Jim Logajan wrote:
If I recall correctly, damage is roughly proportional to energy of
impact, not momentum. (Based on the theory of spring deflection, I
believe: Suppose the object (goose or large weight) strikes a
compression spring. The spring would compress to about the same amount
because the spring equation, E_spring = k_spring_constant *
X_deflection, shows the linear proportionality between energy and
compression.)

Oops! What I wrote here is wrong. The equation E = k*X is only true for a
rare breed of springs known as constant force springs[*]. For conventional
Hook's law springs (F = k*X), the equation is of course E = 0.5*k*X^2.

So if E_kinetic = 0.5*m*V^2 and E_spring = 0.5*k*X^2, and the two energies
are set equal, after a little algebra the deflection is found:

X = V*sqrt(m/k)

So by the spring theory, damage WOULD be linearly propotional to the speed
while proportional to the square root of the mass - i.e. doesn't rise as
fast. Given the earlier example:

X_goose = 120*sqrt(14/k) ~= 449 * sqrt(1/k)
X_wt = 15*sqrt(1000/k) ~= 474 * sqrt(1/k)

Hmmm - interesting that they are still comparable with this selection of
weights and speeds!
[*] A spring loaded measuring tape is the most commonly known household
example of an item that has a constant-force spring in it. The restoring
force is the same no matter how far you pull the tape out.