A nucleus ruptures into two nuclear parts, which have their velocity ratio equal to $$2:1.$$  What will be the ratio of their nuclear size (nuclear radius)?

Solution :

From law of conservation of momentum
$$\eqalign{ & 0 = {m_1}{v_1} + {m_2}{v_2} \cr & \therefore {m_1}{v_1} = - {m_2}{v_2} \cr & {\text{or}}\,\,\frac{{{v_1}}}{{{v_2}}} = - \frac{{{m_1}}}{{{m_2}}} = \frac{2}{1} \cr} $$
One sign indicates that velocity is in opposite direction
As nucleus is assumed to be spherical of radius $$r,$$ density $$\rho .$$
$$\therefore m = \frac{4}{3}\pi {r^3}\rho \Rightarrow m \propto {r^3}$$
So for two different parts of nuclei,
$$\eqalign{ & \frac{{{m_2}}}{{{m_1}}} = \frac{{r_2^3}}{{r_1^3}} \cr & \therefore \frac{{{v_1}}}{{{v_2}}} = \frac{{r_2^3}}{{r_1^3}}\,\,{\text{or}}\,\,\frac{{{r_1}}}{{{r_2}}} = {\left( {\frac{{{v_2}}}{{{v_1}}}} \right)^{\frac{1}{3}}} = {\left( {\frac{1}{2}} \right)^{\frac{1}{3}}} \cr & \Rightarrow {r_1}:{r_2} = 1:{2^{\frac{1}{3}}} \cr} $$