2 cars crashing head on at 50 mph is like.....
Chuck Jackson
brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
stop, the impact is less than if you were doing 100 mph and come to a complete
stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
want to add to this?
Rick Colombo
So let me get this straight, if you're going 50 and hit another car,
their speed is irrelevant? What if the other car was going 5,000mph.
You don't think that will matter to you because you were only going
50mph and came to a sudden stop?
--
Rick Col...@fnal.gov
Christopher M. Ripp
>
> A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
> brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
> stop, the impact is less than if you were doing 100 mph and come to a complete
> stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
> want to add to this?
Sure, your friend is right.
It's got to do with kinetic energy.
Simply, with two cars going 50 mph hitting head on, each car has
a kinetic energy that gets 'dissipated' when they collide (boom)
so the total kinetic energy is twice that of one car going 50 mph
or the same as one car going 100mph.
Hitting a brick wall at 50mph results in having to dissipate only
half of the above kinetic energy, since the wall has *zero* kinetic
energy.
Of course it has to do with the mass of the cars, the collision angle,
elasticity constants and all.
put simply
2x50=1x100
CTERENZI
colliding objects are of similar mass. For instance, 2 Buick Roadmasters
colliding head on at 50 mph will result in the same force as 1 B.R.M.
hitting a wall at 100mph but A buick Roadmaster and a Civic would not
collide with the same force. I believe the formula is Force=Mass *
Acceleration
Chris Terenzi T'was a woman who drove me to
drink
Amber Waves Brewing Supplies and I never had the courtesy to thank
Smyrna, Ga her..............W.C. Fields
(770)384-1448
Kev
: A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
: brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
: stop, the impact is less than if you were doing 100 mph and come to a complete
: stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
: want to add to this?
There's this little part of physics called momentum. It is defined as an
object's velocity times its mass. It is an inertial quantity. It means if
a twig hits you at 50mph, it doesn't hurt as much as a two by four hitting you
at 50mph. It also means that if you hit a Yugo, it won't feel like a semi.
If two cars of equal mass travelling at the same speed collide head on, both
cars will stop. This is, essentially, like hitting a wall (that doesn't break
or move) at a hundred miles per hour (total change in momentum is the same).
Now, if you are in your Ferrari going fifty and you hit a new Dodge V-10
pickup, his car has more momentum than yours does. This means you not only
stop instantaneously, you begin to move backwards. Ouch. Also, getting in
an accident with a one ton vehicle moving sixty miles per hour would feel the
same as getting into a wreck with a two ton vehicle moving thirty.
Oh, for the wall analogy, if you hit a wall, you hit a wall. If you
hit a moving car, a moving car hits you. If you hit a wall when you're going
a hundred, it's the same as hitting a wall while you're going fifty and the
wall is moving toward you at fifty (relative to a stationary point on Earth).
The moral of this lesson is that if you hit somthing going fifty, whether it's
a wall or a car, it's going to hurt. How much depends on mass and velocity
of BOTH objects. Basically, avoid hitting things in your car.
Kev
--
-----------------------------------------------------------------------------
Kevin Mather
POD Engineering Dept.
Mail: gt4...@prism.gatech.edu
-----------------------------------------------------------------------------
Adour Vahe Kabakian
Christopher M. Ripp <chris_...@amoco.com> wrote:
>Chuck Jackson wrote:
>>
>> A friend of mine says that 2 cars hitting head on at 50 mph is like
>> hitting a > brick wall at 100 mph. I disagree.
>Sure, your friend is right.
>It's got to do with kinetic energy.
>Simply, with two cars going 50 mph hitting head on, each car has
>a kinetic energy that gets 'dissipated' when they collide (boom)
>so the total kinetic energy is twice that of one car going 50 mph
>or the same as one car going 100mph.
Except that kinetic energy is a function of the _square_ of velocity.
But there is still another flaw in your logic. The total energy in
the frontal collision is dissipated by TWO cars instead of ONE for the
wall collision. This means that regardless of how you compute the
kinetic energy, the total energy dissipated in the wall collision
should be half of that for the frontal collision to make things
equivalent.
>put simply
>2x50=1x100
Put even more simply:
2 x 50^2 != 100^2 where != means not equal.
Assuming perfect frontal collision of the two cars and an infinitely
rigid wall, smashing into a wall at 50mph is is equivalent to the
collision of two vehicles each going at 50mph. This can actually be
deduced from kinematics. In both cases, the vehicle is being
decelerated with the bumper coming to a stop at the contact point of
the collision.
The original poster was right in disagreeing with his friend. In the
frontal collision, the other car going at the same speed is serving as
a brick wall substitute :-)
-adour
Matthew Rader
On 18 Apr 1996, Chuck Jackson wrote:
> A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
> brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
> stop, the impact is less than if you were doing 100 mph and come to a complete
> stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
> want to add to this?
>
>
>
Yeah, I'll add to this...
it is worse to hit the brick wall. More specifically about 2 times
worse. The energy dispersed is proportional to the velocity squared.
Matthew
Dick Wells
many can start with the right idea (equal kinetic energy) and then make
the great illogical jump from 50 to 100 miles per hour. The variables are
speed and mass. If you are sitting still in a car and are hit at 50 miles
per hour by a equal car the impact you receive is a 25 mph bw (brick wall)
impact. Of course the two cars would have to hit so squarely that it
would never happen in real life. I notice from time to time that people
who didn't pay attention in High School physics make some inuitive
statements that are way off.
Ed Hahn
Chuck Jackson wrote:
>
> A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
> brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
> stop, the impact is less than if you were doing 100 mph and come to a complete
> stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
> want to add to this?
> Sure, your friend is right.
> It's got to do with kinetic energy.
> Simply, with two cars going 50 mph hitting head on, each car has
> a kinetic energy that gets 'dissipated' when they collide (boom)
> so the total kinetic energy is twice that of one car going 50 mph
> or the same as one car going 100mph.
Hmm. Almost right, except that kinetic energy = 0.5 * Mass *
Velocity^2. (The square of the speed is important.)
Case 1: 2 cars @ 50mph
KE = 2 * (0.5*Mass*50mph^2)
= Mass * 2500 mi^2/hr^2
If we assume that each car dissapates the same amount of energy, one
car will be half this value, or Mass * 1250 mi^2/hr^2. (Forgive the
units - I'm not in a handy place to do some conversions to more
standard kinds of units).
In this case, it may be a reasonable assumption that each car would
dissapate the same amount of energy, as the collision would probably
by fairly inelastic (i.e. the hulks would stick together - not bounce
off each other.) Note that this is actually the most important factor
in the severity of the collision, as it defines which body get the
majority of the energy to dissapate.
Case 2: 1 car @ 100 mph
KE = 0.5 * Mass * 100mph^2
= 0.5 * Mass * 10000 mi^2/hr^2
= Mass * 5000 mi^2/hr^2
If we assume that the single car dissapates all of the energy (which
may not be a very good assumption, by the way - the collision would be
more elastic than Case 1, but not totally elastic.), this would be FOUR
TIMES the amount dissapated in Case 1.
Even if we assume that the wall absorbs half of the energy, the car is
still dissapating TWICE the energy in case one.
In any case, since there is four times the energy to be dissapated in
the second case vs. the first, in all likelihood the result would be
more severe in the second case.
Any volunteers out there willing to find out the difference in
collision elasticity, which would properly allocate the amount of
energy dissapated by each car / wall?
ed
-------- Ed Hahn | eh...@mitre.org | (703) 883-5988 --------
The above comment reflects the opinions of the author, and does not
constitute endorsement or implied warranty by the MITRE Corporation.
Really, I wouldn't kid you about a thing like this.
Nathan J. Nagel
at .. by Chuck Jackson@???
> brick wall at 100 mph. I disagree. If you are doing 50 mph and come to
a sudde
> n
> stop, the impact is less than if you were doing 100 mph and come to a
complete
>
> stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
> want to add to this?
It's the same as crashing into a wall at 50 mph. Both cars will deform
in the head-on crash but if you hit a wall only one car will deform.
The math works out rather nicely, but I'm too lazy to do it right now,
maybe someone else will?
later,
Nate
Allen M. Ashley
>Your friend is correct, though the assumption must be made that the
>colliding objects are of similar mass. For instance, 2 Buick Roadmasters
>colliding head on at 50 mph will result in the same force as 1 B.R.M.
>hitting a wall at 100mph but A buick Roadmaster and a Civic would not
>collide with the same force. I believe the formula is Force=Mass *
>Acceleration
Wrong: it is the same as 1 BRM colliding at 50mph with a brick wall.
Imagine a thin vertical film between the two cars at the moment of
impact. By symmetry neither of the cars would penetrate that film
if the two cars are identical. Then replace that film with a brick
wall.
--
==============================================
Allen Ashley ash...@alumni.caltech.edu
=============================================
Frank Mallory
J>A friend of mine says that 2 cars hitting head on at 50 mph is like
J>hitting a
J>brick wall at 100 mph. I disagree. If you are doing 50 mph and come to
J>a sudden
J>stop, the impact is less than if you were doing 100 mph and come to a
J>complete
J>stop. The fact that the other car is doing 50 mph also is irrelevant.
J>Anyone
J>want to add to this?
It happened here yesteday, on the George Washington Parkway in No. Virginia.
Truly devestating.
Kevin Geraghty
>Chuck Jackson wrote:
>>
>> A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
>> brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
>> stop, the impact is less than if you were doing 100 mph and come to a complete
>> stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
>> want to add to this?
>So let me get this straight, if you're going 50 and hit another car,
>their speed is irrelevant? What if the other car was going 5,000mph.
>You don't think that will matter to you because you were only going
>50mph and came to a sudden stop?
>--
Well, obviously it's complex. If a passenger car going 50 runs head
on into a semi going 50, the car is going to go from moving 50 mph
forward to (almost) 50 mph backward, which strikes me as more or less
equivalent to running into a wall at 100 mph. Or if a car going 50
runs into a car of equal mass going, say 100 mph, the car going 50
will have some backward momentum imparted to it, so it receives more
shock than a car hitting a brick wall at 50; but if two cars of equal
mass and traveling at equal speed (say 50 mph) hit each other head on,
they will both stop dead (no pun intented) which, as the original
poster asserted, is exactly like hitting a brick wall at 50.
So it depends on the speed and mass of the colliding objects.
--
Kevin Geraghty
Kevin Geraghty
>If two cars of equal mass travelling at the same speed collide head on, both
>cars will stop. This is, essentially, like hitting a wall (that doesn't break
>or move) at a hundred miles per hour (total change in momentum is the same).
Well, yes, but it's shared between two cars; so the total change in
momentum per car is like hitting a wall at 50, that is, each car stops
instantly, more or less.
--
Kevin Geraghty
Yaakov Eisenberg
* Chuck Jackson wrote:
* >
* > A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
* > brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
* > stop, the impact is less than if you were doing 100 mph and come to a complete
* > stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
* > want to add to this?
*
* So let me get this straight, if you're going 50 and hit another car,
* their speed is irrelevant? What if the other car was going 5,000mph.
* You don't think that will matter to you because you were only going
* 50mph and came to a sudden stop?
* --
If the other car was going 5,000mph, you would not just come to a sudden
stop. You would come to a stop and then start going backwards very
quickly. That's why that collision would be worse.
The severity of a collision as perceived by person A depends on the
force exerted on A during the collision, which depends on A's rate
of deceleration. What causes that deceleration is irrelevant.
--
-Yaakov Eisenberg (yaa...@cc.gatech.edu)
Steve Barnes
>
>It's got to do with kinetic energy.
>Simply, with two cars going 50 mph hitting head on, each car has
>a kinetic energy that gets 'dissipated' when they collide (boom)
>so the total kinetic energy is twice that of one car going 50 mph
>or the same as one car going 100mph.
>
>half of the above kinetic energy, since the wall has *zero* kinetic
>energy.
>
>Of course it has to do with the mass of the cars, the collision angle,
>elasticity constants and all.
>
>put simply
>2x50=1x100
Completely incorrect. Kinetic energy=0.5*mass*velocity^2. Therefore one car doing
100mph has more kinetic energy than two doing 50mph, and a wall will have less give than
a car. So the car doing 100 will have much more kinetic energy and will be hitting a
much less compliant object, so the damage will be much greater than the two cars at
50 mph, where the kinetic energy is less, and it is shared between the two cars in the event
of a collision.
Steve
Barry Kashar
writes:
What about the inertia, or the force to overcome the body at rest
wanting to stay at rest, also the affect of brake drag even if they are
not on?
What you have is an initial impact that is like hitting a brick wall
at aprox 50 mph (allowing for some parasitic forces) that decreases
exponentially at a rate dependent on the weight and crush
characteristics of the cars.
