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