Answer:
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
is the Planck constant. In our problem, the frequency of the photon is 
, and by using these numbers we can find the energy of the photon: