Respuesta :
From the Law of Conservation of Linear Momentum, we have:
[tex]m_1v_1+m_2v_2=m_1v_1^{\prime}+m_2v_2^{\prime}[/tex]If the two particles have the same velocity after the collision, then v₁'=v₂'.
Let v be equal to the final velocity of the particles. Then:
[tex]\begin{gathered} m_1v_1+m_2v_2=m_1v+m_2v \\ \\ \Rightarrow m_1v_1+m_2v_2=(m_1+m_2)v \end{gathered}[/tex]Since v is unknown, isolate it from the equation:
[tex]v=\frac{m_1v_1+m_2v_2}{m_1+m_2}[/tex]Replace the data to find the value of v:
[tex]\begin{gathered} m_1=5kg \\ v_1=5\frac{m}{s} \\ \\ m_2=2kg \\ v_2=0 \\ \\ \Rightarrow v=\frac{(5kg)(5\frac{m}{s})+(2kg)(0)}{5kg+7kg}=\frac{25kg\frac{m}{s}}{12kg}=2.08333...\frac{m}{s} \end{gathered}[/tex]Therefore, the combined final velocity of both train cars is approximately 2.1m/s.
