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Location:
Perth
Registered:
June 2002
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Re: Physics problem.
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Sat, 26 March 2005 02:59
 
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| FWDboy wrote on Sat, 26 March 2005 09:47 |
And the next person who says 'I wanted to consult my Year 12 physics book' or 'I passed year 11 physics' please be careful with you 'argument'.
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I hope this is not directed at me.
If so what is wrong with my argument?
And why wouldn't I consult my year 12 Physics book, it is, after all, what we are talking about here.
After reading back through your difficult to understand posts, I still am not quite sure what you are trying to say.
Are you saying that hitting a wall at 100 km/h is the same as 2 cars, each traveling at 50km/h, having a head on collision? This is after all what the original question was about.
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Location:
South Australia
Registered:
July 2002
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Re: Physics problem.
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Sat, 26 March 2005 03:21
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Force is linearly related to the mass and the acceleration - correct? F = ma -> one of Newtons basic formula (I think it's Newton's)
Acceleration is a result of the rate of change of velocity, not the actual velocity itself. In this case, one car will change it's velocity from 50km/h to 0km/h in some period of time - that period of time is determined by the rate at which the car's body can convert the kinetic energy of it's travel into heat/sound/whatever via crumpling.
So the rate at which the car's body can convert kinetic energy is the key factor in the 'force' related to an accident. Thus the total kinetic energy is what is important in determining an accident's destructiveness...as the more kinetic energy there is, the more the car has to absorb.
We can assume for this example that loss of mass (ie bits of glass flying off etc) is negligible and won't affect the argument/result. So we are basically focussing on the acceleration and thus the kinetic energy of the problem.
The formula for kinetic energy is KE = 0.5 * m * v^2
Or as I stated above, kinetic energy is linearly related to mass, and linearly related to the square of velocity.
In the two car collision, both cars have equal mass, and equal velocity. Whatever energy both cars contribute to the accident (ie kinetic) it will be divided between the two cars evenly (assuming that both cars are identical, generally even identical cars won't crumple exactly the same in any accident - but this is all theoretical anyway). So for one car, it will absorb it's own value of kinetic energy and convert it to heat via crumpling etc.
As it absorbs it's own amount of kinetic energy, it accelerates (or 'decelerates') from it's original speed to stationary based solely on how it converts the energy - and how much energy it has to convert, and because mass isn't a factor here (it is 'constant') we can assume that the force applied to the one car is solely related to it's own original kinetic energy.
If you were to look at the other car, the same result would apply, it will absorb kinetic energy equivalent to it's original value of kinetic energy, and come to a rest. The force it experiences will also be related to whatever it's original kinetic energy was.
Now - we will look at a car hitting a wall. The wall has no kinetic energy at all, and will not absorb kinetic energy upon impact ie no deformation to it's structure, ho heat/friction generated. It will basically sit there and just reflect whatever energy is thrown at it because it (in this theoretical example) doesn't change it's state in any way during the collision.
So whatever happens to the car (whatever energy it absorbs) must come from it's own stockpile of kinetic energy. The impact causes a force on the wall which is reflected back onto the car and it (the car) will absorb the energy related to that force (via crumpling). So you can see that whatever crumpling occurs to the car is done by it's own original kinetic energy. A similar argument follows on from the previous accident, the force of the accident will be determined mainly by the amount of original kinetic energy and (of course) the rate at which the car absorbs it.
Now if you wanted to have equivalence between the two accident set-ups (ie car-car, and car-wall) in terms of 'destructiveness' you would merely need to work out the amount of kinetic energy you want the car to absorb (and convert). In terms of kinetic energy only ONE car is involved in the equation of the car-wall accident, so we have half the mass of the two-car collision.
The mass of the wall? Not important, it's velocity is zero - thus it has no kinetic energy.
So if you crunch the numbers, you will find that you have to times the velocity by 1.404 (the square root of two) in order to come up with an equivalent level of kinetic energy for an accident with half the mass.
