A fission reaction is given by $$_{92}^{236}U \rightarrow _{54}^{140}Xe + _{38}^{94}Sr + x + y$$ where x and y are two particles. Considering $$_{92}^{236}U$$ to be at rest, the kinetic energies of the products are denoted by $$K_{Xe}, K_{Sr}, K_x (2 MeV)$$ and $$K_y (2 MeV),$$ respectively. Let the binding energies per nucleon of $$_{92}^{236}U, _{54}^{140}Xe$$ and $$_{38}^{94}Sr$$ be 7.5 MeV, 8.5 MeV and 8.5 MeV,respectively. Considering different conservation laws, the correct option(s) is(are)