Bob Anderson
Christopher M. Ripp <chris_...@amoco.com> wrote:
>Chuck Jackson wrote:
>>
>> A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
>> brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
>> stop, the impact is less than if you were doing 100 mph and come to a complete
>> stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
>> want to add to this?
>
>Sure, your friend is right.
>It's got to do with kinetic energy.
>Simply, with two cars going 50 mph hitting head on, each car has
>a kinetic energy that gets 'dissipated' when they collide (boom)
>so the total kinetic energy is twice that of one car going 50 mph
>or the same as one car going 100mph.
>
You sure you don't want to think this over a bit more??
If you're going 50 MPH and hit an immoveable object(brick wall)
You will dissipate all your energy. IF you hit another car, of
the same size and weight, head on, you'll dissipate your energy,
the other car will dissipated his energy. Granted, it's twice
the energy, but it won't be like YOU hitting the wall at 100 MPH.
>Hitting a brick wall at 50mph results in having to dissipate only
>half of the above kinetic energy, since the wall has *zero* kinetic
>energy.
>
>Of course it has to do with the mass of the cars, the collision angle,
>elasticity constants and all.
>
>put simply
>2x50=1x100
--
"I have been told that men are natural warriors and killers and that
women are naturally kind, natural mothers, the protectors of stray
cats and waifs." - Jack O'Connor - Complete Book of Shooting - 1965
George Gogis
> A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
> brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
> stop, the impact is less than if you were doing 100 mph and come to a complete
> stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
> want to add to this?
>
Well, you are mostly wrong . If the vehicle you hit head on weighs the same or
more than your vehicle and then it is like you hit a brick wall at 100 mph or
even worse . If the vehicle you hit weighs less then yours then it is like
hitting a brick wall at less than 100 mph .
The relationship is as follows :
Force = Mass(vehi#1) x Velocity(veh#1) + Mass(veh#2) x Velocity(veh#2)
This holds true if the collision occurs head on only .
George Gogis
Christopher M. Ripp
>
> In article <317672...@amoco.com> "Christopher M. Ripp" <chris_...@amoco.com> writes:
>
> Chuck Jackson wrote:
> >
> > brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
> > stop, the impact is less than if you were doing 100 mph and come to a complete
> > stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
> > want to add to this?
>
> > It's got to do with kinetic energy.
> > Simply, with two cars going 50 mph hitting head on, each car has
> > a kinetic energy that gets 'dissipated' when they collide (boom)
> > so the total kinetic energy is twice that of one car going 50 mph
> > or the same as one car going 100mph.
>
OK OK OK :)
I admit it, in my rush to reply I got the fundamental concepts of
energy and momentum screwed up. I'm an idiot! :)
but anyhow, the idea is still the same, apologies to everyone I
confused and ticked off.
--
---Christopher M. Ripp--------------------------------------------------------
chris_...@amoco.com "The only thing to fear is fea...OMIGOD!
ri...@galstar.com What the h@ll is that!! AAAAAAGHHH!
------------------------------------------------------------------------------The views expressed here do not necessarily represent the views of Amoco corp.
...or my Mom.
(GUEST USER)
>>Sure, your friend is right.
>>It's got to do with kinetic energy.
>>Simply, with two cars going 50 mph hitting head on, each car has
>>a kinetic energy that gets 'dissipated' when they collide (boom)
>>so the total kinetic energy is twice that of one car going 50 mph
>>or the same as one car going 100mph.
>>
>
>
>If you're going 50 MPH and hit an immoveable object(brick wall)
>You will dissipate all your energy. IF you hit another car, of
>the same size and weight, head on, you'll dissipate your energy,
>the other car will dissipated his energy. Granted, it's twice
>the energy, but it won't be like YOU hitting the wall at 100 MPH.
No, I seriously think you are wrong. Think about it this way. Hit your
forhead against the wall. It will probably hurt. Now get a friend and
simultaneously head butt eachother with the same force as when you hit
your head on the wall. It will hurt a lot more.
But think about your own example: the car and the immovable brick wall.
If you're going 50 and hit the wall, you will disipate x amount of energy
in the form of crushed bodies and magled metal. Now what happens if the
wall were on a track moving at you at 50 mph as well as you headed toward
it at 50....are you saying that that wont increase the amount of energy
that needs to be dissipated?? I think not.
------------------------Roy
Dick Wells
any energy like a car will.
Lance Cheney
: Chuck Jackson wrote:
: >
: > A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
: > brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
: > stop, the impact is less than if you were doing 100 mph and come to a complete
: > stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
: > want to add to this?
If you had two cars of the same weight that crashed head on at
50mph there will be no momentum after the collision (they add to 0). In
this case it is like hitting a [very thick] brick wall at 50mph, yes. It
is technically irrelevant what speed the other car is going -- just that
the mass of the car x the velocity has to equal the same as yours. You
could hit a Geo metro that's going 80 and it would be the same as hitting
a heavier car that's going 50.
It most certainly is not the same as hitting a brick wall at
100mph (not even the Kinetic energy is the same -- let alone the momentum).
-Lance
Dan Vandeputte
> > brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
> > stop, the impact is less than if you were doing 100 mph and come to a complete
> > stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
> > want to add to this?
> >
>
> more than your vehicle and then it is like you hit a brick wall at 100 mph or
> hitting a brick wall at less than 100 mph .
> The relationship is as follows :
>
> Force = Mass(vehi#1) x Velocity(veh#1) + Mass(veh#2) x Velocity(veh#2)
>
> This holds true if the collision occurs head on only .
>
another car, the car itself will absorb some energy as the body crumples, but a wall isn't going
to give, so the car will feel the full impact. That's why most cars incorporate crumple zones
today.
Richard Bell
Allen M. Ashley <ash...@alumnae.caltech.edu> wrote:
>cter...@aol.com (CTERENZI) writes:
>
>>Your friend is correct, though the assumption must be made that the
>>colliding objects are of similar mass. For instance, 2 Buick Roadmasters
>>colliding head on at 50 mph will result in the same force as 1 B.R.M.
>>hitting a wall at 100mph but A buick Roadmaster and a Civic would not
>>collide with the same force. I believe the formula is Force=Mass *
>>Acceleration
>
>Wrong: it is the same as 1 BRM colliding at 50mph with a brick wall.
>Imagine a thin vertical film between the two cars at the moment of
>impact. By symmetry neither of the cars would penetrate that film
>if the two cars are identical. Then replace that film with a brick
>wall.
Actually, by newtonian relativity, being involved in a head on
collission where bothe cars have equal mass and are travelling at
50mph is the same as hit by a fully loaded semi while parked (the
semi is doing 50 mph.
Rich Lapham
>==========Kevin Geraghty, 4/18/96==========>>> A friend of mine says that 2 cars hitting head on at 50 mph
>is like hitting a
>>> brick wall at 100 mph. I disagree. If you are doing 50 mph
>and come to a sudden
>>> stop, the impact is less than if you were doing 100 mph and
>come to a complete
>>> stop. The fact that the other car is doing 50 mph also is
>irrelevant. Anyone
>>> want to add to this?
>
>>their speed is irrelevant? What if the other car was going 5,000mph.
>
>
>Well, obviously it's complex. If a passenger car going 50 runs head
>on into a semi going 50, the car is going to go from moving 50 mph
>forward to (almost) 50 mph backward, which strikes me as more or less
>equivalent to running into a wall at 100 mph. Or if a car going 50
>runs into a car of equal mass going, say 100 mph, the car going 50
>will have some backward momentum imparted to it, so it receives more
>shock than a car hitting a brick wall at 50; but if two cars of equal
>mass and traveling at equal speed (say 50 mph) hit each other head on,
>they will both stop dead (no pun intented) which, as the original
>poster asserted, is exactly like hitting a brick wall at 50.
>
>So it depends on the speed and mass of the colliding objects.
>
>--
>Kevin Geraghty
>
>kger...@wolfenet.com
Now how about if the car is going 50 MPH and the Wall is also
going 50 MPH?
Or think about - The first car is going 50, the second is stopped?
Yes mass has a lot to do with the outcome, but two cars hitting
head on is much
worse than one hitting a solid object, buy 3 identical cars and
try it out..
it will probably come close to the wall and 100 MPH.
--
/----------------------\ SCSI
|--------------------| | \ BUS Richard...@AtlantaGA.ATTGIS.COM
| __ CyberCorps _ | | | DRIVER
[|-/OO\------------/O\----|]
Or going back to NCR Again (either works)
Richard...@AtlantaGA.NCR.COM
Ludis Langens
>Case 2: 1 car @ 100 mph
>KE = 0.5 * Mass * 100mph^2
> = 0.5 * Mass * 10000 mi^2/hr^2
> = Mass * 5000 mi^2/hr^2
>
>If we assume that the single car dissapates all of the energy (which
>may not be a very good assumption, by the way - the collision would be
>more elastic than Case 1, but not totally elastic.), this would be FOUR
>TIMES the amount dissapated in Case 1.
>
>Even if we assume that the wall absorbs half of the energy, the car is
>still dissapating TWICE the energy in case one.
If the wall didn't get cracked, chipped, gouged, moved, or otherwise
damaged, then it absorbed none of the energy. (Assuming it also
didn't have any audio frequency oscilations and so on...)
--
unsigned long BinToBCD(unsigned long i) {unsigned long t;
Ludis Langens return i ? (t = BinToBCD(i >> 1), (t << 1) + (i & 1) +
lu...@netcom.com (t + 858993459 >> 2 & 572662306) * 3) : 0;}
Kboetzer
> J>hitting a
SNIP
>It happened here yesteday, on the George Washington Parkway in No. Virginia.
>Truly devestating.
>
Let's do a little mental physics experiment.
1. We have two cars comming at each other at 50 MPH. They are in perfect alignment, going
exactly the same speed.
2. We hang a piece of paper between them at the exact location where they will meet.
3. We can stop time to look at the collision in tiny time fragments.
The the exact time the bumpers touch the forces between the cars is equal so the paper sits
there and is just squeezed between the two bumpers. As the cars continue to decelerate due to
the force generated by the distortion of the metal/plastic/glass/flesh, in an ideal straight
on collision, the paper will just be squeezed more and more between the two collapsing cars.
the paper in this ideal collision would not be bent but just squeezed more and more as the
collapse continues until the cars finally stop and any energy left over due to the final bit
of elastic distortion would propel the cars back a little until they come to rest. This
assumes identical cars etc., you get the picture, Balanced forces.
So now we put the same piece of paper against a brick wall and run into it with the third
identical car. Actually we would probably want to use a fresh piece of paper. ;-)
The paper will have a similar experience of being crushed between the car and the wall.
Again breaking the collision down into very small segments we will find teh same balanced
forces with the wall and the car.
So we are lead to the conclusion that the 50MPH head on is the same as runing into the wall
at 50 MPH.
The math one can do will substantiate this. At least the math that I can do, You may not be
able to do the math.
Believe me or not I don't care. I make a good living at engineering!
Have a good one,
Ken
tbea...@utdallas.edu
Equal and opposite forces. You have to take in the equation the force of
both cars. The brick wall being stationary has no force.
--
_________________________
& I still haven't found
what I'm looking for - U2
Stan
>Chuck Jackson (jacksonc) wrote:
>: A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
>: brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
>
>object's velocity times its mass. It is an inertial quantity. It means if
>a twig hits you at 50mph, it doesn't hurt as much as a two by four hitting you
>at 50mph. It also means that if you hit a Yugo, it won't feel like a semi.
So far, so good. He was paying attention at the beginning of Physics 101.
>If two cars of equal mass travelling at the same speed collide head on, both
>cars will stop. This is, essentially, like hitting a wall (that doesn't break
>or move) at a hundred miles per hour (total change in momentum is the same).
No. Now it becomes apparent that he fell asleep later on in that lecture.
>Kevin Mather
>POD Engineering Dept.
^^^^^^^^^^^^^^^^
In short, the original poster was correct. Two identical cars hitting head
on at 50 is the same as one hitting a wall at *50*.
Note that the other person from Caltech who posted on this subject was the
only other one who got it right. ;-)
--
-------------------------------------------------------------------------------
Stan Schwarz | "I just want to live like Yogi Bear
st...@bombay.gps.caltech.edu | He kicks ass on the average bear."
---------------------------------------------------- -Stukas Over Bedrock -----
Robert Cole
towards a pillow that is being hurled at you. The difference is the
pillow will hurt a lot less because it is a deformable body just like a
the crumple zones of a car. The wall will not crumple but the two cars
will. The cars will decelerate over a larger distance than the one
car-wall collision thus less force on the the people inside. So your
friend is misinformed.