Now - finally - you are mostly going wrong with your interpretation of relative velocity. If you are travelling at 50km/h and someone is coming in the opposite direction also travelling at 50km/h, then yes, you are approaching them at 100km/h, but they are also approaching you at 100km/h! So when you 'collide' you are trying to say that there is suddenly 2 * accidents that are 100km/h? Definitely not! There is only 2 * 50km/h accidents.
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Location:
South Australia
Registered:
July 2002
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Re: Physics problem.
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Sat, 26 March 2005 03:29
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hahah - the only reason my posts are difficult to understand is cause I think I'm trying to explain something that usually takes a few weeks to teach at high school level  Whereas just quoting some formula, without explanation - is a very clear statement. But it doesn't mean alot without an explanation of what it is supposed to mean in relation to the problem at hand.
And yes i realise my communication skills aren't top notch and it takes ages to actually figure out what the hell I'm trying to say  Sorry 'bout that! I'm only trying to help 
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Club Member
Location:
sydney
Registered:
May 2002
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Re: Physics problem.
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Sat, 26 March 2005 03:29
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very nicely worded, but i think all this was summed up back at page one
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PE= 0.5 x mass x velocity^2
so lets take the first scenario of two cars traveling at 50kmh towards each other and assume they both weight 1000kg
PE= 0.5 x 1000 x 50^2 x 2
PE= 500 x 2500 x 2
PE= 1.25M x 2
PE= 2.5Mj of energy
scenario 2 the same car of 1000kg travelling at 100kmh to a stationary object that weighs 1000kg
PE= (0.5 x 1000 x 100^2) + (0.5 x 1000 x 0^2)
PE= (500 x 10000) + (500 x 0)
PE= 5Mj + 0j
PE= 5Mj of energy
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please by all means carry on the banter between the two of you as im finding it quite interesting to read the two opinions and arguments presented for each 
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Location:
South Australia
Registered:
July 2002
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Re: Physics problem.
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Sat, 26 March 2005 03:33
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yeah - those equations (provided the background knowledge that go with them is at hand) pretty much show what is going on. 2 * the speed = 4 * the energy.
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Location:
Coffs Harbour, NSW
Registered:
November 2004
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Re: Physics problem.
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Sat, 26 March 2005 03:41
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ok, now im sure we're actually on different levels:
you're talking about energy, im talking about relative velocity
we're trying to argue about two different things 
im basically saying that the two accidents will have approximately the same collective damage
but let me get this straight: you're saying occupants will experience the same forces, right?
well that seems to make sense to me, in real life, as there are two cars two absorb twice the amount of energy put into the collision
we're talking about different perspectives, and frankly, i cant be bothered any more 
go read this thread, it will offer some relief
http://forums.toymods.org.au/index.php?t=msg&t h=63703&rid=8333&S=2a78b8e670d6681c37b1f0c f983e140c&pl_view=&start=0#msg_589354
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Location:
South Australia
Registered:
July 2002
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Re: Physics problem.
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Sat, 26 March 2005 04:13
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Look - KINETIC ENERGY IS RELATED TO THE SQUARE OF VELOCITY! IF YOU DOUBLE THE SPEED, THEN YOU QUADRUPLE THE ENERGY, WHICH MEANS IT'S LIKE HAVING FOUR CARS COLLIDE AT 50KM/H, NOT TWO CARS! IT IS ALL ABOUT THE ENERGY AND NOTHING TO DO WITH 'RELATIVE VELOCITY'. IF YOU TAKE ONE CAR AND SMASH IT INTO A WALL WITH 1.414 TIMES THE VELOCITY OF SMASHING TWO CARS TOGETHER HEAD ON (BOTH TRAVELLING AT THE SAME MAGNITUDE OF VELOCITY, IN OPPOSITE DIRECTIONS) THEN YOU WILL HAVE EQUIVALENT 'FORCES'. EXAMPLE, ONE CAR IS GOING 50KM/H TO THE RIGHT, ONE IS GOING 50KM/H TO THE LEFT, WHEN THEY COLLIDE IT IS EQUIVALENT ENERGY TO ONE CAR HITTING A WALL AT 70.7KM/H HOUR.
