Explanation:
Given that,
Mass of a freight car, [tex]m_1=30,000-kg[/tex]
Speed of a freight car, [tex]u_1=0.85\ m/s[/tex]
Mass of a scrap metal, [tex]m_2=110,000\ kg[/tex]
(a) Let us assume that the final velocity of the loaded freight car is V. The momentum of the system will remain conserved as follows :
[tex]30000\times 0.85=(30000+110000)V\\\\V=\dfrac{30000\times 0.85}{30000+110000}\\\\=0.182\ m/s[/tex]
So, the final velocity of the loaded freight car is 0.182 m/s.
(b) Lost on kinetic energy = final kinetic energy - initial kinetic energy
[tex]\Delta K=\dfrac{1}{2}[(m_1+m_2)V^2-m_1u_1^2)]\\\\=\dfrac{1}{2}\times [(30,000+110,000 )0.182^2-30000(0.85)^2]\\\\=-8518.82\ J[/tex]
Lost in kinetic energy is 8518.82. Negative sign shows loss.