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

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

$Mv_i=4m(v_{mc}+v_{ce})+mv_{ce}$

$5mv_i=4mv_{mc}+4mv_{ce}+mv_{ce}$

$5mv_i=4mv_{mc}+5mv_{ce}$

$v_{ce}=v_i-\frac{4}{5}\times v_{mc}$

$v_{ce}=4350-\frac{4}{5}\times 64$

$v_{ce}=4350-51.2$

$v_{ce}=4298.8\ km/h$