------------------------------------------------
Robert J. Cole
rc...@chat.carleton.ca rc...@engsoc.carleton.ca
http://www.engsoc.carleton.ca/~rcole
Mechanical Engineering, Carleton University
BHM
| In short, the original poster was correct. Two identical cars hitting head
| on at 50 is the same as one hitting a wall at *50*.
|
Okay, but would this mean that hitting a wall at 50 mph is the same as
one car, going 100mph, hitting a stationary car? Or one car, going 75 mph,
hitting another car which is going 25 mph? It seems like the speeds in
the head-on collision would just be relative to any certain "fixed" point
of reference.
If both cars, doing 50 mph each, were to stop dead in a head-on
collision, then a 100 mph car hitting an identical, stationary car would
decelerate to 50 mph and accelerate the stationary car to 50 mph, right?
If that's right, then the total deceleration of the car would be 50 mph in
all three cases: hitting head-on at 50 mph, hitting a wall at 50 mph, or
hitting a stationary car at 100 mph.
Then again, if the stationary car had more mass (big lard-butt driver),
then the car that was originally going 100 mph would decelerate to below
50 mph. That would be more deceleration than hitting a wall at 50 mph
(right?) and would be worse.
Of course, this neglects the differences in the amount of energy
absorbed by the wall or cars, crumple zones, or any of the other ten
million variables that could affect a car accident, but it is pretty dang
interesting...
Speaking of crumple zones, is their purpose to absorb the energy of an
impact, slow the rate of deceleration, both, or what?
bmu...@mednet.swmed.edu
>
> In article <317672...@amoco.com> "Christopher M. Ripp" <chris_...@amoco.com> writes:
>
> >
> > A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
> > brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
> > stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
> > want to add to this?
>
> > It's got to do with kinetic energy.
> > Simply, with two cars going 50 mph hitting head on, each car has
> > a kinetic energy that gets 'dissipated' when they collide (boom)
> > so the total kinetic energy is twice that of one car going 50 mph
> > or the same as one car going 100mph.
>
> Velocity^2. (The square of the speed is important.)
>
> Case 1: 2 cars @ 50mph
> KE = 2 * (0.5*Mass*50mph^2)
> = Mass * 2500 mi^2/hr^2
>
> If we assume that each car dissapates the same amount of energy, one
> car will be half this value, or Mass * 1250 mi^2/hr^2. (Forgive the
> units - I'm not in a handy place to do some conversions to more
> standard kinds of units).
>
> In this case, it may be a reasonable assumption that each car would
> dissapate the same amount of energy, as the collision would probably
> by fairly inelastic (i.e. the hulks would stick together - not bounce
> off each other.) Note that this is actually the most important factor
> in the severity of the collision, as it defines which body get the
> majority of the energy to dissapate.
>
> KE = 0.5 * Mass * 100mph^2
> = 0.5 * Mass * 10000 mi^2/hr^2
> = Mass * 5000 mi^2/hr^2
>
> If we assume that the single car dissapates all of the energy (which
> may not be a very good assumption, by the way - the collision would be
> more elastic than Case 1, but not totally elastic.), this would be FOUR
> TIMES the amount dissapated in Case 1.
>
> Even if we assume that the wall absorbs half of the energy, the car is
> still dissapating TWICE the energy in case one.
>
> the second case vs. the first, in all likelihood the result would be
> more severe in the second case.
>
> Any volunteers out there willing to find out the difference in
> collision elasticity, which would properly allocate the amount of
> energy dissapated by each car / wall?
>
> ed
>
> -------- Ed Hahn | eh...@mitre.org | (703) 883-5988 --------
> The above comment reflects the opinions of the author, and does not
> constitute endorsement or implied warranty by the MITRE Corporation.
> Really, I wouldn't kid you about a thing like this.
Thank you for saving me the trouble of writing this same thing.
has anyone here taken basic high school physics? geez.
--
Bruce Musgrove
bmu...@mednet.swmed.edu
"Always reach for new heights. Use the drapes, that is what they are
there for."
from the musings of Master Meow
bmu...@mednet.swmed.edu
> > brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
> > stop, the impact is less than if you were doing 100 mph and come to a complete
> > stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
> > want to add to this?
> >
>
> more than your vehicle and then it is like you hit a brick wall at 100 mph or
> even worse . If the vehicle you hit weighs less then yours then it is like
> hitting a brick wall at less than 100 mph .
> The relationship is as follows :
>
> Force = Mass(vehi#1) x Velocity(veh#1) + Mass(veh#2) x Velocity(veh#2)
>
> This holds true if the collision occurs head on only .
>
> George Gogis
Assuming a totally inelastic collision and a 10 pound car,
Force = Mass(vehi#1) x Velocity(veh#1) + Mass(veh#2) x Velocity(veh#2)
Force = (10lbs x 50 mph) + (10 lbs x50 mph) = 1,000 lbsMPH
So if you now hit the brick wall instead (assume a brick wall weight of
100 lbs. Actual weight is irrelevant)
Force = (100lbs x 0 mph) + (10 lbs x 50 mph) = 500 lbs mph
the speed of the car would have to double to get the same force generated
in the 2 car crash.
bmu...@mednet.swmed.edu
at around 70 MPH. Of course that is assuming an elastic collision.
Paul Patricio
>
>Finally...... someone came up with the right answer.. I dont see how so
>many can start with the right idea (equal kinetic energy) and then make
>the great illogical jump from 50 to 100 miles per hour. The variables are
>speed and mass. If you are sitting still in a car and are hit at 50 miles
>per hour by a equal car the impact you receive is a 25 mph bw (brick wall)
>impact. Of course the two cars would have to hit so squarely that it
>would never happen in real life. I notice from time to time that people
>who didn't pay attention in High School physics make some inuitive
>statements that are way off.
But there are TWO cars travelling at 50 mph. Twice as much energy to
dissipate between two cars. Two vehicles with the same mass hitting each
other at 50mph would absorb the same amount of energy as hitting a wall at
50mph. (good luck on YOUR next physics test)
Paul Patricio
David Gersic
>Your friend is correct, though the assumption must be made that the
>colliding objects are of similar mass. For instance, 2 Buick Roadmasters
>colliding head on at 50 mph will result in the same force as 1 B.R.M.
>hitting a wall at 100mph but A buick Roadmaster and a Civic would not
>collide with the same force.
Or, the way it was simply shown to me back in Driver's Ed class was a
"crash test" film. Take two cars, run them head - on into each other at 60
mph. First they did two big old nasty 70's land yachts. Both cars stopped
pretty much where they collided, with roughly the same ammount of damage to
each. Then they did two sub-compact import cars, and again there was about
the same damage to each, and both just stopped where they hit with only a
little bit of bounce. Then they ran the land yacht into the sub-compact.
The land yacht was almost undamaged, and the dummies in the front seat
barely moved from their positions, as the car kept moving forward. The
sub-compact, OTOH, went airborn in the other direction, the dummies went
through the windshield, and the car *landed* about thirty yards from the
point of impact.
It's not the car's impact that kills you, it's how fast it decelerates when
it hits (or, more specifically, how fast *you* decelerate).
======================================================================
The packet goes out the card, into the copper, out the router,
onto the fiber, across the world, thru the copper............
NOTHING BUT NET.
David Gersic dge...@niu.edu
Systems Programmer Northern Illinois University
Gene Morrow
>Chuck Jackson wrote:
>>
>> A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
>> brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
>> stop, the impact is less than if you were doing 100 mph and come to a complete
>> stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
>> want to add to this?
>So let me get this straight, if you're going 50 and hit another car,
>their speed is irrelevant? What if the other car was going 5,000mph.
>You don't think that will matter to you because you were only going
>50mph and came to a sudden stop?
>--
>Rick Col...@fnal.gov
Einstein's Special Theory of Relativity says speeds add, up to the speed of
light. The variable is a brick wall vs a collapsible car.
tle...@ior.com
>On 18 Apr 1996, Chuck Jackson wrote:
>
>> A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
>> brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
>> stop, the impact is less than if you were doing 100 mph and come to a complete
>> stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
>> want to add to this?
>>
is not a good one. When hitting a brick wall, all of the kinetic energy will be
expended on your car. When hitting another (we'll assume identical) car,
only half would be applied to your car, and half to the other.
Since kinetic energy is proportional to the square of the velocity, a car going
100mph has 4x the kinetic energy that a 50mph car does. Now, replace the
brick wall with a parked car for a more accurate comparison and you'll find
if you work out the math that two cars going head on at 50mph releases the
same kinetic energy as one car going 70.71mph hitting a parked car.
Hitting a brick wall at 50mph would be identical to hitting another car also
going 50mph. When hitting the wall, all the energy is applied to your car. In
the two car scenario, the amount of energy would be doubled, but there's
two cars to absorb it, so each car gets the same amount of "work" done on
it. On the other hand, one car going 50mph hitting a parked car is another
matter entirely and is equivalent to two cars hitting head on at 35.35mph.
All assumptions above assume identical mass and construction of vehicles
involved and does not take into account the difference between static and
rolling friction which may slightly alter the result of hitting a parked car,
and also does not take relativistic effects into account which makes the
apparent speed differential of two objects infinitesimally less that you'd
expect. So there.
Tom.
As a matter of fact, Yes I do have college degrees in math and physics.
Adour Vahe Kabakian
George Gogis <dp...@ceco.ceco.com> wrote:
>
>Force = Mass(vehi#1) x Velocity(veh#1) + Mass(veh#2) x Velocity(veh#2)
>
I thought that even those who haven't taken physics for poets would at
least know that
Force = Mass x Acceleration
-adour
P. J. Remner
In a previous article, rc...@chat.carleton.ca (Robert Cole) says:
>Take your head and bash it against the wall. Now take your head and run
>towards a pillow that is being hurled at you. The difference is the
>pillow will hurt a lot less because it is a deformable body just like a
>the crumple zones of a car. The wall will not crumple but the two cars
>will. The cars will decelerate over a larger distance than the one
>car-wall collision thus less force on the the people inside. So your
>friend is misinformed.
>
So, the correct analogy is that two identical cars crashing head-on
at 50mph is like one car traveling 100mph and hitting an identical
parked car.
But it would still hurt like a SOB.
- Pete (That's why cab-forward makes no sense)
--
'72 Thunderbird, 429 4bbl, lots o' looks, the "larger hammer"
"Beat them gently until you get bored, then use progressively larger hammers."
- Andy Dingley, din...@codesmth.demon.co.uk
Jerry Whittle
>
> A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
> brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
> stop, the impact is less than if you were doing 100 mph and come to a complete
> stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
> want to add to this?
Your friend is more correct than you as the relative impact speed is 100 mph. However, I
tend to somewhat disagree with your friend as the other car will absorb some impact while
crushing whereas a brick wall (unless very flimsy) will not. Maybe it would be more like
hitting the wall at 80 mph. The net result would be the same in either situation for an
occupant - body bag.
Jerry Whittle
Belleville, Illinois, USA
whit...@apci.net
My minivan is a Voyager and bicycle is a Trek.
April 21, 1996
Jerry Whittle
Yaakov Eisenberg
* > A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
* > brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
* > stop, the impact is less than if you were doing 100 mph and come to a complete
* > stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
* > want to add to this?
* >
*
*
* Well, you are mostly wrong . If the vehicle you hit head on weighs the same or
* more than your vehicle and then it is like you hit a brick wall at 100 mph or
* even worse . If the vehicle you hit weighs less then yours then it is like
* hitting a brick wall at less than 100 mph .
You are correct in that the relative mass of the vehicles needs to be
taken into account. However, assuming the two are of equal mass,
a 50 mph head-on is like 50 mph into a brick wall, not 100 mph into
a brick wall.
* The relationship is as follows :
*
* Force = Mass(vehi#1) x Velocity(veh#1) + Mass(veh#2) x Velocity(veh#2)
This formula doesn't look right. What if Mass(veh#2) = 0, i.e. veh#1
hasn't hit anything? Certainly, the force in this case should be
zero, but that's not what your formula gives. Or, suppose veh#2 is
not moving, so Velocity(veh#2) = 0. Then, your formula says that
the mass of veh#2 is irrelevant, which is not correct.
*
* This holds true if the collision occurs head on only .
*
*
* George Gogis
*
*
--
-Yaakov Eisenberg (yaa...@cc.gatech.edu)
Yaakov Eisenberg
* >
* >On 18 Apr 1996, Chuck Jackson wrote:
* >
* >> A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
* >> brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
* >> stop, the impact is less than if you were doing 100 mph and come to a complete
* >> stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
* >> want to add to this?
