JEE Advanced 2015 · Paper 2, Physics question 16

A fission reaction is given by $^{236}_{92}\mathrm U\to{}^{140}_{54}\mathrm{Xe}+{}^{94}_{38}\mathrm{Sr}+x+y$, where $x$ and $y$ are two particles. Considering $^{236}_{92}\mathrm U$ 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 $^{236}_{92}\mathrm U$, $^{140}_{54}\mathrm{Xe}$ and $^{94}_{38}\mathrm{Sr}$ be 7.5 MeV, 8.5 MeV and 8.5 MeV, respectively. Considering different conservation laws, the correct option(s) is(are)