Your physical intuition is correct. A resistive force arising from motion in a viscous medium should not depend on the mass of the object. See, for example, Stokes drag for a common model of this kind of resistive force. So it is likely that the force is defined this way to make the equation of motion look nice. If you used a different object with a different mass, $k$ would have to change accordingly.
The fastest point of sail depends on the boat (both its hull shape and its sail plan), the wind strength, and the sea state. In general, a beam reach is not the fastest point of sail.
For instance, in very light wind some boats will go fastest on a close reach due to the increased apparent wind from going toward the wind. For boats that sail faster than the wind, the limiting factor for speed is how close they can sail to the apparent wind; when they are going faster than the wind speed the apparent wind is always forward of the beam. A broader angle to the true wind allows them to go faster before their sails are sheeted all the way in, so a broad reach is the fastest.
Check out the polar diagrams on this site:
![](https://76trombones.files.wordpress.com/2009/10/solpolar1.jpg?w=510)
https://76trombones.wordpress.com/2009/10/17/polar-diagrams-vmg/
90 degrees to the true wind is not usually the fastest.
As for why some angles are faster than others, that's a bit complicated and beyond the scope of this answer. Suffice it to say that as you sail closer to the wind the component of lift in the direction of travel is less, plus there's increased drag. So you slow down as you approach a close-hauled course. As you bear away to a run, the sails eventually become stalled and less efficient. Somewhere in between a close-hauled course and a run you'll find the angle of maximum speed, which is sometimes approximately 90 degrees to the true wind but not always.
BTW, I should add that the term "beam reach" doesn't have a precise definition. Some sources say a beam reach is 90 degrees to the true wind, others say its 90 degrees to the apparent wind. Most sources introduce the points of sail before the concept of apparent wind and elide over the difference entirely. For boats that only go a fraction of the wind speed (i.e. the vast majority of them) its not that important a distinction. In the answer above, I've used the true wind definition, but even if you choose the apparent wind definition a beam reach is not always fastest. Boats that go faster than the wind have their sails trimmed to close hauled regardless of their point of sail.
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On the most basic level I would say that both your suspicion and your reasoning are correct, although some caveats should follow.
First I should address this on the basis of simple kinematics. Yes, the energy of an air molecule is $m v^2$ but we need to formalize when and if this will translate into pressure. Imagine a pipe that spews a fluid onto a surface, and the fluid comes to a complete rest after impacting the surface. As speed changes the impulse from a molecule in the fluid increases (and impulse is the more true analog to force/pressure than energy), but so does the flow rate, $\dot{m}$. We say that area and density stays the same.
$$\dot{m} = \rho V A$$
$$P=\frac{F}{A}$$
$$F=\dot{m} \Delta V = \rho A V^2$$
$$P = \rho V^2$$
If you study thermal hydraulics, this will not be the last time you see that final expression. Not by a long shot. I should point out that it's not exactly energy per se that leads to the $V^2$ nature, I would say that it's the above math that leads to it.
The example I provide (much like a garden hose spraying someone holding a shield) is a similar system to that which you describe, which is wind in a hurricane hitting the wall of a house. There are several ways that these two systems are not identical, and science has done a lot of work to develop empirical correlations and representative models for fluid pressure losses.
We can look at the system as either a forms loss (on a human-sized scale) or as a boundary layer friction force (on the larger climate level). Equations for the change in pressure of a fluid due to friction or due to a obstacle ("forms" loss) are as follows.
$$\Delta P_{friction} = f \frac{L}{D} \frac{\rho V^2}{2}$$ $$\Delta P_{forms} = K \frac{\rho V^2}{2}$$
In the above equations, both $f$ and $K$ represent constants that we multiply the $V^2$ term by to account for all of the real-life complexity of fluid flow. Yes, these constants will change with the wind speed, so the answer is that the pressure is not perfectly proportional to $V^2$, but we wouldn't be using the above format if this wasn't the closest form that it does follow.