One mass on, one half off, attached by a string

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When mass B is dropped, it swings downward while attached to mass A by a taut string. As mass B falls, the radius of its arc increases, preventing it from reaching the edge of the cliff before mass A does. Mass A, starting L/2 away from the edge, will reach the cliff's edge first due to its direct vertical drop. The reasoning is sound, as the increasing arc radius of mass B means it cannot hit the edge in time. The explanation aligns with the principles of physics regarding motion and tension in the string.
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Homework Statement


Two equal masses (A and B) are attaced by a massless string of length L. Mass A is on a cliff and B is being held even with the cliff but a length L/2 away from the edge. If you drop mass B from this position, would mass B swing down and hit the edge of the cliff before mass A reaches the edge of the cliff?


Homework Equations


No equations are necessary according to the problem



The Attempt at a Solution


I'm just supposed to explain my reasoning for my answer. So my answer goes...

Mass A would reach the edge of the cliff because as mass B falls and pulls on A, the radius of the arc that B would need to make to reach the edge of the cliff would continually increase, thus not allowing B to hit the edge of the cliff.

Does this make sense or am I just making stuff up? Is there a better "logical" explanation that would seem to fit better with physics?
 
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Addition... as for the actual question, mass A is L/2 away from the edge of the cliff. So the string is taut (the string won't stretch as mass B falls either).
 
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