A space vehicle is traveling at 4140 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The relative speed between the motor and the command module is then 99 km/h. The mass of the motor is four times the mass of the module. What is the speed of the command module relative to Earth just after the separation

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

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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im doing peak rn so sry if its wrong