A nuclide at rest emits an $\mathrm{\alpha }$-particle. In this process: 

As ${\mathrm{F}}_{\mathrm{ext}} = 0, \mathrm{hence} \overrightarrow{{\mathrm{P}}_{\mathrm{s}}} = \mathrm{constant}$

or $\overrightarrow{{\mathrm{P}}_{\mathrm{\alpha }}}+\overrightarrow{{\mathrm{P}}_{\mathrm{N}}} = \mathrm{constant}$

Because nuclide was at rest before emission of $\mathrm{\alpha }$ particle, hence according to law of conservation of momentum, $\overrightarrow{{\mathrm{P}}_{\mathrm{\alpha }}}+\overrightarrow{{\mathrm{P}}_{\mathrm{N}}} = 0$

or ${\mathrm{m}}_{\mathrm{\alpha }}\overrightarrow{{\mathrm{v}}_{\mathrm{\alpha }}}+ {\mathrm{m}}_{\mathrm{N}}\overrightarrow{{\mathrm{v}}_{\mathrm{N}}} = 0 \mathrm{or} \stackrel{\to }{{\mathrm{v}}_{\mathrm{\alpha }} } = -\frac{{\mathrm{m}}_{\mathrm{N}}}{{\mathrm{m}}_{\mathrm{\alpha }}}\overrightarrow{{\mathrm{v}}_{\mathrm{N}}}$

Thus, $\mathrm{\alpha }$-particle and the nucleus move in opposite directions. Further, as ${\mathrm{m}}_{\mathrm{N}} >> {\mathrm{m}}_{\mathrm{\alpha }}, \mathrm{hence} {\mathrm{v}}_{\mathrm{\alpha }}>>{\mathrm{v}}_{\mathrm{N}}$