The radioactive core decayed into three fragments of mass m_1, m_2, and m_3, having mutually perpendicular velocities v_1, v_2, and v_3, respectively. What was the speed of the nucleus before the decay?
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<h3> Statement </h3>
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$2.2.16$
The radioactive core decayed into three fragments of mass $m_1$, $m_2$, and $m_3$, having mutually perpendicular velocities $v_1$, $v_2$, and $v_3$, respectively. What was the speed of the nucleus before the decay?
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<h3>Solution</h3>
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Let's consider the following figure
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Velocity's analysis
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From Momentum Conservation Law (MCL)
$$(m_1+m_2+m_3)\~\vec{V} = m_1\vec{v_1}+m_2\vec{v_2}+m_3\vec{v_3}$$
Since velocity's vectors are mutually perpendicular, their directions can be fundamental for establishing the spatial coordinated system. So, let be $\hat{v_1}$, $\hat{v_2}$ and $\hat{v_3}$ the unitary vectors for each direction. Applying MCL for each direction,
$$(m_1+m_2+m_3)V_1 = m_1v_1 \\;(1)$$
$$(m_1+m_2+m_3)V_2 = m_2v_2 \\;(2)$$
$$(m_1+m_2+m_3)V_3 = m_3v_3 \\;(3)$$
where $V_1$, $V_2$ and $V_3$ are the components of velocity $\vec{V}$,
$$V=\sqrt{V_1^2+V_2^2+V_3^2} \\;(4)$$
Substituting $(1)$, $(2)$ and $(3)$ into $(4)$
$$\boxed{V=\frac{\sqrt{m_1^2v_1^2+m_2^2v_2^2+m_3^2v_3^2}}{m_1+m_2+m_3}}$$
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<p style="text-align: right; font-style: italic; font-size: 14;">
BSc. Luis Daniel Fernández Quintana<br>
Physics Department (FCNE)<br>
Universidad de Oriente, Cuba<br>
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