Question

A 790-kg two-stage rocket is traveling at a speed of 6.90×103 m/s away from Earth when a predesigned explosion separates the rocket into two sections of equal mass that then move with a speed of 2.80×103 m/s relative to each other along the original line of motion.How much energy was supplied by the explosion? [Hint: What is the change in kinetic energy as a result of the explosion?]

Answer #1

initial KE = 0.5*790*(6900)^2 = 1.88*10^10 J

initial momentum MiVi = 790*6900 = 5.45*10^6 kg-m/s

final masses are m , m and final speeds are v1 , v2 , and 2m = 790 kg , v1- v2 = 2800 m/s

according to momentum conservation

mv = mv1 + mv2

=> 5.45*10^6 = m(v1+v2) =395*(v1 + v2)

=> v1 + v2 = 13.79*10^3 m/s ........eq1

and v1 - v2 = 2.8*10^3 m/s ........eq2

from both equations we get v1 = 8.29*10^3 m/s , v2 = 5.49*10^3 m/s

final KE = 0.5*m*( v1^2 + v2^2 ) = 0.5*395*( 8290^2 + 5490^2) = 1.95*10^10 J

so energy supplied by explosion

KE = KEf - KEi = (1.95 - 1.88) = 0.07*10^10 = 700 MJ

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