Let m = mass of the nucleus at rest. $\vec{u}$ = its initial velocity = 0 as it is at rest.Let $m_1, m_2$ be the masses of the two smaller nuclei also called product nuclei and be their respective velocities.If $\vec{p}_i$ and $\vec{p}_f$ be the initial and final momentum of the nucleus and the two nuclei respectively, then$\vec{p}_i = m \vec{u} = 0$...(i)$\vec{p}_f = m_1 \vec{v}_1 + m_2 \vec{v}_2 \dots (ii)$According to the law of conservation of linear momentum,$\vec{p}_i = \vec{p}_f$ or $0 = m_1 \vec{v}_1 + m_2 \vec{v}_2$or $m_2 \vec{v}_2 = -m_1 \vec{v}_1$ or $\vec{v}_2 = -\frac{m_1 \vec{v}_1}{m_2}$...(iii)The negative sign in equation (iii) shows that $\vec{v}_1$ and $\vec{v}_2$ are in oppositedirections i.e. the two smaller nuclei are moved in opposite directions.