A block of mass M = 4.0 kg rests on a smooth horizontal floor…

Step 1. Speeds. No horizontal force acts from outside, and the pair starts at rest, so the blocks move in opposite directions with $mv = MV$. Every surface is smooth, so $mgR = \tfrac12 mv^2 + \tfrac12 MV^2$. Together:

So $v = 2.68\ \text{m/s}$, and $V = \dfrac{m}{M}v = 0.671\ \text{m/s}$, the other way.

Step 2. The big block's frame at the bottom. At the bottom the groove is horizontal, so the push between the blocks is vertical. No horizontal force then acts on the big block, and its acceleration is zero at that instant. So, for that instant, its frame is an inertial frame, and Newton's second law works there without any pseudo force.

Step 3. Normal force. In that frame the small block moves on a circle of radius $R$, with the relative speed $v + V = 3.354\ \text{m/s}$. At the lowest point of that circle:

So $N = 10.0\ \text{N} + 25.0\ \text{N} = 35.0\ \text{N}$.

Step 4. Distance. The centre of mass of the pair does not move sideways. Relative to the big block, the small block moves $R$ sideways. If the big block moves $X$, then $m(R - X) = MX$, so $X = \dfrac{mR}{M + m} = 0.090\ \text{m}$.