Question

Car A was traveling east at high speed when it collided at point with car B, which was traveling north at 46 mi/h. Car C, which was traveling west at 61 mi/h, was 32 ft east and 10 ft north of point at the time of the collision. Because the pavement was wet, the driver of car C could not prevent his car from sliding into the other two cars, and the three cars, stuck together, kept sliding until they hit the utility pole P. Knowing that the weights of cars A, B, and Care, respectively, 3000 lb, 2600 lb, and 2400 lb, and neglecting the forces exerted on the cars by the wet pavement, solve the problems indicated. Knowing that the speed of car A was 77 mi/h and that the time elapsed from the first collision to the stop at P was 2.4 s, determine the coordinates of the utility pole P. (Round the final answer to two decimal places.) УР MU A 0 B The coordinates of pole P are: Xp = ft ура ft

Answer 1

m MA = 3000 lb BE 2600Lb mc= 2400 lb NA= 112.93 î ft/s VB= 67.47 Ĵ ft/s Ve=89.47 (i) fils Уr cons nserved As the momentum will be after collision of A and B C MATA IMBUB=(MAX MB) W (3000x112.93) i +(2600X67.47) ſ 5600 vi -

> ✓ = 60.58 +31-333 Now, the car c was at position Be= 32 i + 10 As A, B, C three cars collide, so they have to collide at Loft north of O. Now, after collision of A and B, the vertical component of the car is 31.33 Û ft/s. 31.33 xt=10 10 > Z = 31-33 = 0.325

cars Now, in 0-32s the A and B will go = (60-58 0:32) It eastwards 19.36 fZ eastwards So, the second collision will happen at To = 19.3.6 ito As after the second collision momentum will be conserved (MAIMB) + marca = (matmot me ve => 5600 (60.5 î +3133 9 )+2400(-89.47) Vé = 15:51 9 +21.93 ĝ

as, the total time ellapsed after first collision is 2.45. 6. total time ellapsed after second collision is t2 (2.4-0.32)s : 2.085 rap + + Jat = So, the final position of the utility pole is ržt = 19.36 +10j +{2.08*(15-51 î +21-93;)} = 51.62 i + 55.61 :. Xp2 51-62 ft, yp=55-61 ft