To all sceptics here in this forum.
======================
Please always keep in mind our post of June 21, 2021, 02:06:31 PM.
Asking my direct and clear question for the 3rd time: Do you have any objections against any of items 1 - 15? Yes or no?
Looking forward to your answer for the 3rd time. (Only one word -- either "yes" or "no"!)

                             click bait
si !

Second and final warning George2
or I'm telling my mom on you !

To JerryVolland.
===================
NO, THIS IS NEITHER A WORM VIRUS NOR A SPAM! YOU ARE SIMPLY A PAID AGENT OF THE OFFICIAL SCIENCE MAFIA, WHO TRIES TO MANIPULATE THE AUDIENCE  AGAIN IN A CLUMSY AND UNSKILLFUL MANNER! TAKE SOME BEGINNER'S MANIPULATION TECHNIQUES GUIDE AND STUDY IT CAREFULLY! BECAUSE OTHERWISE YOU SIMPLY RESEMBLE A CLOWN!
===================
Asking my direct and clear question for the 7th consecutive time: Do you have any objections against any of items 1 - 16 below? Yes or no? (If yes, then specify exactly which item you do not agree with and why.)
Looking forward to your answer for the 7th consecutive time. (Only one word -- either "yes" or "no"!)
============================
============================
1) Look again at PART 3 of the link https://www.youtube.com/watch?v=xX14NK8GrDY&t=24s  . Focus on the “upper” zigzag case.
2) Ma = 1 kg.
3) Mb = 4 kg.
------------------------------------
4) Va’ = pre-zig-zag velocity of the blue component = 1 m/s = const.; Va’ = V1.
5) Vb’ = pre-zig-zag velocity of the black component = 0 m/s; the black component is motionless.
------------------------------------
6) Va” = during-zig-zag velocity of the blue component = variable and comparatively difficult (but not impossible) to calculate.
7) Vb” = during-zig-zag velocity of the black component = variable and comparatively difficult (but not impossible) to calculate.
8/ Vy = during-zig-zag velocity of each couple blue rod-blue ball along the Y-axis = variable and comparatively difficult (but not impossible) to calculate.
------------------------------------
9) Va”’ = post-zig-zag velocity of the blue component = 0.6 m/s = const.; Va”’ = V2.
10) Vb”’ = post-zig-zag velocity of the black component = 0.1 m/s = const.; Vb”’ = V3.
------------------------------------
11) According to the third Newton’s law and to the related law of conservation of linear momentum we can write down the equalities
((Ma) x (Va’)) +  ((Mb) x (Vb’)) =  ((Ma) x (Va”’)) +  ((Mb) x (Vb”’))      <=>
<=>  ((Ma) x (Va’)) + 0 =  ((Ma) x (Va”’)) +  ((Mb) x (Vb”’))       <=>
<=>  (Ma) x (Va’) =  ((Ma) x (Va”’)) +  ((Mb) x (Vb”’))       <=>
<=>  (1 kg) x (1 m/s) = ((1 kg) x (0.6 m/s)) +  ((4 kg) x (0.1 m/s))     <=>
<=>  1 kg.m/s = 1 kg.m/s.
12) In one word, the values of Va”, Vb” and Vy are actually of no interest to us.  Actually only the values of Va’, Va”’ and Vb”’ are of interest to us as these three values determine the validity of the third Newton’s law and the related law of conservation of linear momentum.
13) The mass of each couple blue rod-blue ball is much smaller than the mass of the blue T-shaped component. For example if Ma=1kg, then the mass of each couple blue rod-blue ball must be equal to, let’s say, 0.0001 kg (and even smaller).
14) In our numerous real experiments we strongly reduce friction and the mean experimental values of  Va”’ and Vb”’ are equal to 0.5999992 m/s and to 0.0999997 m/s, respectively, that is, Va”’ = 0.5999992 m/s and  Vb”’ =  0.0999997 m/s. The latter clearly shows that the experimental error (due to friction) is much smaller than 1 % and this experimental error is perfectly acceptable.
