Answer:
23.0760769 %
Explanation:
[tex]m_1[/tex] = Mass of freight car = 15000 kg
[tex]m_2[/tex] = Mass of second car = 50000 kg
[tex]v_1[/tex] = Velocity of freight car = 2 m/s
[tex]v_2[/tex] = Velocity of second car = 0
v = Combined mass velocity
As the linear momentum of the system is conserved we have
[tex]m_1v_1+m_2v_2=(m_1+m_2)v\\\Rightarrow v=\dfrac{m_1v_1+m_2v_2}{m_1+m_2}\\\Rightarrow v=\dfrac{15000\times 2+50000\times 0}{15000+50000}\\\Rightarrow v=0.46153\ m/s[/tex]
The initial kinetic energy
[tex]K_i=\dfrac{1}{2}15000\times 2^2\\\Rightarrow K_i=30000\ J[/tex]
Final kinetic energy
[tex]K_f=\dfrac{1}{2}(15000+50000)\times 0.46153^2\\\Rightarrow K_f=6922.82307\ J[/tex]
The percentage is given by
[tex]\dfrac{6922.82307}{30000}\times 100=23.0760769\ \%[/tex]
The change in percentage of initial kinetic energy is 23.0760769 %
