Time of travel along different paths

[brotherbobby]

TL;DR

Refer to diagram below.
Two balls are released simultaneously from rest at the left end of equal-length tracks A and B as shown. Which ball reaches the end of its track first?
While I have an answer to the question, namely $\text{Ball 2}$, the reasoning is surprisingly tricky, given the elementary nature of the problem. Namely, that ball 2 travels a "significant distance in between" at a higher speed than ball 1 does for the same distance. I hope this is correct.

Problem statement : I copy and paste the problem as it appears in the text, Conceptual Physics (12th Edition) by Hewitt, P.

Two balls are released simultaneously from rest at the
left end of equal-length tracks A and B as shown. Which
ball reaches the end of its track first?

Attempt : The answer for velocity (or speed) is easier here. Both the balls have the same velocity upon arrival : $v_1 = v_2$. To argue for this however, I can refer to energy. Since both balls start and end at the same height, they both have the same loss of potential energies and therefore gain equal kinetic energies. Can there be a "kinematic" argument for the answer? Gaining velocity $\delta v=g\sin\theta\Delta t$ for ball 2 between points C' and D' is exactly counterbalanced by its equal loss between points E' and F' (see figure drawn again below). This results in equal speeds at the end. (1)

As for times, which is what the question is about, the answer itself is ambiguous, let alone the argument.

Both balls 1 and 2 have the same speeds at C and C' as shown. Clearly $v_1(\text D) < v_2 (\text D')$. Moving uniformly again, this implies the time of travel $t_1(\text {DE}) > t_2 (\text {D'E'})$. Since velocities equate at the points F and F', so do times of travel following those points till P.

Hence, ball 2 "saves" time by moving at a faster speed between points D'E' than ball 1 does between the same two of its points, DE. Thus for the total time of flight $T$, we can say that $\boxed{T_1 > T_2\;\Rightarrow\text{Ball 2 reaches the end of the track first}}$.

Doubt : The shorter time of travel of ball 2 rests crucially on the length of the track D'E'. The shorter the track, the closer the two times. However short the track though, it seems always true that $T_1>T_2$, since even if D'E' became infinitesimal, ball 2 would move earlier along incline C'D' at a higher speed than ball 1 for the same length of track CD horizontally. Is this correct? If it is, then the length of the track DE becomes unimportant. (2)

I hope I am correct at points (1) and (2) above, shown in blue, for the arguments regarding final speeds and journey times. Thanks for your time.

 

[anuttarasammyak]

Let consuming time to infinitesimal part of pass $dl$ be $dt$ with speed $v$.
$$dt=\frac{dl}{v}$$
v depends on where on the path the ball is, $v=v(l)$.
Integrating it along the path L, consuming time to go through the path L is
$$T =\int_L \frac{dl}{v}$$
Conservation of energy
$$E=mgz+\frac{1}{2}mv^2=const.$$
where m is mass of ball rotational energy of which is neglected.
$$v=\sqrt{\frac{2}{m}(E-mgz)}$$
Thus
$$T =\int_L \frac{dl}{\sqrt{\frac{2}{m}(E-mgz)}}$$
where $z=z(l)$ is height of the path .

For short journey time, the additional slopes C'D' and E'F' are disadvntageous due to longer path. However, the ball goes horizontal path D'E' in shorter time than it goes DE thanks to greater speed gained by going down the slope C'D'. The integral gives us their total.

Let us introduce y=-z , x be coordinate of horizontal direction and let E=0 for starting point x=0, y=0.
As $$dl=\sqrt{dx^2+dy^2}=\sqrt{1+y'^2}dx$$
$$T =\int_0^X \frac{\sqrt{1+y'^2}}{\sqrt{2gy}} dx$$
It is a familiar formula to get function y(x) on which journey time is minimum i.e., to solve "brachistochrone problem" by variation method.