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(D) According to the law of conservation of linear momentum,the initial momentum is zero,so $m_1v_1 = m_2v_2$.This implies $\frac{m_1}{m_2} = \frac{v_

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$1/2$

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(D) According to the law of conservation of linear momentum,the initial momentum is zero,so $m_1v_1 = m_2v_2$.This implies $\frac{m_1}{m_2} = \frac{v_2}{v_1}$.Given the ratio of velocities $\frac{v_1}{v_2} = \frac{8}{1}$,so $\frac{v_2}{v_1} = \frac{1}{8}$.Assuming the density $\rho$ of the nucleus is constant,the mass $m$ is proportional to the volume,$m = \rho \cdot \frac{4}{3}\pi r^3$.Therefore,$\frac{m_1}{m_2} = \frac{r_1^3}{r_2^3}$.Equating the two expressions: $\frac{r_1^3}{r_2^3} = \frac{v_2}{v_1} = \frac{1}{8}$.Taking the cube root on both sides: $\frac{r_1}{r_2} = \sqrt[3]{\frac{1}{8}} = \frac{1}{2}$.

$A$ heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio $8 : 1$. Find the ratio of the radii of the fragments.

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