There I'm done now Back to my little shell
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Location:
Coffs Harbour, NSW
Registered:
November 2004
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Re: Physics problem.
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Sat, 26 March 2005 04:24
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| FWDboy wrote on Sat, 26 March 2005 15:13 |
Look - KINETIC ENERGY IS RELATED TO THE SQUARE OF VELOCITY! IF YOU DOUBLE THE SPEED, THEN YOU QUADRUPLE THE ENERGY, WHICH MEANS IT'S LIKE HAVING FOUR CARS COLLIDE AT 50KM/H, NOT TWO CARS! IT IS ALL ABOUT THE ENERGY AND NOTHING TO DO WITH 'RELATIVE VELOCITY'. IF YOU TAKE ONE CAR AND SMASH IT INTO A WALL WITH 1.414 TIMES THE VELOCITY OF SMASHING TWO CARS TOGETHER HEAD ON (BOTH TRAVELLING AT THE SAME MAGNITUDE OF VELOCITY, IN OPPOSITE DIRECTIONS) THEN YOU WILL HAVE EQUIVALENT 'FORCES'. EXAMPLE, ONE CAR IS GOING 50KM/H TO THE RIGHT, ONE IS GOING 50KM/H TO THE LEFT, WHEN THEY COLLIDE IT IS EQUIVALENT ENERGY TO ONE CAR HITTING A WALL AT 70.7KM/H HOUR.
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can you see the button on your keyboard that says "caps lock"?
PRESS IT
haha, j/k
as i said, i cant be bothered, feel free to take up the argument with someone else tho. any takers?
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Location:
South Australia
Registered:
July 2002
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Re: Physics problem.
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Sat, 26 March 2005 04:47
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Okay we want 'equivalent' two-car and car-car accidents, and we both agree that kinetic energy is what determines an accidents 'strength'.
Both cars have mass 'm1' and both cars are travelling at velocity 'v1' relative to the road - and are heading towards each other. And the car hitting the wall has mass 'm1' and velocity 'v2'
So the car-car accident has KE1 = 0.5 * (m1 + m1) * (v1)^2
The car-wall accident has KE2 (== KE1) = 0.5 * (m1) * (v2)^2
So
0.5 * (m1 + m1) * (v1)^2 = 0.5 * (m1) * (v2)^2
Cancel out the 0.5
(2 * m1) * (v1)^2 = m1 * (v2)^2
Now take out the factor of m1
2 * (v1)^2 = (v2)^2
Now take the square root of both sides
sqr(2 * (v1)^2) = sqr((v2)^2)
Square root and square cancel out on the RHS, and on the left hand side we separate it into the product of two sqr roots
so
sqr(2) * sqr((v1)^2) = v2
And using the same square root & square cancelling out we get
sqr(2) * v1 = v2
Ie
1.414 * v1 = v2
Ie in the one car accident it must travel 1.414 times the speed of one of the cars in the two car accident to have an accident with overall equiavlent energy.
Anyway, whatever, I'm sure that you understand what's going on by now but you are too lazy to re-arrange the KE formula and substitute stuff around to prove it to yourself. Now I'm going to go do some fun stuff with my Easter saturday instead of whoring on various forums
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Club Member
Location:
sydney
Registered:
May 2002
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Location:
Coffs Harbour, NSW
Registered:
November 2004
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Re: Physics problem.
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Sat, 26 March 2005 05:26
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for the third time:
I CANT BE BOTHERED ANY MORE
I DONT GIVE A FUCK
IM OVER IT
there, got it now?
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Location:
Coffs Harbour, NSW
Registered:
November 2004
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Re: Physics problem.
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Sat, 26 March 2005 06:06
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haha, just found this pic
i officially surrender from this argument, turning the victory over to you
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Location:
Launceston
Registered:
March 2005
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Re: Physics problem.
