Answer:
A   u = 0.36c      B u = 0.961c
Explanation:
In special relativity the transformation of velocities is carried out using the Lorentz equations, if the movement in the x direction remains
      u ’= (u-v) / (1- uv / c²)
Where u’ is the speed with respect to the mobile system, in this case the initial nucleus of uranium, u the speed with respect to the fixed system (the observer in the laboratory) and v the speed of the mobile system with respect to the laboratory
The data give is u ’= 0.43c and the initial core velocity v = 0.94c
Let's clear the speed with respect to the observer (u)
       u’ (1- u v / c²) = u -v
       u + u ’uv / c² = v - u’
       u (1 + u ’v / c²) = v - u’
       u = (v-u ’) / (1+ u’ v / c²)
Let's calculate
       u = (0.94 c - 0.43c) / (1+ 0.43c 0.94 c / c²)
       u = 0.51c / (1 + 0.4042)
       u = 0.36c
We repeat the calculation for the other piece
In this case u ’= - 0.35c
We calculate
        u = (0.94c + 0.35c) / (1 - 0.35c 0.94c / c²)
        u = 1.29c / (1- 0.329)
        u = 0.961c