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?

Explanation:

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

fgravity=10x10=100n case 1

fgravity=30x20=600n case 2

6 times more

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