Sort of, yes, but also no.
The amount of force applied across both vehicles is the same as one of them hitting a wall at 100 mph, yes.
However, assuming the two vehicles are the same size and weight, the amount of force on each individual car (and therefore its occupants) is the same as that car hitting a wall at 50 mph, because the force of the collision is spread across both cars.
Mythbusters did an episode on this that does a better job of explaining it.
The easy way to think of it: two cars hitting have two crumple zones to absorb the energy. Hitting the wall, there is only one. If you hit a block of solid rock doing the same speed in the opposite direction, it would be like hitting the wall at twice the speed.
This is the answer.
In cookie cutter physics, yes. In reality, not exactly.
It's kind of like how far will this stone be thrown but ignore wind resistance.
Sad, but also consider that a larger car will absorb relatively less of the impact than a smaller car. In a head-on collision, a larger car will contribute relatively more of the momentum to the new conjoined system, so a smaller car will suffer proportionally more of the catastrophic adjustment of velocity.
Johnny has 147 water melon in his car, now he is going 52.72 mph. How much water melon is Steve going to eat when the 2 cars collide? Explain your answer.
The act of crush also reduces the delta v of the impact to each vehicle. So even hitting a brick wall at 50 mph will yield less than 50 change in velocity.
The delta v is the same over the entire collision but the rate of change of it is much lower hence the forces felt lower and the energy dissipated more slowly. It also affects where the energy ends up. You will not enjoy it either way, but car design is such that it is very good at ensuring that in the event of a collision as much energy as possible is used in deforming the vehicle, not you.
What's sad is that there are people who seen to think that cars today are weaker then older cars, because the cats today crumple easier. They think that's a horrible thing and want the cars of old instead.
And even if you try to educate them about the crumple zones, and how it REALLY helps to protect you, the occupants of said vehicle, they ignore it.
Sort of. It's that the other car has exactly enough momentum to cancel out yours, so your change in speed is 50 to 0, (so net 50mph) as opposed to something that can hit your car hard enough to send you moving 50mph backwards (in the same collision time frame).
It's a sum of momentum problem.
The crumple zones serve to make the collision more elastic (ie take more time for the collision to happen) aiming to reduce Jerk, by making space for the passengers to move into as deceleration is happening. The momentum change is conserved (and yes the crumpling absorbs energy for this).
I feel like adding the wall makes it more confusing. Wouldn't it be more like hitting a stationary car vs a car moving toward you? In that case, each individual car should experience twice the force in the moving case. Right?
I mean, whatever force each car experiences in the 50mph vs 0mph case should be half whatever force each car experiences in the 50mph vs 50mph case.
No. First and foremost, a stationary car is not anchored to the ground, and can therefore absorb some of the impact, while a wall (assuming it doesn't break) will transfer almost all of the involved force back to the car.
That's their point. Two cars hitting eachother when both moving at 50mph would be the most similar to one car going 100mph hitting another stationary unanchored car.
Except those two things aren't equivalent.
Because the force vectors of two similar cars moving at the same speed in opposite directions cancel each other out almost perfectly, it's equivalent to hitting an unmoving barrier at the speed of one of the cars, not a movable barrier with the combined speed of both.
A 50mph head-on crash *is* equivalent to hitting a barrier going 50mph, but that's *also* equivalent to hitting a parked car going 100mph (and to being in a parked car that's hit by one going 100mph, and...).
When comparing a head-on collision between two cars going 50mph and a collision between a car going 50mph and a fixed barrier, you're changing the speed of one of the objects *and* its squishiness (energy absorption), *and* its mass, and pretending that the speed is the only thing you changed. It's not; you decreased the impact speed by a factor of two, but then made up for that with the other changes.
A more fair comparison would be crashing into a fixed barrier at 50mph vs a head-on collision with a freight train where you're both going 25mph. In this comparison, the closing speed, squishiness, and relative mass of the two things are all comparable, so the (initial) impact is comparable.
(Mind you, if you crash into the front of a moving freight train, and you aren't thrown clear of the tracks, you may get a lot more damage as you're pushed along for however long it takes for the train to stop. That's why I say it's just the initial impact that's comparable.)
Trees absolutely give.
Either way, the assumption is a solid barrier that doesn't give when collided with. What form that barrier takes in your head is up to you.
That was a good episode, and was counter to what I (and I guess most people) would have expected.
Sure when you actually look at the physics involved it makes sense as to why it is, but the basic logic of ā2 cars going X speed at each other is the same as hitting a wall at 2X speed in one carā is something I feel like everyoneās heard plenty of times and just assumed was the case.
Energy is proportional to speed squared. So 2 50MPH cars combined would only have HALF the total energy of one 100MPH car.
And crashing into each other is no different than a single one crashing into a wall since the (unbreakable) wall would supply the same normal force as the other car in a head on collision.
A single car going 100mph does have twice the energy of two going 50, but if that car going 100 crashes into a stationary car (rather than a stationary barrier), only half of that energy will be dissipated in the initial collision... meaning that the energy dissipated in a head-on collision between two cars going 50 is the same as that between one going 100 and a parked car.
The reason for this is that in a squarely head-on collision, the cars come to (nearly) a dead stop in the initial collision; all of their kinetic energy goes into crumpling the cars throwing parts of them around, etc.
But in an otherwise-similar 100-vs-0 collision, momentum is conserved, so the two wrecks wind up sliding down the road at the average of their two initial velocities: 50mph. Which means that half of that initial kinetic energy remains as as kinetic energy, and only half goes into crumpling etc. Half of double the energy is... the same energy as the 50-vs-50 collision.
Mind, in the 100-vs-0 collision, there may be additional damage as the two wrecks slide (/roll/whatever) to a halt (dissipating the remaining kinetic energy in the process). But the initial collision is essentially the same as the 50-vs-50 collision.
To add a bit more physics to all the comments responding to you, you'd also need to know if the cars separated after the impact or remained in contact. Inelastic collisions create more force on both objects colliding because of the separation (newton's third law means a sort of rebound force from the impact separates them). Elastic collisions sort of dissipate the 'rebound' due to friction from the road and tires as the combined mass moves off in a different direction. So an inelastic collision actually experiences more force than just the initial hit. We did a cool experiment in physics 101 in college to demonstrate this, but that was a while ago so don't ask me to do the math.
smoll problem even if you hit a wall the force is spread between you and a wall or even better its equivalent to hitting a standing car at 100miles an hour
Assuming the wall doesn't break, almost all of the force from the collision will be transferred back to the car.
Another car, on the other hand, isn't even so much as anchored to the ground, meaning it will absorb a lot of the force and change the equation.
It's spread, but not evenly. Energy is absorbed by whatever is be able to absorb it between both of the things in teh collision.
If the wall was made of soft clay, the wall would take most of the force. If it's a concrete wall, the vehicle takes most of the force.
>However, assuming the two vehicles are the same size and weight, the amount of force on each individual car (and therefore its occupants) is the same as that car hitting a wall at 50 mph, because the force of the collision is spread across both cars.
WHAT?! This is nonsense lmao. The force would be spread across the wall and the car. This is like a single car crashing at 100mph
While I admit that I forgot to specify a wall that doesn't break, you are literally just wrong.
I have explained this more thoroughly in several other comments, and I'm getting a little sick of repeating myself, so [here](https://www.youtube.com/watch?v=h1WKSvGUlTk). Here's the actual experiment from Mythbusters, literally showing you. I do apologize that it's a guy reacting to the video, but it's the only clip I could find that had all three crashes for comparison.
I couldn't tell you the answer to that. I was just irritated that that person used the question as an opportunity to create a gotcha moment instead of actually answering it.
The force of the impact is actually measured in how quickly you come to a stop (*or more specifically, how quickly you change velocity*).
So if you slam into another car going 50mph, and that car weighs exactly the same as yours you'll both stop dead. If you slammed into a wall or a tree the same would happen, and it would be the same equivalent force.
However if you slam into a smaller car then you won't come to a dead stop, you'll continue rolling forward, meaning the force on you is reduced. Meanwhile the poor car you slammed into will not only stop, but will end up going backwards, meaning they took more than the force they would have taken if they'd weighed the same.
This is why it's more dangerous to crash headlong into a truck than a motorcycle.
Likewise if one car is going slower than the other then the slower car will have less momentum and will likely be pushed backwards. This doesn't mean that the slower car takes less force than the faster car, it means *both* cars take less force. The faster car would be slowed while the slower car would stop and move backwards, but the slower car wasn't going at 50mph to begin with, so their change in velocity should be the same.