15) Let me remind only again (it is written in the explanatory text of the link https://www.youtube.com/watch?v=xX14NK8GrDY  ) that the zigzags generate a mechanical effect (let us call this mechanical effect the "X effect"), (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat. (We take gravity and friction out of equation and consideration.) And really even if the mean experimental value of force of friction inside the zigzag channels is equal to 0.0000001 N (our last experimental result), then the "X effect" still remains and can be clearly observed as in PART 3 of the link above.
16) And now we can easily calculate the pre-zig-zag and post-zig-zag kinetic energies of the bodies, taking part in the experiment. That is, we can write down the inequalities
(0.5) x (Ma) x (Va’) x (Va’) >  ((0.5) x (Ma) x (Va”’) x (Va”’)) +  ((0.5) x (Mb) x (Vb”’) x (Vb”’))       <=>
<=>  (0.5) x (1 kg) x (1 m/s) x (1 m/s) > ((0.5) x (1 kg) x (0.6 m/s)x (0.6 m/s)) +  ((0.5) x (4 kg) x (0.1 m/s) x (0.1 m/s))     <=>   0.5 J > 0.2 J
The last three inequalities unambiguously show a severe violation of the law of conservation of mechanical energy.
============================
============================
Asking my direct and clear question for the 7th consecutive time: Do you have any objections against any of the above items 1 - 16? Yes or no? (If yes, then specify exactly which item you do not agree with and why.)
Looking forward to your answer for the 7th consecutive time. (Only one word -- either "yes" or "no"!)

To JerryVolland and to Floor.
============================
Asking my direct and clear question for the 9th consecutive time: Do you have any objections against any of items 1 - 16 below? Yes or no? (If yes, then specify exactly which item you do not agree with and why.)
Looking forward to your answer for the 9th consecutive time. (Only one word -- either "yes" or "no"!)
============================
============================
1) Look again at PART 3 of the link https://www.youtube.com/watch?v=xX14NK8GrDY&t=24s  . Focus on the “upper” zigzag case.
2) Ma = 1 kg.
3) Mb = 4 kg.
------------------------------------
4) Va’ = pre-zig-zag velocity of the blue component = 1 m/s = const.; Va’ = V1.
5) Vb’ = pre-zig-zag velocity of the black component = 0 m/s; the black component is motionless.
------------------------------------
6) Va” = during-zig-zag velocity of the blue component = variable and comparatively difficult (but not impossible) to calculate.
7) Vb” = during-zig-zag velocity of the black component = variable and comparatively difficult (but not impossible) to calculate.
8/ Vy = during-zig-zag velocity of each couple blue rod-blue ball along the Y-axis = variable and comparatively difficult (but not impossible) to calculate.
------------------------------------
9) Va”’ = post-zig-zag velocity of the blue component = 0.6 m/s = const.; Va”’ = V2.
10) Vb”’ = post-zig-zag velocity of the black component = 0.1 m/s = const.; Vb”’ = V3.
------------------------------------
11) According to the third Newton’s law and to the related law of conservation of linear momentum we can write down the equalities
((Ma) x (Va’)) +  ((Mb) x (Vb’)) =  ((Ma) x (Va”’)) +  ((Mb) x (Vb”’))      <=>
<=>  ((Ma) x (Va’)) + 0 =  ((Ma) x (Va”’)) +  ((Mb) x (Vb”’))       <=>
<=>  (Ma) x (Va’) =  ((Ma) x (Va”’)) +  ((Mb) x (Vb”’))       <=>
<=>  (1 kg) x (1 m/s) = ((1 kg) x (0.6 m/s)) +  ((4 kg) x (0.1 m/s))     <=>
<=>  1 kg.m/s = 1 kg.m/s.
12) In one word, the values of Va”, Vb” and Vy are actually of no interest to us.  Actually only the values of Va’, Va”’ and Vb”’ are of interest to us as these three values determine the validity of the third Newton’s law and the related law of conservation of linear momentum.
