A couple of physics questions
Discussion
2 cars head on.... 1000kg each with velocity 10m/s
0.5*1000*10^2=50,000j total collision energy is 100,000j
After collision ke=0
2 cars one stationary... 1000kg. Car one 0m/s car two 20m/s
0.5*1000*0^2=0 0.5*1000*20^2=200,000j
After the collision there is still kinetic energy. Presuming there is zero friction, there is now half the speed but twice the mass.
0.5*2000*10^2=100,000j. So the moving car has lost 100,000j but the stationary car has gained 100,000j
Not sure what that means but I think the numbers are correct?
0.5*1000*10^2=50,000j total collision energy is 100,000j
After collision ke=0
2 cars one stationary... 1000kg. Car one 0m/s car two 20m/s
0.5*1000*0^2=0 0.5*1000*20^2=200,000j
After the collision there is still kinetic energy. Presuming there is zero friction, there is now half the speed but twice the mass.
0.5*2000*10^2=100,000j. So the moving car has lost 100,000j but the stationary car has gained 100,000j
Not sure what that means but I think the numbers are correct?
brilliant
So, after the collision, in both cases 100kj has been spent ruining the drivers' days, meaning that the two scenarios are equally severe in the first impact, but where one was doing 120 and the other 0, there's still another 100jk shared between the two cars to potentially add to the misery, i.e. possible secondary collision with something else like a wall or a tree?
So, after the collision, in both cases 100kj has been spent ruining the drivers' days, meaning that the two scenarios are equally severe in the first impact, but where one was doing 120 and the other 0, there's still another 100jk shared between the two cars to potentially add to the misery, i.e. possible secondary collision with something else like a wall or a tree?
kiseca said:
Only if the ground starts moving relative to you. My problem is everyone seems to be treating velocity along the ground as absolute zero. I'm saying if you're talking about two moving objects, then the ground isn't the correct reference point. What if it were two aircraft instead? Or two satellites? When two spaceships dock, they're both doing maybe 15,000mph, but because they are doing maybe 1mph relative to eachother, the docking doesn't end in a catastrophic fireball. Why does having wheels on the ground make that different? If a car doing 120 rear ends a car doing 119 in the same direction, you might not even have a mark to show for it.
The ground is a correct reference for road vehicles simply because, ignoring other vehicles, everything you want to avoid crashing into is attached to it and travelling at the same speed as it, so if your ground speed in a car is zero then you won’t be crashing into anything either.Aircraft use the ground as a reference point for navigation, mainly because ultimately if you’re flying a plane then you’re interested in getting to one point on the ground to another. In terms of actually flying then the aircraft’s speed compared to the mass of air around it is the correct reference as that’s what’s required to keep it airborne.
Spacecraft the ground is arguably irrelevant, except for perhaps communications and timing for launch or landing. In terms of two spacecraft intercepting then their speed and direction relative to eachother is the only relevant metric.
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.
I also think you've nailed it there. The TV program is comparing a car that collides at 100mph with an object that isn't designed to crumple to absorb energy with two cars traveling at the same speed difference, but are both designed to absorb energy. I cannot fathom why that would be comparable without calculating in how much energy the wall absorbs, and how much energy one of these cars absorb. 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.
Bodo said:
I also think you've nailed it there. The TV program is comparing a car that collides at 100mph with an object that isn't designed to crumple to absorb energy with two cars traveling at the same speed difference, but are both designed to absorb energy. I cannot fathom why that would be comparable without calculating in how much energy the wall absorbs, and how much energy one of these cars absorb.
In basic physics (which is all I know) if the wall doesn’t move, then it doesn’t absorb any energy.Think of it as two identical cars hitting a solid wall at 60 mph from both sides at once. The damage to both cars will be the same. If we neglect the energy absorbed by the wall then it doesn’t matter how thick the wall is. So the wall can be infinitely thin. Or not there at all.
Edit to emphasise the if. How much energy the wall absorbs compared to the crumple zone gives the difference in what happens to the car. Crumple zones are designed to absorb the energy. Doubling the speed of a car and hit the wall means the car and wall have to absorb four times the energy.
The amount of energy the immovable, incompressible wall absorbs is tiny.
Edit to emphasise the if. How much energy the wall absorbs compared to the crumple zone gives the difference in what happens to the car. Crumple zones are designed to absorb the energy. Doubling the speed of a car and hit the wall means the car and wall have to absorb four times the energy.
The amount of energy the immovable, incompressible wall absorbs is tiny.
Edited by V8LM on Saturday 1st August 09:32
Bodo said:
I also think you've nailed it there. The TV program is comparing a car that collides at 100mph with an object that isn't designed to crumple to absorb energy with two cars traveling at the same speed difference, but are both designed to absorb energy. I cannot fathom why that would be comparable without calculating in how much energy the wall absorbs, and how much energy one of these cars absorb.
Did they also try a car hitting the concrete block at 50 mph?Bodo said:
I also think you've nailed it there. The TV program is comparing a car that collides at 100mph with an object that isn't designed to crumple to absorb energy with two cars traveling at the same speed difference, but are both designed to absorb energy. I cannot fathom why that would be comparable without calculating in how much energy the wall absorbs, and how much energy one of these cars absorb.
Did they also try a car hitting the concrete block at 50 mph?Kawasicki said:
In basic physics (which is all I know) if the wall doesn’t move, then it doesn’t absorb any energy.
Correct. (Virtually all of) The energy lost (0.5 m v^2) in the collision is proportional to the amount the crumple zone compresses by (distance over which the car decelerates) plus the amount the wall compresses/moves. For the concrete block in the 60 mph case this latter distance is negligible. Kawasicki said:
In basic physics (which is all I know) if the wall doesn’t move, then it doesn’t absorb any energy.
Well if it ends up cracked or damaged then it does absorb some, but I suppose in that situation the wall itself hasn't moved but not all of it ends up in quite the same place as it started 
Imagine hitting a giant spring, it will absorb the energy and then throw it back into the car, the stiffer the spring the more force it will require to compress, the higher force means the car will crumple more (thus the higher the peak deceleration of the car will be and the energy absorbed) and subsequently the less energy will be absorbed by the spring. If it's a hypothetical very soft, very long spring, softer than the front bumper of the car and long enough to absorb all the energy of the car without becoming coil bound, then the car will end up bouncing back in the opposite direction at a slightly lower speed to how it started with no damage.
The slightly lower speed due to the energy consumed by hysteresis of the spring converting a small amount of energy into heat
Add in a conveyor belt and the car will end up taking off backwards at twice the speed it was going to hit the wall at
ETA for clarity
Edited by The Wookie on Wednesday 5th August 17:56
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