is quite clear.
And the second one also seems to be untrue.
But can an F1 car turn instantaneously? Maybe there would be a transition between the opposite G forces? I don’t know.
Cheers
The first quote was from an uncle of mine who was a police officer and he was referring to crashes he had seen, but the Myth Busters clip is quite clear.
And the second one also seems to be untrue.
But can an F1 car turn instantaneously? Maybe there would be a transition between the opposite G forces? I don’t know.
Cheers
Nimby said:
Arnd said:
There are 2 things I’ve heard over the years, that I would like clearing up. Both of which I’ve thought, ‘“nah not sure about that” but haven’t been able to verify one way or the other.
1. 2 vehicles travelling at 60 mph crashing head-on is the equivalent, for each vehicle, of hitting a stationary immovable object at 120 mph.
This came up on PH quite recently. See Mythbusters1. 2 vehicles travelling at 60 mph crashing head-on is the equivalent, for each vehicle, of hitting a stationary immovable object at 120 mph.
On 1) it's the opposite, assuming identical cars in all cases. If an officer is used to seeing cars hit other cars, then a head-on with both cars travelling at 60mph will naturally look (and be) much worse than a car travelling at 60mph and hitting a stationary car.
Two cars hitting eachother at 60mph each would be about the same as one car doing 120mph hitting a stationary car. In that impact, the stationary car will deform and move backwards, taking some of the energy away from the moving car, which itself won't come to a dead stop immediately. It will carry on travelling forward past the point of impact.
Compare this to a car at 120mph hitting an immovable wall, and it should be clear that the damage would be much more severe, since it will stop immediately (or bounce backwards).
When both cars are doing 60mph, they cancel eachother out. One doesn't push the other one backwards, so neither of them manage to proceed forwards past the point of impact. This makes the collision equivalent to hitting an immovable wall at 60mph.
Two cars hitting eachother at 60mph each would be about the same as one car doing 120mph hitting a stationary car. In that impact, the stationary car will deform and move backwards, taking some of the energy away from the moving car, which itself won't come to a dead stop immediately. It will carry on travelling forward past the point of impact.
Compare this to a car at 120mph hitting an immovable wall, and it should be clear that the damage would be much more severe, since it will stop immediately (or bounce backwards).
When both cars are doing 60mph, they cancel eachother out. One doesn't push the other one backwards, so neither of them manage to proceed forwards past the point of impact. This makes the collision equivalent to hitting an immovable wall at 60mph.
kiseca said:
On 1) it's the opposite, assuming identical cars in all cases. If an officer is used to seeing cars hit other cars, then a head-on with both cars travelling at 60mph will naturally look (and be) much worse than a car travelling at 60mph and hitting a stationary car.
Two cars hitting eachother at 60mph each would be about the same as one car doing 120mph hitting a stationary car. In that impact, the stationary car will deform and move backwards, taking some of the energy away from the moving car, which itself won't come to a dead stop immediately. It will carry on travelling forward past the point of impact.
Compare this to a car at 120mph hitting an immovable wall, and it should be clear that the damage would be much more severe, since it will stop immediately (or bounce backwards).
When both cars are doing 60mph, they cancel eachother out. One doesn't push the other one backwards, so neither of them manage to proceed forwards past the point of impact. This makes the collision equivalent to hitting an immovable wall at 60mph.
No it won't - a car doing 60mph is doing 60mph - it can't suddenly gain the extra energy and momentum a car doing 120mph would have. Kinetic energy is 0.5 x mass x velocity squared, and does not change just because you hit a moving object. On top of this, hitting another car travelling in the opposite direction is better because you have two crumple zones instead of one, so the deceleration is less. Hitting an immovable wall means the crumple zone in the car has to absorb all the energy of the impact.Two cars hitting eachother at 60mph each would be about the same as one car doing 120mph hitting a stationary car. In that impact, the stationary car will deform and move backwards, taking some of the energy away from the moving car, which itself won't come to a dead stop immediately. It will carry on travelling forward past the point of impact.
Compare this to a car at 120mph hitting an immovable wall, and it should be clear that the damage would be much more severe, since it will stop immediately (or bounce backwards).
When both cars are doing 60mph, they cancel eachother out. One doesn't push the other one backwards, so neither of them manage to proceed forwards past the point of impact. This makes the collision equivalent to hitting an immovable wall at 60mph.
Monty Python said:
kiseca said:
On 1) it's the opposite, assuming identical cars in all cases. If an officer is used to seeing cars hit other cars, then a head-on with both cars travelling at 60mph will naturally look (and be) much worse than a car travelling at 60mph and hitting a stationary car.
Two cars hitting eachother at 60mph each would be about the same as one car doing 120mph hitting a stationary car. In that impact, the stationary car will deform and move backwards, taking some of the energy away from the moving car, which itself won't come to a dead stop immediately. It will carry on travelling forward past the point of impact.
Compare this to a car at 120mph hitting an immovable wall, and it should be clear that the damage would be much more severe, since it will stop immediately (or bounce backwards).