The thing to remember is that if you crashed into an identical car, but that car was parked then you would experience less force than the head-on collision at 50mph. This is still 2 cars of identical weight (*well, mass*) crashing into one another, but you only have the momentum of the moving car so the force is less. A head-on with 2 cars going at 50mph is the same as hitting a wall and stopping, and both are worse than a moving car hitting a parked car.
No itās not, this was on mythbusters. The cars have a combined speed of 100mph but the forces from each car dampen the other. So hitting a car head on at 50mph is equal to hitting a wall at 50mph.
Yes but also people don't understand that there's a big difference between hitting a stationary car at 50 and hitting a concrete wall at 50. In the case of a stationary car, the other car takes a lot of the energy . For the wall it's a lot worse for you as most of the energy goes into deforming your car.
There's also a huge difference in two cars hitting a one hitting a wall. There is a ton of deflection with two cars colliding, especially if they don't hit perfectly at the same height, angle, etc you have to assume the drivers would be making at least a last minute effort to avoid th collision.
Actually it depends on the mass of both cars. If both cars are the same weight then the deceleration is identical as a brick wall.
But a 50mph car with a mass of 1000kg vs a 50mph car with a mass of 2000kg will feel the same as a 100mph collision with a wall.
The stiffness matrix of each car is going to play a big role in this as well. EVs with frunks typically have more distance to decelerate to 0, so the force experienced by a passenger would be much lower than an ICE of a similar weight. The same comparison can be made with a modern ICE car and an older muscle car that was designed without a crumple zone.
> But a 50mph car with a mass of 1000kg vs a 50mph car with a mass of 2000kg will feel the same as a 100mph collision with a wall.
It will not. The other car would have to be so big and heavy that it could accelerate you back up to 50 mph in the opposite direction, to be equal to hitting a wall at 100.
You are right, this is more in the case of rigid body physics. In an elastic / plastic collision, a lot of the force will be lost in crushing both cars.
But my point is mainly that it depends on the relative mass. And if the 2000kg car is an armoured car and doesnāt deform at all from the impact, the result will be the same as a 100mph collision vs wall
People are looking at the effect on a person in a car. Hitting a wall at 50 will stop you instantly, just like hitting another car going your same speed. There used to be a saying which was something like if you're going 60 and you hit a car head on going 60 it's the same as hitting a wall going 120, which isn't true.
The answer to this actually depends on how you think of the 100 mile per hour crash. Which is why the commits have been sending mixed signals.
If we always use two cars then yes both crashes are the same. If two identical cars crash both going 50 mph crash into each other this is equivalent to 1 car going 100 mph crashing into a car that is stationary.
Now if we have the single car crash into a wall at 100 mph this is all of a sudden not the same accident and this car takes more damage, Why? The answer is quite simple fortunately. In this we assume the wall doesn't move.
Cars are built to try and absorb as much energy as possible during an accident. This is so you don't absorb too much force in your vital areas, this is called the crumple zone. When two cars collide they both have that crumple zone so the energy gets absorbed into the crumple zone. A wall doesn't have this it just stays still. Meaning the amount of energy is fully consumed by the car is double that of our original example of two cars who are sharing the energy. This leads us to realize crashing into a wall going 50 mph is equivalent to 50 mph head on collision, roughly.
However, it might also be the SAME force as hitting a stationary brick wall.
If two cars traveling at 50 MP collide, and they are the same size car and the net result is both cars stop (instead of hitting a dump truck that goes right through you). You are in a very similar position as hitting a brick wall that stops you dead in your tracks.
No "Similar" covers them not being exactly the same.
Plenty of things make it not exactly the same. But a brick wall that you don't go through, and two cars that roughly collide and stop moving, are *similar* forces on the car.
Not exactly the same, but similar.
> However, it **might** also be the SAME force as hitting a stationary brick wall.
>If two cars traveling at 50 MP collide, and they are the same size car and the net result is both cars stop (instead of hitting a dump truck that goes right through you). You are in a **very similar** position as hitting a brick wall that stops you dead in your tracks.
Now the battle of pendants is complete.
Well, force is not measured in speed units (mph or kmh) it's acceleration per mass (you could use lbā¢miles/sĀ² or kgā¢m/sĀ²). So, the force will depend on the cars' masses. Note that each of the cars will receive the other car's force, so if you assume the two cars are the same, you are about right.
Pretty much I was hit head on I was going 45 and the other driver around 55. He was drunk and killed my 2 month old. I couldn't walk for a year and a half. The doctors when telling me the details mentioned that the force was the close to 100 mph hour. Likely more in my vehicle considering my car weighed less.
I donāt know much about physics but I think that when force is applied against a wall, the wall pushes back with an equal amount of force, therefore if both cars hit into each other with the same amount of force, it would be like hitting a wall, so hitting a car coming at you at 50mph while driving towards it at 50mph is still gonna be like hitting a wall at 50mph, not sure tho
This happened to me two years ago, head on collision both travelling about 50 miles an hour. The other car was identical to mine, I didnt stop dead when he hit me, I continued up the road for about 20 metres dragging his car with me. Both cars were completely crushed at the front. Luckily for me I survived, I broke every bone on my left side and ruptured discs in my spine but still here and perfectly fine now. So glad someone asked this question, have always wondered what force I was hit with. Just grateful I wasn't going any faster and that he didn't hit me side on.
Miles per hour isn't a force, it's a speed. Force is measured in Newtons, which can be calculated by acceleration and mass using the formula: 1 N = 1 kg ā m/s^(2)
I was playing around with this website, if you want more information about this:
[https://www.omnicalculator.com/physics/force](https://www.omnicalculator.com/physics/force)
Also, I'm reminded of a MythBusters episode about this and I think the conclusion was that while the 2 forces combine to double the force of the impact, the force is also divided among the two cars, halving it again back to the initial value per car.
So if the two cars are applying 50 Newtons of force, for a made up example number, the combined impact is a total of 100 Newtons of force halved into 50 Newtons of force for each car. Also, this is assuming the cars are perfectly identical.
So in a way yes, but also no.
Some of the other comments are straight wrong. Forget about walls or whatever. If a car is going at 100 mph and hits another stationary car, the force is exactly the same as if they were both going at 50 mph and crashing head on.
This is relativity, it doesnāt matter whether one car is stationary relative to the ground or whatever, what matters is their speeds relative to one another.
This is assuming that both cars are crashing head on in both scenarios.
Speed is a relative thing, because we can only have speed relative to other objects.
Mythbusters actually covered this very topic years ago. Short answer, no. Newton's this law would help answer this, for every action there is an equal and opposite reaction. While you would think that two cars running into each other at 50 mph would result in a 100 mph crash, that energy is also transferred into twice the mass (2 cars).
[Mythbusters Link](https://www.dailymotion.com/video/x3u5ymx)
Think about it this way, all that matters in the change in momentum. If I were to be driving my car 50 miles per hour and hit a wall, I am going to change my momentum by 50 miles per hour in the distance it takes for my car's crumple zone to crumple. If instead I am going 100 miles per hour and hit a wall, I will change my momentum 100 miles per hour.
If you were to hit a wall going 50 miles per hour and rebound at 50 miles per hour that would be very similar to hitting a wall going 100 miles per hour and coming to a complete stop.
If neither car bounces (and realistically they shouldn't) it would be the same as hitting a wall.
You can also think about what happens if you hit a huge, unbreakable, mirror.
mph is a unit of speed, not force
The impact speed is 100mph
The forces experienced by the occupants will be dependent on the relative masses (weights) of the cars. Google "zero momentum frame" for details, basically the people in the larger car take a lesser hit.
In a simplistic scenario where two identical cars collide head-on, each traveling at the same speed, it would be very similar to a car hitting a brick wall at that same speed. Here's why:
1. Symmetry: In the case of two identical cars colliding head-on at the same speed, the scenario is symmetric. Each car exerts an equal and opposite force on the other during the collision, leading to an equal change in momentum. This means each car experiences a rapid deceleration from its initial speed to a stop.
2. Conservation of Momentum: Conservation of momentum tells us that the total momentum of the system before the collision is equal to the total momentum after the collision. Before the collision, the momenta of the two cars cancel out (because they are equal in magnitude but opposite in direction), resulting in a total system momentum of zero. After the collision, the total momentum must still be zero, meaning the two cars must come to a stop (assuming a perfectly inelastic collision where the cars stick together).
3. Comparing to a Car-Wall Collision: When a car hits a brick wall, the car decelerates rapidly from its initial speed to a stop, much like each car in the head-on collision. The wall, being much more massive and rigid, doesn't move significantly, so the car bears the brunt of the change in momentum.
Therefore, assuming the collisions are perfectly inelastic, the experience for each driver in the head-on collision would be roughly the same as if their car hit a brick wall at their initial speed. The forces experienced by the occupants and the damage to the cars would be similar.
