What formula can give "the" impulse and lead to an answer of 16 without any units?
For the ball you can write linear momentum before = 10 m/s x 3 kg = 30 kgm/s
and momentum after -6 m/s x 3 kg = -18 kgm/s, so a momentum change of 48 kgm/s.
So during the collision it has experienced a force F = Δp/Δt = 48 kgm/s / Δt.
The analogue of ##\vec F = d\vec p/dt## for rotations is ## \vec \tau = d\vec L/dt## in vector notation, with
##\vec \tau ## the torque and ##\vec L## the angular momentum
And if things aren't too complicated , like in F = Δp/Δt this becomes ##\tau = \Delta L/\Delta t##.
In the absence of external forces, and making use of action = -reaction, one can state for a system of e.g. two colliding masses ## \vec F=0 \Rightarrow d\vec p/dt = 0 \Rightarrow \vec p## is conserved.
Likewise, in the absence of external torques, we get ## \tau \equiv \vec r \times \vec F=0 \Rightarrow d(\vec r \times \vec p)/dt \equiv d\vec L/dt = 0 \Rightarrow \vec L## is conserved (r is constant during the collision).
If we can combine: FΔt and ##\tau##Δt via ##\tau## = rF we have rΔp = ΔL and no more Δt!
Now take the pivot point of the bar as the center for rotation. And, just for the fun of it, consider the change in angular momentum of the ball:
before: ##\vec L = \vec r \times \vec p ## and since ## \vec v \bot \vec r## this becomes L = rp
after: idem, so ΔL = r Δp
We have all we need to calculate it. Do so, you need it later on.
Wrt this same point the bar initially has zero angular momentum and for the system of bar + ball there are no external torques wrt this center of rotation, which means that ΔL
ball + ΔL
bar=0 !
(for the ball r Δp = I Δω :
## \vec v = \vec \omega \times \vec r## and since ## \vec v \bot \vec r## this becomes ## |\vec \omega | = {|\vec v|\over | \vec r|}## so ω = v/r and Δω = Δ(v/r) = Δv /r. Moment of inertia: I = mr2, so Angular momentum L = I ω= mr2 v/r = mvr and ΔL = r Δp). You already calculated it.
For the bar we now have ΔL = ΔIω = I Δω = I ω
after, which is the one we were after !
Reason I drew attention to b was that you have to realize that (horizontal) force of ball on bar is not the only (horizontal) force on the bar, but the torque from the collsion with the ball ball is the only non-zero torque wrt the pivot point!
Your questions [quote The linear momentum of the ball is NOT conserved (where does it go, though? Isn't this an elastic collision?) [/quote] can now be answered with ease, right ? Try me!