@ponderable said
Track B wins (because the length in the depths is run through faster than in Track A)
Ok. Correct! ( unless deepthoughts "second guessing" is justified in some subtle way )
The tracks are identical before and after the second dip in track B. So focus on the common sections of track across the second dip.
Put a coordinate system across with respect to Track A.
x - direction parallel to track A at the flat length, y direction normal to it.
Where the two tracks deviate the initial velocity of both balls is "v" in the x-direction.
Track A maintains a constant velocity "v" over the sections in question.
The ball on track B once on the second slope experiences accelerations in both the y and x directions.
If we first analyze the forces acting on the ball in a coordinate system x' -y' aligned with the slope we find that the magnitude of the Normal Force ( the reaction force perpendicular to the slope ) is given by:
N = m*g*cosΦ
Where Φ is the angle of the slope with respect to the original coordinate x.
Now, I wish to compare velocities with respect to the original coordinate system in the "x" direction. The sum of the forces in x direction is as follows:
ΣF_x = N*sinΦ = m*a_x
m*g*cosΦ*sinΦ = m*a_x
a_x = g *cosΦ *sinΦ
Checking extremes to verify it makes sence let Φ = 0, we expect a_x = 0
a_x = g *cos0 *sin0 = g*1*0 = 0
And at Φ = π/2 we again expect the acceleration in x direction to be zero. ( the ball is in free fall )
a_x = g *cos(π/2) *sin(π/2) = g*0*1 = 0
what we have is that for:
0 ≤ Φ ≤ π/2
0 ≤ N ≤ mg
a_x ≥ 0
As a result, the velocity of the ball on track B in the "x-direction" grows from v to v' = v + δv as it continues down the slope. In otherwords for the simplest comparison the average velocity of the ball on B in the x-direction is greater than the velocity of the ball on track A over the same interval and same direction.
( 2v + δv )/2 > v
Next, the flat section the ball will again experience no acceleration in the x-direction, but it will maintain its velocity of v' = v + δv which is again greater than the velocity of the ball on track A over the same interval and direction.
v' = v + δv > v
Finally, on the ascent the ball will experience the mirror acceleration of the first section. the velocity of the ball on B will decrease from v' to v. Again, the average velocity of the ball on track B is greater than that of the ball on track A over the same interval and direction.
( 2v + δv )/2 > v
So in every section of the track B across the that second "dipped" region ball B experiences a higher average velocity than the ball on track A in the x-direction ( over the same distance in that direction ). So it arrives at the top of the trough with the same velocity as Track A, but it does so in a lesser amount of time.
Ball B wins the race.
I think that is a sufficient explanation, but if Deepthought justifies his "second guessing" with some subtle argument about angular momentum I'll be glad to hear it.