Answer:2.5 m/s
37.5KJ
Explanation:
Let [tex]u_1, u_2 , v_f[/tex] be the initial velocity of rail road car ,coupled cars & Final velocity of system respectively.
[tex]m=2.50\times 10^{4}[/tex]
Conserving momentum
[tex]mu_1+3mu_2=4mv_f[/tex]
[tex]4m+6m=4mv_f[/tex]
[tex]v_f=2.5 m/s[/tex]
Therefore Final velocity of system is 2.5m/s
(b)Mechanical Energy lost =Initial Kinetic Energy -Final Kinetic Energy
[tex]Initial Kinetic Energy=\frac{1}{2}m\left ( 4^2\right )+\frac{1}{2}m\left ( 2^2\right )=14m J[/tex]
[tex]Final Kinetic Energy=\frac{1}{2}4m\left ( 2.5^2\right )=12.5m J[/tex]
[tex]Mechanical Energy lost=14m-12.5m=3.75\times 10^4=37.5 KJ[/tex]