(a) The final velocity if the trucks lock together after the collision is 7.66 m/s.
(b) The decrease in the kinetic energy after the collision is 11,165.1 J.
The final velocity of the two truck system after the collision is determined by applying the principle of conservation of linear momentum as shown below;
m₁u₁ + m₂u₂ = v(m₁ + m₂)
4285(9.35) + 5495(6.35) = v(4285 + 5495)
74,958 = 9,780v
v = 74,958/9780
v = 7.66 m/s
K.E = ¹/₂mv²
K.Ei = ¹/₂(4285)(9.35)² + ¹/₂(5495)(6.35)²
K.Ei = 298,088.78 J
K.Ef = ¹/₂(m₁ + m₂)v²
K.Ef = ¹/₂(4285 + 5495)(7.66)²
K.Ef = 286,923.68 J
ΔK.E = K.Ef - K.Ei
ΔK.E = 286,923.68 J - 298,088.78 J
ΔK.E = - 11,165.1 J
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