A heavy nucleus at rest breaks into two fragm | vedclass

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(A) According to the law of conservation of linear momentum,for a nucleus initially at rest,the magnitudes of momenta of the two fragments must be equ

Vedclass · Accepted Answer

$1: 3^{1/3}$

Vedclass · Answer

(A) According to the law of conservation of linear momentum,for a nucleus initially at rest,the magnitudes of momenta of the two fragments must be equal: $m_1 v_1 = m_2 v_2$. This implies $\frac{m_1}{m_2} = \frac{v_2}{v_1}$. Given the velocity ratio $\frac{v_1}{v_2} = \frac{3}{1}$,we have $\frac{m_1}{m_2} = \frac{1}{3}$. Assuming the density $\rho$ of the nucleus is constant,the mass $m$ is proportional to the volume,so $m = \rho \cdot \frac{4}{3} \pi R^3$. Therefore,$\frac{m_1}{m_2} = \frac{R_1^3}{R_2^3}$. Equating the two expressions 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,we get $\frac{R_1}{R_2} = (\frac{1}{3})^{1/3} = 1 : 3^{1/3}$.

$A$ 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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