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A nucleus breaks into two nuclear parts, which have their velocity ratio $2: 1$. The ratio of their nuclear radii will be
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Verified Answer
The correct answer is:
$\frac{1}{2^{1 / 3}}$
By law of conservation of momentum, the two parts will have equal and opposite momentum, assuming the nucleus was initially at rest.
$$
\begin{aligned}
& \therefore \mathrm{m}_1 \mathrm{v}_1=\mathrm{m}_2 \mathrm{v}_2 \text { or } \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{\mathrm{v}_2}{\mathrm{v}_1}=\frac{1}{2} \\
& \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^3=\frac{1}{2} \\
& \therefore \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{1}{2^{1 / 3}}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \mathrm{m}_1 \mathrm{v}_1=\mathrm{m}_2 \mathrm{v}_2 \text { or } \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\frac{\mathrm{v}_2}{\mathrm{v}_1}=\frac{1}{2} \\
& \frac{\mathrm{m}_1}{\mathrm{~m}_2}=\left(\frac{\mathrm{r}_1}{\mathrm{r}_2}\right)^3=\frac{1}{2} \\
& \therefore \frac{\mathrm{r}_1}{\mathrm{r}_2}=\frac{1}{2^{1 / 3}}
\end{aligned}
$$
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