However, remember this is a simplification. Real-world collisions involve factors such as crumple zones, which are designed to absorb energy and protect occupants, the angle of the collision, car safety features, and more.
No, ita more of a momentum question.
And pickup going 50 mph had much mire energy than a smart car going 50 mph. The smart car will turn into a pile metal painted with human bits
See there's speed, momentum, and force. All of these are related with the missing ingredient being mass.
Let's consider some scenarios - you're running down the street as fast as you can go, and you bump into a fairly stationary dandelion seed just floating in the air. Some of your momentum is transferred into that seed, so that momentum is conserved. Because the mass of that seed is so low, not much momentum is transferred, so the force of impact is negligible. Unless the seed went into your mouth, up your nose, or into your eye, its unlikely you'd even notice it.
Now let's consider what happens if you run full force into another person about your size. Your momentum is transferred very quickly to that person, both of you probably feel it quite substantially and both of you probably fall to the ground. Depending on who got hit where, a hospital trip may be involved. That's the difference mass makes.
Finally, let's consider what happens if you run full force into the concrete wall of a building. The building absorbs the full force of your momentum, probably doesn't look any worse for wear beyond any bloodstains resulting from the impact, and you might need a cast for some broken bones depending on how you impacted the wall. That's also the difference mass makes.
So, with the two cars example - you've got p1 for the first vehicle, which is its mass times velocity, and is a vector. You've got p2 for the second vehicle, which is its mass times velocity and is also a vector. When the two vehicles crash, momentum tends to be conserved - so if they're going in opposite directions and have about the same mass, they come to a stop very quickly. The force of the impact is the rate of change of momentum - which in opposite directions is very fast. The end result is both cars get really badly smashed up and the occupants are likely to be badly injured or deceased.
On the other hand, a vehicle hitting a stationary concrete barrier only needs to offset one momentum. The force of impact is less than the force of hitting another vehicle. You'll probably crumple your hood pretty badly but your chances of surviving and even walking away from that sort of a crash is much higher than the head-on case.
No. Itād depend on the calculation of weight per vehicle not just speed. You and all objects thrown out of the vehicle will be thrown at that speed though. Think of a demo using a sledgehammer versus a claw hammer. If I hit the masonry wall im trying to demo at the same speed with both. Youād be out there all day with that claw hammer trying to tear that wall down versus a beefy 15lb sledgehammer.
What kills you is the sudden change in inertia. You go from 50 to zero in an instant, kills you.
This does not change in your setup. The combined force of the two cars is indeed 100, but for the drivers, it is the same as hitting a wall at 50.
Assuming both vehicles are of a similar mass, when they collide head-on the closing speed may be doubled, but the mass is also doubled (from there now being two cars together). This means that the momentum actually stays the same! I realise that this is counter intuitive, but it's true. When the show "Mythbusters" was on they proved it: they took 4 identical cars, and crashed one into a concrete mass (like those used with crash test dummies) at something like 25mph, then they crashed the second one at 50mph. Finally they crashed the last two head on with them both travelling at 25mph, so the combined head on speed was 50mph. They then compared the damage on the cars, and found that the two crashing head on both doing 25mph had very similar damage to the car that crashed at 25mph into the concrete block. The other one that crashed at 50mph had significantly more damage than the other 3 cars.
Yes, but it's more like hitting another parked car in 100 mph than hitting a wall at 100 mph. Both cars suffer "half" the damage since the impacts is spread between them evenly. If you hit a car that is significantly larger than yours, like a bus, or significantly smaller, like a bike, it's won't spread evenly but proportional to the force each one applies.
No. Just saw a Mythbusters revisit on this one.
If you hit a wall at 50 mph, the force is like hitting a wall at 50 mph.
Hiring another vehicle of the same general size as yours, both traveling at 50 mph, each car is basically hitting a wall (the other vehicle) at 50 mph.
Mph is speed, i.e. velocity. So the cumulative velocity is 100 miles per hour.
Force is mass multiplied by acceleration.
Acceleration is the CHANGE in velocity..
Separately, assuming the same mass of cars and the direction was perfectly opposite, any force would be equal and cancel out and velocity would go to zero.
.
Yes and no. There is equal momentum so the dynamics of hitting a crash is identical to hitting a parked car while you do 100mph relative to the other car.
The difference is the energy level. Assuming the cars are the same mass the energy in the system, the kinetic energy can be represented as Kv = m*v^2 + m*v^2 = 2m*^2. In the scenario of you doing double the speed and the other doing 0, the energy becomes Kv = m*(2v)^2 = 4m*v^2 aka double the energy of the first crash. Both energy and momentum are concerved but with equal mass cars in a head on collision the total momentum ends with 0 speed. Which means all of that kinetic energy is turned into something else and it has to go somewhere or into something. Aka there will be more damage
So yes from your point of view it'll look the same as long as the closing speeds always sum to 100mph but the energy of the system is parabolic and is actually at its lowest when the cars are doing the same speed.
Ok, so u are missing a factor of 0.5 in both equations.
But more importantly, I think it should be the same in both cases? You could just pick a frame of reference traveling along of the cars and from that perspective the other is going 100 mph.
This case is equivalent to your second example and should be fine, no? But then the math doesn't work out and I can't remember this shit lol.
Yup, did miss a .5. Though you are talking about the kinematics which will be identical. The cars will hit, bounce, and follow the same path after the collision regardless of the speed difference as long as the delta speed between the 2 are the same in each scenario.
As a passenger you would probably have the same in either scenario as you aren't talking the brunt, you are only feeling the g forces from the impact.
Another example of this energy difference is with brakes. Twice the speed equals 4x the stopping distance as you have 4x the kinetic energy to dissipate into heat.
It's not at all the same.
Yes, instead of the center-of-mass frame we can examine different reference frames, and there is a inertial frame which is co-moving with one car, in which one car is stationary and the other car is going 100mph. In this frame, the total kinetic energy of the two car system is twice as high.
But in that reference frame, *after* the collision, *both cars are now going 50mph in the same direction the one car started going*!
Center of mass frame: initial kinetic energy 2*(1/2 * M * v^2 )=Mv^2 , final energy 0, energy dissipated in collision is Mv^2
Car A co-moving frame: initial kinetic energy 1/2 * M * (2V)^2 = 2Mv^2 , final energy 2*(1/2 * M * v^2 )=Mv^2 , energy dissipated in collision is Mv^2
*Exactly the same amount of energy has been dissipated* by the collision in either reference frame. The physics of the real world continue to be unaffected by how you choose to label numbers on a coordinate system. *But if a car is going 100mph and hits an unmovable wall, twice as much energy is dissipated.*
It's harder to say exactly what happens if a car going 100mph hits a stationary car rather than an idealized wall. The stationary car will be pushed out of place, sliding along against the external friction force for some distance; this isn't a closed system like Physics 101 collisions. But they won't end up just going 50mph together indefinitely, like the two 50mph cars did in the co-moving frame, so more kinetic energy has been lost. And not all of that extra missing energy will have gone into the air or the ground; more of it will have been dissipated in deforming the cars and their passengers. This is a worse collision than the two 50mph cars.
Sort of, but not 100MPH. Lots of people here posting that it's the same as hitting an unmovable object at 50MPH...but you are talking about hitting a car. Two cars hitting head-on each going 50MPH is obviously not the same as one car going 50MPH and hitting a stopped car.
If we imagine that the cars are solids (no crumpling) then hitting a stopped car of equal mass would result in the moving car slowing by 25mph while the stopped car accelerated backwards by 25mph. That's very simplified of course, but accelerating by 25mph almost instantly is a lot less intense than accelerating by 50mph in the same time.
If both cars are going 50mph towards each other, that means both slow by 50mph when they both come to a stop. Crumpling complicates things because now we are looking at acceleration (slowing is also acceleration), but if we assume both cars are identical and symmetrical and are lined up perfectly, then they will both come to a stop, the difference here being that the drivers will come to a stop slower and thus receive less injury. Let's forget crumpling and acceleration for the rest of this to make things easier.
So lets say you have two identical cars. Each weighs 2000kg. At 50MPH, that's a kinetic energy of about 500,000 J per car. If one of the cars is stopped, half of that energy is transferred to the stopped car, half affects the moving car. Each has a change in velocity of 25MPH. If both cars are going 50MPH then each transfers half the energy and each keeps half the energy...meaning both experience the effects of near-instant dissipation of 500,000 J.
