Can't understand derivation of relativistic mass in Kleppner

In Kleppner and Kolenkow chapter 13, they derive the expression of relativistic mass by considering a symmetric glancing elastic collision.

It was analyzed from two reference frames. One in which the velocity of A in the x direction was zero and another one in which the velocity of B in the x direction was zero.

Here is how the derivation goes in the book:

Our task is to find a conserved quantity analogous to classical momentum. We suppose that the momentum of a particle moving with velocity $mathbf{w}$ is $$mathbf{p} = m(w) mathbf{w}$$ where $m(w)$ is a scalar quantity yet to be determined, analogous to Newtonian mass but which could depend on the speed $w$.

The x momentum in A’s frame is due entirely to particle B. Before the collision B’s speed is
$w = sqrt{V^2 + u_0^2/gamma^2}$ and after the collision it is $w’ = sqrt{V^2 + u’^2/gamma^2}$. Imposing conservation of momentum in the x direction yields $$m(w)V = m(w’)V$$ It follows that $w=w’$, so that $$u’ = u_0$$ In other words, y motion is reversed in the A frame.

Next we write the statement of the conservation of momentum in the y direction as evaluated in A’s frame. Equating the y momentum before and after the collision gives
$$-m_0 u_0 + m(w) frac{u_0}{gamma} = m_0 u_0 – m(w) frac{u_0}{gamma}$$
which gives $$m(w) = gamma m_0$$ In the limit $u_0 rightarrow 0$, $m(u_0) rightarrow m(0)$, which we take to be the Newtonian mass, or "rest mass" $m_0$, of the particle. In this limit, $w = V$. Hence $$m(V) = gamma m_0 = frac{m_0}{sqrt{1 – V^2/c^2}}$$
Consequently, momentum is preserved in the collision provided we define the momentum of a particle moving with velocity $mathbf{v}$ to be $$mathbf{p} = m mathbf{v}$$ where
$$m = frac{m_0}{sqrt{1 – v^2/c^2}} = gamma m_0$$

Now I have a few problems with this derivation. They are:

1. They assumed that both A and B has the same mass. Unless I’m mistaken, the momentum equation in the $x$ direction should remain unchanged because during the collision the impulse is in the $y$ direction. So suppose the masses were different, namely $m_A$ and $m_B(w)$. Then since
   $$m_B(w)V = m_B(w’)V$$
   it follows that $$u’ = u_0$$ But then the y equation becomes
   $$-m_A u_0 + m_B(w) frac{u_0}{gamma} = m_0 u_0 – m_B(w) frac{u_0}{gamma}$$
   or
   $$m_B(w) = gamma m_A$$
   which is weird because B’s mass shouldn’t depend on A’s. Personally I think that their argument for $u’ = u_0$ is flawed. Because it doesn’t matter whether A and B’s masses are different but intuitively I think it should. I don’t see how the collision could be elastic and symmetrical without the two particles having the same mass. Because I could always take extreme cases when one is much more massive than the other and following their argument we would still have $u’ = u_0$. Or maybe I misunderstood their argument and the masses really do matter. This is all very confusing to me.
2. It seems that when writing the momentum equation in the $y$ direction, the author represented $m(u_0)$ as $m_0$ while in the final equation they meant $m_0$ to be the rest mass which makes sense because A was also moving in the y direction in A’s frame so it’s mass can’t be just the rest mass $m_0$. However before taking the limit $u_0 rightarrow 0$, $m(u_0) rightarrow m(0)$, the equation for $m(w)$ was
   $$m(w) = frac{m(u_0)}{sqrt{1 – V^2/c^2}}$$
   and after taking the limit it became
   $$m(V) = frac{m_0}{sqrt{1 – V^2/c^2}}$$
   However both equations are supposed to be true and using the final result we should have
   $$m(w) = frac{m_0}{sqrt{1 – w^2/c^2}}$$
   and similarly for A in A’s frame,
   $$m(u_0) = frac{m_0}{sqrt{1 – u_0^2/c^2}}$$
   Substituting this in the first equation
   $$m(w) = frac{m(u_0)}{sqrt{1 – V^2/c^2}}$$
   $$ = frac{m_0}{sqrt{(1 – V^2/c^2)(1 – u_0^2/c^2)}}$$
   $$ = frac{m_0}{sqrt{1 – (u_0^2/c^2 + V^2/c^2) + (Vu_0)^2/c^4}}$$
   $$ = frac{m_0}{sqrt{1 – w^2/c^2 + (Vu_0)^2/c^4}}$$
   $$ neq frac{m_0}{sqrt{1 – w^2/c^2}} $$
   I’m probably missing something but I can’t figure out what.