* >>
* is not a good one. When hitting a brick wall, all of the kinetic energy will be
* expended on your car. When hitting another (we'll assume identical) car,
* only half would be applied to your car, and half to the other.
*
* Since kinetic energy is proportional to the square of the velocity, a car going
* 100mph has 4x the kinetic energy that a 50mph car does. Now, replace the
* brick wall with a parked car for a more accurate comparison and you'll find
* if you work out the math that two cars going head on at 50mph releases the
* same kinetic energy as one car going 70.71mph hitting a parked car.
This is incorrect. In one case, the relative speed of the cars is 100 mph;
in the other, it is 70.71 mph. How can the two collisions be equally
severe?
I don't believe that analyzing the kinetic energy dissipated is an
especially good way to understand this problem. However, if done
correctly, it certainly should not yield wrong answers. The point
that you're missing is this: After the 50 mph head-on, both cars
have come to a stop (let's assume all our collisions are inelastic),
so all the kinetic energy that the cars initially had (2 * 50^2 = 5000)
has been dissipated. On the other hand, after the 70.71 mph car hits
the stationary one, the two cars are now moving at 35.36 mph, so
the kinetic energy dissipated during the crash is the difference
between the initial kinetic energy (70.71^2 = 5000) and the final
kinetic energy (2 * 35.36^2 = 2500), which is only 2500, or half
that of the 50 mph head-on.
* Hitting a brick wall at 50mph would be identical to hitting another car also
* going 50mph. When hitting the wall, all the energy is applied to your car. In
* the two car scenario, the amount of energy would be doubled, but there's
* two cars to absorb it, so each car gets the same amount of "work" done on
* it.
Yes.
* On the other hand, one car going 50mph hitting a parked car is another
* matter entirely and is equivalent to two cars hitting head on at 35.35mph.
No.
*
* All assumptions above assume identical mass and construction of vehicles
* involved and does not take into account the difference between static and
* rolling friction which may slightly alter the result of hitting a parked car,
* and also does not take relativistic effects into account which makes the
* apparent speed differential of two objects infinitesimally less that you'd
* expect. So there.
*
* Tom.
*
* As a matter of fact, Yes I do have college degrees in math and physics.
ao...@lehigh.edu
requires 2500 for each of the two. Thus, hitting a brick wall at 50 mph
will cause dissipation of the same 2500 units and the same damage.
But hitting the wall at 100 mph will dissipate 10000, that is 4 times
as much! Looks like only some lurking atoms will be left!
Alex
In article <MSGID_1=3A109=2F417_3...@fidonet.org>, Frank_...@srs.blkcat.
com (Frank Mallory) writes:
> J>From: Chuck Jackson <jacksonc>
>
> J>A friend of mine says that 2 cars hitting head on at 50 mph is like
> J>hitting a
> J>brick wall at 100 mph. I disagree. If you are doing 50 mph and come to
> J>a sudden
> J>stop, the impact is less than if you were doing 100 mph and come to a
> J>complete
> J>stop. The fact that the other car is doing 50 mph also is irrelevant.
> J>Anyone
> J>want to add to this?
>
>It happened here yesteday, on the George Washington Parkway in No. Virginia.
>Truly devestating.
>
>
Alexis Ostapenko
610-758-3517 ao...@Lehigh.edu
Gerald N. Cameron
>
> Finally...... someone came up with the right answer.. I dont see how so
> many can start with the right idea (equal kinetic energy) and then make
> the great illogical jump from 50 to 100 miles per hour. The variables are
> speed and mass. If you are sitting still in a car and are hit at 50 miles
> per hour by a equal car the impact you receive is a 25 mph bw (brick wall)
> impact. Of course the two cars would have to hit so squarely that it
> would never happen in real life. I notice from time to time that people
> who didn't pay attention in High School physics make some inuitive
> statements that are way off.
What a fascinating conversation. It does indeed remind me of high school physics. I've
heard a lot of correct explanations, and some not-so correct. Because this is a
simplified problem (i.e. identical collision cars which interact as simple point
particles [like billiards], exactly head on with no couple induced motion in any plane,
etc.), we need to also clearly state that the WALL in question must also be assumed to be
IMMOVABLE. Given this case, the collision of two identical cars each traveling 50mph
with a relative vel. of 100mph is the exact equivalent of a single car traveling at 50mph
colliding with an IMMOVABLE wall.
--GNC, Ga. Tech. ChE '85 (What do chemical engineers know about particle physics anyway!)
Dick Wells
of the total. One car on the brick wall generates half the energy with
the single car absorbing all of it. Fairly simple.
RW
Prof Calculus
>A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
>brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
>stop, the impact is less than if you were doing 100 mph and come to a complete
>stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
>want to add to this?
Might as well jump in.
Assuming that both BOTH cars come to rest (velocity=0) after
the collision, 2 identical cars crashing into each other at 50mph
is equivalent to one car hitting an infinite mass wall at
50mph (not 100mph). This is because the wall absorbs no energy. The
deformation of the car is identical in both cases since the energy
absorbed by the car is the same in both cases (viz. 0.5*m*v*v) where
v=50mph. In case 1 the other car dissipates the identical
amount of energy. The impact of the collision is determined
by the equation E=F.d (E=total energy) and d is the "distance"
travelled after impact (the crumpling of the car). The more
the distance travelled, lower is the force experienced by
the car. The deceleration then of course can be calulated
as F/m (and related to "g's" if necessary).
If the cars are of unequal mass but same speed, the final
velocity of the now cojoined mass of both cars is NOT 0.
Similar calculations will apply but in real life this
will be affected by crumpling properties of different
cars. m1*v1+m2*v2 = (m1+m2)*v. If m1 is larger than
m2, the effect of impact on car 1 is less since the
final velocity of both cars is in the same direction as
that of car 1 before the impact. For car2 its velocity
reverses direction and so it will experience a much
worse impact (assuming both cars are built to similar
safety standards).
Sriram
--
nar...@netcom.com
Saleem Ahmad
>Take your head and bash it against the wall. Now take your head and run
>towards a pillow that is being hurled at you. The difference is the
>pillow will hurt a lot less because it is a deformable body just like a
>the crumple zones of a car. The wall will not crumple but the two cars
>will. The cars will decelerate over a larger distance than the one
>car-wall collision thus less force on the the people inside. So your
>friend is misinformed.
Right, energy dissapated over a few seconds (as the cars fold upon
themselves) should cause less havoc than when it is released almost
instantly (as in the case of the collision with an inelastic wall).
That's a real world answer.
The paper and pencil answer could go something like this:
Remember: every action has an equal and opposite reaction.
Hit a wall (stationary object) at 50 mph, it hits you back at 50mph.
But now, take away the wall and substitute it with a another car (B)
traveling towards the first (A) at 50mph. At first it would seem that
you are going to get much more of a headache. However, it will take
all the energy of car B travelling at 50 mph to stop the original car
A, as effectively as the wall.
So without formulas, but through logic, I am led to deduce that two
identical cars rushing at each other 50mph will be less catastophic
than one car hitting a wall at 100mph.
Oh Lord, I wish I could remember my physics 101, but then it was 16
years ago.
Dennis Jensen
head
>|
>
> Okay, but would this mean that hitting a wall at 50 mph is the same as
>one car, going 100mph, hitting a stationary car? Or one car, going 75 mph,
>hitting another car which is going 25 mph? It seems like the speeds in
>the head-on collision would just be relative to any certain "fixed" point
>of reference.
This is correct. In crash dynamics, it is said that if you hit into a
stationary vehicle of the same mass, the effect is the same as hitting a solid
object at twice the speed (i.e. wall = 100mph, stationary vehicle = 50mph).
Dennis
Dennis Nesseth
In article <klRjblm00iWTAGMmg=@andrew.cmu.edu>, "Nathan J. Nagel" <gear...@andrew.cmu.edu> writes:
> Excerpts from netnews.rec.autos.tech: 18-Apr-96 2 cars crashing head on
> at .. by Chuck Jackson@???
> > brick wall at 100 mph. I disagree. If you are doing 50 mph and come to
> a sudde
> > n
> > stop, the impact is less than if you were doing 100 mph and come to a
> complete
> >
> > stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
> > want to add to this?
>
> in the head-on crash but if you hit a wall only one car will deform.
> The math works out rather nicely, but I'm too lazy to do it right now,
> maybe someone else will?
>
> later,
>
> Nate
Welllll I'll put it this way. If I was running towards
someone who was the same size as me and we were going
to bump into each other, I would much prefer the other
person were standing still rather than running towards me at
the same speed. Guess I'm a little confused as to how
the speed of the oncoming vehicle is not to be considered
as part of the impact calculation???
Dennis
Humberto Fernando Garcia Figueiredo
Chuck Jackson <jacksonc> wrote:
>A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
>brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a
sudden
>stop, the impact is less than if you were doing 100 mph and come to a
complete
>stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
>want to add to this?
>
>
2 equal cars hitting at 50 mph each is the same as 1 car hitting a perfect
wall at 50 mph! Also, is the same as 1 car at 100 mph hitting the other equal
car while stoped.
Any doubts?
Humberto
Kev
: Wrong... IF the wall is moving at you at 50mph it still will not absorb
: any energy like a car will.
To keep this conversation less confusing, let's assume rigid bodies. Just
ignore crumple zones; I don't think the original poster meant to consider
these things, although they are interesting.
--
-----------------------------------------------------------------------------
Kevin Mather
POD Engineering Dept.
Mail: gt4...@prism.gatech.edu
-----------------------------------------------------------------------------
Kev
: 1. We have two cars comming at each other at 50 MPH. They are in perfect alignment, going
: exactly the same speed.
: 2. We hang a piece of paper between them at the exact location where they will meet.
: 3. We can stop time to look at the collision in tiny time fragments.
: The the exact time the bumpers touch the forces between the cars is equal so the paper sits
: there and is just squeezed between the two bumpers. As the cars continue to decelerate due to
: the force generated by the distortion of the metal/plastic/glass/flesh, in an ideal straight
: on collision, the paper will just be squeezed more and more between the two collapsing cars.
: the paper in this ideal collision would not be bent but just squeezed more and more as the
: collapse continues until the cars finally stop and any energy left over due to the final bit
: of elastic distortion would propel the cars back a little until they come to rest. This
: assumes identical cars etc., you get the picture, Balanced forces.
: So now we put the same piece of paper against a brick wall and run into it with the third
: identical car. Actually we would probably want to use a fresh piece of paper. ;-)
: The paper will have a similar experience of being crushed between the car and the wall.
: Again breaking the collision down into very small segments we will find teh same balanced
: forces with the wall and the car.
: So we are lead to the conclusion that the 50MPH head on is the same as runing into the wall
: at 50 MPH.
: The math one can do will substantiate this. At least the math that I can do, You may not be
: able to do the math.
: Believe me or not I don't care. I make a good living at engineering!
: Have a good one,
: Ken
How about this. Stand against a wall and let a car hit you at 5 mph. Now
stand in the middle of the road and let two cars hit you from opposite
directions; each going 5 mph. Bet you wish you were a piece of paper.
Kev
Kev
: >Sure, your friend is right.
: >It's got to do with kinetic energy.
: >Simply, with two cars going 50 mph hitting head on, each car has
: >a kinetic energy that gets 'dissipated' when they collide (boom)
: >so the total kinetic energy is twice that of one car going 50 mph
: >or the same as one car going 100mph.
: Except that kinetic energy is a function of the _square_ of velocity.
: But there is still another flaw in your logic. The total energy in
: the frontal collision is dissipated by TWO cars instead of ONE for the
: wall collision. This means that regardless of how you compute the
: kinetic energy, the total energy dissipated in the wall collision
: should be half of that for the frontal collision to make things
: equivalent.
: >put simply
: >2x50=1x100
: Put even more simply:
: 2 x 50^2 != 100^2 where != means not equal.
: Assuming perfect frontal collision of the two cars and an infinitely
: rigid wall, smashing into a wall at 50mph is is equivalent to the
: collision of two vehicles each going at 50mph. This can actually be
: deduced from kinematics. In both cases, the vehicle is being
: decelerated with the bumper coming to a stop at the contact point of
: the collision.
: The original poster was right in disagreeing with his friend. In the
: frontal collision, the other car going at the same speed is serving as
: a brick wall substitute :-)
: -adour
You are completely full of shit. It is quite apparent that you have *NO* clue
what you're talking about. I am a mechanical engineer, and I'm telling you
first off, in your equation you forgot to account for the change in KE of
EACH vehicle. Number two, we're talking about changes in inertia right now,
not necissarily energy (although it can be used). Proper quantities in
question are called MOMENTUM (P=M*V) and IMPULSE (Change in P/time). Hitting
an oncoming car of equal mass and equal velocity will result in FAR (double,
actually) greater impulse and thus be a lot more violent than hitting a
stationary wall at that speed. If you still don't believe me, get another
degree.
