A 9300 kg boxcar traveling at 15.0 m/s strikes a second boxcar at rest. The two...

Question

Question

Answer 1

Answer #1

m1 = 9300 kg, u1 = 15 m/s, u2 =0 , v =6 m/s

(1) From conservation of momentum

m1u1 +m2u2 = (m1+m2)v

(9300*15) + 0 = (9200+ m2)*6

m2 = 14050 kg

(2) K.Ef = (1/2)(m1+m2)v^2

K.Ef = 0.5*(9300+14050)*6^2

K.Ef = 420300 J

K.Ei = (1/2)mu1^2 = 0.5*9300*15^2

K.Ei = 1046250 J

Thermal energy = K.Ei - K.Ef

= 1046250 - 420300 = 625950 J

Answer 2

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