An airplane of mass 30 Mg is flying due east at 700 kph at an altitude of 12,000
ID: 2995446 • Letter: A
Question
An airplane of mass 30 Mg is flying due east at 700 kph at an altitude of 12,000 m. When is directly above point P, it suddenly explodes into 3 fragments (A, B and C) which then land in a dense forest. NOT TO SCALE Assuming that all three fragments hit the ground at the same time, how long did it take for the fragments to fall? Neglect any air resistance. The locations where the fragments land are shown below relative to point Q 10 km directly east of point P. Complete the missing entries in bold. Fragment Mass Location north/south of point Q Location east/west of point Q A 15 kg 95 m south 125 m east B 6 kg 135 m north 170 m west C kg m of Q north/south m of Q east/westExplanation / Answer
a) since same time for all fragments to fall, they are split horizontally.
Hence time taken = sqrt(2h/g) = sqrt(2*12000/9.8) = 49.487 seconds
b) Total momentum in north-south direction was 0 intially.
Hence final momentum must also be 0
6*135 - 15*95 + s1*m = 0 (here since time taken is same, distances are proportional to speeds)
m*s1 = 615
Total mass = 30 Mg = 30,000 Kg
Hence m = 30000-6-15 = 29979 kg
s1 = 615/29979 = 0.0205 m north of Q
Total momentum in East-West direction = 0
30000*(700*5/18)*49.487 = 29979*s2 + 15*10125 + 6*9830
s2 = 9622.18m
Hence mass of C = 29979 kg
Location is 0.0205m North of Q and 377.82m West of Q