If it hits a funnel like target, it should go up and then overboard

A one kilogram mass that is on the end of a string draped over a 99 kilogram rim will produce 44.29 units of momentum after the one kilogram has dropped only one meter.

A one kilogram mass with 44.29 units of momentum will rise 100 meters.

Transfer the momentum cause by an unbalancing mass to the mass that caused the imbalance and you create very large amounts of energy.

http://garydgordon.com/Despin/    this is interesting and it goes back in time (RCA).   But do you see a slow down when the smaller disks open up and have all the motion. There is an air resistance slowdown but if the disks where left attached they would have restarted the larger disk.

https://www.youtube.com/watch?v=w-7d66JscI8

In the 1.320 kg model the original energy is about .66 joules.

The final energy of the .132 kg moving ten times as fast is 6.6 joules.

That is a production of 5.94 joules.

The final energy of the .132 kg balls in the free-falling frame of reference is also about .66J.
The balls does not move ten time as fast. They simply move with constant velocity, while initially unwinding the wire then winding it back.There are no forces to accelerate the balls, hence their speed remain constant.

There are forces acting within the closed system. (132 g / 1320 g)

The original motion of the 1188 (1188g + 132g = 1320g) gram cylinder is about one meter per second. In a short period of time (around .1 second) the cylinder's rotation has stopped. This stop requires the application of 11.88 newtons of force; applied for .1 seconds. This force is applied by the tension in the string; and the force is applied equally to the cylinder and to the spheres. (Third Law)

When you apply 11.88 newtons of forces to a .132 kg mass for .1 seconds you will get a velocity change of 9 m/sec.        F = ma and V = at

Like the cylinder: the original velocity of the spheres was one meter per second; when the additional 9 m/sec is added we have 10 m/sec.

After the cylinder has stopped: and the strings and spheres are left attached the 11.88 newtons kicks in again and it reverses the process. The cylinder is accelerated up to one m/sec and the spheres slow back down to one meter per second. This 11.88 newtons is of course the average force. And even to my surprise; the spheres do not crash into the cylinder. They simply fly in formation just above its surface.

The final energy of the .132 kg balls in the free-falling frame of reference is also about .66J.
The balls does not move ten time as fast. They simply move with constant velocity, while initially unwinding the wire then winding it back.There are no forces to accelerate the balls, hence their speed remain constant.

This is a well known physics experiment, which is best done by a machine.
the basic level college demonstration done by hand is too inconsistent to take accurate measurements.


The rotational velocity of the balls does exceed that of the initial system,
as the cylinder loses momentum to the string tension.
The system as a whole loses net energy rapidly, in air, however in a vacuum in the absence of
gravity and other effects, it transfers momentum rather efficiently and momentum depletes more slowly.


There are also desirable conditions in terms of diameters, proportionate masses, length and strength of the strings, etc. to make it work best.

You are correct Tinu; the final energy and the initial energy are the same. But this is because the final linear momentum (arc velocity * mass) and the initial linear momentum are the same.  The initial and final configurations are the same so everything is the same at these two points in time. But what is the linear momentum and energy in the middle when the spheres have all the motion?

The initial momentum is 1.320 units: 1.320 kg * 1m/sec.  The initial kinetic energy is .66 joules; ½ 1.320 kg *1 m/sec * 1 m/sec.

The middle linear momentum is 1.320 units: .132 kg * 10 m/sec. The middle kinetic energy is 6.6 joules; ½ .1320 kg *10 m/sec * 10 m/sec.

The final momentum is 1.320 units: 1.320 kg * 1m/sec.  The final kinetic energy is .66 joules; ½ 1.320 kg *1 m/sec * 1 m/sec.

When a 132 g projectile moving 10 m/sec collides with a 1188 gram block at rest we will get a combined mass of 1.320 kg moving 1 m/sec. This is consistent with the loss of energy between middle and final.

But what if the spheres are not left attached, when they are moving 10 m/sec. Lets direct them upward and cut them loose; at 10 m/sec they will rise 5.10 meters.      d = ½ v²/a

Initially we started with 1320 grams moving 1 m/sec. This one meter per second velocity is equal to dropping this mass .0510 meters.        square root of (d * 2 * a) = v

1320 gram at a height of 5.1 cm could lift 132 grams to 51 cm.

132 grams moving 10 m/sec is equal to 132 grams at a height of 510 cm.

Dynamic application: take a 1188 gram mounted rim and wrap a weighted string around it; let 132 grams drop .51 meters. This will accelerate the entire 1320 grams to 1 m/sec. Transfer all this momentum to the 132 grams and throw it upwards. Only one tenth of the distance up is needed to reload the system.

Place 1188 grams on a frictionless plane. Tie a string to the mass and drape the string over a pulley. On the suspended end of the string place 132 grams. Allow the 132 grams to drop until the total mass of 1320 grams reached a velocity of one meter per second. Upon measurement you will find that the 132 grams has dropped .50968 meters. One formula for energy is the newton/meter and this would be .132 kg * 9.81 N / kg = 1.2949 newtons; which would be multiplied by .50968 meters for .65999 joules.

Any mass moving one meter per second can rise .050968 meters. We start with 1320 grams moving 1 m/sec.  So that is 1.320 kg * 9.81 N / kg * .050968 meters = .65999 joules

Take a thin wall tube with a mass of 1188g.  Place the tube in dry ice; then accelerate it with a 132 g weighted string wrapped around its surface, it will have a near perfect F = ma acceleration. It is F = ma  for any radius tube. It would work for a 10 cm tube or a 2 meter tube.  You will have a velocity of one meter per second after the 132 grams has dropped .50968 meters.       And .132 kg * 9.81 N / kg = 1.2949 newtons; which would be multiplied by .50968 meters for .65999 joules.

But is it .65999 joules in the middle when the spheres have all the motion.  ½ * .132 kg * v * v = .65999 joule ?; solving for v we have 3.16 m/sec. Can .132 kg * 3.16 m/sec = .417 units of momentum give you 1.320 (1.320 kg * 1 m/sec) units of momentum?

The main point is that only linear momentum (arc velocity  * mass) conservation can give you a middle velocity that would be sufficient for the end velocity to equal the beginning velocity.