3. a space vehicle is traveling at 4300 km/h relative to earth when the exhausted rocket motor (mass 4m) is disengaged and sent backward with a speed of 82 km/h relative to the command module (mass m). what is the speed of the command module relative to earth just after the separation?

Vce= 4234.4km/hr or 1180m/s

Explanation:

Mass of rocket motor Mr= 4m

Mass of control module Mc= m

Initial velocity of spaceship relative to earth Vs= 4300km/hr

Final velocity of command module relative to the rocket motor Vrc = 82km/hr

Total mass of spaceship M= 4m+m= 5m

Using law of conservation of momentum;

MVs= MrVre + McVce eqn1

Vre = Vrc+Vce eqn2

Where Vce is velocity of control room relative to the earth and Vre is velocity of rocket motor relative to the earth

Substituting eqn2 to eqn1, we have.

MVs= Mr(Vrc+Vce) +McVce

MVs= MrVrc+MrVce+McVce

Making Vce the subject of formula.

Vce = (MVs-MrVrc)/(Mc+Mr)

Substituting the values we have,

Vce = (5m×4300km/hr - 4m×82km/hr)/ (m+4m)

Vce = 21172m/5m

Vce= 4234.4km/hr or 1180m/s