Kev
: Finally...... someone came up with the right answer.. I dont see how so
: many can start with the right idea (equal kinetic energy) and then make
: the great illogical jump from 50 to 100 miles per hour. The variables are
: speed and mass. If you are sitting still in a car and are hit at 50 miles
: per hour by a equal car the impact you receive is a 25 mph bw (brick wall)
: impact. Of course the two cars would have to hit so squarely that it
: would never happen in real life. I notice from time to time that people
: who didn't pay attention in High School physics make some inuitive
: statements that are way off.
Dick, I see a funny flaw in your reasoning. Do brick walls have physical
properties that somehow halve the momentum of objects they hit? Here it is
in simple math. The momentum of the moving car is (We'll say it's going
ten m/s and mass=1000kg to make it REAL easy) p=(10)*(1000)=10,000 kgm/s.
Now, assuming an IDEAL collision, this car transfers ALL its momentum to your
car. Your car's change in momentum is 10,000 kgm/s. Now, if you're driving
your 1000 kg car at 10 m/s towards a wall, your momentum is 10,000 kgm/s.
When you hit the wall, your momentum is suddenly zero. The change in momentum
is still 10,000 kgm/s. In real life, another variable would be how LONG it
takes for that momentum to change. It is not instantaneous due to crush and
give of materials and some other variables. This quantity I'm talking about
is impulse. It's the difference between me pushing on you with my fist slowly
until you reach one mile per hour, or punching you, instantly accelerating
you to one mile per hour. The momentum change is the same, but it happens much
quicker in the latter case, and therefore, hurts.
Kev
you.
Kev
: In article <317672...@amoco.com> "Christopher M. Ripp" <chris_...@amoco.com> writes:
: Chuck Jackson wrote:
: >
: > A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
: > brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
: > stop, the impact is less than if you were doing 100 mph and come to a complete
: > stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
: > want to add to this?
: > Sure, your friend is right.
: > It's got to do with kinetic energy.
: > Simply, with two cars going 50 mph hitting head on, each car has
: > a kinetic energy that gets 'dissipated' when they collide (boom)
: > so the total kinetic energy is twice that of one car going 50 mph
: > or the same as one car going 100mph.
: Hmm. Almost right, except that kinetic energy = 0.5 * Mass *
: the second case vs. the first, in all likelihood the result would be
: more severe in the second case.
: Any volunteers out there willing to find out the difference in
: collision elasticity, which would properly allocate the amount of
: energy dissapated by each car / wall?
: ed
: -------- Ed Hahn | eh...@mitre.org | (703) 883-5988 --------
: The above comment reflects the opinions of the author, and does not
: constitute endorsement or implied warranty by the MITRE Corporation.
: Really, I wouldn't kid you about a thing like this.
Some of your reasoning is pretty good, however I think it would probably be
easier to use momentum/impulse to solve these problems rather than energy.
It takes a bit less effort. My previous posts describe a perfectly inelastic
collision in which the cars do not stick together. Whether they stick or not
is irrelevent if the cars are equal mass/velocity. However, in real life this
is pretty much never the case. If a heavy car hits a light one, it negates
(brings to zero) momentarily the light car's momentum. Then the light car
begins to accelerate BACKWARDS, with the heavy car stuck to it. The two cars
move together at the same velocity. The momentum of the heavy car was not
changed anywhere near as much as that of the light car, therefore the light car
got screwed. In real life cars are designed to prolong the duration of the
impulse using crumple zones. This means a change in momentum is over a longer
period of time, making the effects of the accident on its victims less severe.
In the case of hitting a rigid wall, momentum is only prolonged by the crush
zones of the car, as the wall does not give. It makes sense, then, that
hitting a car of equal mass and an equal velocity of, say, thirty miles per
hour, would be a bit less severe than hitting a wall at sixty. This is because
there are two energy absorbing systems rather than just one. Anyway, that's
pretty much it.
Kev
Kev
: Wrong: it is the same as 1 BRM colliding at 50mph with a brick wall.
: Imagine a thin vertical film between the two cars at the moment of
: impact. By symmetry neither of the cars would penetrate that film
: if the two cars are identical. Then replace that film with a brick
: wall.
My congrats to really making me think. This analogy made me questoion what
makes perfect sense to me. Here's another one. The approach velocity of
car one relative to car two (if both are travelling 50 mph) is 100mph. Now,
to keep total momentum the same, stop car one and accelerate car two to 100mph.
When car two strikes car one, car two stops completely, transferring all of
its momentum to car one. This change in momentum for car two is the same
as the change in momentum if car two were moving at 100mph towards a stationary
wall. It goes from 100mph to zero mph. P=MV, and if we consider equal impulse
time, we realize that car two striking the wall at a hundred is the same as
car one and car two striking each other at fifty. Your little paragraph really
screwed with my head. Thanks for making me think.
Yaakov Eisenberg
Kev <gt4...@acmey.gatech.edu> wrote:
*
* : Assuming perfect frontal collision of the two cars and an infinitely
* : rigid wall, smashing into a wall at 50mph is is equivalent to the
* : collision of two vehicles each going at 50mph. This can actually be
* : deduced from kinematics. In both cases, the vehicle is being
* : decelerated with the bumper coming to a stop at the contact point of
* : the collision.
*
* : The original poster was right in disagreeing with his friend. In the
* : frontal collision, the other car going at the same speed is serving as
* : a brick wall substitute :-)
*
* : -adour
*
* You are completely full of shit. It is quite apparent that you have *NO* clue
* what you're talking about. I am a mechanical engineer, and I'm telling you
* first off, in your equation you forgot to account for the change in KE of
* EACH vehicle. Number two, we're talking about changes in inertia right now,
* not necissarily energy (although it can be used). Proper quantities in
* question are called MOMENTUM (P=M*V) and IMPULSE (Change in P/time). Hitting
* an oncoming car of equal mass and equal velocity will result in FAR (double,
* actually) greater impulse and thus be a lot more violent than hitting a
* stationary wall at that speed. If you still don't believe me, get another
* degree.
*
* --
* -----------------------------------------------------------------------------
* Kevin Mather
* POD Engineering Dept.
* Mail: gt4...@prism.gatech.edu
* -----------------------------------------------------------------------------
I agree that, in this problem, it is more useful to deal with momentum than
with kinetic energy. Since the masses of the cars and brick walls aren't
changing, we can think of momentum as proportional to velocity and impulse
as proportional to acceleration. Reread the message that you responded to;
it's exactly right. The deceleration in both cases is identical, since
in both cases, the very front of the car stops instantly, with the rest
of the car stopping after a bit of crunching. What causes the front to stop
instantly is irrelevant.
--
-Yaakov Eisenberg (yaa...@cc.gatech.edu)
Yaakov Eisenberg
In article <4lirdn$d...@catapult.gatech.edu>,
Kev <gt4...@acmey.gatech.edu> wrote:
*
* Some of your reasoning is pretty good, however I think it would probably be
* easier to use momentum/impulse to solve these problems rather than energy.
* It takes a bit less effort. My previous posts describe a perfectly inelastic
* collision in which the cars do not stick together.
An elastic collision is where they bounce back; an inelastic one is where they
stick together.
* Whether they stick or not
* is irrelevent if the cars are equal mass/velocity. However, in real life this
* is pretty much never the case. If a heavy car hits a light one, it negates
* (brings to zero) momentarily the light car's momentum. Then the light car
* begins to accelerate BACKWARDS, with the heavy car stuck to it.
The light car (and the heavy one, too) was always accelerating backwards. You
mean "Then the light car begins to move backwards."
* The two cars
* move together at the same velocity. The momentum of the heavy car was not
* changed anywhere near as much as that of the light car, therefore the light car
* got screwed.
The momentum of the heavy car changed exactly the same amount as that of the
light car (but in the opposite direction). Since the heavy car is more massive,
this equal change in momentum required a smaller change in velocity.
* In real life cars are designed to prolong the duration of the
* impulse using crumple zones. This means a change in momentum is over a longer
* period of time, making the effects of the accident on its victims less severe.
* In the case of hitting a rigid wall, momentum is only prolonged by the crush
* zones of the car, as the wall does not give. It makes sense, then, that
* hitting a car of equal mass and an equal velocity of, say, thirty miles per
* hour, would be a bit less severe than hitting a wall at sixty.
Quite a bit less, actually :-) In fact, it would be exactly the same
as hitting the brick wall at thirty.
Think of it this way: Shouldn't a head-on collision where each car is going
thirty be the same as a car going sixty hitting a stationary car? After
all, the relative speed of the cars is the same in both cases. Now, shouldn't
hitting a brick wall at sixty be much worse than hitting a stationary car
at sixty?
* This is because
* there are two energy absorbing systems rather than just one. Anyway, that's
* pretty much it.
* Kev
--
-Yaakov Eisenberg (yaa...@cc.gatech.edu)
Yaakov Eisenberg
Kev <gt4...@acmey.gatech.edu> wrote:
* Dick Wells (dick...@aol.com) wrote:
* : Wrong... IF the wall is moving at you at 50mph it still will not absorb
* : any energy like a car will.
* To keep this conversation less confusing, let's assume rigid bodies. Just
* ignore crumple zones; I don't think the original poster meant to consider
* these things, although they are interesting.
* --
* -----------------------------------------------------------------------------
* Kevin Mather
* POD Engineering Dept.
* Mail: gt4...@prism.gatech.edu
* -----------------------------------------------------------------------------
Actually, I think it's pretty important to not assume rigid bodies. Otherwise,
any crash, no matter how low the speed, would cause infinite acceleration, thus
infinite force on the occupant, which could hurt quite a bit. :-)
--
-Yaakov Eisenberg (yaa...@cc.gatech.edu)
Dick Wells
things we know that aren't so." What a chuckle. You can lead the
engineer to reason but you cant make him think. RW
BHM
| [...] The approach velocity of car one relative to car
| two (if both are travelling 50 mph) is 100mph. Now, to keep total
| momentum the same, stop car one and accelerate car two to 100mph. When car
| two strikes car one, car two stops completely, transferring all of its
| momentum to car one. This change in momentum for car two is the same as the
| change in momentum if car two were moving at 100mph towards a stationary
| wall. It goes from 100mph to zero mph. P=MV, and if we consider equal impulse
| time, we realize that car two striking the wall at a hundred is the same as
| car one and car two striking each other at fifty. [...]
Hang on a sec, I'm confused. If two cars, each going 50, hit head-on and
stop dead, then wouldn't a car going 100 just decelerate to 50 when it
hits a stationary car? It seems like both cases are the same, just with
different points of reference. My "C" in Physics 231 certainly doesn't
qualify me to say "you're wrong", but what you said doesn't make any sense
to me.
Car A is going 50 mph in a westward direction
Car B is going 50 mph in an eastward direction
Point of reference is "stationary"
Both cars "stop" on impact (Car B's acceleration is -50 mph) (inelastic
collision)
Now make the point of reference move at 50 mph in a westward direction and
we have:
Car A is stationary
Car B is going 100 mph in an eastward direction
Upon impact, both cars are now moving at 50 mph in an eastward direction
(Car B's deceleration is -50 mph)
If, as you said, the 100 mph car stops when it hits a stationary car, then
in the original 50 mph head-on collision both cars would have to bounce
back from the collision at 50 mph in the opposite direction. (Car B's
deceleration is -100 mph) (elastic collision)
I assume that real-life wrecks are inelastic collisions.
Like I said, I got a "C" in Physics, but I wonder why the important factor
isn't change in velocity over time (acceleration/deceleration).
Decelerating to 0 from 100 mph in .001 second is liable to be much worse
than decelerating to 0 from 50 mph in .001 second. Hitting a brick wall
at 50 or a oncoming car (of equal size and speed) at 50 will both result
in the same deceleration in theory (although, not in the real world).
Thus, in theory, they would be the same. The velocity and mass of the
oncoming car WOULD be important (as common sense tells us), in that
changes in these variables would change the amount of deceleration by the
car in question. (Hitting an oncoming car who is 1 mph faster or 1 pound
heavier at 50 mph would be worse, in theory, than hitting a wall at 50
mph)
In real life, my guess would be that crumple zones would allow the change
in velocity to be spread over a longer time, decreasing the deceleration.
change in velocity = -50 m/s time = .001 second acceleration = -50,000 m/s2
change in velocity = -50 m/s time = .002 second acceleration = -25,000 m/s2
Anyway, those are my thoughts, be they right or wrong...
