Physics Question

[Chalkie]

Nope, you're missing the fact that when the two cars collide, they both take THE SAME time to decelerate, because each of them takes up half of the total distance that they travel in the impact.

Ohhhhhh, yeah sorry, fucked up there. The car will accelerate in a negative direction as if hit a solid wall, correct. But doesn't that mean it's just like asking which will experience less force... a car hitting a wall at 50km/h or a car hitting a wall at 100km/h.

(btw - impact of 1.0 seconds? You have a 14 metre long crumple zone? Consider how far a car travels in one second at 100km/h). You don't add the two times together.

lets say that the crumple zone of a car lets it stop in 1second to make things easy

Haha, I was lazy, I admit i should of made it ALOT smaller.

 

[PINT of Stella. mate!]

My counter to his arguement is that although the impact speed will remain the same, there will be greater damage between two cars than a single car hitting a brick wall.

Who's right? (BTW, theres beer riding on this)

Send your mate to the bottle-o NeBos!

With the brick wall hypothesis, one car gets f***ed up.

With a head-on, two cars get f***ed up.

Twice as much damage!

 

[tu plang]

Send your mate to the bottle-o NeBos!

With the brick wall hypothesis, one car gets f***ed up.

With a head-on, two cars get f***ed up.

Twice as much damage!

i have a friend who could tell you how much worse it is to drive a car into a wall.... his car.... into the front of his house = lotsa damage = lotsa laughs

 

[Chalkie]

I felt that this thread was missing the POSM factor. BTW i just did some more calculations. Hitting a car head on at 50km/h is the same as hitting a brick wall at 50km/h. Therefore if you are hitting the wall at 100km/h the force will be greater.

 

[Ben-e]

Send your mate to the bottle-o NeBos!

With the brick wall hypothesis, one car gets f***ed up.

With a head-on, two cars get f***ed up.

Twice as much damage!

Hahaha, i like your thinking! at the end of the day, we will probably finish the case together.

 

[tu plang]

I felt that this thread was missing the POSM factor. BTW i just did some more calculations. Hitting a car head on at 50km/h is the same as hitting a brick wall at 50km/h. Therefore if you are hitting the wall at 100km/h the force will be greater.

i think we can all disagree on this one, but lets see the maths all the same.

 

[Chalkie]

i think we can all disagree on this one, but lets see the maths all the same.

Oh this means more typing doesnt it....

 

[S.]

I felt that this thread was missing the POSM factor. BTW i just did some more calculations. Hitting a car head on at 50km/h is the same as hitting a brick wall at 50km/h. Therefore if you are hitting the wall at 100km/h the force will be greater.

Actually yeah, that is correct (thus my previous statements were wrong - sorry Ajay, you were actually right there, I misinterpreted it). While the impact speed is the same, you are effectively doubling the size of the crumple zone using two cars. Ajay's original analogy was correct.

 

[S.]

i think we can all disagree on this one, but lets see the maths all the same.

Nah, he's actually right. If you hit an identical car head on and you're both doing 50km/h, then it will provide exactly the right force and deformation to keep the impact point fixed (as I said before), and you'll effectively be running into a solid wall at 50km/h also. What I said about relative velocity before stands, but it's relative TO THE IMPACT FRONT.

 

[Ben-e]

Oh this means more typing doesnt it....

Yes but you just love numbers, dont you? Come on, just admit it, your a maths-whore!!

 

[Octane_Matty]

its really a question of the stength..brittleness youngs modulus or whatver of the wall

id say it would be the same as a car driving at 100km/h into a car which has a wall behind it

hah

apples and oranges boys

 

[Ben-e]

Nah, he's actually right. If you hit an identical car head on and you're both doing 50km/h, then it will provide exactly the right force and deformation to keep the impact point fixed (as I said before), and you'll effectively be running into a solid wall at 50km/h also. What I said about relative velocity before stands, but it's relative TO THE IMPACT FRONT.

So are you saying that two cars doing 50km/h in a head on collision will provide the same force (ie damage) as a car doing 50km/h running into a brick wall

 

[Chalkie]

Getting very lazy so for simplicities sake (well my sake really) lets do Car vs. Car both going 50m.s^-1, mass of 1000kg, and a car of 1000kg travel at 100m.s^-1.

Car(A) vs. Car(B)

ill show it using momentum (p=momentum sN=units)

PB = m.v
= 1000 . 50
= 50000 sN to the right

PB = m.v
= 1000 . 50
= 50,000 sN to the left

Simple vector addition shows that the total momentum is 50,000 sN + (-50,000 sN). Therfore total mometum = 0 before the collision.

Law of conservation of momentum states that mometum of system before a collision must = the momentum after a collision. Hence the total momentum after collision also =0, so 0=1,000.v, both cars come to a complete stop. Therefore the car decelerates as if it hits a solid wall.

With that in mind consider the formula F = change in momentum / change in time.

Say the collision takes 0.1 second to occur, and the change in mometum is as already stated 50,000-0 = 50,000sN.

Therefore force:
= 50,000N/0.1s
= 500,000N

Car vs. Wall

Pcar = m.v
= 1000 . 100
= 100,000 sN to the right

Pcar = m.v
= m . 0
= 0 sN

Hence, Ptotal and also P after the collision
= 100,000 sN to the right

Now although it would change with increase in velocity, say that impact time again is 0.1s.

F = change in momentum / change in time.

F = 100,000/0.1
= 1,000,000 sN

Force of hitting a wall at double the speed on a head-on collision is doubled in magnitude. The same as I previously worked out, if I simply changed both impact times to be the same (which they should be)

 

[S.]

So are you saying that two cars doing 50km/h in a head on collision will provide the same force (ie damage) as a car doing 50km/h running into a brick wall

Yep. See chalkie's explanation.

 

[Chalkie]

Yes but you just love numbers, dont you? Come on, just admit it, your a maths-whore!!

I prefer: intellectual one who accepts monetary goods to undergo sexual intercourse with maths problems, thank you very much.

So are you saying that two cars doing 50km/h in a head on collision will provide the same force (ie damage) as a car doing 50km/h running into a brick wall

As i showed above, yes that is correct. If the collision is perfect (cars stick together, absorb equal kinetic energy, do not spray to the side, which never happens, but lets say it does) then yes, head on collision is the same as driving into a brick wall at equal speed.

 

[Ben-e]

Well im pretty surprised at the results! Thanks guys for lending us your brains for ego's sake.

 

[t]

So are you saying that two cars doing 50km/h in a head on collision will provide the same force (ie damage) as a car doing 50km/h running into a brick wall

so a simplified formulea would be damage = speed / crumple zone

 

[FuTAnT]

Haha, I can't believe this thread went on so far with no one bothering to read the original link at http://mathforum.org/library/drmath/view/60747.html - which I basically just worked out on a piece of paper albeit not as long winded.

1. If you run object A into object B and they both have the same mass, then the collision is the same for either a head on at speed x or ... object b stationary and object A speed 2x.

2. If the mass of object B is very large (assume a really huge concrete reinforced wall) then the it twice as bad running object A into object B as all the energy must now be dissipated within object A.

Using a simple Kinetic energy analysis will sort this out. Anyway, the link above tells it all pretty well.

 

[johnny]

Irrisistable force + Immovable object = ?

 

[FuTAnT]

Irrisistable force + Immovable object = ?

Big bang!!!