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Tags breakup, linear mometum, mechanics, momentum conservation law, physics Difficulty Points 2 Language Type Quantity Last modified 2 months, 2 weeks ago Initial Version by Urs Zellweger (urs)	AG - Radioaktive Emission (Impuls) (urs) Unelastischer Stoss (patrik) Unelastischer Stoss (urs)	0

Exercise

An atomic nucleus initially moving at $\SI{420}{\meter\per\second}$ emits an $\upalpha$-particle in the direction of its velocity, and the remaining nucleus slows to $\SI{350}{\meter\per\second}$. If the $\upalpha$-particle has a mass of $\SI{4.0}{u}$ and the original nucleus has a mass of $\SI{222}{u}$, what speed does the alpha particle have when it is emitted?

Solution

\newqty{v}{420}{\mps} \newqty{we}{350}{\mps} \newqty{mz}{4.0}{u} \newqty{M}{222}{u} We can determine the speed using momentum conservation during the emission. \ImpulsSchritte \PGleichung{2}{\ssc{p}{N} + p_\alpha &= \ssc{p}{N}' + p_\alpha'} \PGleichung{3}{\ssc{m}{N}\ssc{v}{N} + m_\alpha v_\alpha &= \ssc{m}{N}\ssc{v}{N}' + m_\alpha v_\alpha'} \PGleichung{4}{(M-m_\alpha)v + m_\alpha v &= (M-m_\alpha)\ssc{v}{N}' + m_\alpha v_\alpha'} \AlgebraSchritte \MGleichung{1}{Mv &= (M-m_\alpha)\ssc{v}{N}' + m_\alpha v_\alpha'} \MGleichung{2}{m_\alpha v_\alpha' &= Mv - (M-m_\alpha)\ssc{v}{N}'} \MGleichung{4}{v_\alpha' &= \frac{Mv - (M-m_\alpha)\ssc{v}{N}'}{m_\alpha}} % \PHYSMATH Inserting the numbers leads to \solqty{wz}{\frac{Mv - (M-m_\alpha)\ssc{v}{N}'}{m_\alpha}}{(\Mn*\vn - (\Mn-\mzn)*\wen)/\mzn}{\mps} \al{ v_\alpha' &= \wzf \\ &= \frac{\M \cdot \v - \qty(\M-\mz)\cdot \we}{\mz}\\ &= \wz = \wzIII. }

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