Why does bullet speed in spring system not depend on gravitational energy?

[haruspex]

Okay so according to your variables:

$Δy_f = d - \frac{Mg}{k}$
$Δy_i = \frac{Mg}{k}$

If those are the height differences from the spring's relaxed position, yes. (I'm assuming d > Mg/k, but the equations probably still work even if not.)

$v_i = \frac{mv_b}{m+M}$

I'm a bit confused about how to interpret the gravitational PE you've given. I'm assuming that you've calculated $mg(y_f - y_i) = (m+M)gd$.

Are you using m differently on the left and right of that equation?
From the impact of the bullet to the max height of the block, block and bullet ascend together a distance d.

So you're claiming $y_f - y_i = d$?

If y_i is the initial position of the block (stretched position of spring) and y_f is the highest point to which the block rises, then yes.

 

[STEMucator]

If those are the height differences from the spring's relaxed position, yes. (I'm assuming d > Mg/k, but the equations probably still work even if not.)

Are you using m differently on the left and right of that equation?
From the impact of the bullet to the max height of the block, block and bullet ascend together a distance d.

If y_i is the initial position of the block (stretched position of spring) and y_f is the highest point to which the block rises, then yes.

Yes I'm assuming up is positive so $d > Mg/k$ holds.

I usually equate generically and simplify as much as possible before I plug things in usually. So on the left I'm using $m$ in general.

I see how you're defining things now though, which also makes a lot of sense. The book I'm using defined things a bit differently for a few example problems, so I was a bit confused about your interpretation at first (even though it also seems straightforward).

 

[haruspex]

Yes I'm assuming up is positive so $d > Mg/k$ holds.

Well, not necessarily. If mv is small and M is large, the bullet might not even lift the block as far as the spring's relaxed position.