Question

(hrw8c9p40) A space vehicle is traveling at 4350 km/h relative to the Earth when the exhausted rocket motor is disengaged and sent backward with a speed of 64 km/h relative to the command module. The mass of the motor is four times the mass of the module. What is the speed (km/h) of the command module relative to Earth after the separation?

Answer 1

Answer:4298.8 km/h

Step-by-step explanation:

Given

velocity of space vehicle
`v_i=4350\ km/h`

Mass of motor is 4 times the mass of module

suppose m is the mass of module therefore

mass of motor is 4 m

velocity of disengaged motor is 64 km/h relative to command module

we can write

`v_(me)=v_(mc)+v_(ce)`

where
`v_(me)`=velocity of motor relative to earth

`v_(mc)`=velocity of motor relative to command module

`v_(ce)`=velocity of command relative to earth

Conserving momentum

`M(v_i)=4m(v_(me))+mv_(ce)`

where M is combined mass of motor and module

`Mv_i=4m(v_(mc)+v_(ce))+mv_(ce)`

`5mv_i=4mv_(mc)+4mv_(ce)+mv_(ce)`

`5mv_i=4mv_(mc)+5mv_(ce)`

`v_(ce)=v_i-(4)/(5)* v_(mc)`

`v_(ce)=4350-(4)/(5)* 64`

`v_(ce)=4350-51.2`

`v_(ce)=4298.8\ km/h`