13) The mass of each couple blue rod-blue ball is much smaller than the mass of the blue T-shaped component. For example if Ma=1kg, then the mass of each couple blue rod-blue ball must be equal to, let’s say, 0.0001 kg (and even smaller).
14) In our numerous real experiments we strongly reduce friction and the mean experimental values of  Va”’ and Vb”’ are equal to 0.5999992 m/s and to 0.0999997 m/s, respectively, that is, Va”’ = 0.5999992 m/s and  Vb”’ =  0.0999997 m/s. The latter clearly shows that the experimental error (due to friction) is much smaller than 1 % and this experimental error is perfectly acceptable.
15) Let me remind only again (it is written in the explanatory text of the link https://www.youtube.com/watch?v=xX14NK8GrDY  ) that the zigzags generate a mechanical effect (let us call this mechanical effect the "X effect"), (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat. (We take gravity and friction out of equation and consideration.) And really even if the mean experimental value of force of friction inside the zigzag channels is equal to 0.0000001 N (our last experimental result), then the "X effect" still remains and can be clearly observed as in PART 3 of the link above.
16) And now we can easily calculate the pre-zig-zag and post-zig-zag kinetic energies of the bodies, taking part in the experiment. That is, we can write down the inequalities
(0.5) x (Ma) x (Va’) x (Va’) >  ((0.5) x (Ma) x (Va”’) x (Va”’)) +  ((0.5) x (Mb) x (Vb”’) x (Vb”’))       <=>
<=>  (0.5) x (1 kg) x (1 m/s) x (1 m/s) > ((0.5) x (1 kg) x (0.6 m/s)x (0.6 m/s)) +  ((0.5) x (4 kg) x (0.1 m/s) x (0.1 m/s))     <=>   0.5 J > 0.2 J
The last three inequalities unambiguously show a severe violation of the law of conservation of mechanical energy in this particular case.
============================
============================
Asking my direct and clear question for the 9th consecutive time: Do you have any objections against any of the above items 1 - 16? Yes or no? (If yes, then specify exactly which item you do not agree with and why.)
Looking forward to your answer for the 9th consecutive time. (Only one word -- either "yes" or "no"!)

To JerryVolland and to Floor.
============================
Asking my direct and clear question for the 11th consecutive time: Do you have any objections against any of items 1 - 16 below? Yes or no? (If yes, then specify exactly which item you do not agree with and why.)
Looking forward to your answer for the 11th consecutive time. (Only one word -- either "yes" or "no"!)
============================
============================
1) Look again at PART 3 of the link https://www.youtube.com/watch?v=xX14NK8GrDY&t=24s  . Focus on the “upper” zigzag case.
2) Ma = 1 kg.
3) Mb = 4 kg.
------------------------------------
4) Va’ = pre-zig-zag velocity of the blue component = 1 m/s = const.; Va’ = V1.
5) Vb’ = pre-zig-zag velocity of the black component = 0 m/s; the black component is motionless.
------------------------------------
6) Va” = during-zig-zag velocity of the blue component = variable and comparatively difficult (but not impossible) to calculate.
7) Vb” = during-zig-zag velocity of the black component = variable and comparatively difficult (but not impossible) to calculate.
8/ Vy = during-zig-zag velocity of each couple blue rod-blue ball along the Y-axis = variable and comparatively difficult (but not impossible) to calculate.
------------------------------------
9) Va”’ = post-zig-zag velocity of the blue component = 0.6 m/s = const.; Va”’ = V2.
10) Vb”’ = post-zig-zag velocity of the black component = 0.1 m/s = const.; Vb”’ = V3.
------------------------------------
11) According to the third Newton’s law and to the related law of conservation of linear momentum we can write down the equalities
((Ma) x (Va’)) +  ((Mb) x (Vb’)) =  ((Ma) x (Va”’)) +  ((Mb) x (Vb”’))      <=>
<=>  ((Ma) x (Va’)) + 0 =  ((Ma) x (Va”’)) +  ((Mb) x (Vb”’))       <=>
<=>  (Ma) x (Va’) =  ((Ma) x (Va”’)) +  ((Mb) x (Vb”’))       <=>
<=>  (1 kg) x (1 m/s) = ((1 kg) x (0.6 m/s)) +  ((4 kg) x (0.1 m/s))     <=>
<=>  1 kg.m/s = 1 kg.m/s.
