A space vehicle is traveling at 3730 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 85 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?

Given that,

Speed of space ship

V = 3730km/hr relative to earth

Motor speed is v (mc) = 85km/hr relative to command nodule

Let mass of command module be m

Motor mass is 4 times module=4m

Let command module be V (ce)

Note, the space vehicle contain the command module and the motor

Applying conservation of linear momentum

Pi = Pf

Momentum is given as p=mv

For command module p = mv (ce)

For motor module p=4mV (me)

Initial momentum before Pi = MV

M is the total mass of both the command module and motor module

M=m+4m = 5m

Pi = Pf

MV = m•v (ce) + 4m•V (me)

v (ce) velocity of command relative to the earth

V (me) velocity of motor relative earth

V (me) = v (mc) + v (ce)

5mV = mv (ce) + 4m (v (mc) + v (ce))

Divide through by m

5V = v (ce) + 4v (mc) + 4v (ce)

5V - 4v (mc) = 5v (ce)

5v (ce) = 5V - 4 v (mc)

5v (ce) = 5 * 3730 - 4 * 85

5v (ce) = 18310

v (ce) = 18310/5

v (ce) = 3662km/hr

To m/s v (ce) = 3662*1000/3600

v (ce) = 1017.22 m/s

command module relative to Earth just after the separation is 3662km/h or 1017.22 m/s

the speed of the command module relative to Earth = 1017.22 m/s

Explanation:

We are given;

Speed space vehicle is traveling at relative to Earth; Vi = 3730 km/h = (3730 x 10) / 36 m/s = 1036.111 m/s

Rocket motor speed; Vmc = 85km/h = (85x10) / 36 = 23.611 m/s

From conservation of linear momentum, we know that;

Initial momentum = final momentum.

Thus,

MV_i = 4m (V_me) + mV_ce

Where;

M = the mass of the space vehicle which is the sum of the motor's mass and command mass

V_i = initial speed

V_me = speed of the motor relative to the earth

V_ce = speed of command relative to the earth

m = the command module mass

4m = the mass of the motor

Now, M = the sum of the motor's mass and command mass

Thus; M = 4m + m = 5m

So, we now have;

5mV_i = 4m (V_me) + mV_ce

Now, the velocity of the motor relative to the earth is;

V_me = V_mc + V_ce

Thus, we now have;

5mV_i = 4m (V_mc + V_ce) + mV_ce

This gives;

5mV_i = 4mV_mc + 5mV_ce

Divide through by m to get;

5V_i = 4V_mc + 5V_ce

Let's make V_ce the subject;

5V_i - 4V_mc = 5V_ce

Divide through by 5;

V_ce = V_i - (4/5) V_mc

Plugging in the relevant values;

V_ce = 1036.111 - (4/5) •23.611 = 1017.22 m/s