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Sat, 26 March 2005 06:52
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is this like if a plane is flying at 300kph and the wind speed(directly head on) is 300kph then wouldnt the plane be theoreticly flying at 0kph ground speed?
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Location:
Coffs Harbour, NSW
Registered:
November 2004
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Re: Physics problem.
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Sat, 26 March 2005 06:56
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| tom210 wrote on Sat, 26 March 2005 17:52 |
is this like if a plane is flying at 300kph and the wind speed(directly head on) is 300kph then wouldnt the plane be theoreticly flying at 0kph ground speed?
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ok, since this doesnt relate (directly) to the huge argument above, ill reply:
depends, what is the plane flying at 300km/h (which is highly unlikely, at that speed it would drop out of the sky ) relative to? the wind? the ground? a passing bird?
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Location:
South Australia
Registered:
July 2002
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Re: Physics problem.
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Sat, 26 March 2005 07:02
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It always sucks when you go to all this effort to explain something to someone, and instead of them figuring out that they are wrong and trying to learn from the whole thing, they just bury their head in the sand and say "they don't give a fuck"
Well it's both your and my loss
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Location:
Coffs Harbour, NSW
Registered:
November 2004
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Re: Physics problem.
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Sat, 26 March 2005 07:05
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| FWDboy wrote on Sat, 26 March 2005 18:02 |
It always sucks when you go to all this effort to explain something to someone, and instead of them figuring out that they are wrong and trying to learn from the whole thing, they just bury their head in the sand and say "they don't give a fuck"
Well it's both your and my loss
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hey, dont think you're the only one who feels like that (without the "dont give a fuck bit, of course )
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Location:
Launceston
Registered:
March 2005
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Re: Physics problem.
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Sat, 26 March 2005 07:19
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the air speed sorry i forgot to mention
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Location:
Perth
Registered:
August 2003
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Location:
Launceston
Registered:
March 2005
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Re: Physics problem.
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Sat, 26 March 2005 07:28
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ok then a car moving at 50kph and a brick wall at 50kph would that be like 2 cars both traveling at 50ks colliding?
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Location:
Coffs Harbour, NSW
Registered:
November 2004
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Re: Physics problem.
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Sat, 26 March 2005 07:39
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| tom210 wrote on Sat, 26 March 2005 18:19 |
the air speed sorry i forgot to mention
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so the plane is moving at 300km/h relative to the 300km/h wind? then yes, the plane would be moving at 0km/h relative to the ground
obviously thats impossible, but anyway
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ok then a car moving at 50kph and a brick wall at 50kph would that be like 2 cars both traveling at 50ks colliding?
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well, in terms of collective destruction, i say yes, but others beg to differ  oh well...
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Location:
Brisbane
Registered:
May 2002
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Re: Physics problem.
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Sat, 26 March 2005 09:47
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| tom210 wrote on Sat, 26 March 2005 17:28 |
ok then a car moving at 50kph and a brick wall at 50kph would that be like 2 cars both traveling at 50ks colliding?
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The brick wall doesn't have to be moving; a stationary wall will exert the same amount of force on the car as another car travelling at 50km/h.
"For every action, there is an equal and opposite reaction."
I think this is the concept Mr Skellator fails to grasp. It sounds weird at first, but that's how Newtonian physics works. In fact it's Newton's Third Law of Motion, something a high school physics student should be familiar with?
Now I think about it, I seem to recall this very problem being used in high school to explain the third law of motion. Just about everyone in the class got it wrong (including me I expect). It's been about 11 years since high school though, so don't blame me if my recollection of the specifics is a bit hazy.
Anyway, here's a good read to refresh your memory:
http://www.glenbrook.k12.il.us/gbssci/phys/Class/n ewtlaws/u2l4a.html
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Location:
Coffs Harbour, NSW
Registered:
November 2004
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Re: Physics problem.
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Sat, 26 March 2005 10:10
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| Norbie wrote on Sat, 26 March 2005 20:47 |
a stationary wall will exert the same amount of force on the car as another car travelling at 50km/h.
