Question

A 460-kg space vehicle traveling with a velocity Vo = ( 400 m/s)i passes through the origin O att = 0. Explosive charges then separate the vehicle into three parts, A, B, and C, with mass, respectively, 230 kg, 100 kg, and 130 kg. Knowing that at the positions of parts A and B are observed to be A (1170 m, -290 m, -585 m) and B( 1975 m, 365 i 800 m), determine the corresponding position of part C. Neglect the effect of gravity The position of part Cisrc =( m) + m) + m)

Answer 1

Som Intial mass (M)= ubokg. v = 400 e After explosion, MA = 230ks Mc = 130kg. m₂ = looky final co-ordinaty of A = (1170m - 290m - 585m) ¢ B - (1975m, 36m, 8oom) in vector form 7 = 1170ê - 290] - 585 F Xo = 1975 mg + 365 ↑ 18ookt we know velocity ū= x = 4 te t = uha t To = 11701 - 290]-S85o, Te = 19750 +385 § +800k 4 Applyly confervation of momentum. MJ = MAVA + mo yo t mcvc. 466x you î = 230/11701-290 ] - 5852) 11975ê +365] +fook 11 6725 ta 130 To î - 2580 0j + 114550 = 180o a V =89788 7 #198.46 +881.150 ml 1 Vol = 127382 majs