Can a small weight lift a large one?

Factual Questions
1

Simple physics problem, using inelastic string and frictionless pulleys.

Up on the moon you have a weight in the ground which is attached to the inelastic string, which runs up to the frictionless pulley.

This pulley is attached to a branch of a moon tree way above the ground, but directly above the weight. The other end of the string is another weight at half the mass of the first, which you are holding above the pulley.

You drop that smaller weight, assuming you brake the pulley or the moon tree the small weight should fall till that inelastic string reaches it’s max extention. Now what happens?

Does the small weight just stop dead, or does the large weight do a little hop off the surface before falling back down?

How does this change if we replace the inelastic string with a elastic one?

2

Here’s my take on it, assuming a perfectly frictionless environment.

In dropping the small weight, you’re converting gravitational potential energy into kinetic energy. When the small weight hits the end of its string, it will have some excess kinetic energy, which it will lose by applying force to the string.

This force will have the effect of lifting the big weight until the system’s energy briefly equalizes, with the big weight up off the ground a bit, and the net potential energy of the system equal to the excess kinetic energy introduced by dropping the small weight.

After this, the big weight will succumb to the force of gravity and reconvert its potential energy into kinetic, until it hits the surface of the moon with a thud. What happens next depends on how elastic the collision with the moon is. Assuming a perfectly inelastic collision, with the big weight settling into the moon dust, the big weight will stop.

Though the big weight has stopped, the small weight was moving upward, causing the string to go slack. The small weight will keep moving up, slow, stop and then drop in a ballistic arc. When it again reaches the end of its string, it will again lift the big weight up, though the effect will be smaller to the extent that the elastic collision with the moon removes energy from the system. This will repeat in smaller and smaller oscillations until the excess energy is entirely damped out of the system.

On the other hand, if the collision between the big weight and the moon is perfectly elastic, the big weight will bounce, there will be two ballistic arcs on the slack string, with what happens next depending on the relevant values of weight, height dropped, string length, etc, developing into a wierd oscillation. A partially elastic collision will also be dependent on the relevant values, but will eventually damp out.

Making the string elastic will also introduce wierd oscillations.

3

If the string is perfectly unstretchable, as you stipulate, then the larger mass is guaranteed to lift somewhat. When the smaller mass reaches the end of the string, it has some kinetic energy which it must lose. As you know, energy is gained or lost through work, which requires a force acting over some distance. Since the tension (the force which is stopping the mass) is finite, and the work we need to do is nonzero, the distance the force acts over must also be nonzero. So the small mass keeps moving for some distance past where we’ve used up all the slack. If the small mass is still moving, and the string isn’t stretching, then the large mass must be lifted by that same distance.

4

And the large weight now has potential energy which it converts to kinetic energy by falling, which lifts the small weight thus giving it potential energy. So it would seem that that when you drop the small weight the big weight bounces once and then everything stops.

But wait, the small weight on being lifted arrives with some velocity which it must lose by going up against gravity slackening the cord. The small weight then falls arriving at the taut cord point with some kinetic energy which it must lose by lifting the large weight just a trifle. The large weight then falls back down lifting the small weight which arrives at a position with some velocity which it must lose by … etc., etc., etc.

5

This explanation felt so intuitively right that I bought it at first. Then I let this thread go thinking someone would pipe up with a correction, but with none forthcoming I’ll give it a go. What really happens is there is a blinding flash of light as the laws of physics are violated!

The trouble here is the larger weight has to undergo an infinitely large acceleration to match the downward speed of the lighter weight. Your solution, Chronos, has the smaller weight moving continuously past the point of first tension. The larger weight’s speed goes from rest to the speed of the falling weight, well, instantaneously. If accelerations are finite then velocities are continuous functions. Thus the speed of the larger weight an infinitesimal amount of time after the string goes taut approaches zero.

Or, since the big weight won’t be budged, the smaller weight comes to a halt, or even rebounds. All this gets you, though, is an infinitely large acceleration of the smaller weight. All of this assumes that the masses are perfectly rigid as well. Otherwise energy could be stored as strain in the masses as they deform under the load.

6

Yeah, I got to thinking the same thing, in the back of my mind, after posting that. I think that what this proves is that there’s no such thing as an unbreakable, unstretchable string (funny, one usually needs to resort to relativity to prove things like that). In actuality, any real string subjected to this experiment would, in fact, stretch, albeit perhaps by a very small amount. But no matter how “stiff” the string, it would inevitably store a non-negligible amount of energy (stiffer strings stretch less, but they also store more energy for a given amount of stretch). If we did the experiment with a real, but almost-unstretchable, string, I think that what would happen would be that the smaller mass would experience a high but finite acceleration while the string stretched by a small amount, and that both masses would rebound as the string re-converted that springy potential energy back into kinetic. Depending on the relative masses and the elasticity of the string, it would be possible, in some cases, for the larger mass not to budge, but I think that if the string is stiff enough, it would (that is to say, given any two masses, there exists some finite but large string stiffness such that the heavier mass would jump by some nonzero amount).

7

I guess the inextensible string factory was shut down the day after they outlawed division by zero…

8

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.

9

Unfortunately, no, at least if you’re considering the system to be the two masses. There’s an outside force from the pully which can act just as impulsively as the masses act on each other, so momentum will not, in general, be conserved here. As an extreme case, suppose the two masses are equal, and you drop both of them at the same time. Before the collision, all of the momentum is going down, and after they bounce at the end of the string, all of it is going up.

10

Sure, but if M1 is dropped from enough of a height to impact a force on M2 greater than the force of gravity, then the pulley does not rotate, acts acts like a teeter-totter.