12) In one word, the values of Va”, Vb” and Vy are actually of no interest to us.  Actually only the values of Va’, Va”’ and Vb”’ are of interest to us as these three values determine the validity of the third Newton’s law and the related law of conservation of linear momentum.
13) The mass of each couple blue rod-blue ball is much smaller than the mass of the blue T-shaped component. For example if Ma=1kg, then the mass of each couple blue rod-blue ball must be equal to, let’s say, 0.0001 kg (and even smaller).
14) In our numerous real experiments we strongly reduce friction and the mean experimental values of  Va”’ and Vb”’ are equal to 0.5999992 m/s and to 0.0999997 m/s, respectively, that is, Va”’ = 0.5999992 m/s and  Vb”’ =  0.0999997 m/s. The latter clearly shows that the experimental error (due to friction) is much smaller than 1 % and this experimental error is perfectly acceptable.
15) Let me remind only again (it is written in the explanatory text of the link https://www.youtube.com/watch?v=xX14NK8GrDY  ) that the zigzags generate a mechanical effect (let us call this mechanical effect the "X effect"), (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat. (We take gravity and friction out of equation and consideration.) And really even if the mean experimental value of force of friction inside the zigzag channels is equal to 0.0000001 N (our last experimental result), then the "X effect" still remains and can be clearly observed as in PART 3 of the link above.
16) And now we can easily calculate the pre-zig-zag and post-zig-zag kinetic energies of the bodies, taking part in the experiment. That is, we can write down the inequalities
(0.5) x (Ma) x (Va’) x (Va’) >  ((0.5) x (Ma) x (Va”’) x (Va”’)) +  ((0.5) x (Mb) x (Vb”’) x (Vb”’))       <=>
<=>  (0.5) x (1 kg) x (1 m/s) x (1 m/s) > ((0.5) x (1 kg) x (0.6 m/s)x (0.6 m/s)) +  ((0.5) x (4 kg) x (0.1 m/s) x (0.1 m/s))     <=>   0.5 J > 0.2 J
The last three inequalities unambiguously show a severe violation of the law of conservation of mechanical energy in this particular case.
============================
============================
Asking my direct and clear question for the 11th consecutive time: Do you have any objections against any of the above items 1 - 16? Yes or no? (If yes, then specify exactly which item you do not agree with and why.)
Looking forward to your answer for the 11th consecutive time. (Only one word -- either "yes" or "no"!)

To JerryVolland and to Floor.
============================
Asking my direct and clear question for the 13th consecutive time: Do you have any objections against any of items 1 - 16 below? Yes or no? (If yes, then specify exactly which item you do not agree with and why.)
Looking forward to your answer for the 13th consecutive time. (Only one word -- either "yes" or "no"!)
============================
============================
1) Look again at PART 3 of the link https://www.youtube.com/watch?v=xX14NK8GrDY&t=24s  . Focus on the “upper” zigzag case.
2) Ma = 1 kg.
3) Mb = 4 kg.
------------------------------------
4) Va’ = pre-zig-zag velocity of the blue component = 1 m/s = const.; Va’ = V1.
5) Vb’ = pre-zig-zag velocity of the black component = 0 m/s; the black component is motionless.
------------------------------------
6) Va” = during-zig-zag velocity of the blue component = variable and comparatively difficult (but not impossible) to calculate.
7) Vb” = during-zig-zag velocity of the black component = variable and comparatively difficult (but not impossible) to calculate.
8/ Vy = during-zig-zag velocity of each couple blue rod-blue ball along the Y-axis = variable and comparatively difficult (but not impossible) to calculate.
------------------------------------
9) Va”’ = post-zig-zag velocity of the blue component = 0.6 m/s = const.; Va”’ = V2.
10) Vb”’ = post-zig-zag velocity of the black component = 0.1 m/s = const.; Vb”’ = V3.
