Respuesta :
The initial momentum of the system is,
[tex]p_i=m_1u_1+m_2u_2[/tex]The final momentum of the system is,
[tex]p_f=m_1v_1+m_2v_2[/tex]According to conservation of momentum,
[tex]p_i=p_f[/tex]Plug in the known expressions,
[tex]\begin{gathered} m_1u_1+m_2u_2=m_1v_1+m_2v_2 \\ m_2v_2=m_1u_1+m_2u_2-m_1v_1 \\ v_2=\frac{m_1u_1+m_2u_2-m_1v_1}{m_2} \end{gathered}[/tex]Substitute the known values,
[tex]\begin{gathered} v_2=\frac{(1381\text{ kg)(14.5 m/s)+(1381 kg)(13.4 m/s)-}(1381\text{ kg)(13.4 m/s)}}{1381\text{ kg}} \\ =\frac{(1381\text{ kg)(14.5 m/s)}}{1381\text{ kg}} \\ =14.5\text{ m/s} \end{gathered}[/tex]The initial kinetic energy of the system is,
[tex]K=\frac{1}{2}m_1u^2_1+\frac{1}{2}m_2u^2_2[/tex]The final kinetic energy of the system is,
[tex]K^{\prime}=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2[/tex]The change in kinetic energy of the system is,
[tex]\Delta K=K^{\prime}-K[/tex]Plug in the known expressions,
[tex]\begin{gathered} \Delta K=\frac{1}{2}m_2v^2_2+\frac{1}{2}m_1v^2_1-(\frac{1}{2}m_2u^2_2+\frac{1}{2}m_1u^2_1) \\ =\frac{1}{2}m_2(v^2_2-u^2_2)+\frac{1}{2}m_1(v^2_1-u^2_1) \end{gathered}[/tex]Substitute the known values,
[tex]\begin{gathered} \Delta K=\frac{1}{2}(1381kg)((14.5m/s)^2-(13.4m/s)^2)- \\ \frac{1}{2}(1381kg)((13.4m/s)^2-(14.5m/s)^2) \\ =(690.5\text{ kg)(}210.25m^2s^{-2}-179.56\text{ }m^2s^{-2})-(690.5\text{ kg)(}179.56\text{ }m^2s^{-2}-210.25\text{ }m^2s^{-2}) \\ =(690.5\text{ kg)(}30.69\text{ }m^2s^{-2})(\frac{1\text{ J}}{1\text{ kg}m^2s^{-2}})-(690.5\text{ kg)(-30.69 }m^2s^{-2}) \\ =21191.4\text{ J+21191.4 J} \\ =42382.8\text{ J} \end{gathered}[/tex]Therefore, the change in kinetic energy of the system is 42382.8 J.
