Respuesta :
From the Law of Conservation of Linear Momentum, we know that:
[tex]m_1v_1+m_2v_2=m_1v_1^{\prime}+m_2v_2^{\prime}[/tex]Where v₁, v₂ are the initial velocities of masses m₁ and m₂ respectively, and v₁', v₂' are their final velocities.
In the given problem, the mass of the second particle is unknown. Isolate m₂ from the equation:
[tex]m_2=\frac{m_1(v_1^{\prime}-v_1)}{(v_2-v_2^{\prime})}[/tex]Replace the data into the equation to find m₂:
[tex]\begin{gathered} m_1=10kg \\ v_1=10\frac{m}{s} \\ v_1^{\prime}=5\frac{m}{s} \\ \\ m_2=\text{ unknown} \\ v_2=0 \\ v_2^{\prime}=25\frac{m}{s} \end{gathered}[/tex]The initial velocity of the second particle is 0 because the problem says it is stationary.
[tex]m_2=\frac{(10kg)(5\frac{m}{s}-10\frac{m}{s})}{(0-25\frac{m}{s})}=2kg[/tex]Therefore, the mass of the second train car is 2kg.