------------------------------------
11) According to the third Newton’s law and to the related law of conservation of linear momentum we can write down the equalities
((Ma) x (Va’)) +  ((Mb) x (Vb’)) =  ((Ma) x (Va”’)) +  ((Mb) x (Vb”’))      <=>
<=>  ((Ma) x (Va’)) + 0 =  ((Ma) x (Va”’)) +  ((Mb) x (Vb”’))       <=>
<=>  (Ma) x (Va’) =  ((Ma) x (Va”’)) +  ((Mb) x (Vb”’))       <=>
<=>  (1 kg) x (1 m/s) = ((1 kg) x (0.6 m/s)) +  ((4 kg) x (0.1 m/s))     <=>
<=>  1 kg.m/s = 1 kg.m/s.
12) In one word, the values of Va”, Vb” and Vy are actually of no interest to us.  Actually only the values of Va’, Va”’ and Vb”’ are of interest to us as these three values determine the validity of the third Newton’s law and the related law of conservation of linear momentum.
13) The mass of each couple blue rod-blue ball is much smaller than the mass of the blue T-shaped component. For example if Ma=1kg, then the mass of each couple blue rod-blue ball must be equal to, let’s say, 0.0001 kg (and even smaller).
14) In our numerous real experiments we strongly reduce friction and the mean experimental values of  Va”’ and Vb”’ are equal to 0.5999992 m/s and to 0.0999997 m/s, respectively, that is, Va”’ = 0.5999992 m/s and  Vb”’ =  0.0999997 m/s. The latter clearly shows that the experimental error (due to friction) is much smaller than 1 % and this experimental error is perfectly acceptable.
15) Let me remind only again (it is written in the explanatory text of the link https://www.youtube.com/watch?v=xX14NK8GrDY  ) that the zigzags generate a mechanical effect (let us call this mechanical effect the "X effect"), (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat. (We take gravity and friction out of equation and consideration.) And really even if the mean experimental value of force of friction inside the zigzag channels is equal to 0.0000001 N (our last experimental result), then the "X effect" still remains and can be clearly observed as in PART 3 of the link above.
16) And now we can easily calculate the pre-zig-zag and post-zig-zag kinetic energies of the bodies, taking part in the experiment. That is, we can write down the inequalities
(0.5) x (Ma) x (Va’) x (Va’) >  ((0.5) x (Ma) x (Va”’) x (Va”’)) +  ((0.5) x (Mb) x (Vb”’) x (Vb”’))       <=>
<=>  (0.5) x (1 kg) x (1 m/s) x (1 m/s) > ((0.5) x (1 kg) x (0.6 m/s)x (0.6 m/s)) +  ((0.5) x (4 kg) x (0.1 m/s) x (0.1 m/s))   <=>     
<=>   0.5 J > 0.2 J
The last three inequalities unambiguously show a severe violation of the law of conservation of mechanical energy in this particular case.
============================
============================
Asking my direct and clear question for the 13th consecutive time: Do you have any objections against any of the above items 1 - 16? Yes or no? (If yes, then specify exactly which item you do not agree with and why.)
Looking forward to your answer for the 13th consecutive time. (Only one word -- either "yes" or "no"!)

To JerryVolland.
============================
DON'T BEAT ABOUT THE BUSH, YOU OLD CHEATER! AND STOP BEHAVING LIKE A CLOWN! ARE YOU ILLITERATE? ANSWER MY QUESTION!
============================
Asking my direct and clear question for the 15th consecutive time: Do you have any objections against any of items 1 - 16 below? Yes or no? (If yes, then specify exactly which item you do not agree with and why.)
Looking forward to your answer for the 15th consecutive time. (Only one word -- either "yes" or "no"!)
============================
============================
1) Look again at PART 3 of the link https://www.youtube.com/watch?v=xX14NK8GrDY&t=24s  . Focus on the “upper” zigzag case.
2) Ma = 1 kg.
3) Mb = 4 kg.
------------------------------------
4) Va’ = pre-zig-zag velocity of the blue component = 1 m/s = const.; Va’ = V1.
