Need a bet settled about classical momentum

[Paige_Turner]

TL;DR

Will a car crash harder into a wall or an oncoming car?

Consider a car slamming into an unyielding wall at 60 mph. Objects in the car will be slammed against the dashboard with a certain amount of force.
Now, instead of slamming into a stationary wall, you slam into another car coming towards you at 60 mph. Relative speed, 120MPH.

QUESTION: Will you be slammed against the dashboard twice as hard in the second case because of the oncoming car's momentum, or will the stuff crash into the dashboard with the same energy because your car decelerates 60-0 MPH in both cases?

I tried to tell this guy that in the 1st case, the static wall absorbs energy, kind of like a rubber bumper, more or less, but in the second case, your momentum is not absorbed, but reflected back at you.

But like my WV mom used to say, "You can't tell nobody nuthin'."

Thus, I pass the question up to the cognoscenti.

 

[PeroK]

You lose!

 

[PeroK]

You lose!

The critical aspect is the stopping distance, which is determined by how much the front of the car can crumple. Hitting a solid wall, this is pretty much all you get (unless your car demolishes the wall). Hitting a second car, the front of your car may become entangled with the front of the other car, so you get crumple plus some added stopping distance. That should lead to a less drastic deceleration.

Although, in practical terms, there may be little difference between the two collisions:

 

[jbriggs444]

Summary:: Will a car crash harder into a wall or an oncoming car?

Consider a car slamming into an unyielding wall at 60 mph. Objects in the car will be slammed against the dashboard with a certain amount of force.
Now, instead of slamming into a stationary wall, you slam into another car coming towards you at 60 mph. Relative speed, 120MPH.

QUESTION: Will you be slammed against the dashboard twice as hard in the second case because of the oncoming car's momentum, or will the stuff crash into the dashboard with the same energy because your car decelerates 60-0 MPH in both cases?

I tried to tell this guy that in the 1st case, the static wall absorbs energy, kind of like a rubber bumper, more or less, but in the second case, your momentum is not absorbed, but reflected back at you.

Which is it? An unyielding wall? A stationary wall? A static wall? Or something "kind of like a rubber bumper"?

An ideal static wall cannot absorb energy precisely because it is immobile. No work can be done on it because no displacement ensues for any finite impact force. [The laws of mechanics are indeterminate with respect to the result of an irresistible force applied to an immovable object, though one can sometimes take limits which suggest a particular result].

The usual idealization (upon which @PeroK has expanded to allow for interpenetration) is that two identical cars colliding will meet each other in a kind of mirror image, each crumpling against but not passing through an imaginary dividing plane. Obviously this is identical to the ideal static wall case.

One can also do an energy accounting to see that the dissipated energy per car is identical in the two cases. Please let us know if you'd like to see that done.

 

[Baluncore]

Conservation of momentum says that; if the speeds are the same, but your car weighs less than the other car, then you will be going backwards after the collision, so you will be subjected to a greater change in velocity than if you hit a fixed wall.

 

[DaveC426913]

Which is it? An unyielding wall? A stationary wall? A static wall? Or something "kind of like a rubber bumper"?

...

The usual idealization (upon which @PeroK has expanded to allow for interpenetration) is that two identical cars colliding will meet each other in a kind of mirror image, each crumpling against but not passing through an imaginary dividing plane. Obviously this is identical to the ideal static wall case.

Yeah. OK, interpreting as a layperson, I see that

1.
A collision with a perfectly rigid wall would be identical to a collision with a second vehicle of identical mass and (opposite) velocity.

But a real world wall will give some arbitrary amount, and in doing so, absorb some arbitrary amount of damage.

So, it hinges, as jb says, on how you define your wall.

2.
It also depends, as PeroK points out, on how you define the collision.

A dead head-on collision will be like hitting a wall. But what if it is an offset collision? Best case (albeit heavily contrived) you each take out the entire passenger side of each car, taking their own length to stop. That would be quite gentle.

 

[Paige_Turner]

Thanx!

My conclusion, modulo your approval: for an ideal, unyielding wall, the forces are equal. But the one-car will experience less crash force as a function of how much energy the wall absorbs by distortion.

 

[DaveC426913]

for an ideal, unyielding wall,

...and an ideal head-on collision...

the forces are equal.