"For every action, there is an equal and opposite reaction."
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*exasperated sigh* not again
yes, you are correct there, im talking about the collective damage
so would you agree with me that the collective damage will be the same for the two car 50km/h collision as for the one car/wall 100km/h collision? please? im getting sick of this
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Location:
Brisbane
Registered:
May 2002
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Re: Physics problem.
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Sat, 26 March 2005 10:12
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Collective damage? WTF is that supposed to mean?
This was the original problem:
| Quote: |
Do 2 cars colliding each travelling at 50Km/H have the same force as 1 car hitting a wall at 100Km/H??
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Location:
Coffs Harbour, NSW
Registered:
November 2004
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Re: Physics problem.
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Sat, 26 March 2005 10:21
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| Norbie wrote on Sat, 26 March 2005 21:12 |
Collective damage? WTF is that supposed to mean?
This was the original problem:
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Do 2 cars colliding each travelling at 50Km/H have the same force as 1 car hitting a wall at 100Km/H??
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ok, i mean the damage done to both cars in the two car 50km/h collision will be the same as the damage done to the one car in the car/wall collision
i interpreted the original question as:
do both collisions result in the same amount of damage
and that's what ive been arguing all along...
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Location:
Perth, WA
Registered:
December 2004
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Re: Physics problem.
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Sat, 26 March 2005 11:37
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the energy expelled by both collisions (sound, heat.. etc. etc.) will be the same.
and yes, the forces will be the same, it cant be different because the wall isnt moving, yet the other car is.
i did physics in year 11, dropped out (of the class), and i know the answer, sheet you people make things compliacated.
Eldar.O.
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Location:
Launceston
Registered:
March 2005
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Re: Physics problem.
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Sat, 26 March 2005 12:46
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My moving brick wall post was to just mix things up a bit. and Corona RT142 "and even deceleration cannot be calcualted as you don't know how long it takes" its called an accelerometer (i cant spell) but it measures g force over time (i think) like a distance over time measures speed g's over time measure acceleration or negative acceleration.
lets say the cars weighed the same.
we'll take out our newtons cradle (swinging balls on a string) grab the first ball, let go... the force travels through the others to give what appears to be exerting the same force onto the last ball... yes..... well now you have to grab the first and last balls drop them and they will either appear to bounce off the next ball or stop dead. thus two objects of the same mass and will exert the same amount of force on one another.
and i like the equation of skellator
v1 + - v2
and v2 = -50 (opposite direction)
so v1 + - v2
= 50 + - -50
= 100
but also the way you had it before
50 + - 50 = 0 it has no leftover force therefore the forces were the same
50 + - 70 = -20
i'd say someone copped more of a beating.
i know that the original post was talking about velocity but this sorta helps i hope.
Professor Tom
thats the most constructive post all day (that being my first day)...
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Location:
South Australia
Registered:
July 2002
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Re: Physics problem.
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Sat, 26 March 2005 12:53
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100km/h is not equivalent.
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Location:
Brisbane
Registered:
May 2002
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Re: Physics problem.
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Sun, 27 March 2005 02:02
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| EldarO wrote on Sat, 26 March 2005 21:37 |
and yes, the forces will be the same, it cant be different because the wall isnt moving, yet the other car is.
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No worries, let's completely ignore Newton's third law. He didn't know what he was on about anyway.
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i did physics in year 11, dropped out (of the class), and i know the answer
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Location:
Coffs Harbour, NSW
Registered:
November 2004
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Re: Physics problem.
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Sun, 27 March 2005 02:36
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can you tell me if you agree with me now? cos i want to forget about this thread
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Location:
South Australia
Registered:
July 2002
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Re: Physics problem.
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Sun, 27 March 2005 03:15
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Who's agreeing on what? Be more specific.
You keep saying that one car hitting a wall at 100km/h is the same destructiveness (overall) as two cars, both travelling at 50km/h in opposite directions, colliding.
This is not the case.
