Answer
$1.187\times10^{-12}\;J$
Work Step by Step
The linear momentum of Particle 3:
$\vec{P_3}=-(1.002\times10^{-19}\hat i+6.68\times10^{-20}\hat j)\;kg.m/s$
Therefore, the magnitude of the linear momentum of Particle 3 is given by
$P_3=\sqrt {(1.002\times10^{-19})^2+(6.68\times10^{-20})^2}\;kg.m/s$
or, $P_3=1.204\times10^{-19}\;kg.m/s$
Now the kinetic energy appears in this transformation is given by
$K.E=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\frac{1}{2}m_3v_3^2$
or, $K.E=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2+\frac{P_3^2}{2m_3}$
or, $K.E=\frac{1}{2}\times16.7 \times10^{-27}\times(6.00 \times10^6)^2+\frac{1}{2}\times8.35 \times10^{-27}\times(8.00 \times10^6)^2+\frac{(1.204\times10^{-19})^2}{2\times11.7 \times10^{-27}}\;J$
or, $K.E=1.187\times10^{-12}\;J$
