On Thu, 22 Nov 2007 05:20:10 -0600, Cecil Moore
were Cecil's dignity was not conserved, wrote:
If we carry out the math of your own answer, it renders units in your
terms to:
(watts/m²)/(m/s)²
or
((kg·m²/s³)/m²)/(m/s)²
or
kg/(s·m²)

One needs to multiply by the area under consideration
to obtain the total momentum which is conserved.

Which, of course, yields:
kg/s
which is not momentum, but oddly enough the units one might use in
describing how fast his bath tub was filling with photons. :-)

Oh well, third time's a charm! Keep fumbling with the conservation of
dignity, and eventually you might bumble into a job with Anderson
Consulting doing Enron's books.