From: fortunati.luigi@gmail.com   
      
   On Thu, 07 Aug 2025 11:46:25 PDT, Luigi Fortunati   
    wrote:   
      
   >[[Mod. note --   
   >Three comments:   
   >   
   >Comment #1:   
   >I think that instead of a dynamometer (which measures torque and power),   
   >what you actually want at x=0 want is a strain gauge, calibrated to measure   
   >the forces between bodies A1 & B1.   
      
   Whatever it is, it remains stationary if it experiences two equal   
   opposing forces and accelerates if it experiences two unequal opposing   
   forces (Newton's first and second laws).   
      
   >Comment #2:   
   >For the A1+A2 collision with B1, this animation seems   
   >to violate conservation of momentum:   
      
   Momentum conservation is perfectly preserved; just read the values of   
   the forces to see it: F1+F2+F3+F5=+1.33-1+0.33-0.67=0.   
      
   >the initial state has a net momentum to the right (A1 & A2 both   
   >moving to the right, B1 moving to the left), but the final state has a net   
   >momentum to the left (A1 & A2 both moving to the left, B1 moving to the   
   right).   
      
   No, even in the final state, the net momentum is to the right because   
   they all move to the right (including the dynamometer): the collision   
   is inelastic!   
      
   That small return to the left is the second compression-decompression   
   phase that also exists in an inelastic collision, during which the   
   forces that slow the opposing momentum of the other body act (on both   
   bodies).   
      
   After these two phases, there are no more forces and no more   
   accelerations: finally, they all move as one body.   
      
   To the right.   
      
   The final velocities (which I reported in the animation itself) are   
   all positive and equal to +1/3 (including the dynamometer).   
      
   >Comment #3:   
   >Newton's 3rd law says nothing about whether   
   >  force exerted on dynamometer by A1   
   >and   
   >  force exerted on dynamometer by B1   
   >are or are not equal.   
      
   The action (the force F1) exerted by body A on the dynamometer (of   
   negligible mass) is exactly equal to the action the dynamometer then   
   transmits directly to body B (why should it transmit less?).   
      
   The reaction (the force F2) exerted by body B on the dynamometer (of   
   negligible mass) is exactly equal to the reaction the dynamometer then   
   transmits directly to body B (why should it transmit less?).   
      
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