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A $\SI{10.0}{t}$ railroad car $\mathcal{A}$ travelling at a speed of $\SI{86.4}{\kilo\meter\per\hour}$ strikes an identical car $\mathcal{B}$ at rest. If the cars lock together as a result of the collision, what is their common speed just afterwards?
\newqty{m}{10.0e3}{kg} \newqty{ve}{24}{\mps} % The linear momentum before and after the collision remains the same, i.e. the linear momentum of \emph{one} vehicle before the collision is the same as the linear momentum of \emph{both together} after the collision. \ImpulsSchritte \PGleichung{2}{p_1 &= p_1' + p_2'} \PGleichung{3}{m_1v_1 &= m_1v_1' + m_2v_2'} \PGleichung{4}{mv_1 &= mv' + mv'} \AlgebraSchritte \MGleichung{2}{2mv' &= mv_1} \MGleichung{4}{v' &= \frac{v_1}{2}} \PHYSMATH Thus the their common speed is \solqty{w}{\frac{v_1}{2}}{\ven/2}{\mps} \al{ v' &= \wf \\ &= \frac{\ve}{2}\\ &= \wTTTT = \wTTT. } We can see that for two identical railroad cars, this speed does not depend on their mass.
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