A 975-kg two-stage rocket is traveling at a speed of 6.85×10^3 m/s away from Ear

Question

A 975-kg two-stage rocket is traveling at a speed of 6.85×10^3 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 2.70×10^3 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? (Enter your answers numerically separated by a comma.)

(Va,Vb) = ___ m/s

How much energy was supplied by the explosion? [Hint: What is the change in kinetic energy as a result of the explosion?]

(delta) K = ___ J

Explanation / Answer

Let A be the upper stage and B be the lower stage

Let the mass of the rocket be M = 975 kg

Let MA and MB represent the masses of stages A and B such that M = MA + MB       ……………. ( 1 )

Now let the velocity of the rocket before separation be v = 6.85 * 103 m/s = 6850 m/s

This is equal to the velocity of both the stages before separation

Let the relative velocity of both the stages be V = 2.70 * 103 m/s = 2700 m/s

Then we can write V = VA + VB …………… ( 2 )

where VA and VB are relative velocities of stages A and B w.r.to earth

Let us take the initial direction before separation as positive direction

A)From the law of conservation of momentum , the initial momentum before separation must be equal to the momentum after separation for both the masses

So                  Pinitial = Pfinal

              M * v = MA * VA + MB * VB

By using ( 2 ) , we can write

             M * v = MA * VA + MB * ( VA – V )

Finally we get , VA = ( M * v + MB * V ) / ( MA + MB )

                               = ( 975 * 6850 + 487.5 * 2700 ) / 975

                                = ( 6678750 + 1316250 ) / 975

                                = 7995000 / 975

                                = 8200 m/s

So the velocity of the stage will be 8200 m/s away from the earth

From ( 2 ) , VB = VA – V = 8200 – 2700 = 5500 m/s away from the earth

B)Initial kinetic energy of the rocket before separation Ki = 1 / 2 * M * v2

                       = 1 / 2 * 975 * 6850 * 6850

                       = 22,874,718,750 J

Final kinetic energy of the rocket after separation is

                  Kf = ½ * MA * VA2   +   ½ * MB * VB2

                       = ( ½ * 487.5 * 8200 * 8200 ) + ½ * 487.5 * 5500 * 5500 )

                       = 16,389,750,000 + 7,373,437,500

                       = 23,763,187,500 J

So change in kinetic energy is given by

                  K = Kf - Ki

                        = 23,763,187,500 - 22,874,718,750

                        = 888,468,750 J

                        = 8.88469 * 108 J

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