For some reason I interpreted the # as "number" and got confused about what it meant. Sorry about that.
Firstly, it's more convenient for me to do this in metric units, I hope that's okay with you. I'll convert the units back at the end

160# = 72.6 kg
9# = 4 kg
10" = 25.4 cm
According to
http://www.roymech.co.uk/Useful_Tables/Tribology/co_of_frict.htm, the coefficient of friction between stainless steel and teflon is around 0.04. Unfortunately, there's no data for sliding friction in dry conditions so we'll just have to assume it's 0.04 under all conditions.
The contact surface area doesn't matter, only the force on the contact surface and the coefficient of friction.
The braking force applied while it's moving is 72.6 kg * 9.8m/s^2 * 0.04 = 28.5 N, which causes a deceleration of 0.392 m/s^2. The braking distance is given by d = v^2/2a where d is the distance, v is the velocity and a is the deceleration. Re-arranging this formula for velocity gives us v = √(d * 2a). If we plug our numbers for distance and deceleration, we get an initial velocity of 0.45 m/s.
Unfortunately, I can't find anything telling me how to calculate the energy required to overcome the static friction, so I'll skip that part and hope that it doesn't change the result by much.
Because you said that an angled hit with a lot of rebound moves the weight farther, I think it's best to model the impact as an elastic collision:
https://en.wikipedia.org/wiki/Elastic_collision
The formula describing the velocities of both objects before and after the collision is as follows:
m1, u1 and v1 are the mass, velocity before impact and velocity after impact of object 1. m2, u2 and v2 are the mass, velocity before impact and velocity after impact of object 2.
Let's say that the hammer is object 1, and the weight is object 2. We know that u2 = 0 and that v2 = 0.45 m/s, as well as the values of m1 and m2. We don't know u1 and v1, which is a bit tricky, because we only have one equation but two unknowns. However, we also know that the collision was elastic, so we have conservation of kinetic energy. This can give us our 2nd equation: 0.5 (m1 * u1^2 + m2 * u2^2) = 0.5 (m1 * v1^2 + m2 * v2^2)
Below is my working for solving this system:
For conservation of momentum:
4 kg * u1 = 4 kg * v1 + 72.6 kg * 0.45 m/s
Conservation of energy:
0.5 (m1 * u1^2 + m2 * u2^2) = 0.5 (m1 * v1^2 + m2 * v2^2)
m1 * u1^2 + m2 * u2^2 = m1 * v1^2 + m2 * v2^2
4 kg * u1^2 = 4 kg + v1^2 + 14.46
Unfortunately, the 2nd equation has square terms. I'll solve them using the substitution method - to do this, first I'll re-arrange the conservation of momentum formula to give me v1:
(4 kg * u1 - 72.6 kg * 0.45 m/s) / 4 kg = v1
Now I'll substitute that for v1 in the formula for conservation of energy:
4 kg * u1^2 = 4 kg + ((4 kg * u1 - 72.6 kg * 0.45 m/s) / 4 kg)^2 + 14.46
The result is a formula with one unknown. I struggled with it for a bit, but ultimately decided a computer might be better at solving it than I am:
https://www.wolframalpha.com/input/?i=solve+4+*+u1^2+=+4+++((4+*+u1+-+72.6+*+0.45)+/+4)^2+++14.46+for+u1
We see 2 solutions there. Of course, because u1 is technically a vector, it must have a positive length - so we choose the solution that is a positive number, and we get 3.26 meters per second. This is the speed the hammer moves before the impact.
That's about 10.7 feet per second. In the video it looks like the hammer is accelerated to that speed in about half a second (may have been different in that demo, I don't know that), which would require a force of 26.08 Newtons or, roughly 5.86 lbs to be applied throughout the entire acceleration.
Hopefully you found this useful
