Question

A 720-kg two-stage rocket is traveling at a speed of 6.90×10^3 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 2.10×10^3 m/s relative to each other along the original line of motion.What is the speed of each section (relative to Earth) after the explosion? R/= v1=5850, v2=7950. How much energy was supplied by the explosion? [Hint: What is the change in kinetic energy as a result of the explosion?] R/= 3.97*10^8 J. Could somebody help me to know why the answer to C is 3.9*10^8J?

Answer 1

Given

Rocket mass M = 720 kg , initial speed is v = 6.90*10^3 m/s

after explosion the speed is 2.10*10^3 m/s

and after the explosion the the speed is (v1-v2) = 2.10*10^3 m/s ==> v1 = v2+2.10*10^3 m/s

the two parts are of equal mass ( half of the initial mass M/2)

from conservation of momentum  

M*V = m1v1+m2*v2

M*V = M/2*v1+M/2*v2

V = 0.5(v1+v2)

v1+v2 = 2*V

substituting v1 in above eq

v2+2.10*10^3 +v2 = 2*6.90*10^3

solving for v2 , v2 = 5.850*10^3 m/s

and substituting the value in v1 = v2+2.10*10^3 m/s

v1 = 5.850*10^3+2.10*10^3 m/s

v1 = 7.950*10^3 m/s

v1 = 7950 m/s, v2 = 5850m/s

now the conservation of kinetic energy is  

the change in kientic energies are

D k.e = k.e _f - k.e_i

= (0.5*m1*v1^2 +0.5*m2*v2^2) - (0.5*M*V^2)

= (0.5*M)(0.5*v1^2+0.5*V2^2 -V^2)

substituting the values

= (0.5*720)(0.5*7950^2+0.5*5850^2-6900^2) J

= 396900000 J

= 3.9*10^8 J

so the energy supplied for the explosion is 3.9*10^8 J