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 values,
[tex]m_1u_{1_{}}+m_2u_2=m_1v_1+m_2v_2[/tex]Substitute the known values,
[tex]\begin{gathered} (1230kg)(14.6m/s)+(1278kg)(13m/s)=(1230kg)(13.6m/s)+(1278kg)v_2 \\ 17958\text{ kgm/s+}16614\text{ kgm/s=}16728\text{ kgm/s+}(1278kg)v_2 \\ v_2=\frac{34572\text{ kgm/s-}16728\text{ kgm/s}}{1278\text{ kg}} \\ =13.96\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-v^2_1)+\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}(1278kg)((13.96m/s)^2-(13m/s)^2)+ \\ \frac{1}{2}(1230kg)((13.6m/s)^2-(14.6m/s)^2) \\ =(639\text{ kg)(}194.88m^2s^{-2}-169m^2s^{-2})+(615\text{ kg)(}184.96m^2s^{-2}-213.16m^2s^{-2}) \\ =(639\text{ kg)(}25.88\text{ }m^2s^{-2})(\frac{1\text{ J}}{1\text{ kg}m^2s^{-2}})+(615\text{ kg)(}-28.2m^2s^{-2})(\frac{1\text{ J}}{1\text{ kg}m^2s^{-2}}) \\ =16537.32\text{ J-}17343\text{ J} \\ =-805.68\text{ J} \end{gathered}[/tex]Therefore, the change in kinetic energy of the system is -805.68 J.