5) Vb’ = pre-zig-zag velocity of the black component = 0 m/s; the black component is motionless.
------------------------------------
6) Va” = during-zig-zag velocity of the blue component = variable and comparatively difficult (but not impossible) to calculate.
7) Vb” = during-zig-zag velocity of the black component = variable and comparatively difficult (but not impossible) to calculate.
8/ Vy = during-zig-zag velocity of each couple blue rod-blue ball along the Y-axis = variable and comparatively difficult (but not impossible) to calculate.
------------------------------------
9) Va”’ = post-zig-zag velocity of the blue component = 0.6 m/s = const.; Va”’ = V2.
10) Vb”’ = post-zig-zag velocity of the black component = 0.1 m/s = const.; Vb”’ = V3.
------------------------------------
11) According to the third Newton’s law and to the related law of conservation of linear momentum we can write down the equalities
((Ma) x (Va’)) +  ((Mb) x (Vb’)) =  ((Ma) x (Va”’)) +  ((Mb) x (Vb”’))      <=>
<=>  ((Ma) x (Va’)) + 0 =  ((Ma) x (Va”’)) +  ((Mb) x (Vb”’))       <=>
<=>  (Ma) x (Va’) =  ((Ma) x (Va”’)) +  ((Mb) x (Vb”’))       <=>
<=>  (1 kg) x (1 m/s) = ((1 kg) x (0.6 m/s)) +  ((4 kg) x (0.1 m/s))     <=>
<=>  1 kg.m/s = 1 kg.m/s.
12) In one word, the values of Va”, Vb” and Vy are actually of no interest to us.  Actually only the values of Va’, Va”’ and Vb”’ are of interest to us as these three values determine the validity of the third Newton’s law and the related law of conservation of linear momentum.
13) The mass of each couple blue rod-blue ball is much smaller than the mass of the blue T-shaped component. For example if Ma=1kg, then the mass of each couple blue rod-blue ball must be equal to, let’s say, 0.0001 kg (and even smaller).
14) In our numerous real experiments we strongly reduce friction and the mean experimental values of  Va”’ and Vb”’ are equal to 0.5999992 m/s and to 0.0999997 m/s, respectively, that is, Va”’ = 0.5999992 m/s and  Vb”’ =  0.0999997 m/s. The latter clearly shows that the experimental error (due to friction) is much smaller than 1 % and this experimental error is perfectly acceptable.
15) Let me remind only again (it is written in the explanatory text of the link https://www.youtube.com/watch?v=xX14NK8GrDY  ) that the zigzags generate a mechanical effect (let us call this mechanical effect the "X effect"), (a) which is absolutely identical and equivalent to friction and (b) which does not generate heat. (We take gravity and friction out of equation and consideration.) And really even if the mean experimental value of force of friction inside the zigzag channels is equal to 0.0000001 N (our last experimental result), then the "X effect" still remains and can be clearly observed as in PART 3 of the link above.
16) And now we can easily calculate the pre-zig-zag and post-zig-zag kinetic energies of the bodies, taking part in the experiment. That is, we can write down the inequalities
(0.5) x (Ma) x (Va’) x (Va’) >  ((0.5) x (Ma) x (Va”’) x (Va”’)) +  ((0.5) x (Mb) x (Vb”’) x (Vb”’))       <=>
<=>  (0.5) x (1 kg) x (1 m/s) x (1 m/s) > ((0.5) x (1 kg) x (0.6 m/s)x (0.6 m/s)) +  ((0.5) x (4 kg) x (0.1 m/s) x (0.1 m/s))   <=>     
<=>   0.5 J > 0.2 J
The last three inequalities unambiguously show a severe violation of the law of conservation of mechanical energy in this particular case.
============================
============================
Asking my direct and clear question for the 15th consecutive time: Do you have any objections against any of the above items 1 - 16? Yes or no? (If yes, then specify exactly which item you do not agree with and why.)
Looking forward to your answer for the 15th consecutive time. (Only one word -- either "yes" or "no"!)