Answer:
V_ce = 1128 m/s
Explanation:
We are given;
Speed of space vehicle relative to the earth; V_i = 4140 km/h = 1150 m/s
relative speed between the motor and the command module; V_mc = 99 km/h = 27.5 m/s
mass of the motor is four times the mass of the module.
If the mass of the module is m
Mass of motor = 4m
From conservation of momentum, since the system is isolated, then;
Initial momentum = final momentum.
Thus;
MV_i = 4m(V_me) + m(V_ce) - - - (eq 1)
Where;
M is mass of space vehicle
V_me is speed of motor relative to the earth
V_ce is speed of command module relative to the earth
Now, relative speed between the motor and the command module is given by the formula;
V_mc = V_me - V_ce
Thus, V_me = V_mc + V_ce
Putting (V_mc + V_ce) for V_me in eq 1, we have;
MV_i = 4m(V_mc + V_ce) + m(V_ce)
Now, the mass of the space vehicle will be the sum of the mass of the module and mass of the motor.
Thus;
M = m + 4m
M = 5m
Thus;
5mV_i = 4m(V_mc + V_ce) + m(V_ce)
Divide through by m to get;
5V_i = 4(V_mc + V_ce) + (V_ce)
5V_i = 4V_mc + 4V_ce + V_ce
5V_i = 4V_mc + 5V_ce
We want to find the speed of the command module relative to Earth just after the separation. Thus, let's make V_ce the subject;
V_ce = (5V_i - 4V_mc)/5
V_ce = ((5 × 1150) - (4 × 27.5))/5
V_ce = 1128 m/s