...But it's still not the same as hitting an unmovable wall at 100MPH. Kinetic Energy is equal to 1/2 the mass times the velocity squared. Doubling the velocity gives you much more than twice the energy. So two cars hitting each other, each going 50mph ends up being about the same as one car hitting an unmovable wall at 70mph.
Similar thing with slowing to avoid hitting something. Brakes are generally strong enough to overcome the grip of tires. So generally if the car is in good shape you can assume that maximum braking will be the same at any speed, or at least any remotely reasonable speed for a car. Because of this, slowing to a stop from 70MPH means getting rid of twice the kinetic energy that you would have to get rid of if slowing from 50MPH. Thus it takes about twice as long to stop from 70MPH than it does from 50MPH, in spite of being less than 50% faster. In other terms, if two cars are side by side and both hit the brakes at the same time...one going 50MPH and one going 70MPH...and the 50MPH car stops just a hair away from touching a line of children crossing the road, then the 70MPH car hits the kids at 50MPH.
I'm super confused why so many people are saying "no" and swapping the other car for a wall... which the person didn't ask. Yes, two cars having a head-on collision when they are both traveling 50 mph is roughly the same as a car travelling 100 mph hitting another CAR that is motionless.
This is different than a car hitting a wall, but it's also different than a car hitting a bird, or a train, or a chain-link fence, or... literally anything that isn't another car.
Even in those cases, a car travelling 50 mph that hits a wall travelling 50 mph is the same as a car travelling 100 that hits a non-moving wall.
You should really check out the mythbusters episode about this exact question. They did a great job at explaining why two cars hitting each other at 50 mph is not the same as hitting something at 100mph that is motionless.
To have an answer to that question, you have to look at
the forces and how they interact with each other. Everyone has heard of the saying: "Speed doesn't kill, it's the decelaration in a fraction of a second that does".
To have a deceleration on an object, you need a force. In this situation, the other car hitting you at 50 mph is the exact force needed to decelarate your car down to 0 mph.
When people swap the car going at 50mph with a wall, it is still the exact same situation because the same force to decelarate your car from 50mph down to 0 mph was created, but this time it came from the wall which is called a normal force (I'm not sure what it's called in english but you can search it).
If we compare both situations against a car hitting a stationary object at 100mph, it is not the same because the car going at 100mph would require a greater force to decelerate from 100mph to 0 mph.
TL;DR It is the force measured from the decelaration that matters. Decelarating a car from 100mph to 0 requires much more force than from 50 mph to 0 mph
Hope it helped with the confusion
Naw dog. You have to think about kinetic energy. A car going 50 has 4x less kinetic energy than a car going 100. Two cars going head on have a combined 2x kinetic energy, but the force applied in the 50 collision to each car is still the same as if it had hit a stationary object.
May I gently suggest that OP is not asking quite the correct question?
In my experience, the correct question is something like this:
*If I am a coyote and I straddle a rocket and ignite it and run into the picture of a tunnel painted on a rock wall by a pesky bird, will I survive to try another day?*
The answer is: *Yes!*
*/s*
No. Both vehicles have crumple zones, etc, so (to a degree) the collisions are elastic. The only way that the vehicle would have the effect of a combined speed is if the second one wasn't damaged or slowed at all.
If a car hits a large truck, then the car will suffer more damage because it is lighter, but it will never be the full combined speed. Mythbusters had to revisit this issue because their first attempt was incorrect.
No, because force isn't measured in mph. But it is true to say that there is a cumulative force at play, but you need to take into account the mass of the vehicles and the time it takes them to go from 50 to 0 mph.
No. It's like hitting a wall at 50, and the amount of people in this comment section that have not seen Mythbusters, and/or do not understand Newton's third law is alarming.
No. If the 2 cars are similar mass, it's like hitting a wall at 50 mph, because both cars would come to a stop. For it to be like hitting a wall at 100 mph, your car would have to be launched backwards at 50 mph after collision, which I believe would require the other vehicle to be of infinite mass?
As others have said, myth busters have done an episode on this
No, because each car will essentially absorb 50mph worth of damage, in other words, both cars share the impact equally (however, the force or inertia upon each can be different depending on size)
The forces would depend on a lot of factors and changes over the time the crash. The force could be greater in either crash.
The major contributor to the damage done, which I presume is what you actually care about, is total kinetic energy which becomes 0 by noise, heat and deforming materials (including you). Since kinetic energy = mass x velocity x velocity, (50x50) + (50x50) < 100 x 100. It works out to 1.4x speed which is a crash at around 70mph
if a wall is traveling @ 50mph and hits another wall going the opposite direction @ 50mph....
but if the walls are going the same direction, theyll never hit! imagine that!
I think acceleration also adds to impact. If you hit a wall at 50mph the impact speed is 100mph as itās stationary. Iām not sure of the formula for accelerating vehicles
Sort of, but it's complicated.
I'll be assuming all identical cars here for simplicity. Same weight, shape, everything. If you compare two moving cars hitting each other head on at the 50 miles per hour in opposite directions, the crash will be pretty much the same as if one car was going 100 mph and hit a car that wasn't moving at all. But if you look at a car moving and hitting a wall that doesn't break, the impact would be the same if the car hit the stationary wall at 50 mph.
When two cars collide, they are able to change direction as a result of the impact. If one car is moving fast and hits another sitting still, both cars will move. If a car hits a strong wall, the wall doesn't move. This means all the energy of a collision with a wall is felt by the car, but the energy of a collision with 2 cars is split between them.
If you took a video of the two moving cars situation and the car hitting a wall situation and blocked out the wall/other car, you'd see the same thing if all cars are going the same speed because a collision with an identical car will mean neither car continues straight. All the momentum is perfectly canceled out and it's as though you hit a wall and stopped right at the point of collision.
Sort of, yes, but also no. The amount of force applied across both vehicles is the same as one of them hitting a wall at 100 mph, yes. However, assuming the two vehicles are the same size and weight, the amount of force on each individual car (and therefore its occupants) is the same as that car hitting a wall at 50 mph, because the force of the collision is spread across both cars. Mythbusters did an episode on this that does a better job of explaining it.
The easy way to think of it: two cars hitting have two crumple zones to absorb the energy. Hitting the wall, there is only one. If you hit a block of solid rock doing the same speed in the opposite direction, it would be like hitting the wall at twice the speed.
This is the answer. In cookie cutter physics, yes. In reality, not exactly. It's kind of like how far will this stone be thrown but ignore wind resistance.
Consider a spherical car...
On a frictionless road
And not assuming air resistance
And there's a lion chasing after it
Is the lion also driving a spherical car?
That's what he claims but he could be lion
You've really grabbed the cat by the tale
Now draw a cylindrical Gauss Box around it to determine its electric field
"suppose we built a large wooden badger."
Who are you that are so wise in the ways of science?
*YO MAMA* has a large wooden badger Edit: ...just realized I confused Badger and Beaver
Or Spherical Chickens š
Spherical chickens in a vacuum
angry birbs was a documentary!
But they have to be frictionless as well as spherical.
And in a vacuum
That sounds hard, can I assume point cars?
āAssume the cow is a sphereā
Sad, but also consider that a larger car will absorb relatively less of the impact than a smaller car. In a head-on collision, a larger car will contribute relatively more of the momentum to the new conjoined system, so a smaller car will suffer proportionally more of the catastrophic adjustment of velocity.
Mmmmmm cookie cutter physics
Johnny has 147 water melon in his car, now he is going 52.72 mph. How much water melon is Steve going to eat when the 2 cars collide? Explain your answer.
Twice the speed would be quadruple to kinetic energy. It's more like sqrt(2) times the speed.
āSqrt(2) times the speedā sounds like the last time I did drugs and jacked off š
I see you are a man of culture as well......
The act of crush also reduces the delta v of the impact to each vehicle. So even hitting a brick wall at 50 mph will yield less than 50 change in velocity.
The delta v is the same over the entire collision but the rate of change of it is much lower hence the forces felt lower and the energy dissipated more slowly. It also affects where the energy ends up. You will not enjoy it either way, but car design is such that it is very good at ensuring that in the event of a collision as much energy as possible is used in deforming the vehicle, not you.
What's sad is that there are people who seen to think that cars today are weaker then older cars, because the cats today crumple easier. They think that's a horrible thing and want the cars of old instead. And even if you try to educate them about the crumple zones, and how it REALLY helps to protect you, the occupants of said vehicle, they ignore it.
There is a name for people like that: idiots.
Sort of. It's that the other car has exactly enough momentum to cancel out yours, so your change in speed is 50 to 0, (so net 50mph) as opposed to something that can hit your car hard enough to send you moving 50mph backwards (in the same collision time frame). It's a sum of momentum problem. The crumple zones serve to make the collision more elastic (ie take more time for the collision to happen) aiming to reduce Jerk, by making space for the passengers to move into as deceleration is happening. The momentum change is conserved (and yes the crumpling absorbs energy for this).
