My grandson asked me to explain the impact dynamics and forces involved in auto collisions. When I attempted to explain using simple vector addition, I screwed it up and got a negative result.
Could one of the SD’s here please provide the correct explanation?
The problem: two cars are travelling in opposite directions; one at 100mph, the other at 150mph.
They collide head on. The common fallacy is that for either car, the force of impact is equivalent to either of them impacting a stationary wall at 250mph. Ie: the velocities and forces are additive.
In fact, for either of them, the dynamics of impact are the equivalent to either of them impacting a stationary wall at their corresponding velocity, regardless of the speed of the other car.
How do I demonstrate this quantitatively using vector addition, and confirm to my grandson that I am all knowing?
There are two types of collisions, elastic and inelastic. Do the cars bounce off of each other after the collision, or do they stick together? Is it perfectly elastic if they bounce off? These different situations will be governed by different equations.
I was a physics major and I would have sworn that a head on collision would absolutely depend on the relative velocity of the two bodies before collision, not just that they were heading into a brick wall with their initial respective velocities. You’re throwing me for a loop calling this a fallacy. The equations that govern collisions definitely involve both velocities of the moving bodies.
The kinetic energies involved of course will be very different compared to a car traveling at 250mph and individual cars traveling at 100mph and 150mph because KE is proportional to velocity squared.
I don’t know the ins and outs of crashes, but clearly you cannot demonstrate this using simple vector addition as that is precisely what suggests that the effect is the same as a 250 mph collision. Indeed if two asteroids A and B collide in space, A travelling at 100 mph and B at 150 mph, the only difference between this scenario and A travelling at 250 mph into a stationary B is a change of frame.
I think the answer is that the ground provides a preferred state of motion for both cars and they will always come approximately to halt shortly after the crash. This is independent of the speed that the object they hit is travelling at.
If the wall is absolutely rigid (a typical physics problem assumption), then running into it is approximately the same as running into an identical car going the same speed as your car. You could imagine inserting an invisible absolutely rigid shield right at the point on impact.
If the wall isn’t rigid, then it will take some of the damage and the damage to the car is less.
The question basically is how much energy is involved and how is it distributed. The energy will depend on the frame of reference, but then so will the kinetic energy of the output of the collision.
If you work in the frame of reference where there is no momentum and have an inelastic collision, then the cars will be stopped after the collision and all the energy will be converted into damage. I see no reason not to assume that identical cars wouldn’t get equal proportions of the damage energy. I don’t think the problem would want you to assume anything else. Of course if one car is an SUV and the other a VW Bug, then the latter is going to take more of the energy.
You have to read a physics homework problem in the right frame of mind. A “rigid wall” means one that’s infinitely massive (so the frame of reference with zero momentum is it standing still and which doesn’t take any of the energy of the collision.
That isn’t true in general. It’s only true if their combined momentum is 0.
Remember, momentum is conserved, so if a car going 100 mph hits another car with equal mass going 150 mph in the opposite direction, they won’t come to a stop. Assuming a perfectly inelastic collision, both cars will end up going 25 mph in the direction the faster car was going before. This differs from either car hitting an immovable wall.
Actually, neither of these is correct for reasons other have pointed out. It is correct to say that it is the same as either car hitting the other stationary car (not a wall) while going 250 mph. That is just shifting one inertial reference frame for another.
Of course, it is only correct for this simplified version where you are not taking into account things like atmosphere and interaction with the ground.
If we postulate two exactly similar cars impacting in a perfectly straight line, then surely the impact speed for each car is the sum of the two speeds. 250mph in the OP’s scenario. If one car was stationary, and the other doing 250mph, the result would be the same (disregarding any outside factors).
This is not observed in practice.
What is observed in practice is that the collision is not so bad as you say.
Because the level of damage is related to energy, the level of damage tells you how much energy… which is that its the same as if each car hit brick walls.
Only (in simplified terms) if the combined momentum of the two cars is in the same direction as you were travelling. If not, the crumple zone of the oncoming car will typically absorb only the energy of the oncoming car.
It is. Kinetic Energy is different in different reference frames, but still conserved (so long as the collision is elastic, which it is not for cars). If you solve the problem using either the one car at the sum of speeds other still or the each car moving methods, you will still get the same answer. This, even though the kinetic energy you are looking at is different. Thus, we see that while kinetic energy has some relationship with damage, it is not direct.
To the extent that “damage” is proportional to any simple physical quantity, it’s mostly proportional to energy. And two cars, each traveling at speed v, hitting head-on will do the same amount of damage to each car as if that car had hit a perfectly rigid, extremely massive wall. If the wall had a crumple zone instead of being perfectly rigid, then the damage from hitting the wall would be less than the head-on collision.
To use a specific example, a car going 150 hitting a car going 100 head-on will be equivalent, damage-wise, to two cars going 125, or a car going 50 hitting one going 200, or a car going 250 hitting a stationary car. It will also be equivalent to a car going 125 hitting an ideal brick wall. The reason that hitting a parked car is more forgiving than hitting a wall is that, in the collision with the wall, everything ends up at rest, and thus all of the kinetic energy ends up going towards wrecking the cars, while in the collision with the other car, after the collision, both are still moving, so there’s some kinetic energy which remained kinetic.
