Answer:
[tex]\Delta KE = 218.375\ kJ[/tex]
Explanation:
Given,
Mass of the truck, M = 1470 Kg
initial speed of the truck, u = 27 m/s
mass of car, m = 1010 Kg
initial speed of car, u' = 0 m/s
Final speed, V = ?
Using conservation of momentum
[tex]Mu = (M+m) V[/tex]
[tex]1470\times 27 = (1470 + 1010 ) V[/tex]
[tex] V = 16\ m/s[/tex]
Change in kinetic energy
[tex]\Delta KE = \dfrac{1}{2} Mu^2 - \dfrac{1}{2}(M+m)V^2[/tex]
[tex]\Delta KE = \dfrac{1}{2}\times 1470\times 27^2 - \dfrac{1}{2}\times (1470 + 1010)\times 16^2[/tex]
[tex]\Delta KE = 218.375\ kJ[/tex]
Change in KE = [tex]\Delta KE = 218.375\ kJ[/tex]