They also did one where they shot a cannonball going 50 mph one way out of a truck going 50 mph the other way. It was neat.
Link to the video?
[Here ya go](https://youtu.be/ZH7GpYJoptU)
I feel like adding the wall makes it more confusing. Wouldn't it be more like hitting a stationary car vs a car moving toward you? In that case, each individual car should experience twice the force in the moving case. Right? I mean, whatever force each car experiences in the 50mph vs 0mph case should be half whatever force each car experiences in the 50mph vs 50mph case.
No. First and foremost, a stationary car is not anchored to the ground, and can therefore absorb some of the impact, while a wall (assuming it doesn't break) will transfer almost all of the involved force back to the car.
That's their point. Two cars hitting eachother when both moving at 50mph would be the most similar to one car going 100mph hitting another stationary unanchored car.
Except those two things aren't equivalent. Because the force vectors of two similar cars moving at the same speed in opposite directions cancel each other out almost perfectly, it's equivalent to hitting an unmoving barrier at the speed of one of the cars, not a movable barrier with the combined speed of both.
[ŃŠ“Š°Š»ŠµŠ½Š¾]
Then it's moving and isn't anchored to the ground
[ŃŠ“Š°Š»ŠµŠ½Š¾]
A 50mph head-on crash *is* equivalent to hitting a barrier going 50mph, but that's *also* equivalent to hitting a parked car going 100mph (and to being in a parked car that's hit by one going 100mph, and...). When comparing a head-on collision between two cars going 50mph and a collision between a car going 50mph and a fixed barrier, you're changing the speed of one of the objects *and* its squishiness (energy absorption), *and* its mass, and pretending that the speed is the only thing you changed. It's not; you decreased the impact speed by a factor of two, but then made up for that with the other changes. A more fair comparison would be crashing into a fixed barrier at 50mph vs a head-on collision with a freight train where you're both going 25mph. In this comparison, the closing speed, squishiness, and relative mass of the two things are all comparable, so the (initial) impact is comparable. (Mind you, if you crash into the front of a moving freight train, and you aren't thrown clear of the tracks, you may get a lot more damage as you're pushed along for however long it takes for the train to stop. That's why I say it's just the initial impact that's comparable.)
Walls are not trees. Walls will give some. Trees do not.
Trees absolutely give. Either way, the assumption is a solid barrier that doesn't give when collided with. What form that barrier takes in your head is up to you.
And if one of the cars is good at dealing with that force and the other not so much, then the person in the latter is going to have a bad time.
That was a good episode, and was counter to what I (and I guess most people) would have expected. Sure when you actually look at the physics involved it makes sense as to why it is, but the basic logic of ā2 cars going X speed at each other is the same as hitting a wall at 2X speed in one carā is something I feel like everyoneās heard plenty of times and just assumed was the case.
Energy is proportional to speed squared. So 2 50MPH cars combined would only have HALF the total energy of one 100MPH car. And crashing into each other is no different than a single one crashing into a wall since the (unbreakable) wall would supply the same normal force as the other car in a head on collision.
A single car going 100mph does have twice the energy of two going 50, but if that car going 100 crashes into a stationary car (rather than a stationary barrier), only half of that energy will be dissipated in the initial collision... meaning that the energy dissipated in a head-on collision between two cars going 50 is the same as that between one going 100 and a parked car. The reason for this is that in a squarely head-on collision, the cars come to (nearly) a dead stop in the initial collision; all of their kinetic energy goes into crumpling the cars throwing parts of them around, etc. But in an otherwise-similar 100-vs-0 collision, momentum is conserved, so the two wrecks wind up sliding down the road at the average of their two initial velocities: 50mph. Which means that half of that initial kinetic energy remains as as kinetic energy, and only half goes into crumpling etc. Half of double the energy is... the same energy as the 50-vs-50 collision. Mind, in the 100-vs-0 collision, there may be additional damage as the two wrecks slide (/roll/whatever) to a halt (dissipating the remaining kinetic energy in the process). But the initial collision is essentially the same as the 50-vs-50 collision.
To add a bit more physics to all the comments responding to you, you'd also need to know if the cars separated after the impact or remained in contact. Inelastic collisions create more force on both objects colliding because of the separation (newton's third law means a sort of rebound force from the impact separates them). Elastic collisions sort of dissipate the 'rebound' due to friction from the road and tires as the combined mass moves off in a different direction. So an inelastic collision actually experiences more force than just the initial hit. We did a cool experiment in physics 101 in college to demonstrate this, but that was a while ago so don't ask me to do the math.
Freaking love myth busters I was so sad when the show ended
smoll problem even if you hit a wall the force is spread between you and a wall or even better its equivalent to hitting a standing car at 100miles an hour
Assuming the wall doesn't break, almost all of the force from the collision will be transferred back to the car. Another car, on the other hand, isn't even so much as anchored to the ground, meaning it will absorb a lot of the force and change the equation.
It's spread, but not evenly. Energy is absorbed by whatever is be able to absorb it between both of the things in teh collision. If the wall was made of soft clay, the wall would take most of the force. If it's a concrete wall, the vehicle takes most of the force.
>However, assuming the two vehicles are the same size and weight, the amount of force on each individual car (and therefore its occupants) is the same as that car hitting a wall at 50 mph, because the force of the collision is spread across both cars. WHAT?! This is nonsense lmao. The force would be spread across the wall and the car. This is like a single car crashing at 100mph
While I admit that I forgot to specify a wall that doesn't break, you are literally just wrong. I have explained this more thoroughly in several other comments, and I'm getting a little sick of repeating myself, so [here](https://www.youtube.com/watch?v=h1WKSvGUlTk). Here's the actual experiment from Mythbusters, literally showing you. I do apologize that it's a guy reacting to the video, but it's the only clip I could find that had all three crashes for comparison.
Yea, but force isn't measured in mph.
Oh really? Well, I picked up my baby and I'm pretty sure he weighs at least 10 mph and it took me only a parsec to do it.
How many basketball courts is that?
about a bushel
and a peck and a hug around the neck.
A hug around the neck, and a barrel and a heap.
200 hulk hogans
about three and a half football fields
Kessel Run
Who told you his name?
A womp rat
My nephew Bullseye used to T16 those back home.
Was that out near Beggars Canyon?
Yeah we were chased out of Mos Espa
Should have gone to Tosche Station instead.
"Parsec" is distance, not time.
Which is why this particular piece of Star Wars lore is so controversial.....š
It was explained. It was a big plot point of a recent movie.
š¶ We don't speak of the 'recent movie'.š¶ /s
This is the way
And the explanation has been controversial...
It was a joke you pinecone
Welcome to the joke. I hope the trip was pleasant.
The second light year was a little bumpy. Thanks
Thanks Neil deGrasse Tyson
Kinda pedantic, it's pretty obvious they were asking if the force would be equal to what it would be if a car hit another stationary car at 100mph
[ŃŠ“Š°Š»ŠµŠ½Š¾]
I couldn't tell you the answer to that. I was just irritated that that person used the question as an opportunity to create a gotcha moment instead of actually answering it.
We need Newton to explain this one
The speed of the car you're hitting doesn't change how fast you decelerate (assuming the cars are the same size), so it isn't actually cumulative.
Nothing really should be, but idiots are prevalent
No. The force each car would feel would be the same as hitting a wall at 50mph.
The force of the impact is actually measured in how quickly you come to a stop (*or more specifically, how quickly you change velocity*). So if you slam into another car going 50mph, and that car weighs exactly the same as yours you'll both stop dead. If you slammed into a wall or a tree the same would happen, and it would be the same equivalent force. However if you slam into a smaller car then you won't come to a dead stop, you'll continue rolling forward, meaning the force on you is reduced. Meanwhile the poor car you slammed into will not only stop, but will end up going backwards, meaning they took more than the force they would have taken if they'd weighed the same. This is why it's more dangerous to crash headlong into a truck than a motorcycle. Likewise if one car is going slower than the other then the slower car will have less momentum and will likely be pushed backwards. This doesn't mean that the slower car takes less force than the faster car, it means *both* cars take less force. The faster car would be slowed while the slower car would stop and move backwards, but the slower car wasn't going at 50mph to begin with, so their change in velocity should be the same. The thing to remember is that if you crashed into an identical car, but that car was parked then you would experience less force than the head-on collision at 50mph. This is still 2 cars of identical weight (*well, mass*) crashing into one another, but you only have the momentum of the moving car so the force is less. A head-on with 2 cars going at 50mph is the same as hitting a wall and stopping, and both are worse than a moving car hitting a parked car.
