Explanation:
Let \({A_1}\) and \({A_2}\) be the mass number of the two nuclear parts. Their radii are given by
\({R_1} = {R_0}A_1^{1/3}\,\,\,\,\,\,(1)\)
and \({R_2} = {R_0}A_2^{1/3}\,\,\,\,\,\,(2)\)
Dividing eqn. (1) by eqn(2), we get
\(\frac{{{R_1}}}{{{R_2}}} = {\left( {\frac{{{A_1}}}{{{A_2}}}} \right)^{1/3}}or\frac{{{A_1}}}{{{A_2}}} = {\left( {\frac{{{R_1}}}{{{R_2}}}} \right)^3}\)
As \(\frac{{{R_1}}}{{{R_2}}} = \frac{1}{2}\) (given)
\(\therefore \frac{{{A_1}}}{{{A_2}}} = {\left( {\frac{1}{2}} \right)^3} = \frac{1}{8}\)
Hence the ratio of their masses is
\(\frac{{{m_1}}}{{{m_2}}} = \frac{1}{8}\) (3)
(\(\therefore \) mass is approximately proprtional to \(A\))
According to law of conservation of linear momentum
magnitude of \({{\vec p}_1} = \) magnitude of \({{\vec p}_2}\)
i.e., \({m_1}{v_1} = {m_2}{v_2}\)
or \(\frac{{{v_1}}}{{{v_2}}} = \frac{{{m_2}}}{{{m_1}}} = \frac{8}{1}\) (using (3))