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A fission reaction is given by${}_{92}{}^{236}U\to {}_{54}{}^{140}Xe+{}_{38}{}^{94}Sr+x+y$, where$x$and$y$are two particles. Considering${}_{92}{}^{236}U$to be at rest, the kinetic energies of the products are denoted by$K_{Xe},\mathrm{ }{K}_{Sr},\mathrm{ }{K}_x\left(2\mathrm{ }MeV\right)$and$K_y\left(2\mathrm{ }MeV\right)$, respectively. Let the binding energies per nucleon of${}_{92}{}^{236}U,\mathrm{ }{}_{54}{}^{140}Xe\mathrm{ }$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)