S.E. Homer
> My congrats to really making me think. This analogy made me questoion
> what makes perfect sense to me. Here's another one. The approach
> velocity of car one relative to car two (if both are travelling 50 mph) is
> 100mph. Now, to keep total momentum the same, stop car one and accelerate
> car two to 100mph. When car two strikes car one, car two stops completely,
> transferring all of its momentum to car one. This change in momentum for
> car two is the same as the change in momentum if car two were moving at
> 100mph towards a stationary wall. It goes from 100mph to zero mph. P=MV,
> and if we consider equal impulse time, we realize that car two striking
> the wall at a hundred is the same as car one and car two striking each
> other at fifty. Your little paragraph really screwed with my head.
> Thanks for making me think.
But what you're describing is a situation where KE is conserved which
means that no energy has been dissipated. i.e. No harm done. That's not
the case in a car wreck.
If you assume the worst (most damage) in terms of energy dissipation
then you minimize KE while conserving P. You end up with both vehicles
traveling in the same direction at 50 mph, again assuming identical
vehicles. In terms of damage (the energy dissipated by the vehicles),
this is the same as a perfectly inelastic 50 mph headon between two
identical vehicles.
Your focus on impulse and momentum is incorrect. Damage is related to
the amount of energy absorbed by an object. This is an energy problem.
--
S.E. Homer
Allen M. Ashley
In the case of two cars of equal mass heading at each other at equal
velocity the total momentum is zero. In any such collision energy is
never conserved, being dissipated in the deformation of the vehicles, the
sound emitted, and the heat released.
Momentum also depends upon the coordinate system used. In the case of
two vehicles approaching each other, a coordinate system mounted on one
of the two vehicles will see zero momentum for that vehicle, and twice
the momentum for the other vehicle. That is the case for a vehicle
traveling at twice the speed impacting a stationary vehicle in an
earth refernce system (parked vehicle). The critical thing in these
impacts is that momentum before the impact=momentum after the impact.
--
==============================================
Allen Ashley ash...@alumni.caltech.edu
=============================================
Dick Wells
to get the same energy on one car if it hits an immovable object?
Humberto Fernando Garcia Figueiredo
>
>: The original poster was right in disagreeing with his friend. In the
>
>
>You are completely full of shit. It is quite apparent that you have *NO*
clue
I don't believe you! And I have enough degrees.
How did you calculate the impulse? You know the change in P, but you don't
know the amount of time in wich P will change. So, you say the impulse is
double for the cars collision because, by INTUITION, you think the crash takes
place (till the end of deformation of the cars) in half the time it takes in
the car-wall collision. Why???????????????
You should learn about INTUITION with the guy you were answering. He does'nt
have to deal with momentum or impulse because he sees what you can't see, and
he put it very well when he said:
"In the frontal collision, the other car going at the same speed is serving as
a brick wall substitute :-)"
Like a mirror, I would add!
Humberto
Humberto Fernando Garcia Figueiredo
If you still did'nt figure it out, I will gladly explain why two cars hitting
at 30 mph each is the same as one car hitiing a wall at 30 mph!
Humberto
Humberto Fernando Garcia Figueiredo
gt4...@acmey.gatech.edu (Kev) wrote:
>
>: Wrong: it is the same as 1 BRM colliding at 50mph with a brick wall.
>: Imagine a thin vertical film between the two cars at the moment of
>: impact. By symmetry neither of the cars would penetrate that film
>: if the two cars are identical. Then replace that film with a brick
>: wall.
>
>makes perfect sense to me. Here's another one. The approach velocity of
>car one relative to car two (if both are travelling 50 mph) is 100mph. Now,
>to keep total momentum the same, stop car one and accelerate car two to
100mph.
>When car two strikes car one, car two stops completely, transferring all of
>its momentum to car one. This change in momentum for car two is the same
>as the change in momentum if car two were moving at 100mph towards a
stationary
>wall. It goes from 100mph to zero mph. P=MV, and if we consider equal
impulse
>time, we realize that car two striking the wall at a hundred is the same as
>car one and car two striking each other at fifty. Your little paragraph
really
>screwed with my head. Thanks for making me think.
Sorry! Following you was not my intention. But you're still wrong. Why do you
say car two will stop? It won't.
Humberto
Kev
: * Well, you are mostly wrong . If the vehicle you hit head on weighs the same or
: * more than your vehicle and then it is like you hit a brick wall at 100 mph or
: * even worse . If the vehicle you hit weighs less then yours then it is like
: * hitting a brick wall at less than 100 mph .
: You are correct in that the relative mass of the vehicles needs to be
: taken into account. However, assuming the two are of equal mass,
: a 50 mph head-on is like 50 mph into a brick wall, not 100 mph into
: a brick wall.
: * The relationship is as follows :
: *
: * Force = Mass(vehi#1) x Velocity(veh#1) + Mass(veh#2) x Velocity(veh#2)
Uh, since when does F=MV? I think it's more like F=MA. The cars are only
accelerating durring the collision. THe quantity you describe, MV, is called
momentum.
: This formula doesn't look right. What if Mass(veh#2) = 0, i.e. veh#1
: hasn't hit anything? Certainly, the force in this case should be
: zero, but that's not what your formula gives. Or, suppose veh#2 is
: not moving, so Velocity(veh#2) = 0. Then, your formula says that
: the mass of veh#2 is irrelevant, which is not correct.
: *
: * This holds true if the collision occurs head on only .
: *
: *
: * George Gogis
: *
: *
: --
:
: -Yaakov Eisenberg (yaa...@cc.gatech.edu)
Kev
: But what you're describing is a situation where KE is conserved which
: means that no energy has been dissipated. i.e. No harm done. That's not
: the case in a car wreck.
: If you assume the worst (most damage) in terms of energy dissipation
: then you minimize KE while conserving P. You end up with both vehicles
: traveling in the same direction at 50 mph, again assuming identical
: vehicles. In terms of damage (the energy dissipated by the vehicles),
: this is the same as a perfectly inelastic 50 mph headon between two
: identical vehicles.
: Your focus on impulse and momentum is incorrect. Damage is related to
: the amount of energy absorbed by an object. This is an energy problem.
Actually, it's both. Magnatude of impulse tells you a lot about damage, and
so does energy. Here's the deal. In real life, the wall HAS to move to
conserve momentum. Energy is absorbed by the wall when this happens, as, in
order for the car to stop, the wall has to stop it. The mass of the wall is
used with the momentum of the car to calculate energy absorbed by the wall.
To stop the car, the portion of the wall that is knocked out or moved must
be equal to or greater than the mass of the car. In all likelyhood, that mass
is going to be very close to that of the car (as the car can only smash a mass
equal to or less than its own). Thus, we can deduce that the wall will absorb
close to the same amount of energy that the car does. Therefore, if the wall
is stationary, the car needs to be moving twice as fast to achieve the same
energy absorbtion by the automobile.
Yaakov Eisenberg
Kev <gt4...@acmex.gatech.edu> wrote:
* S.E. Homer (ho...@owens.ridgecrest.ca.us) wrote:
*
* : But what you're describing is a situation where KE is conserved which
* : means that no energy has been dissipated. i.e. No harm done. That's not
* : the case in a car wreck.
*
* : If you assume the worst (most damage) in terms of energy dissipation
* : then you minimize KE while conserving P. You end up with both vehicles
* : traveling in the same direction at 50 mph, again assuming identical
* : vehicles. In terms of damage (the energy dissipated by the vehicles),
* : this is the same as a perfectly inelastic 50 mph headon between two
* : identical vehicles.
*
* : Your focus on impulse and momentum is incorrect. Damage is related to
* : the amount of energy absorbed by an object. This is an energy problem.
*
* Actually, it's both. Magnatude of impulse tells you a lot about damage, and
* so does energy. Here's the deal. In real life, the wall HAS to move to
* conserve momentum. Energy is absorbed by the wall when this happens, as, in
* order for the car to stop, the wall has to stop it. The mass of the wall is
* used with the momentum of the car to calculate energy absorbed by the wall.
* To stop the car, the portion of the wall that is knocked out or moved must
* be equal to or greater than the mass of the car. In all likelyhood, that mass
* is going to be very close to that of the car (as the car can only smash a mass
* equal to or less than its own). Thus, we can deduce that the wall will absorb
* close to the same amount of energy that the car does. Therefore, if the wall
* is stationary, the car needs to be moving twice as fast to achieve the same
* energy absorbtion by the automobile.
* --
* -----------------------------------------------------------------------------
* Kevin Mather
* POD Engineering Dept.
* Mail: gt4...@prism.gatech.edu
* -----------------------------------------------------------------------------
Momentum does need to be conserved; however, we are assuming a very strong
wall rigidly attached to the earth, so that the wall and earth can be
considered a single body. The momentum change of the wall-earth system
is exactly equal to the momentum change of the car (in the opposite direction).
This does not mean that the wall-earth system dissipates the same amount of
energy as the car does.
If the wall-earth's change in velocity is such that momentum is conserved
(i.e. very small compared to the change in the car's velocity), the change
in the energy of the wall-earth, which is proportional to the square of its
velocity, will be miniscule.
--
-Yaakov Eisenberg (yaa...@cc.gatech.edu)
Kev
: Sorry to nitpick, but...
: * Some of your reasoning is pretty good, however I think it would probably be
: * easier to use momentum/impulse to solve these problems rather than energy.
: * It takes a bit less effort. My previous posts describe a perfectly inelastic
: * collision in which the cars do not stick together.
: An elastic collision is where they bounce back; an inelastic one is where they
: stick together.
Sorry for that one.
: * is irrelevent if the cars are equal mass/velocity. However, in real life this
: * is pretty much never the case. If a heavy car hits a light one, it negates
: * (brings to zero) momentarily the light car's momentum. Then the light car
: * begins to accelerate BACKWARDS, with the heavy car stuck to it.
: The light car (and the heavy one, too) was always accelerating backwards. You
: mean "Then the light car begins to move backwards."
That, too.
: * The two cars
: * move together at the same velocity. The momentum of the heavy car was not
: * changed anywhere near as much as that of the light car, therefore the light car
: * got screwed.
: The momentum of the heavy car changed exactly the same amount as that of the
: light car (but in the opposite direction). Since the heavy car is more massive,
: this equal change in momentum required a smaller change in velocity.
Sorry. I meant the momentum of the passengers.
: * In real life cars are designed to prolong the duration of the
: * impulse using crumple zones. This means a change in momentum is over a longer
: * period of time, making the effects of the accident on its victims less severe.
: * In the case of hitting a rigid wall, momentum is only prolonged by the crush
: * zones of the car, as the wall does not give. It makes sense, then, that
: * hitting a car of equal mass and an equal velocity of, say, thirty miles per
: * hour, would be a bit less severe than hitting a wall at sixty.
: Quite a bit less, actually :-) In fact, it would be exactly the same
: as hitting the brick wall at thirty.
I still disagree. This is only true with a wall of infinite mass. In real life the wall HAS to move to conserve momentum.
: Think of it this way: Shouldn't a head-on collision where each car is going
: thirty be the same as a car going sixty hitting a stationary car? After
: all, the relative speed of the cars is the same in both cases. Now, shouldn't
: hitting a brick wall at sixty be much worse than hitting a stationary car
: at sixty?
Correct, if the wall doesn't move. See earlier posts which describe what
happens to a real wall.
: * This is because
Kev
: wall rigidly attached to the earth, so that the wall and earth can be
: considered a single body. The momentum change of the wall-earth system
: is exactly equal to the momentum change of the car (in the opposite direction).
: This does not mean that the wall-earth system dissipates the same amount of
: energy as the car does.
: If the wall-earth's change in velocity is such that momentum is conserved
: (i.e. very small compared to the change in the car's velocity), the change
: in the energy of the wall-earth, which is proportional to the square of its
: velocity, will be miniscule.
: --
This is true, because your wall has effectively got infinite mass. However,
I'd think that most walls one would collide with would probably be damaged
quite a bit by an auto travelling at fifty miles per hour. In this event, the
wall absorbs a very significant amount of energy, making it similar (but
not exactly) to hitting a stationary car of the same mass.
Kev
Kev
previous posts. It is very untypical of me to behave this way. I was rather
upset at the time (and a bit drunk) and took it out on the newsgroup. Just
wanted to let everyone know that I am not, in fact, a dick. Apologies to
those offended.
Kev
: Why is it so hard to take the two car energy and divide it by the two cars
: to get the same energy on one car if it hits an immovable object?
Defbods, my friend, defbods.