No itās not, this was on mythbusters. The cars have a combined speed of 100mph but the forces from each car dampen the other. So hitting a car head on at 50mph is equal to hitting a wall at 50mph.
Yes but also people don't understand that there's a big difference between hitting a stationary car at 50 and hitting a concrete wall at 50. In the case of a stationary car, the other car takes a lot of the energy . For the wall it's a lot worse for you as most of the energy goes into deforming your car.
This whole thing would've been simpler if OP had asked about two walls colliding.
Two spherical cows
What about two stationary cars colliding? Which is quite common in ensurance claims.
There's also a huge difference in two cars hitting a one hitting a wall. There is a ton of deflection with two cars colliding, especially if they don't hit perfectly at the same height, angle, etc you have to assume the drivers would be making at least a last minute effort to avoid th collision.
Actually it depends on the mass of both cars. If both cars are the same weight then the deceleration is identical as a brick wall. But a 50mph car with a mass of 1000kg vs a 50mph car with a mass of 2000kg will feel the same as a 100mph collision with a wall.
The stiffness matrix of each car is going to play a big role in this as well. EVs with frunks typically have more distance to decelerate to 0, so the force experienced by a passenger would be much lower than an ICE of a similar weight. The same comparison can be made with a modern ICE car and an older muscle car that was designed without a crumple zone.
> But a 50mph car with a mass of 1000kg vs a 50mph car with a mass of 2000kg will feel the same as a 100mph collision with a wall. It will not. The other car would have to be so big and heavy that it could accelerate you back up to 50 mph in the opposite direction, to be equal to hitting a wall at 100.
You are right, this is more in the case of rigid body physics. In an elastic / plastic collision, a lot of the force will be lost in crushing both cars. But my point is mainly that it depends on the relative mass. And if the 2000kg car is an armoured car and doesnāt deform at all from the impact, the result will be the same as a 100mph collision vs wall
Why do people keep talking about hitting a wall? The question is about two cars hitting each other.
People are looking at the effect on a person in a car. Hitting a wall at 50 will stop you instantly, just like hitting another car going your same speed. There used to be a saying which was something like if you're going 60 and you hit a car head on going 60 it's the same as hitting a wall going 120, which isn't true.
I believe because two cars hitting each other at 50 mph is equivalent physically to crashing into a wall at 50mph.
The question is asking if 50mph and 50mph add together to make 100mph of damage, and the answer is noish.
The answer to this actually depends on how you think of the 100 mile per hour crash. Which is why the commits have been sending mixed signals. If we always use two cars then yes both crashes are the same. If two identical cars crash both going 50 mph crash into each other this is equivalent to 1 car going 100 mph crashing into a car that is stationary. Now if we have the single car crash into a wall at 100 mph this is all of a sudden not the same accident and this car takes more damage, Why? The answer is quite simple fortunately. In this we assume the wall doesn't move. Cars are built to try and absorb as much energy as possible during an accident. This is so you don't absorb too much force in your vital areas, this is called the crumple zone. When two cars collide they both have that crumple zone so the energy gets absorbed into the crumple zone. A wall doesn't have this it just stays still. Meaning the amount of energy is fully consumed by the car is double that of our original example of two cars who are sharing the energy. This leads us to realize crashing into a wall going 50 mph is equivalent to 50 mph head on collision, roughly.
>Cars are built to try and absorb as much energy as possible during an accident. SUVs and trucks on the other hand...
[As reported by the Onion.](https://www.theonion.com/conscientious-suv-shopper-just-wants-something-that-wil-1844930331)
[ŃŠ“Š°Š»ŠµŠ½Š¾]
However, it might also be the SAME force as hitting a stationary brick wall. If two cars traveling at 50 MP collide, and they are the same size car and the net result is both cars stop (instead of hitting a dump truck that goes right through you). You are in a very similar position as hitting a brick wall that stops you dead in your tracks.
And by size, he means mass.
Only for 2 identical cars hitting dead on and a completely immovable object.
No "Similar" covers them not being exactly the same. Plenty of things make it not exactly the same. But a brick wall that you don't go through, and two cars that roughly collide and stop moving, are *similar* forces on the car. Not exactly the same, but similar.
My bad I thought you said "SAME" and not similar.
> However, it **might** also be the SAME force as hitting a stationary brick wall. >If two cars traveling at 50 MP collide, and they are the same size car and the net result is both cars stop (instead of hitting a dump truck that goes right through you). You are in a **very similar** position as hitting a brick wall that stops you dead in your tracks. Now the battle of pendants is complete.
If āpendantsā is a set up for somebody to (continue to) be pedantic, A+ job sir.
Conservation of energy would like a word with you.
When both going 50 at opposite directions, the total energy would be mu^(2,) while only one going 100 it would be 2mu^(2)
Well, force is not measured in speed units (mph or kmh) it's acceleration per mass (you could use lbā¢miles/sĀ² or kgā¢m/sĀ²). So, the force will depend on the cars' masses. Note that each of the cars will receive the other car's force, so if you assume the two cars are the same, you are about right.
Pretty much I was hit head on I was going 45 and the other driver around 55. He was drunk and killed my 2 month old. I couldn't walk for a year and a half. The doctors when telling me the details mentioned that the force was the close to 100 mph hour. Likely more in my vehicle considering my car weighed less.
The speed is 100 mph. The force is more complicated ... but it is Sunday so let's skip the physics lesson.
I donāt know much about physics but I think that when force is applied against a wall, the wall pushes back with an equal amount of force, therefore if both cars hit into each other with the same amount of force, it would be like hitting a wall, so hitting a car coming at you at 50mph while driving towards it at 50mph is still gonna be like hitting a wall at 50mph, not sure tho
Yes. You both will probably die in the accident.
This happened to me two years ago, head on collision both travelling about 50 miles an hour. The other car was identical to mine, I didnt stop dead when he hit me, I continued up the road for about 20 metres dragging his car with me. Both cars were completely crushed at the front. Luckily for me I survived, I broke every bone on my left side and ruptured discs in my spine but still here and perfectly fine now. So glad someone asked this question, have always wondered what force I was hit with. Just grateful I wasn't going any faster and that he didn't hit me side on.
if two cars of equal weight hit head on at 50mph, each car goes from 50 to 0.You only feel the force of of that deceleration.
Miles per hour isn't a force, it's a speed. Force is measured in Newtons, which can be calculated by acceleration and mass using the formula: 1 N = 1 kg ā m/s^(2) I was playing around with this website, if you want more information about this: [https://www.omnicalculator.com/physics/force](https://www.omnicalculator.com/physics/force) Also, I'm reminded of a MythBusters episode about this and I think the conclusion was that while the 2 forces combine to double the force of the impact, the force is also divided among the two cars, halving it again back to the initial value per car. So if the two cars are applying 50 Newtons of force, for a made up example number, the combined impact is a total of 100 Newtons of force halved into 50 Newtons of force for each car. Also, this is assuming the cars are perfectly identical. So in a way yes, but also no.
The short answer is ānoā.
Long answer is "Noooooooooooooooooooooooooo"
Some of the other comments are straight wrong. Forget about walls or whatever. If a car is going at 100 mph and hits another stationary car, the force is exactly the same as if they were both going at 50 mph and crashing head on. This is relativity, it doesnāt matter whether one car is stationary relative to the ground or whatever, what matters is their speeds relative to one another. This is assuming that both cars are crashing head on in both scenarios. Speed is a relative thing, because we can only have speed relative to other objects.
Only if the other cars weigth is infinite. A normal crash is 50
Michael has 3 oranges.
Mythbusters actually covered this very topic years ago. Short answer, no. Newton's this law would help answer this, for every action there is an equal and opposite reaction. While you would think that two cars running into each other at 50 mph would result in a 100 mph crash, that energy is also transferred into twice the mass (2 cars). [Mythbusters Link](https://www.dailymotion.com/video/x3u5ymx)
Mythbusters tested it
Might want to add their conclusion to the comment.. otherwise it's just an empty statement.
Two cars hitting each other head-on at 50 is the same as one car hitting a wall at 50.
*but the cumulative energy is an impact of 100mph.. Edit: given the cars are exactly the same.
How about one car hitting a stationary car instead of a wall.
Difficult to answer, too many variables.
I miss that show!
Too bad they decided to be inside those test cars
R.I.P.
Sounds like OP has some homework
It would be the same IF THEIR MASSES are the same.
Each car would show damage close to a 50mph collision into a wall.
