Aspace vehicle is traveling at 5320 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 98 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 ?

5398.4km/h

Explanation:

IN THIS CASE THE MOMENTUM IS CONSERVED. THE VALUE OF MOMENTUM OF ONE COMBINED ROCKET WILL BE SAME AS OF TWO COMBINED.

Let mass of module be m

then

mass of motor  = 4m  (four times the mass of rocket module)

total mass = m + 4m = 5m

combined velocity = V = 5320kph

Let

absolute (relative to earth)motor velocity after disengagement = v

then

rocket module velocity (relative to earth) after disengagement = v+98  (relative velocity = 98)

momentum conservation equation

combined momentum = module momentum + motor momentum

(m+4m)V = m(v+98) + 4m*v

5mV = 98m+mv + 4mv

5V = 98+v + 4v (m cancels out)

5V - 98 = 5v

((5*5320)-98)/5 = v

v = 5300.4 km/h

velocity of rocket module relative to earth = v +98

                                                                       = 5300.4 + 98

                                                                       = 5398.4km/h  

i think the answer is a, hope this : )

the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this