When both cars are doing 60mph, they cancel eachother out. One doesn't push the other one backwards, so neither of them manage to proceed forwards past the point of impact. This makes the collision equivalent to hitting an immovable wall at 60mph.
No it won't - a car doing 60mph is doing 60mph - it can't suddenly gain the extra energy and momentum a car doing 120mph would have. Kinetic energy is 0.5 x mass x velocity squared, and does not change just because you hit a moving object. On top of this, hitting another car travelling in the opposite direction is better because you have two crumple zones instead of one, so the deceleration is less. Hitting an immovable wall means the crumple zone in the car has to absorb all the energy of the impact.Two cars hitting eachother at 60mph each would be about the same as one car doing 120mph hitting a stationary car. In that impact, the stationary car will deform and move backwards, taking some of the energy away from the moving car, which itself won't come to a dead stop immediately. It will carry on travelling forward past the point of impact.
Compare this to a car at 120mph hitting an immovable wall, and it should be clear that the damage would be much more severe, since it will stop immediately (or bounce backwards).
When both cars are doing 60mph, they cancel eachother out. One doesn't push the other one backwards, so neither of them manage to proceed forwards past the point of impact. This makes the collision equivalent to hitting an immovable wall at 60mph.
kiseca said:
Monty Python said:
kiseca said:
On 1) it's the opposite, assuming identical cars in all cases. If an officer is used to seeing cars hit other cars, then a head-on with both cars travelling at 60mph will naturally look (and be) much worse than a car travelling at 60mph and hitting a stationary car.
Two cars hitting eachother at 60mph each would be about the same as one car doing 120mph hitting a stationary car. In that impact, the stationary car will deform and move backwards, taking some of the energy away from the moving car, which itself won't come to a dead stop immediately. It will carry on travelling forward past the point of impact.
Compare this to a car at 120mph hitting an immovable wall, and it should be clear that the damage would be much more severe, since it will stop immediately (or bounce backwards).
When both cars are doing 60mph, they cancel eachother out. One doesn't push the other one backwards, so neither of them manage to proceed forwards past the point of impact. This makes the collision equivalent to hitting an immovable wall at 60mph.
No it won't - a car doing 60mph is doing 60mph - it can't suddenly gain the extra energy and momentum a car doing 120mph would have. Kinetic energy is 0.5 x mass x velocity squared, and does not change just because you hit a moving object. On top of this, hitting another car travelling in the opposite direction is better because you have two crumple zones instead of one, so the deceleration is less. Hitting an immovable wall means the crumple zone in the car has to absorb all the energy of the impact.Two cars hitting eachother at 60mph each would be about the same as one car doing 120mph hitting a stationary car. In that impact, the stationary car will deform and move backwards, taking some of the energy away from the moving car, which itself won't come to a dead stop immediately. It will carry on travelling forward past the point of impact.
Compare this to a car at 120mph hitting an immovable wall, and it should be clear that the damage would be much more severe, since it will stop immediately (or bounce backwards).
When both cars are doing 60mph, they cancel eachother out. One doesn't push the other one backwards, so neither of them manage to proceed forwards past the point of impact. This makes the collision equivalent to hitting an immovable wall at 60mph.
Anyway I've thought about it a bit more, and to explain what I am saying, if two cars colliding with eachother with a closing speed of 120mph, the differences in speeds between each of the two cars is a relatively minor influence on the outcome and that the energy in the crash would be the same whether both were doing 60, or one was doing 30 and the other 90, or one doing 120 and the other doing 0. The speed portion of the kinetic energy equation, IMO, needs to be measured using the relative speed of the car compared to whatever it is crashing into, not between the car and the road, since the road is not involved in the collision.
Happy to discuss the pros and cons of that viewpoint, but that's what I was getting at. Still, my point was that if I a car doing 120mph can crash either into a massive concrete block, or into another stationary identical car, I'm sure it will hurt a lot more if it hits the wall, because the stationary car will get pushed out of the way to some extent. Maybe the difference between an open coffin and a closed one..
kiseca said:
since the road is not involved in the collision.
As an interesting (possibly) aside, these days with modern cars the road actually IS involved, because modern cars channel crash forces downwards through controlled deformation of the front crash members, subframe and suspension and actually expend a significiant proportion of the impact energy into the road on purpose!1.what matters is how much engery needs to be obsorbed. As already said, energy held by a moving object is proportional to the square of its speed. So doubling the speed results in 4 x the energy. Hence one car at 120 has double the energy of two car at 60. So a one car accident at 120 into a immovable object has to obsorb twice the energy that will be shared by 2 cars in a head-on collision where both cars are at 60.
Glosphil said:
1.what matters is how much engery needs to be obsorbed. As already said, energy held by a moving object is proportional to the square of its speed. So doubling the speed results in 4 x the energy. Hence one car at 120 has double the energy of two car at 60. So a one car accident at 120 into a immovable object has to obsorb twice the energy that will be shared by 2 cars in a head-on collision where both cars are at 60.
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