Question

A 780-kg two-stage rocket is traveling at a speed of 6.90×103 m/s away from Earth when...

A 780-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 2.50×103 m/s relative to each other along the original line of motion. a.)What is the speed of each section (relative to Earth) after the explosion? Express your answers using three significant figures separated by a comma. b.)What are the direction of each section (relative to Earth) after the explosion? c.)How much energy was supplied by the explosion? [Hint: What is the change in kinetic energy as a result of the explosion?]

Homework Answers

Answer #1

let M = 780 kg
v = 6.9*10^3 m/s

let m1 = m2 = M/2 = 780/2 = 390 kg

let v1 is the speed of m1.

then speed of m2, v2 = v1 + 2.5*10^3

a) Apply conservation of momentum

final momentum = initial momentum

m1*v1 + m2*v2 = M*v

390*v1 + 390*(v1 + 2.5*10^3) = 780*6.9*10^3

==> v1 = 5.65*10^3 m/s <<<<<<<<<<---------------Answer

so, v2 = v1 + 2.5*10^3

= 5.65*10^3 + 2.5*10^3

= 8.15*10^3 m/s <<<<<<<<<<---------------Answer


b) Both travel away from the Earth

c) KEi = (1/2)*M*v^2

= (1/2)*780*(6.9*10^3)^2

= 1.857*10^10 J

KEf = (1/2)*m1*v1^2 + (1/2)*m2*v2^2

= (1/2)*390*(5.65*10^3)^2 + (1/2)*390*(8.15*10^3)^2

= 1.918*10^10 J


Energy supplied by the explosion, E = KEf - KEi

= 1.918*10^10 - 1.857*10^10

= 6.10*10^8 J  <<<<<<<<<<---------------Answer

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