John Kennedy
of 100 mph, it doesn't matter if they are both doing 50, one is doing 75 and
the other 25, or one is doing 100 and the other 0, or one is doing 110 and
the other is backing away at 10 mph. The only thing different in each of the
4 scenarios is the relative speed of the soil under the cars relative to the
point of impact.
If the above scenarios weren't equivalent, think of what the rotational
speed of the earth would do to the math!
--
John Kennedy jo...@secondsource.com
Second Source, Inc.
Annapolis, MD
C & K Claxton
50 is really goofy thinking. You are making a big assumpion here - that the car you are
hitting will absorb exactly 1/2 of the energy of the impact! Uhh - Nope!
The only way to put this whole thing into proper perspective would be to say
that 2 cars of given types hitting head on at 50 mph each would be the same as if the
one car driving at 100mph hit the other, stationary, car head on.
Stan
>S.E. Homer (ho...@owens.ridgecrest.ca.us) wrote:
>
>: Your focus on impulse and momentum is incorrect. Damage is related to
>
Kev really needs to go back and take freshman Physics again.
In *any* impact, momentum is conserved. Yes, that means that the wall will
move a bit when the car hits it. In this case, the collision is inelastic,
so energy is *not* conserved. Note also that momentum is a *vector*
quantity, so it has both magnitude and direction.
>order for the car to stop, the wall has to stop it. The mass of the wall is
>used with the momentum of the car to calculate energy absorbed by the wall.
>To stop the car, the portion of the wall that is knocked out or moved must
>be equal to or greater than the mass of the car. In all likelyhood, that mass
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Equal or *greater*. Make note of this. Now watch as he contradicts
himself:
>is going to be very close to that of the car (as the car can only smash a mass
>equal to or less than its own). Thus, we can deduce that the wall will absorb
^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
Um, where do you get this? If I walk into a brick wall, am I only
going to knock out 170lbs of bricks? I don't think so.
>close to the same amount of energy that the car does. Therefore, if the wall
>is stationary, the car needs to be moving twice as fast to achieve the same
>energy absorbtion by the automobile.
No. I repeat, No. The wall in this case can be modeled as an a car of
infinite mass. The relevant equations are:
Momentum: P = m*v
Force: F = m*a = dP/dt (the first-order derivative of momentum with
respect to time)
Energy: K = 1/2 m*v^2
In the original scenario, if two cars of equal mass traveling at equal
speeds collide, the momentum of each goes to zero. This is the *same* as
if one car hits an immovable wall.
--
-------------------------------------------------------------------------------
Stan Schwarz | "I just want to live like Yogi Bear
st...@bombay.gps.caltech.edu | He kicks ass on the average bear."
---------------------------------------------------- -Stukas Over Bedrock -----
Adour Vahe Kabakian
>what you're talking about. I am a mechanical engineer, and I'm telling you
I will try to use this line in my next paper: "What I wrote is right,
because I'm an engineer." Better yet, if you still take classes, try
it on your exams.
>question are called MOMENTUM (P=M*V) and IMPULSE (Change in P/time). Hitting
>an oncoming car of equal mass and equal velocity will result in FAR (double,
>actually) greater impulse and thus be a lot more violent than hitting a
>stationary wall at that speed.
In both cases, the momentum is going from (50mph)*M to 0 in the same
fashion, ie. by the front bumper coming to a sudden stop. Therefore,
the impulses are the same.
>If you still don't believe me, get another degree.
What do you suggest, O wise one?
-adour
Adour Vahe Kabakian
Kev <gt4...@acmey.gatech.edu> wrote:
>
> The approach velocity of car one relative to car two (if both are
> travelling 50 mph) is 100mph. Now, >to keep total momentum the
> same, stop car one and accelerate car two to 100mph.
That doesn't keep the the total momentum the same! You either don't
know that momentum is a vector quantity or you can't do very simple
algebra. You have to go back to freshman physics.
> When car two strikes car one, car two stops completely, transferring
> all of its momentum to car one. This change in momentum for car two
> is the same as the change in momentum if car two were moving at
> 100mph towards a stationary wall. It goes from 100mph to zero mph.
> P=MV, and if we consider equal impulse time.....
Because of your previous mistake, this argument becomes invalid.
> Your little paragraph really screwed with my head.
You mean the paragraph screwed with your little head.
> Thanks for making me think.
You really need to do a lot more of it. The contrast between the
arrogant "your full of shit" post and your complete inaptitude
displayed here is truly farcical.
-adour
Martyn Uttley
(immovable) wall at 50 MPH is considered a 50 MPH impact. Two identical
cars travelling towards each other at 50 MPH each (a closing speed of 100
mph) is still a 50 mph impact. The damage suffered by the first car will
be the same as 50mph into the wall. Only in the second case you have two
damaged cars.
This is simple physics and the basis of every accident reconstruction
process. BUT it is impossible to simulate in a real situation.
--
!!!!!
(O O)
+-----------------oOO------(_)-------------------------+
! Martyn C. Uttley Hong Kong !
! Fax (852) 2873 3342 e-mail mut...@hk.super.net !
+-----------------------------------oOO----------------+
|__|__|
|| ||
ooO Ooo
P. J. Remner
In a previous article, dge...@niu.edu (David Gersic) says:
>
>Or, the way it was simply shown to me back in Driver's Ed class was a
>"crash test" film. Take two cars, run them head - on into each other at 60
>mph. First they did two big old nasty 70's land yachts. Both cars stopped
>pretty much where they collided, with roughly the same ammount of damage to
>each. Then they did two sub-compact import cars, and again there was about
>the same damage to each, and both just stopped where they hit with only a
>little bit of bounce. Then they ran the land yacht into the sub-compact.
>The land yacht was almost undamaged, and the dummies in the front seat
>barely moved from their positions, as the car kept moving forward. The
>sub-compact, OTOH, went airborn in the other direction, the dummies went
>through the windshield, and the car *landed* about thirty yards from the
>point of impact.
>
>It's not the car's impact that kills you, it's how fast it decelerates when
>it hits (or, more specifically, how fast *you* decelerate).
>
Heh heh heh heh heh heh... heeeeeeeere Suzuki Suzuki Suzuki Suzuki...
But seriously... :-) I saw a video in a tech class with a mid-Eighties
front-drive LeSabre hitting a Geo Storm frontally but off-center.
The Buick was munched up, to be sure, but the Geo arched upwards,
the driver's door opened, the "driver" slipped to the left, bypassing
the airbag, and its head wedged tightly between the A-pillar and the
opened door's front pillar! Not good! I will do my best to never
ride in a car where the seatbelts mount to the doors, and I recommend
the same to everyone else.
For an extreme example, remember in the movie "Mad Max" where the guy
on the motorcycle hits a tanker truck head-on? Yeah, it's a movie,
but I'm sure it's pretty accurate.
--
'72 Thunderbird, 429 4bbl, lots o' looks, the "larger hammer"
"Beat them gently until you get bored, then use progressively larger hammers."
- Andy Dingley, din...@codesmth.demon.co.uk
P. J. Remner
In a previous article, yaa...@cc.gatech.edu (Yaakov Eisenberg) says:
>
>Think of it this way: Shouldn't a head-on collision where each car is going
>thirty be the same as a car going sixty hitting a stationary car? After
>all, the relative speed of the cars is the same in both cases. Now, shouldn't
>hitting a brick wall at sixty be much worse than hitting a stationary car
>at sixty?
Yes! This makes the most sense. I was wrong before.
(At least, right now, I believe I was wrong before. This may change. :-) )
Two *identical* cars, having *identical* velocity, mass, and crush
characteristics, should have a stationary point of impact. By that,
I mean the cars just stop, neither one getting "pushed". In this
situation, each car is absorbing its own energy, just like an
individual car hitting an immovable object, say, a good brick wall.
So, two cars hitting each other at 50mph is like one car hitting
a brick wall ay 50mph, assuming completely identical cars. Of course,
this is never the case, so try to drive a large car, preferably one
with most of its mass low and in front of you so it doesn't try to
munch your body.
P. J. Remner
In a previous article, ash...@alumnae.caltech.edu (Allen M. Ashley) says:
>cter...@aol.com (CTERENZI) writes:
>
>>Your friend is correct, though the assumption must be made that the
>>colliding objects are of similar mass. For instance, 2 Buick Roadmasters
>>colliding head on at 50 mph will result in the same force as 1 B.R.M.
>>hitting a wall at 100mph but A buick Roadmaster and a Civic would not
>>collide with the same force. I believe the formula is Force=Mass *
>>Acceleration
>
>Wrong: it is the same as 1 BRM colliding at 50mph with a brick wall.
>Imagine a thin vertical film between the two cars at the moment of
>impact. By symmetry neither of the cars would penetrate that film
>if the two cars are identical. Then replace that film with a brick
>wall.
>
Yes! That's it!
This all assumes, of course, that the brick wall is a truly stationary
device. In reality, the brick wall will either deform or transfer its
forces to its point of contact with the ground. If the point of contact
is completely stationary, then the energy will be transferred to the
tectonic plates! Do you know what this means? We can get about ten
thousand 18-wheelers, load them to the ceilings with lead and concrete,
build some extremely sturdy brick walls near the San Andreas fault,
and hold Southern California for ransom!! "Give us the money, or we'll
run these suckers at those walls at 80mph. What do we want the money
for? To pay for all these trucks, that's what for."
- Pete (I *really* need some caffeine)
S.E. Homer
> S.E. Homer (ho...@owens.ridgecrest.ca.us) wrote:
>
>
> : Your focus on impulse and momentum is incorrect. Damage is related to
> : the amount of energy absorbed by an object. This is an energy problem.
>
> Actually, it's both. Magnatude of impulse tells you a lot about damage, and
> so does energy.
The impulse is equal to the change in momentum. A car coming to rest
from 50 mph has been subjected to the same impulse whether it coasts
safely to a stop or smashes into a wall.
For a qualitative answer to the "which is worse" question just look at
the energies involved.
--
S.E. Homer
Humberto Fernando Garcia Figueiredo
>By saying that 2 cars hitting each other at 50 (each) is the same as 1
hitting a wall at
>50 is really goofy thinking. You are making a big assumpion here - that the
car you are
>hitting will absorb exactly 1/2 of the energy of the impact! Uhh - Nope!
I guess you did'nt pay attention. I mentioned two IDENTICAL cars in a perfect
head-on collision. Obviously they absorve the same energy. They are equal,
there's no reason for a different energy distribution between them. In other
words, the problem is simmetrical!
> The only way to put this whole thing into proper perspective would be
to say
>that 2 cars of given types hitting head on at 50 mph each would be the same
as if the
>one car driving at 100mph hit the other, stationary, car head on.
Again you did'nt pay attention. I also said that in the post you replyed. But
there must be no friction between tyres and the floor in order to obtain equal
results. Because there is friction, I can only say that the results will be
almost the same.
Hope I was useful
Humberto
Humberto Fernando Garcia Figueiredo
>I agree that, in this problem, it is more useful to deal with momentum than
>with kinetic energy. Since the masses of the cars and brick walls aren't
>changing, we can think of momentum as proportional to velocity and impulse
>as proportional to acceleration. Reread the message that you responded to;
>it's exactly right. The deceleration in both cases is identical, since
>in both cases, the very front of the car stops instantly, with the rest
>of the car stopping after a bit of crunching. What causes the front to stop
>instantly is irrelevant.
Well, you shoud'nt agree. You see, you are dealing with a perfect wall, wich
means infinite mass. Therefore you can not deal with momentum. This is a
concept experience. Do it with energy.
Even if you consider an almost perfect wall (Big mass compared to the car
mass) you'll get nowhere.
Humberto
Nate
>
> George Gogis wrote:
> >
> > In article 3...@bbs.ug.eds.com, Chuck Jackson <jacksonc> writes:
> > > A friend of mine says that 2 cars hitting head on at 50 mph is like hitting a
> > > brick wall at 100 mph. I disagree. If you are doing 50 mph and come to a sudden
> > > stop, the impact is less than if you were doing 100 mph and come to a complete
> > > stop. The fact that the other car is doing 50 mph also is irrelevant. Anyone
> > > want to add to this?
> > >
> >
> > Well, you are mostly wrong . If the vehicle you hit head on weighs the same or
> > more than your vehicle and then it is like you hit a brick wall at 100 mph or
> >
> >
> >
>
> Assuming a totally inelastic collision and a 10 pound car,
>
> Force = Mass(vehi#1) x Velocity(veh#1) + Mass(veh#2) x Velocity(veh#2)
>
>
> So if you now hit the brick wall instead (assume a brick wall weight of
> 100 lbs. Actual weight is irrelevant)
>
> Force = (100lbs x 0 mph) + (10 lbs x 50 mph) = 500 lbs mph
>
> the speed of the car would have to double to get the same force generated
> in the 2 car crash.
