A nucleus at rest splits into two nuclear par | vedclass

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(A) According to the law of conservation of linear momentum,for a nucleus initially at rest,the total momentum after splitting must be zero.$m_1 v_1 =

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$8$

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(A) According to the law of conservation of linear momentum,for a nucleus initially at rest,the total momentum after splitting must be zero.$m_1 v_1 = m_2 v_2 \implies \frac{v_1}{v_2} = \frac{m_2}{m_1}$Since the density of nuclear matter is constant,the mass $m$ of a nucleus is proportional to its volume,which is proportional to the cube of its radius $R$.$m \propto R^3 \implies \frac{m_2}{m_1} = \left(\frac{R_2}{R_1}\right)^3$Given the ratio of radii $\frac{R_1}{R_2} = \frac{1}{2}$,we have $\frac{R_2}{R_1} = \frac{2}{1}$.Substituting this into the velocity ratio equation:$\frac{v_1}{v_2} = \left(\frac{2}{1}\right)^3 = \frac{8}{1}$.Thus,the ratio of their velocities is $8:1$.

$A$ nucleus at rest splits into two nuclear parts having radii in the ratio $1:2$. Their velocities are in the ratio: (in $:1$)

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