OK. It seems that this problem is a lot more complicated than I thought.
So let’s try reformulating the problem
Assume both cars are identical; the other parameters stay the same. Assume the wall is of infinite mass, non elastic, and immobile.
Think of a tennis ball hitting a racquet: the ball is moving in direction “a”, on contact it decelerates to a complete stop, then changes velocity to that imparted by the racquet. Its initial velocity is irrelevant to the final velocity.
Now, looking at the cars, at the moment of contact, would it not be the case that each individual car would decelerate, then accelerate in the direction in which there was excess energy? So the slower car would be pushed backwards; the faster car would just slow down a bit. The faster car would sustain less damage.
So how does this relate to the cars colliding with a wall? Are the velocities additive? If they are additive, is the effect the same as running into the wall at the equivalent of a collision occurring at 50mph?
Crudely put, is it better to have two cars collide when they are travelling at 150mph and 100mph respectively; or is it better for one of them to run into the wall at 250mph?
I realise that my thinking here is somewhat confused, but I am having a hard time trying to establish an oranges and oranges equivalence scenario in quantifying the velocities, forces and the effects. Any ideas on this?
This isn’t true. The ball may gain energy from the racket, but it also has its own energy that contributes to its final speed.
Here’s a thought experiment: a tennis ball that bounces off a stationary tennis racket will come to a complete stop before rebounding in the opposite direction. The final velocity is a function of the initial velocity and the elasticity of the collision (I’m assuming here that no energy is imparted to the racket, which of course won’t happen in reality). The higher the ball’s initial speed, the higher its final speed.
Now imagine the same situation, except that the racket isn’t stationary: it’s moving toward the ball at one mile per hour (i.e. slowly). The racket will add a little energy to the ball, so the rebound speed will be higher than if the racket were stationary. If the racket speed is ten miles per hour, it will add more energy to the ball, and if it’s a hundred miles an hour it will add even more speed. But this speed is added to the speed the ball would have had if the racket weren’t moving.
Are you saying that if 1 car is stationary and the other is moving, only the moving car takes damage?
I can tell you from personal experience that this is not true.
Right, but we’re talking about the difference in hitting a car vs a wall. If the car is moving, it is contributing kinetic energy, but if the car crumples, that relieves kinetic energy. A stationary car will tend to crumple and roll. A rigid wall will do neither.
Look at Chronos’s descriptions. A car going 150 mph hitting a car going 100 mph head on will be equivalent to hitting a stationary car at 250 mph or a wall at 125 mph. That is essentially Polar Iceman’s original description.
The difference with Polar Iceman’s description is that he’s saying each car sees the energy from only it’s own velocity, whereas Chronos is saying you have to take the total velocity sum and divide it by the mass of each car (i.e. momentum balance).
I always get confused by the terms “elastic” and “inelastic”. I have to go back to materials science definitions where elastic is recoverable and inelastic is plastic, i.e. nonrecoverable.
A bucket of sand is inelastic - the impact will disperse into movement of all the sand grains. A car is inelastic - the car will crumple and change permanent shape. The rigid wall should be elastic, because we are saying the wall does not deform or dent. Right?
No, the ball is elastic - it hits the racquet, momentarily deforms, then restores to original shape. That rebound takes the momentum and energy from the impact and redirects it away from the racquet. The racquet itself is also largely elastic - i.e. rigid. That’s why the wires are under tension, to make them solid, not spongy.
And there you go wrong. First off, we are ideal collision, so we are restricted to one line of motion, so no bouncing off at an angle. Kinetic energy is shared between both cars. It has to be, or there is a defined reference frame. For relative motion to be true, you can’t separate the damage and say “this car gets more damage because it was traveling faster” - faster with regards to what? Change reference frames, and the “faster” car is now stationary. How does it get less damage than the other car?
Also, the final velocity will depend upon the amount of inelasticity of the crash, i.e. how much energy is absorbed by the crumples. If the cars were rigid objects but instantly fused, you’d be able to say they would both end up leaving the collision at 25 mph in the direction the faster car was going. But because the cars aren’t rigid, momentum is absorbed by the crumples, and the final velocity will be reduced.
The energy will be conserved, it is just transferred to damage instead of final motion of the cars as a whole.
Chronos just ran down the equivalent scenarios. The kinetic energy from both cars has to be conserved. The velocities add for the momentum balance.
I think maybe you need to go back and look at momentum balance equations.
Sum the momentum of both cars before the collision and the momentum of both cars after the collision, and the amount of loss due to crumple zones.
mc1b = mass of car 1 before collision
vc1b = velocity of car 1 before collision
mc2b = mass of car 2 before collision
vc2b = velocity of car 2 before collision
mc = mc1b = mc2b = mass of a car (identical cars)
va = velocity after collision
Cr = coefficient of restitution (how much energy is returned; 1 = perfect elastic, 0 = perfect inelastic, i.e. all energy absorbed in the collision)
mc1b * vc1b + mc2b * vc2b = 2 * mc * va * Cr
Car 1 traveling 150 mph hits car 2 traveling 100 mph head on = both cars fused into one crumpled block, velocity determined by how much energy is lost to crumple zones.
And then of course the wheels are damaged and won’t roll, etc, so friction comes into play.
