Answer:
a) 1.67 m/s
b) 23kJ
Explanation:
We need to apply the linear momentum conservation formula, that states:
[tex]m1*v_{o1}+m2*v_{o2}=m1*v_{f1}+m2*v_{f2}[/tex]
in this case:
[tex]3.45*10^4kg*2.60m/s+2*3.45*10^4kg*1.20m/s=3*m1*v_{f}\\v_f=1.67m/s[/tex]
the initital kinetic energy is:
[tex]K_i=\frac{1}{2}*3.45*10^4kg*(2.60m/s)^2+2(\frac{1}{2}*3.45*10^4kg*(1.20m/s)^2\\K_i=167kJ[/tex]
and the final:
[tex]K_f=3*\frac{1}{2}*3.45*10^4kg*(1.67m/s)^2\\K_f=144kJ[/tex]
The energy lost is given by:
[tex]E_l=|K_f-K_i|\\E_l=23kJ[/tex]
