Car A has a mass of 1400 kg and is approaching the intersection shown at the sam

Question

Car A has a mass of 1400 kg and is approaching the intersection shown at the sam time Car B with a mass of 1100 kg is approaching the same intersection. If the two cars are involved in a collision such that they become entangled and the ensuing wreckage moves with velocity v = 10i + 7j m/s, then determine the speed of each car prior to impact. Also determine the percentage of energy lost in the process of the collision. v_A,1 = 22.7 m/s, v_B,1 = 12.5 m/s v_A,1 = 12.5 m/s, v_B,1 = 22.7 m/s v_A,1 = 10.5 m/s, v_B,1 = 20.7 m/s v_A,1 = 20.7 m/s, v_B,1 = 10.5 m/s

Explanation / Answer

given that

ma = 1400 kg

mb = 1100 kg

V = 10i+7j m/s

using conservation of momentum

P before collision = P after collision

mb*vb i + ma*va j = (ma+mb) * V

1100*vbi + 1400*vbj = (1400+1100) * (10i+7j)

1100*vbi + 1400*vaj = 25000i + 17500j

1100*vbi = 25000i            .......eq1

1400*vaj = 17500j             ........eq2

from eq1

vbi = 25000/1100 = 22.72 i m/s

from eq2

vaj = 17500/1400 = 12.5 j m/s

so answer is (b)

part (b)

initial kinetic energy

KEi = 1/2*ma*va^2 + 1/2*mb*vb^2

KEi = 1/2*1400*(12.5)^2 + 1/2*1100*(22.72)^2

KEi = 393284.12 J

final kinetic energy after collision

KEf = 1/2*(ma+mb)*V^2

magnitude of V = sqrt (Vi^2+Vj^2)

KEf = 1/2 * (1400+1100) * ((10)2 + (7)2)

KEf = 1/2 * 2500 * 149

KEf = 186250 J

loss in kinetic energy = kEi - KEf / KEi

loss in kinetic energy = 393284.12 - 186250 / 393284.12 = 0.52

percentage loss = 52 %

answer

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