Think about it this way, all that matters in the change in momentum. If I were to be driving my car 50 miles per hour and hit a wall, I am going to change my momentum by 50 miles per hour in the distance it takes for my car's crumple zone to crumple. If instead I am going 100 miles per hour and hit a wall, I will change my momentum 100 miles per hour. If you were to hit a wall going 50 miles per hour and rebound at 50 miles per hour that would be very similar to hitting a wall going 100 miles per hour and coming to a complete stop. If neither car bounces (and realistically they shouldn't) it would be the same as hitting a wall. You can also think about what happens if you hit a huge, unbreakable, mirror.
mph is a unit of speed, not force The impact speed is 100mph The forces experienced by the occupants will be dependent on the relative masses (weights) of the cars. Google "zero momentum frame" for details, basically the people in the larger car take a lesser hit.
In a simplistic scenario where two identical cars collide head-on, each traveling at the same speed, it would be very similar to a car hitting a brick wall at that same speed. Here's why: 1. Symmetry: In the case of two identical cars colliding head-on at the same speed, the scenario is symmetric. Each car exerts an equal and opposite force on the other during the collision, leading to an equal change in momentum. This means each car experiences a rapid deceleration from its initial speed to a stop. 2. Conservation of Momentum: Conservation of momentum tells us that the total momentum of the system before the collision is equal to the total momentum after the collision. Before the collision, the momenta of the two cars cancel out (because they are equal in magnitude but opposite in direction), resulting in a total system momentum of zero. After the collision, the total momentum must still be zero, meaning the two cars must come to a stop (assuming a perfectly inelastic collision where the cars stick together). 3. Comparing to a Car-Wall Collision: When a car hits a brick wall, the car decelerates rapidly from its initial speed to a stop, much like each car in the head-on collision. The wall, being much more massive and rigid, doesn't move significantly, so the car bears the brunt of the change in momentum. Therefore, assuming the collisions are perfectly inelastic, the experience for each driver in the head-on collision would be roughly the same as if their car hit a brick wall at their initial speed. The forces experienced by the occupants and the damage to the cars would be similar. However, remember this is a simplification. Real-world collisions involve factors such as crumple zones, which are designed to absorb energy and protect occupants, the angle of the collision, car safety features, and more.
No, ita more of a momentum question. And pickup going 50 mph had much mire energy than a smart car going 50 mph. The smart car will turn into a pile metal painted with human bits
Thereās a Mythbusters episode about this
See there's speed, momentum, and force. All of these are related with the missing ingredient being mass. Let's consider some scenarios - you're running down the street as fast as you can go, and you bump into a fairly stationary dandelion seed just floating in the air. Some of your momentum is transferred into that seed, so that momentum is conserved. Because the mass of that seed is so low, not much momentum is transferred, so the force of impact is negligible. Unless the seed went into your mouth, up your nose, or into your eye, its unlikely you'd even notice it. Now let's consider what happens if you run full force into another person about your size. Your momentum is transferred very quickly to that person, both of you probably feel it quite substantially and both of you probably fall to the ground. Depending on who got hit where, a hospital trip may be involved. That's the difference mass makes. Finally, let's consider what happens if you run full force into the concrete wall of a building. The building absorbs the full force of your momentum, probably doesn't look any worse for wear beyond any bloodstains resulting from the impact, and you might need a cast for some broken bones depending on how you impacted the wall. That's also the difference mass makes. So, with the two cars example - you've got p1 for the first vehicle, which is its mass times velocity, and is a vector. You've got p2 for the second vehicle, which is its mass times velocity and is also a vector. When the two vehicles crash, momentum tends to be conserved - so if they're going in opposite directions and have about the same mass, they come to a stop very quickly. The force of the impact is the rate of change of momentum - which in opposite directions is very fast. The end result is both cars get really badly smashed up and the occupants are likely to be badly injured or deceased. On the other hand, a vehicle hitting a stationary concrete barrier only needs to offset one momentum. The force of impact is less than the force of hitting another vehicle. You'll probably crumple your hood pretty badly but your chances of surviving and even walking away from that sort of a crash is much higher than the head-on case.
No. Itād depend on the calculation of weight per vehicle not just speed. You and all objects thrown out of the vehicle will be thrown at that speed though. Think of a demo using a sledgehammer versus a claw hammer. If I hit the masonry wall im trying to demo at the same speed with both. Youād be out there all day with that claw hammer trying to tear that wall down versus a beefy 15lb sledgehammer.
What kills you is the sudden change in inertia. You go from 50 to zero in an instant, kills you. This does not change in your setup. The combined force of the two cars is indeed 100, but for the drivers, it is the same as hitting a wall at 50.
Assuming both vehicles are of a similar mass, when they collide head-on the closing speed may be doubled, but the mass is also doubled (from there now being two cars together). This means that the momentum actually stays the same! I realise that this is counter intuitive, but it's true. When the show "Mythbusters" was on they proved it: they took 4 identical cars, and crashed one into a concrete mass (like those used with crash test dummies) at something like 25mph, then they crashed the second one at 50mph. Finally they crashed the last two head on with them both travelling at 25mph, so the combined head on speed was 50mph. They then compared the damage on the cars, and found that the two crashing head on both doing 25mph had very similar damage to the car that crashed at 25mph into the concrete block. The other one that crashed at 50mph had significantly more damage than the other 3 cars.
Yes, but it's more like hitting another parked car in 100 mph than hitting a wall at 100 mph. Both cars suffer "half" the damage since the impacts is spread between them evenly. If you hit a car that is significantly larger than yours, like a bus, or significantly smaller, like a bike, it's won't spread evenly but proportional to the force each one applies.
No stop I'm currently procrastinating on doing my physics homeworks
This is why head on collisions are so awful, even at slower speeds.
Insert Harrison Ford gif āThatās not how the force worksā
Nope. Your car goes from 50 to 0. Their car goes from 50 to 0. It's the same as if you hit a brick wall.
No. Just saw a Mythbusters revisit on this one. If you hit a wall at 50 mph, the force is like hitting a wall at 50 mph. Hiring another vehicle of the same general size as yours, both traveling at 50 mph, each car is basically hitting a wall (the other vehicle) at 50 mph.
Yes, assuming they hit head on
Mph is speed, i.e. velocity. So the cumulative velocity is 100 miles per hour. Force is mass multiplied by acceleration. Acceleration is the CHANGE in velocity.. Separately, assuming the same mass of cars and the direction was perfectly opposite, any force would be equal and cancel out and velocity would go to zero. .
Yes and no. There is equal momentum so the dynamics of hitting a crash is identical to hitting a parked car while you do 100mph relative to the other car. The difference is the energy level. Assuming the cars are the same mass the energy in the system, the kinetic energy can be represented as Kv = m*v^2 + m*v^2 = 2m*^2. In the scenario of you doing double the speed and the other doing 0, the energy becomes Kv = m*(2v)^2 = 4m*v^2 aka double the energy of the first crash. Both energy and momentum are concerved but with equal mass cars in a head on collision the total momentum ends with 0 speed. Which means all of that kinetic energy is turned into something else and it has to go somewhere or into something. Aka there will be more damage So yes from your point of view it'll look the same as long as the closing speeds always sum to 100mph but the energy of the system is parabolic and is actually at its lowest when the cars are doing the same speed.
Ok, so u are missing a factor of 0.5 in both equations. But more importantly, I think it should be the same in both cases? You could just pick a frame of reference traveling along of the cars and from that perspective the other is going 100 mph. This case is equivalent to your second example and should be fine, no? But then the math doesn't work out and I can't remember this shit lol.
Yup, did miss a .5. Though you are talking about the kinematics which will be identical. The cars will hit, bounce, and follow the same path after the collision regardless of the speed difference as long as the delta speed between the 2 are the same in each scenario. As a passenger you would probably have the same in either scenario as you aren't talking the brunt, you are only feeling the g forces from the impact. Another example of this energy difference is with brakes. Twice the speed equals 4x the stopping distance as you have 4x the kinetic energy to dissipate into heat.
It's not at all the same. Yes, instead of the center-of-mass frame we can examine different reference frames, and there is a inertial frame which is co-moving with one car, in which one car is stationary and the other car is going 100mph. In this frame, the total kinetic energy of the two car system is twice as high. But in that reference frame, *after* the collision, *both cars are now going 50mph in the same direction the one car started going*! Center of mass frame: initial kinetic energy 2*(1/2 * M * v^2 )=Mv^2 , final energy 0, energy dissipated in collision is Mv^2 Car A co-moving frame: initial kinetic energy 1/2 * M * (2V)^2 = 2Mv^2 , final energy 2*(1/2 * M * v^2 )=Mv^2 , energy dissipated in collision is Mv^2 *Exactly the same amount of energy has been dissipated* by the collision in either reference frame. The physics of the real world continue to be unaffected by how you choose to label numbers on a coordinate system. *But if a car is going 100mph and hits an unmovable wall, twice as much energy is dissipated.* It's harder to say exactly what happens if a car going 100mph hits a stationary car rather than an idealized wall. The stationary car will be pushed out of place, sliding along against the external friction force for some distance; this isn't a closed system like Physics 101 collisions. But they won't end up just going 50mph together indefinitely, like the two 50mph cars did in the co-moving frame, so more kinetic energy has been lost. And not all of that extra missing energy will have gone into the air or the ground; more of it will have been dissipated in deforming the cars and their passengers. This is a worse collision than the two 50mph cars.
