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I am considering a situation where two identical masses are side by side and slide down a hill, ignoring friction. But one mass dips into a trench and comes back up, while the other one simply finishes rolling down the hill and continues at constant velocity. I would like to know which ball first passes a "finish line" somewhere after the trench. I drew a crude diagram to explain:

enter image description here

The way I tried to solve was to assume they have the same speed right before and after the trench, therefore they won't pass each other after the trench, so the one that is ahead once they both are past the trench is automatically the winner. Then I tried to figure out if the added velocity for the mass in the trench is enough to overcome the added distance for travelling in the trench. But I am stuck because I don't know how to formulate this properly mathematically. I considered a special case for the trench being a triangle (ignoring any speed loss from impacting a sharp corner) but the result I came up with is that the winner seems to depend on both the depth of the trench and the height fallen before the trench, which seems wrong to me.

My main question is how to conceptually figure out which mass will win the race? As a secondary question to show mathematically whether or not it depends on the heights, shape of the trench, etc.

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    $\begingroup$ it will depend on the shape. $\endgroup$ – Lelouch Nov 1 '16 at 4:10
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Assuming the crudity of your diagram to be quite large,and my understanding of it to be correct, this is the brachistochrone problem. The one that dips below will make it to the GOAL faster, provided the function of the curve is a particular cycloid in shape. I suggest you read up on variational calculus, and when you're done, read THIS.

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  • $\begingroup$ This is helpful, but I don't fully understand the justifications given on wikipedia for the Lagrangian and "virtual gravity" solutions. In particular, it says "In order to be an isochrone, the Lagrangian must be that of a simple harmonic oscillator: the height of the curve must be proportional to the arclength squared." Why is that the case and how is it related to the SHO? Also, why does it say the "virtual gravity" required for the tautochrone is proportional to the distance remaining to be traveled? What is the physical justification for that? $\endgroup$ – AAC Nov 1 '16 at 5:40
  • $\begingroup$ @JohnForkosh : I would have agreed, but in fact the brachistochrone (cycloid) dips below the lower point if the ratio of horizontal to vertical separation is less than $\pi/2$. (BTW I think you mean "turning point" not "inflection point".) $\endgroup$ – sammy gerbil Nov 3 '16 at 2:34

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