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Subject: A physics teacher begs for his subject back
Armchair Private    9/22/2007 11:23:16 AM
h*tp://www.civitas.org.uk/blog/2007/06/a_physics_teacher_begs_for_his.html Some of you may have seen this elsewhere, it's been around a bit. The teacher argues that physics has become a media studies class at GCSE with pupils taught to debate about issues of the day, based on their opinions. The actual science has been removed. You have to read it to get the full gist, so please do. If this is true it's horrific, with all the Government's talk of a knowledge economy, and the tractor-production-like incessant increase in education results too.
 
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Armchair Private       9/24/2007 4:21:15 PM




Sorry Nichevo... You had ought to read Asymmetrics answer it is correct.



 



From this simple diagram:



 



      V1                                                   V1



__________a                               d_________  



                      \        V2               /



                        \b___________/c



 



 



V1 is the constant Velocity the ball is released at.



V2 is V1 plus the potential energy of a-b (which is equal to the loss of potential energy c-d in this case as friction is NEGLIGIBLE)



 



And of course this is assuming that the increase in speed makes up for the increase in distance.



 



 



 



That's what I SAID!!!


No - v0=v0b=v0a ...then drops, +v to v1b and it covers the low track at v1 then decelerates to v0...


i.e. they both start at v0, b (v0b) accels to v1b along the trough...then reverts to v0b, the same speed as a!


John G was referring to your comment to tissapherenes, which he mis-understood to mean you disagreed with  tissapherenes, when in fact you were just pointing out that tissapherenes was late to the party.

Armchair "let's talk about our problems" Private
 
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EW3    Armchair Private    9/24/2007 6:56:57 PM
 
Still not on board with you on this....

As you said -
 
"In a 'V' shape both balls arrive at the same time as although distance is higher in (b) so is speed."
I agree with this, but the implication of this is that C is correct....
 
Both balls are equal before the dip. 
After the dip both have the same speed and distance. 
So if a V shaped dip is introduced then C is correct.
 
 
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Asymmetric    Replies   9/25/2007 12:23:56 AM
 

(Takes a long deep breath)

To EW3 and John G

There seems to be a little confusion between you 2 about the distance travelled. The increase in distance (scalar quantity) is only on the vertical axis but there is no increase in the vertical displacement (vector quantity) or vertical velocity (vector quantity) therefore they are irrelevant. In other words the net horizontal velocity can only increase with a “dip” in the track if the horizontal displacement between the start and finish remains the same as it does in this case.

Ok just pause to think about this for a minute. From the start to the edge the ball is travelling along then when it gets to the edge of the “dip” its horizontal velocity starts to increase, even when its decelerating coming up the other side it is still going faster than if it hadn’t fallen down only when it reaches to the top of the other side of the dip has it returned to its initial velocity.

Notice that the short-time uses more track than the shortest time. This should help to explain why we don’t use little indentations on the road since you would end up losing more kinetic energy to friction.

Longest-time

___________________________________________________________________

 

Long-time

________                           ____________________________________________
               \                         /
                 \                      /
                   \_________/

 

Short-time

 ________                             ________                           _________________
                 \                         /                 \                        /
                   \                      /                    \                    /   
                    \__________/                       \________/

 

Shortest-time

_                                                                                                                ____
  \                                              
 
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flamingknives       9/25/2007 4:42:03 AM
Smooth curves, like a parabola.

If we assume that the ball is constrained to the rails and that they can be any geometry, then the answer is not always true. Since this experiment takes place in theory space, I can set anything I like as a constant in order to test my theory, so gravity can be constant over the infinite dip.
If that's a problem, and I don't see why it should be, the infinite dip could also be expressed as one that is large compared to the horizontal distance.

So, say the dip is 1 kilometre compared to a horizontal distance of 1 metre. Gravity is essentially the same across 1km of vertical displacement, unless the gravitational field is very small. But if your rail conforms to those specs, I guarantee that the one crossing the dip will not reach the far side before the straight line, unless the initial velocity is very low.

A parabola would constrain the depth of the dip relative to the length and make no alteration to the horizontal velocity. As soon as you start redirecting things and having to constrain the ball to the rail, you alter horizontal velocity.
 
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Asymmetric    flammingknives reply and clarifacation   9/25/2007 7:34:57 AM

Ah I see what you’re implying now, didn’t realise your were altering the scale of the drawing.

Yes if the ball is constrained to the track this would alter it’s allowed horizontal velocity noticeably at high speeds. And yes if you were to make extreme limits, different to the scale shown in my rail track in my pictures you could get different results but to resolve that I could increase g and there is no physical maximum on how big you can make “g” in the universe (unlike an entirely theoretical infinite dip) i.e. it is possible to make it have infinite acceleration on an infinitely dense object i.e. the singularity of a black hole, ignoring relativistic effects as is common for a 14-15 year old group. So you would have to increase g proportionally the more distorted you made the depth to gap ratio. In other words if g is set to a massive number compared to the depth or the initial speed track b) will always win regardless of the dimensions of the parabola unless the parabola was set to infinity in which case the ball would never get to the other side, even if it had an infinite velocity and infinite acceleration. But that’s just one of the anomalies of using a value like infinity in mathematics. i.e. infinity minus infinite is till infinity not zero.

I can see why there was some confusion over this; it’s probably something I should have clarified that I was implying this track was on a macroscopic human scale, since I unfortunately didn’t specify its limits.

 
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Nichevo       9/25/2007 1:45:51 PM
Hmm.  As a wrinkle, reflect that if both rail horizontals are straight then gravity which is curved in Terran reference will speed both of the balls towards the center and slow them coming out...i.e. the altitude wrt surface of the earth is lower in the middle than at the ends, thus gravity is stronger.
 
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Herald1234    The point is...........    9/25/2007 1:58:10 PM
that a gravity influenced accleration is additional to the vector sums here and that you have to change the way you visualize processes to understand what goes on.
 
As long as you look at vector process as lineal and not as directional you make mistakes in mechanics.
 
In other words, sometimes it makes sense to move into a parabola to get from A to B to cut the straight liner bastard off and kill him.
 
Herald
 
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