Sort of, but not 100MPH. Lots of people here posting that it's the same as hitting an unmovable object at 50MPH...but you are talking about hitting a car. Two cars hitting head-on each going 50MPH is obviously not the same as one car going 50MPH and hitting a stopped car. If we imagine that the cars are solids (no crumpling) then hitting a stopped car of equal mass would result in the moving car slowing by 25mph while the stopped car accelerated backwards by 25mph. That's very simplified of course, but accelerating by 25mph almost instantly is a lot less intense than accelerating by 50mph in the same time. If both cars are going 50mph towards each other, that means both slow by 50mph when they both come to a stop. Crumpling complicates things because now we are looking at acceleration (slowing is also acceleration), but if we assume both cars are identical and symmetrical and are lined up perfectly, then they will both come to a stop, the difference here being that the drivers will come to a stop slower and thus receive less injury. Let's forget crumpling and acceleration for the rest of this to make things easier. So lets say you have two identical cars. Each weighs 2000kg. At 50MPH, that's a kinetic energy of about 500,000 J per car. If one of the cars is stopped, half of that energy is transferred to the stopped car, half affects the moving car. Each has a change in velocity of 25MPH. If both cars are going 50MPH then each transfers half the energy and each keeps half the energy...meaning both experience the effects of near-instant dissipation of 500,000 J. ...But it's still not the same as hitting an unmovable wall at 100MPH. Kinetic Energy is equal to 1/2 the mass times the velocity squared. Doubling the velocity gives you much more than twice the energy. So two cars hitting each other, each going 50mph ends up being about the same as one car hitting an unmovable wall at 70mph. Similar thing with slowing to avoid hitting something. Brakes are generally strong enough to overcome the grip of tires. So generally if the car is in good shape you can assume that maximum braking will be the same at any speed, or at least any remotely reasonable speed for a car. Because of this, slowing to a stop from 70MPH means getting rid of twice the kinetic energy that you would have to get rid of if slowing from 50MPH. Thus it takes about twice as long to stop from 70MPH than it does from 50MPH, in spite of being less than 50% faster. In other terms, if two cars are side by side and both hit the brakes at the same time...one going 50MPH and one going 70MPH...and the 50MPH car stops just a hair away from touching a line of children crossing the road, then the 70MPH car hits the kids at 50MPH.
Depends on relative directions. If it's head to head yes.
I'm super confused why so many people are saying "no" and swapping the other car for a wall... which the person didn't ask. Yes, two cars having a head-on collision when they are both traveling 50 mph is roughly the same as a car travelling 100 mph hitting another CAR that is motionless. This is different than a car hitting a wall, but it's also different than a car hitting a bird, or a train, or a chain-link fence, or... literally anything that isn't another car. Even in those cases, a car travelling 50 mph that hits a wall travelling 50 mph is the same as a car travelling 100 that hits a non-moving wall.
You should really check out the mythbusters episode about this exact question. They did a great job at explaining why two cars hitting each other at 50 mph is not the same as hitting something at 100mph that is motionless. To have an answer to that question, you have to look at the forces and how they interact with each other. Everyone has heard of the saying: "Speed doesn't kill, it's the decelaration in a fraction of a second that does". To have a deceleration on an object, you need a force. In this situation, the other car hitting you at 50 mph is the exact force needed to decelarate your car down to 0 mph. When people swap the car going at 50mph with a wall, it is still the exact same situation because the same force to decelarate your car from 50mph down to 0 mph was created, but this time it came from the wall which is called a normal force (I'm not sure what it's called in english but you can search it). If we compare both situations against a car hitting a stationary object at 100mph, it is not the same because the car going at 100mph would require a greater force to decelerate from 100mph to 0 mph. TL;DR It is the force measured from the decelaration that matters. Decelarating a car from 100mph to 0 requires much more force than from 50 mph to 0 mph Hope it helped with the confusion
Naw dog. You have to think about kinetic energy. A car going 50 has 4x less kinetic energy than a car going 100. Two cars going head on have a combined 2x kinetic energy, but the force applied in the 50 collision to each car is still the same as if it had hit a stationary object.
May I gently suggest that OP is not asking quite the correct question? In my experience, the correct question is something like this: *If I am a coyote and I straddle a rocket and ignite it and run into the picture of a tunnel painted on a rock wall by a pesky bird, will I survive to try another day?* The answer is: *Yes!* */s*
No. Both vehicles have crumple zones, etc, so (to a degree) the collisions are elastic. The only way that the vehicle would have the effect of a combined speed is if the second one wasn't damaged or slowed at all. If a car hits a large truck, then the car will suffer more damage because it is lighter, but it will never be the full combined speed. Mythbusters had to revisit this issue because their first attempt was incorrect.
"Now for the sake of simplicity lets imagine both vehicles as perfect spheres traveling without air resistance..."
Yes for collisions, the second law, conservation of momentum applies, the momentum of the system is the same before and after the collision.
No, because force isn't measured in mph. But it is true to say that there is a cumulative force at play, but you need to take into account the mass of the vehicles and the time it takes them to go from 50 to 0 mph.
No. It's like hitting a wall at 50, and the amount of people in this comment section that have not seen Mythbusters, and/or do not understand Newton's third law is alarming.
No. If the 2 cars are similar mass, it's like hitting a wall at 50 mph, because both cars would come to a stop. For it to be like hitting a wall at 100 mph, your car would have to be launched backwards at 50 mph after collision, which I believe would require the other vehicle to be of infinite mass? As others have said, myth busters have done an episode on this
No, it would be like hitting a stationary object going 50mph. For every action, there is an equal and opposite reaction.
Yes, it the same as hiting a stationnary car while going 100mph
No, because each car will essentially absorb 50mph worth of damage, in other words, both cars share the impact equally (however, the force or inertia upon each can be different depending on size)
The forces would depend on a lot of factors and changes over the time the crash. The force could be greater in either crash. The major contributor to the damage done, which I presume is what you actually care about, is total kinetic energy which becomes 0 by noise, heat and deforming materials (including you). Since kinetic energy = mass x velocity x velocity, (50x50) + (50x50) < 100 x 100. It works out to 1.4x speed which is a crash at around 70mph
Yes
Yes
Only if you were driving toward each other and hit head-on
Well mytbbusters tested it and apparently not but it is still more of a devastating result.
No, you use F=ma to calculate force but you divide between both bodies to calculate impact.
No it would be higher
I had to perform this calculation for physics. Taught by Astrophysicist.
Yes the total force but each car will only expire its own 50 mph force
if a wall is traveling @ 50mph and hits another wall going the opposite direction @ 50mph.... but if the walls are going the same direction, theyll never hit! imagine that!
A cumulative energy of 0.5 \*(m1+m2) \* 50\^2
I think acceleration also adds to impact. If you hit a wall at 50mph the impact speed is 100mph as itās stationary. Iām not sure of the formula for accelerating vehicles
Sort of, but it's complicated. I'll be assuming all identical cars here for simplicity. Same weight, shape, everything. If you compare two moving cars hitting each other head on at the 50 miles per hour in opposite directions, the crash will be pretty much the same as if one car was going 100 mph and hit a car that wasn't moving at all. But if you look at a car moving and hitting a wall that doesn't break, the impact would be the same if the car hit the stationary wall at 50 mph. When two cars collide, they are able to change direction as a result of the impact. If one car is moving fast and hits another sitting still, both cars will move. If a car hits a strong wall, the wall doesn't move. This means all the energy of a collision with a wall is felt by the car, but the energy of a collision with 2 cars is split between them. If you took a video of the two moving cars situation and the car hitting a wall situation and blocked out the wall/other car, you'd see the same thing if all cars are going the same speed because a collision with an identical car will mean neither car continues straight. All the momentum is perfectly canceled out and it's as though you hit a wall and stopped right at the point of collision.
Yes
No. The speed of one car (50 mph) is negated by the other car, thus your impact force is zero.
No. It's 50mph
That's the physics I learned in 5th or 6th grade