A nucleus ruptures into two nuclear parts whi | vedclass

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(B) According to the law of conservation of linear momentum,$m_1v_1 = m_2v_2$.Given the velocity ratio $v_1 : v_2 = 2 : 1$,we have $\frac{v_1}{v_2} =

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$1 : 2^{1/3}$

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(B) According to the law of conservation of linear momentum,$m_1v_1 = m_2v_2$.Given the velocity ratio $v_1 : v_2 = 2 : 1$,we have $\frac{v_1}{v_2} = 2$.From the momentum conservation equation,the mass ratio is $\frac{m_1}{m_2} = \frac{v_2}{v_1} = \frac{1}{2}$.Since the mass number $A$ is proportional to the mass of the nucleus,$\frac{A_1}{A_2} = \frac{1}{2}$.The nuclear radius is given by $R = R_0 A^{1/3}$.Therefore,the ratio of their nuclear radii is $\frac{R_1}{R_2} = \left(\frac{A_1}{A_2}\right)^{1/3} = \left(\frac{1}{2}\right)^{1/3} = 1 : 2^{1/3}$.

$A$ 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)?

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