The results confirmed that momentum is always conserved

What i am talking about is that you can use
the momentum of the large weight to
throw the smaller weight more than it 'should'
but you are missing the point that you putt that
into the larger weight to begin with

Quote
Re: re: energy producing experiments
« Reply #241 on: November 16, 2021, 05:45:57 PM
The Administrator is welcome to remove the kolbacict post.

You are some kind of angry ...

I know this is a recent repeat but I want to cover the first paragraph.

And the 'MIT Atwood's machine' has an input momentum of .0443 units and an output of .4624 units of momentum.    .01 kg * 4.429 m/sec            1.11 kg * .41666 m/sec

If this MIT Atwood's was a modified Atwood's you would have 1.100 kg moving .4166 m/sec (on a plane) and, one meter away, there would also be .01 kg moving .4166 m/sec. If these two masses were placed on the ends of a balanced beam; and the beam were rotating about its center of mass, then the 1.100 kg would be moving .2102 m/sec and the .01 kg would be moving 23.135 m/sec. At 23.135 m/sec the .01 kg will rise 27.255 m.

The .01 kg will rise 27.255 m and it was dropped 1 meter.

This 27.25 times more energy is done without transfer all the motion to the small mass. If you transferred all the motion to the small mass (as in the cylinder and spheres) the energy increase would be 110 times (ideal).

"And the 'MIT Atwood's machine' has an input momentum of .0443 units and an output of .4624 units of momentum.    .01 kg * 4.429 m/sec    1.11 kg * .41666 m/sec"

The input momentum (.0443) is the mass times the velocity achieved by dropping a .01 kg mass 1 meter.

Any mass dropped one meter will develop a velocity of 4.429 m/sec. Therefore the momentum needed to throw an object up one meter is 4.429 units of momentum for each kilogram. Ten grams is .01 kg so it only needs .0443 units of momentum in order to rise 1 meter.

The .01 kg was dropped one meter; so to return the system to it original starting configuration .0443 kg m/sec is all that is needed.

The MIT Atwood's covers one meter in 4.80 seconds for and average speed of .20833 m/sec, and a final speed of .41666 m/sec. The center of mass of the 1.100 kg does not move. But the 1.100 kg is moving .4166 m/sec at the end of the drop of the .01 kg mass.

So the output momentum is 1.110 kg * .41666 m/sec = .4625 kg m/sec

This is an increase of .4625 - .0443 = .4182 kg m/sec

We can restart the MIT Atwood's ten times. That is 1000% of the energy we put in.

smOky2: Could you post links or an address for the Atwood's experiment conducted in space? Thanks

m1*g*h = 1/2 m1 * v(final)^2 + 1/2 m2 * v(final)^2 + m2 * g * h
———————————————————————————————————


m1*g*h - m2*g*h = 1/2(m1+m2)*v(final)^2
————————————————————————-
(m1-m2)gh=1/2(m1+m2)*v(final)^2
———————————————————-


2[ (m1-m2)g/(m1+m2)]h=v(final)^2


it doesnt matter when or where you stop it
when or where you transfer the energy between the 2 masses