## A space vehicle is traveling at 3760 km/h relative to Earth when the exhausted rocket motor is disengaged and sent backward. The relative sp

Question

A space vehicle is traveling at 3760 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 90 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?

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2 months 2021-07-31T08:02:33+00:00 2 Answers 2 views 0

The command module is traveling at 3832 km/h after the separation.

Explanation:

Let’s call the mass of the module M, and the mass of the motor is 4M. The total mass of the vehicle, then, is M + 4M = 5M.

Now, momentum = force x distance.

Thus, momentum of the vehicle before separation 3760 x 5M = 18800M kg-km/h.

Let’s call the velocity of the motor v. So the velocity of the module must be v + 93.

Now, after the separation, the momentum of the motor will be 4Mv, and the momentum of the module is M(v + 90) = Mv + 90M. The total momentum is 4MV + Mv + 90M = 5Mv + 90M.

From conservation of momentum that says initial momentum equals final momentum, we have;

18800M = 5Mv + 90M

Divide both sides by M to get;

18800 = 5v + 90

18800 – 90 = 5v

18710 = 5v

v = 18710/5

v = 3742 km/h

But v is the velocity of the motor, so we have to add 90 to get the velocity of the module, thus, velocity of module = 3742 + 90 = 3832

So the command module is traveling at 3832 km/h after the separation.

3688 km/h

Explanation:

Given:-

– The speed of vehicle relative to earth, vs_e = 3760 km/h

– The relative speed of command and motor, v_c/m = 90 km/h

– The mass of command = m

– The mass of motor = 4m

Find:-

What is the speed of the command module relative to Earth just after the separation?

Solution:-

– Consider the space vehicle as a system that detaches itself into two parts ( command and motor ). We will assume that the gravitational pull due to Earth on the space vehicle is negligible. With that assumption we have our system in isolation. We will apply the principle of conservation of linear momentum on the system as follows:

Initial momentum = Final momentum

Pi = Pf

M*vs_e = m*vc_e + 4m*vm_e

Where,

M = m + 4m = 5m

vc_e = Velocity of command relative to earth

vm_e = Velocity of motor relative to earth

– We will develop a relation of velocities of command and motor in the frame of earth as follows:

vm_e =  v_c/m + vc_e

– Substituting (vm_e) from Equation 2 into Equation 1, we have:

5m*vs_e = m*vc_e + 4m*(v_c/m + vc_e)

5m*vs_e = 5m*vc_e + 4m*(v_c/m)

Solve for vc_e:

5m*vs_e –  4m*(v_c/m) = 5m*vc_e

vs_e – 0.8*(v_c/m) = vc_e

Plug in values and evaluate vc_e:

vc_e = 3760 – 0.8*(90)

vc_e = 3,688 km/h