The car hitting the wall would have to be going 70.7km/h for the accidents to have similar levels of overall damage. I've said this at least three times now and argued it in both logic and mathematical terms yet someone keeps saying '100km/h' and it's not me.
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Location:
Canberra
Registered:
August 2003
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Re: Physics problem.
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Tue, 29 March 2005 03:48
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| FWDboy wrote on Sat, 26 March 2005 14:21 |
The formula for kinetic energy is KE = 0.5 * m * v^2
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We have a winner...
That is all I was looking for!! The rest of your answer is correct (and quite good!) but this is the equation proved it for me, and won the argument!
As for the rest of the argument - quoting your qualifications doesn't make you right, empirical proof makes you right. I have a university qualification in physics, but forgot the equation I needed. I could have said i have a uni qualification believe me, but without remembering the proof that is a shallow statement. Hence I asked for help.
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Location:
South Australia
Registered:
July 2002
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Re: Physics problem.
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Tue, 29 March 2005 04:41
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hahah - yeah I'm still retarded though - damn running in special olympics.
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Location:
Coffs Harbour, NSW
Registered:
November 2004
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Re: Physics problem.
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Tue, 29 March 2005 05:50
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| FWDboy wrote on Tue, 29 March 2005 14:41 |
hahah - yeah I'm still retarded though - damn running in special olympics.
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THANKYOU for not being a dickhead about it
i still think we were arguing different things, though
i asked my physics teacher today and he said i was right, so we must of just had a misunderstanding
congratulations on the win, tho!
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Location:
South Australia
Registered:
July 2002
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Re: Physics problem.
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Tue, 29 March 2005 12:07
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Haha - gimme your physics teachers number! I will set him straight
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Location:
Perth, WA
Registered:
December 2004
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Re: Physics problem.
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Tue, 29 March 2005 16:41
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| Norbie wrote on Sun, 27 March 2005 10:02 |
| EldarO wrote on Sat, 26 March 2005 21:37 |
and yes, the forces will be the same, it cant be different because the wall isnt moving, yet the other car is.
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No worries, let's completely ignore Newton's third law. He didn't know what he was on about anyway.
| Quote: |
i did physics in year 11, dropped out (of the class), and i know the answer
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objects in motion want to stay in motion, basically.
and objects will continue doing their own thing unless acted on by an external force.
so, if a wall is stationary, obviously it wants to stay stationary this force is called intertia.
both 50km/h cars also have the same inertia. being that they are both moving at the same speed.
the wall is not relevant at all in the second instand, seeing as the 100 km/h has twice the inertia.
ass.
edit: i cant be fucked editing my spelling,its too late.
[Updated on: Tue, 29 March 2005 16:43]
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Location:
Coffs Harbour, NSW
Registered:
November 2004
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Re: Physics problem.
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Tue, 29 March 2005 21:05
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| FWDboy wrote on Tue, 29 March 2005 22:07 |
Haha - gimme your physics teachers number! I will set him straight
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haha, no you wont, he was in the army for a few years, knows a few tricks apparently
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Location:
NSW Engadine
Registered:
June 2003
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Re: Physics problem.
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Tue, 29 March 2005 21:20
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If your breaking the speed of sound then you hit teh wall. you die...
Of course your hitting at the same speed, just you'll have two crumple zones when the cars collide, but with 1 car into a wall you'll only have the crumple zone of your car...
So you'll get less injured when your colliding into a car...
Simple
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Location:
Adelaide
Registered:
May 2002
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Location:
South Australia
Registered:
July 2002
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Re: Physics problem.
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Wed, 30 March 2005 00:42
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Skellator -> You are asking about how much overall damage is done. Ie in the two car accident, twice as much damage is done (as a total of two cars get trashed) compared to one car hitting a wall (all cars travelling at the same speed). Then he's saying that if one car were to hit a wall at double the speed, it would do an equivalent amount of damage (ie twice the damage to one car) as the two-car accident.
Is this what you are saying? If not, please clarify as to what you are trying to argue!
