A heavy nucleus at rest breaks into two fragm | vedclass

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(A) According to the law of conservation of linear momentum,the initial momentum is zero,so the magnitudes of the momenta of the two fragments must be

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$1: 3^{1/3}$

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(A) According to the law of conservation of linear momentum,the initial momentum is zero,so the magnitudes of the momenta of the two fragments must be equal:$m_1 v_1 = m_2 v_2$$\frac{m_1}{m_2} = \frac{v_2}{v_1} = \frac{1}{3}$Assuming the density $\rho$ of the nucleus is constant,the mass $m$ is proportional to the volume $V = \frac{4}{3} \pi R^3$,where $R$ is the radius.$\frac{m_1}{m_2} = \frac{R_1^3}{R_2^3}$Equating the two expressions for the mass ratio:$\frac{R_1^3}{R_2^3} = \frac{v_2}{v_1} = \frac{1}{3}$Taking the cube root on both sides:$\frac{R_1}{R_2} = \left(\frac{1}{3}\right)^{1/3} = 1 : 3^{1/3}$

$A$ heavy nucleus at rest breaks into two fragments which fly off with velocities in the ratio $3: 1$. The ratio of radii of the fragments is

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