A "slick solution" to the kinetic energy lost when two pucks collide if they are pulled by the string that connects them From David Morin's Classical Mechanics, problem 5.4

A massless string of length $2L$ connects two hockey pucks that lie on frictionless ice. A constant horizontal force F is applied to the midpoint of the string, perpendicular to it (see Fig. 5.20). By calculating the work done in the transverse direction, find how much kinetic energy is lost when the pucks collide, assuming they stick together. The answer you obtained above should be very clean and nice. Find the slick solution that makes it transparent why the answer is so nice.


I have included the first part of the problem for context, however, I am more interested in the slick solution. Morin provides a great answer to the problem, which turns out to be $Fl$. The slick answer is not what I would expect.
Question:
Is it a coincidence that the KE lost is $Fl$? Could I argue that regardless of whether there are masses attached to the string or not, applying a force $F$ through $l$ requires work $Fl$. We can see the midpoint travels a distance $l$ before the collision. So we know that whether it is converted to KE or not along the way to the end of the motion, all useful energy has left the system
 A: (a) Suppose that just before the pucks collide the midpoint of the string has moved forward a distance $x$. The pucks have moved forward (that is in the $\vec F$ direction) by the smaller distance $(x-l)$.
The forward component of the force that the string exerts on each puck is $F/2$ in magnitude, so the contribution of the acquired forward velocity to each puck's kinetic energy just before they collide is $\tfrac12 F(x-l)$, so for both pucks together it is $F(x-l)$.
However the work input to the system, that is the work done by the point of application of $\vec F$, is $Fx$. This is greater by $Fl$ than the work finishing up as the contribution of forward motion to the pucks' kinetic energy. Since there are no dissipative effects until the pucks collide, $Fl$ must be the contribution of their transverse motion to their pre-collision kinetic energy. And this contribution is the kinetic energy lost when the pucks collide. That's it. Is this slick enough? Is it Morin's method?
(b) "Could I argue that regardless of whether there are masses attached to the string or not, applying a force  through  requires work  [?]" I'm not sure that this is even possible. With no masses attached to a massless string the acceleration would be infinite.
A: The argument applies only in the centre of mass frame of the two pucks. I think this is the slick solution (otherwise you have to take into account the acceleration of the system to the right).
I think it is not trivial to convince yourself that the argument is valid, because the centre of mass frame is not an inertial frame. It means introducing varying inertial (fictional) forces on the pucks to cancel the $x$-component of tension in the string. I think Morin's argument is that these inertial forces do no work, because there is no movement in the $x$-direction in the COM frame. Then he argues that the work done by $F$ in this frame must be equal to KE due to motion in the $y$-directions. This is the energy lost when the pucks collide and stop in the COM frame. The argument also depends on Pythagoras theorem in calculating KE.