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Location:
Coffs Harbour, NSW
Registered:
November 2004
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Re: Physics problem.
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Wed, 30 March 2005 04:40
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| FWDboy wrote on Wed, 30 March 2005 10:42 |
Skellator -> You are asking about how much overall damage is done. Ie in the two car accident, twice as much damage is done (as a total of two cars get trashed) compared to one car hitting a wall (all cars travelling at the same speed). Then he's saying that if one car were to hit a wall at double the speed, it would do an equivalent amount of damage (ie twice the damage to one car) as the two-car accident.
Is this what you are saying? If not, please clarify as to what you are trying to argue!
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yeah, im saying that in the two car collision X amount of damage is done
and in the car/wall collision, X amount of damage is also done
hope that clears it up a bit
and robertox, i accidentally said that 50 + - 50 = 100, but then someone pointed it out, so i corrected myself, saying that 50 + - - 50 = 100
anyway, i thought this thread was over
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Location:
Adelaide
Registered:
May 2002
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Re: Physics problem.
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Wed, 30 March 2005 04:49
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| Quote: |
yeah, im saying that in the two car collision X amount of damage is done
and in the car/wall collision, X amount of damage is also done
hope that clears it up a bit
and robertox, i accidentally said that 50 + - 50 = 100, but then someone pointed it out, so i corrected myself, saying that 50 + - - 50 = 100
anyway, i thought this thread was over
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Yeh, I know that 50+--50=100 but I just found it amusing that people would just throw in an extra "-" to 'prove' something
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Location:
Coffs Harbour, NSW
Registered:
November 2004
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Re: Physics problem.
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Wed, 30 March 2005 04:57
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| RobertoX wrote on Wed, 30 March 2005 14:49 |
| Quote: |
yeah, im saying that in the two car collision X amount of damage is done
and in the car/wall collision, X amount of damage is also done
hope that clears it up a bit
and robertox, i accidentally said that 50 + - 50 = 100, but then someone pointed it out, so i corrected myself, saying that 50 + - - 50 = 100
anyway, i thought this thread was over
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Yeh, I know that 50+--50=100 but I just found it amusing that people would just throw in an extra "-" to 'prove' something
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no, you dont get it:
i made an honest mistake!
i didnt simply "throw in an extra" one
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Location:
Adelaide
Registered:
May 2002
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Location:
Coffs Harbour, NSW
Registered:
November 2004
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Re: Physics problem.
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Wed, 30 March 2005 06:35
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| RobertoX wrote on Wed, 30 March 2005 15:40 |
| skellator wrote on Fri, 25 March 2005 20:22 |
| FWDboy wrote on Fri, 25 March 2005 20:10 |
| skellator wrote on Fri, 25 March 2005 19:10 |
50km/h of your car + - 50km/h of the other car (opposite direction) = 100km/h in your direction
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Velocity is a vectors too.... and 50 + -50 = 0 .... not 100!!!
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whoops! i meant, use the formula v1 + - v2
and v2 = -50 (opposite direction)
so v1 + - v2
= 50 + - -50
= 100
there that's better!
but i still cant see why you think like you do
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I'm talking about this
It's wrong because you have just chucked in a "-" in
"v1 + -v2" (it should be "v1 + v2")
... this is wrong for explaining what you are trying to explain because the sum of 2 vectors is simply that, the sum (with no fiddling around with putting in extra bits to make it say what you want it to say).
But anyway I think you are right, this thread has died O.o
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ok, so now i know what you're talking about, but that equation is still correct for relative velocities, which is what i was talking about there
i think you might be thinking about adding vector components to get the final vector, as in calculating the initial velocity of a projectile, for instance?
btw: DIE THREAD, DIE!!
[Updated on: Wed, 30 March 2005 06:41]
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Location:
South Australia
Registered:
July 2002
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Re: Physics problem.
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Wed, 30 March 2005 13:47
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Yeah this thread should die - ppl can make their own conclusions about what they want to 'believe'.
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