This explanation seems wrong to me. Please let me know if this reasoning is better. You have to excuse me, it’s been light years since I had physics so I don’t remember all of the nomenclature and terms.
The flaw is not recognizing this for what it is – a collision problem. At the point when the stack is taken up, we’ve got two objects, each with their own momentums, the product of their masses and velocities. The sum of the momentums must remain the same before and after the collision.
For the moment, let’s simplify the problem by saying that the masses are equal. Afterwards we can look at what happens when they are unequal.
Let’s call the falling object mass 1 and the stationary object mass 2. At the time of the collision, M1V1 + M2V2 must be equal to the sum after the collision M1V1’ + M2V2’
Since M2 has zero velocity before the collusion, V2 = zero. Thus M1V1 = M1V1’ + M2V2’
As one can see from billiard ball collisions, it is possible for one object to transfer energy to another instantaneously. The moving ball will stop and the stationary ball will start moving. This does not violate laws of physics.
The falling object will transfer energy to the stationary object. We can think of our perfectly non-elastic rope and frictionless pulley as a “mass-less” teeter-totter. Drop M1 from a height onto one side of the teeter-totter. The result of the collusion is that M2 will now have a V2’ in the upward (negative) direction. This is similar to our billiard ball collision. If this were on a horizontal plane, then M1 would stop, and actually, it does, however momentarily. Then, with the presence of gravity, it resumes its downward trajectory, but at a reduced speed. This pushes the teeter-totter down. The rising M2 flies above the teeter-totter since V2’ is faster than the moon’s gravity is accelerating M1 downward, and M1 is the driving force on the teeter-totter. M2 fights the downward force of gravity, and eventually stops its upward climb. It then starts its journey back to the surface.
M2 then falls until – and let’s assume for the moment that M1 is still falling and hasn’t bottomed out – it hits up rising teeter-totter (the rope loses slack) and we have a second collision which must be solved by M1V2 + M2V2 = M1V2’ + M2V2’. Note that gravity has been accelerating both objects in the same direction so the net difference in velocities will be less the second go around. Gravity prevents endless oscillations and eventually the teeter-totter is balanced.
Let’s clean up some of the assumptions. Going backward, if M1 has bottomed out, if the collision is elastic, then it will be propelled upward at the same velocity as it had when it hit. This will create slack on the rope, but eventually M2 will catch it and we’re back to the above. If the collision in non-elastic then M2 will catch it that much quicker and the collisions will end sooner.
For objects of non-equal masses, the velocities after the collisions change, but the principles remain the same. The end result is that M2 will be on the moon surface after a little joy ride which depends on how far above the moon tree you drop M1, and if the collisions with the surface are elastic or non-elastic.