Asked by Jay
On an icy road, a 2100-kg car moving at 52 km/h strikes a 410-kg truck moving in the same direction at 35 km/h. The pair is soon hit from behind by a 1800-kg car speeding at 45 km/h, and all three vehicles stick together. Find the speed of the wreckage.
Answers
Answered by
henry2,
Given: M1 = 2100 kg, V1 = 52 km/h.
M2 = 410 kg, V2 = 35km/h.
M3 = 1800 kg, V3 = 45 km/h.
V4 = velocity of M1 and M2 after collision.
V5 = velocity of M1, M2, and M3 after collision.
Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V4 + M2*V4.
2100*52 + 410*35 = 2100*V4 + 410*V4,
2510V4 = 123,550,
V4 = 49.22 km/h.
M3 Joins Wreckage:
M1*V4 + M2*V4 + M3*V3 = M1*V5 + M2*V5 + M3*V5.
2100*49.22 + 410*49.22 + 1800*45 = 2100V5 + 410V5 + 1800V5,
4310V5 = 204,542,
V5 = 47.5 km/h.
M2 = 410 kg, V2 = 35km/h.
M3 = 1800 kg, V3 = 45 km/h.
V4 = velocity of M1 and M2 after collision.
V5 = velocity of M1, M2, and M3 after collision.
Momentum before = Momentum after.
M1*V1 + M2*V2 = M1*V4 + M2*V4.
2100*52 + 410*35 = 2100*V4 + 410*V4,
2510V4 = 123,550,
V4 = 49.22 km/h.
M3 Joins Wreckage:
M1*V4 + M2*V4 + M3*V3 = M1*V5 + M2*V5 + M3*V5.
2100*49.22 + 410*49.22 + 1800*45 = 2100V5 + 410V5 + 1800V5,
4310V5 = 204,542,
V5 = 47.5 km/h.
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