Imagine an immovable plane, and on that plane at a specific point collides the full perpendicular force of a 1984 Chevy Corolla going 70 mph.
Imagine the same point above a road where a 1984 Chevy Corolla going 70 mph collides with another Chevy Corolla going 70 in the opposite direction.
At the point of impact, are the forces the same?
The way I see it, and I have told many wrongs about physics here, it depends on what you mean by the point of impact and force. The first Chevy will be totalled in the same way in both cases, so the force the first Chevy experiences is the same. But in the second case, the point of impact will experience itself the double amount of force, as it will be compressed from both sides. And the second Chevy will be totalled in the same way as the first one, only symmetrically. In the second case the force is enough to crush two Chevys, in the first example only one, as the plane is immovable and therefore, I reckon, indestructible.
Fortunatelly immovable planes do not exist. It is always rocks that cannot be moved.
Which way did you bet?
I think the answer to the bet is that I can’t imagine Chevy building a Toyota Corolla. They badged theirs as a Prizm.
Did you mean to post that the speed of the first Chevy Corolla (it’s a cut-and-shut, so I hope no-one is inside) is 140mph? Because if not, it’s going to take some pretty clever physics for me to entertain the idea that a collision with a closing speed of 70mph has the same force as one with a closing speed of 140mph. Then again, my intuitions about physics have often been wrong, so hopefully an actual physicist will soon roll up in their Toyota Camaro.
What exactly do you mean by “are the forces the same”? Mythbusters actually tested this and yes, while two cars hitting each other means the forces are doubled, it’s now spread between the two cars, so the overall impact is the same:
Yeah, 140 mph force is the idea. So is it?
Just to clarify, is this what you intended?
Imagine an immovable plane, and on that plane at a specific point collides the full perpendicular force of a 1984 Chevy Corolla going 140 mph.
Imagine the same point above a road where a 1984 Chevy Corolla going 70 mph collides with another Chevy Corolla going 70 in the opposite direction.
Think of it like this:
You have two identical cars. In one scenario, Car A and Car B are each going 70 mph and they collide head on. This impact is equal to Car A going 140 mph and crashing into a stationary (0 mph) Car B. At the end of that perfect collision, Car A has instantly decelerated to 70 mph, and Car B is now going 70 mph in the same direction as Car A.
However, a “brick wall” or “immovable plane” or whatever, is going to have infinite mass. The car crashing into it will decelerate to 0 instantly. If you have one car and the brick wall, then the car needs to be going 70 mph to be equivalent to your first scenario of two cars crashing head on at 70 mph each.
So, if Car A is going 140mph and crashes into a brick wall at 140 mph, that is equivalent of crashing into another car going 140mph in the opposite direction.
You only add the velocities like you’re thinking when the masses are the same and they are free to move. If it has infinite mass, its mass at impact will be equal and opposite of the force that impacts it. That means it will be like itself is a car doing the same speed as the car that hit it.
Sure, that’s the other expected answer. Either way, which is correct? Or is there something in the middle?
That’s right. The momentum in each kilogram of car is the same, but the total amount of energy released is doubled. Each kilogram-sized lump of debris is thrown out of the impact with the same amount of energy as the collision of the single car, but there are twice as many of them, and they are thrown in all directions (instead of covering a 180° half circle).
Imagine there’s a giant curtain hanging in the air. A car drives at it, hits it, and stops dead. What was on the other side of the curtain? A wall, or a car driving the opposite direction? If it’s impossible to tell, how can the force be different?
If you put a pressure sensor in the curtain, in one case the pressure would be P, in the other 2P. If the sensor survives the crash and can be read out, or transmits the data before breaking, you can know.
ETA: I am still in the original scenario. One car, 70 MPH, and a wall or two cars, 70 MPH each in opposite directions. This 140 MPH nonsense was not in the original post by the OP and is confusing me.
It’s relevant in that we can always pick a reference frame, and there’s a frame in which one car is traveling 140 mph and the other car is stationary.
The physics works out the same, with the exception of how the cars interact with the road surface–but at these speeds and accelerations, we can pretty much ignore that in the short term.
Kinetic energy scales with the square of speed. If we double the speed of one car (140 vs. 70), and spread it between both, doesn’t that leave another factor of two to consider?
Yes, but in a frame where one car is stationary, the final wreck will still be moving at 70 mph (and have the mass of both cars). That accounts for the remaining energy.
This remaining energy is one explanation for why it’s much better (for you) to hit a stationary car of the same mass than a brick wall. Some of the energy that would have gone into your car instead goes into the wreck sliding along the ground.
You’re claiming a car reacts the exact same way to being hit with force P and 2P?
Can you draw a free body diagram of the sensor and how it differs in the two scenarios?
Assuming symmetry (either the car itself is symmetrical, or the two cars are mirror images of each other), a car hitting an immovable, indestructible wall at 70 MPH will cause exactly the same damage to the car, as if that car hit another car head-on that was also going 70 MPH in the other direction. Though of course, in the two-car case, you end up with two cars wrecked by that amount, not just one.
A car would react the same being hit by a force P as two cars would react being hit by a force 2P equally shared between both cars (in opposite directions and so on).
You are right about the frame of reference, but I am confused by post one (one car, 70 MPH vs two cars, 70 MPH each) and post five, where all of a sudden the lonely car hits 140 MPH. That is no longer the same problem, but both posts are by the OP.
For the damage absorb by the car you need to consider the state of affairs after the collision. In all cases below, the rigid plane has infinite mass, and it does not absorb any of the energy of the collision. The plane does not change velocity after the collision.
(1) Car moving East @ 70mph vs stationary plane.
After the collision, both plane and car are stationary.
The car has absorbed 70mph-worth of kinetic energy.
(2) Car moving East @ 140mph vs stationary plane.
After the collision, both plane and car are stationary.
The car has absorbed 140mph-worth of kinetic energy (four times as much as 1).
(3) Car 1 moving East @ 70mph vs Car 2 moving West @ 70mph.
After the collision, both cars are stationary.
Each car has absorbed 70mph-worth of kinetic energy.
Damage to each car will be the same as the car in scenario (1).
(4) Car moving East @ 70mph vs plane moving West @ 70mph.
After the collision, the car and plane are BOTH MOVING WEST @70mph.
The car has absorbed 140mph-worth of kinetic energy.
Damage to the car will be the same as scenario (2).
The critical part when considering relative velocity is to see the difference between scenarios (3) and (4). In each case, the amount of kinetic energy absorbed by the car is proportional to the change in velocity of the car. (The damage absorbed by the car in (2) and (4) is four times as great rather than double because kinetic energy goes with velocity squared.)
That’s just a car being hit by P in both scenarios. I’m not sure why a sensor would read differently in one vs the other.
Imagine the sensor not being suspended in the air (or the curtain), but mounted on the front of one of the two cars. In the frame of reference of that car the sensor will be hit by the second car with 140 MPH, while when hitting the wall, it will only hit it (the wall) with 70 MPH, the apparent speed of the wall from that point of reference. A fly caught in the unfortunate spot will be double as crushed in the two car scenario compared with the fly that was sitting on the wall. But the cars will be crushed the same, one car in one scenario, two in the other. To crush two cars you need double the force compared with crushing just one car.
I admit that for the fly the difference is negligible. And a sensor would have to be sturdy. But if you put a piece of the right material in there, perhaps some quarz, the way it shatters may give a clue as to whether there was a wall behind the hypothetical curtain or another car coming the other way. But that is over-engineering it: after the crash it will be evident from the debris which scenario was happenig.
And the sound will be different. Louder in the second scenario.