I can't believe no one has seen this. Get a clue! F=MA, not F=MV!!!
MV is POTENTIAL ENERGY, and is NOT being applied correctly here.
Your talking Bull!
Not thing in this post make ANY sense!!
> from the musings of Master Meow
More like mental masturbations...
--
Nathanial Beckwith
http://www.itlabs.umn.edu/~beck0238/
Kev
: >I agree that, in this problem, it is more useful to deal with momentum than
In real life walls get screwed up when you hit them with cars. When they do,
they absorb energy. Momentum needs to be dealt with so one can calculate
that amount of energy. The guy's original post probably was intended to
figure out effects on the passengers, in which case you'll have to include
the fact that I've never heard of someone running into a perfect wall.
Kevin Mather
POD Engineering
gt4...@prism.gatech.edu
-----------------------------------------------------------------------------
----------------------------------------------------------------------------
Kevin Mathere
Kev
: I can't believe no one has seen this. Get a clue! F=MA, not F=MV!!!
: MV is POTENTIAL ENERGY, and is NOT being applied correctly here.
MV is momentum.
--
Malcolm G. Costello
one car traveling at 100 mph crashing head on into a parked car ?
Wouldn't the damage to the moving car then be worse if it hit a brick
wall instead of the parked car? Thus 2 cars crashing head on at 50 mph
is not as bad as a single car crashing at 100 mph into a brick wall.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Mack Costello <mcos...@oasys.dt.navy.mil> Code 652 ?
David Taylor Model Basin, Carderock Division Hq. NSWC ___/-\_____
Bethesda, MD 20084-5000 Phone (301) 227-2431 (___________>8
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Kev
: >You are completely full of shit. It is quite apparent that you have *NO* clue
: >what you're talking about. I am a mechanical engineer, and I'm telling you
: >first off....
: I will try to use this line in my next paper: "What I wrote is right,
: because I'm an engineer." Better yet, if you still take classes, try
: it on your exams.
I already apologized for this outburst.
: >question are called MOMENTUM (P=M*V) and IMPULSE (Change in P/time). Hitting
: >an oncoming car of equal mass and equal velocity will result in FAR (double,
: >actually) greater impulse and thus be a lot more violent than hitting a
: >stationary wall at that speed.
: In both cases, the momentum is going from (50mph)*M to 0 in the same
: fashion, ie. by the front bumper coming to a sudden stop. Therefore,
: the impulses are the same.
This is true, however, the amounts of energy absorbed in either situation are
different if you include a wall that can break (ie a real one) and a car that
can crush (ie a real one).
: >If you still don't believe me, get another degree.
: What do you suggest, O wise one?
I apologized for that one, too.
Chuck Jackson
the same injuries if he hit a block wall at 100 mph vs hitting another
vehicle (same size and weight) head on at 50 mph.
You've probably heard the old expression...
"Wow! those cars hit head on at 50 mph?....that's like hitting a block
wall at 100 mph!"
I disagree. The amount of force on your body exerted when you go from
50 mph to zero is less than the amount of force on your body exerted
when you go from 100 mph to zero. I still contend that the "other" car
has no effect if it is the same quanity of mass. The injuries you would
sustain would be based on "your" bodies deceleration. I do agree that if
the other vehicles weight varied up or down, the results WILL vary.
Hitting a semi at 50 mph is a lot worse than hitting a pillow at 50mph!
Kev
: >
: >: Your focus on impulse and momentum is incorrect. Damage is related to
: >: the amount of energy absorbed by an object. This is an energy problem.
: >
: >Actually, it's both. Magnatude of impulse tells you a lot about damage, and
: Kev really needs to go back and take freshman Physics again.
: In *any* impact, momentum is conserved. Yes, that means that the wall will
: move a bit when the car hits it. In this case, the collision is inelastic,
: so energy is *not* conserved. Note also that momentum is a *vector*
: quantity, so it has both magnitude and direction.
Now a vector quantity depends on your reference frame, right?
: >order for the car to stop, the wall has to stop it. The mass of the wall is
: >used with the momentum of the car to calculate energy absorbed by the wall.
: >To stop the car, the portion of the wall that is knocked out or moved must
: >be equal to or greater than the mass of the car. In all likelyhood, that mass
: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
: Equal or *greater*. Make note of this. Now watch as he contradicts
: himself:
I didn't contradict myself. I said in order to stop the car, the mass of bricks
knocked out needs to be nearly equal (subtract the force it took to break the
bricks) or greater than the mass of the car. The next paragrapgh is saying
that in reality, the car cannot knock out a mass greater than itself, so it
needs to be smaller. I'm saying that the mass of the bricks is going to be
somewhere near the mass of the car.
: >is going to be very close to that of the car (as the car can only smash a mass
: >equal to or less than its own). Thus, we can deduce that the wall will absorb
: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
: Um, where do you get this? If I walk into a brick wall, am I only
: going to knock out 170lbs of bricks? I don't think so.
No, because you aren't going to break the wall (unless you're a hell of a fast
walker).
: >close to the same amount of energy that the car does. Therefore, if the wall
: >is stationary, the car needs to be moving twice as fast to achieve the same
: >energy absorbtion by the automobile.
: No. I repeat, No. The wall in this case can be modeled as an a car of
: infinite mass. The relevant equations are:
: Momentum: P = m*v
: Force: F = m*a = dP/dt (the first-order derivative of momentum with
: respect to time)
: Energy: K = 1/2 m*v^2
: In the original scenario, if two cars of equal mass traveling at equal
: speeds collide, the momentum of each goes to zero. This is the *same* as
: if one car hits an immovable wall.
Missed point: If p = m*v is going to be a relevent equation, the wall cannot
be immovable. Momentum is conserved (unless you have a wall of infinite mass,
which would severely alter the problem from real life, now wouldn't it?).
Kev
: Kev <gt4...@acmey.gatech.edu> wrote:
: >
: > The approach velocity of car one relative to car two (if both are
: > travelling 50 mph) is 100mph. Now, >to keep total momentum the
: > same, stop car one and accelerate car two to 100mph.
: That doesn't keep the the total momentum the same! You either don't
: know that momentum is a vector quantity or you can't do very simple
: algebra. You have to go back to freshman physics.
The direction of a vector depends on your reference frame.
: > When car two strikes car one, car two stops completely, transferring
: > all of its momentum to car one. This change in momentum for car two
: > is the same as the change in momentum if car two were moving at
: > 100mph towards a stationary wall. It goes from 100mph to zero mph.
: > P=MV, and if we consider equal impulse time.....
: > Your little paragraph really screwed with my head.
: You mean the paragraph screwed with your little head.
: > Thanks for making me think.
: You really need to do a lot more of it. The contrast between the
: arrogant "your full of shit" post and your complete inaptitude
: displayed here is truly farcical.
I already apologized for my full of shit post, but fuck you anyway.
Yaakov Eisenberg
Humberto Fernando Garcia Figueiredo <fern...@cfn.ist.utl.pt> wrote:
*
* >I agree that, in this problem, it is more useful to deal with momentum than
* >with kinetic energy. Since the masses of the cars and brick walls aren't
* >changing, we can think of momentum as proportional to velocity and impulse
* >as proportional to acceleration. Reread the message that you responded to;
* >it's exactly right. The deceleration in both cases is identical, since
* >in both cases, the very front of the car stops instantly, with the rest
* >of the car stopping after a bit of crunching. What causes the front to stop
* >instantly is irrelevant.
*
* Well, you shoud'nt agree. You see, you are dealing with a perfect wall, wich
* means infinite mass. Therefore you can not deal with momentum. This is a
* concept experience. Do it with energy.
* Even if you consider an almost perfect wall (Big mass compared to the car
* mass) you'll get nowhere.
*
* Humberto
I'm not sure what you mean. Can you elaborate?
Why can't you deal with momentum if the wall's mass is infinite? It just
means that the car will come to a complete stop, without violating conservation
of momentum. I assume this is the essential characteristic of hypothetical
brick walls, i.e., that they can stop cars instantly without themselves moving
at all. Since this is exactly what happens when two cars of equal momentum
hit head-on (each one stops instantly, besides the crumpling), it is correct
to say that the two situations are equivalent.
Another point about dealing with energy dissipated: Why is energy dissipated a
good measure of the severity of a crash? Isn't it true that a car going at
a given speed dissipates the same amount of energy whether it stops by
crashing into a brick wall or by having its brakes gently applied?
--
-Yaakov Eisenberg (yaa...@cc.gatech.edu)
Adour Vahe Kabakian
In article <4lo1th$f...@catapult.gatech.edu>,
Kev <gt4...@acmey.gatech.edu> wrote:
>Adour Vahe Kabakian (ad...@leland.Stanford.EDU) wrote:
>: In article <4lis90$d...@catapult.gatech.edu>,
>: Kev <gt4...@acmey.gatech.edu> wrote:
>: >
>: > The approach velocity of car one relative to car two (if both are
>: > travelling 50 mph) is 100mph. Now, >to keep total momentum the
>: > same, stop car one and accelerate car two to 100mph.
>
>: That doesn't keep the the total momentum the same! You either don't
>: know that momentum is a vector quantity or you can't do very simple
>: algebra. You have to go back to freshman physics.
>
>The direction of a vector depends on your reference frame.
Wow, are we again repeating "big words" without having a clue as to
what they mean? Regardless of what Newtonian reference one uses, the
momentum of the system formed by the two cars is not conserved in
your analogy.
>I already apologized for my full of shit post, but fuck you anyway.
Are we "upset" and drunk again? You had already entertained us enough
by displaying your complete and utter lack of understanding for the
most basic concepts of Newtonian mechanics and vectors. But you had
to go further and show that you were also too stupid to read the dates
of the posts you are referring to, and realize that my article was
posted hours before you had "already" apologized. But I understand
that frames of references pose a particularly difficult challenge to
your cognitive abilities. It seems like time frames are no exception.
You did well as a buffoon. You may retire now.
-adour
Prof Calculus
ho...@owens.ridgecrest.ca.us (S.E. Homer) writes:
>Kev <gt4...@acmey.gatech.edu> wrote:
>> My congrats to really making me think. This analogy made me questoion
>> what makes perfect sense to me. Here's another one. The approach
>> velocity of car one relative to car two (if both are travelling 50 mph) is
>> 100mph. Now, to keep total momentum the same, stop car one and accelerate
>> car two to 100mph. When car two strikes car one, car two stops completely,
>> transferring all of its momentum to car one. This change in momentum for
>> car two is the same as the change in momentum if car two were moving at
>> 100mph towards a stationary wall. It goes from 100mph to zero mph. P=MV,
>> and if we consider equal impulse time, we realize that car two striking
>> the wall at a hundred is the same as car one and car two striking each
>> other at fifty. Your little paragraph really screwed with my head.
>> Thanks for making me think.
>But what you're describing is a situation where KE is conserved which
>means that no energy has been dissipated. i.e. No harm done. That's not
>the case in a car wreck.
>If you assume the worst (most damage) in terms of energy dissipation
>then you minimize KE while conserving P. You end up with both vehicles
>traveling in the same direction at 50 mph, again assuming identical
>vehicles. In terms of damage (the energy dissipated by the vehicles),
>this is the same as a perfectly inelastic 50 mph headon between two
>identical vehicles.
>Your focus on impulse and momentum is incorrect. Damage is related to
>the amount of energy absorbed by an object. This is an energy problem.
>--
>S.E. Homer
I think you are the only poster who makes sense. I don't think
momentum has any place in this discussion! The total momentum of
both cars before and after the collision is IDENTICAL
whether it is an elastic or inelastic collision. This is because
momentum is a vector (and we are assuming 0 deg angle of incidence).
In this case with identical cars and same speeds (velocities are
opposite though), total momentum is 0 before and after the
collision. So what information does momentum give us ? Nothing.
If the cars were of different velocities and mass it would be
a nice tool to figure out post-collision velocty(ies) depending
on whether it is an elastic/inelastic collision.
Here, we have an inelastic collision between 2 cars running head-on
versus one car against a infinite mass wall at identical "speeds"
(50 mph). These two from an energy point of view are identical.
Same damage. Here K.E. energy is converted to heat (all of it).
Compare the energies of the bodies involved (see my last post),
and assume an infinte wall will not absorb energy. If the last
part is not true, then the wall impact is less severe than the
2 cars crashing head-on. If you plan to bring real-world into
this, I can bring other factors in as well!
End of story (hopefully) :-).
Sriram