Respuesta :
Answer:
a) use the law of conservation of the moment where the system is formed by the rocket plus the ejected fuel
c)  Δm v ’= m Δv,  v = v’ ln (m₀ / m)
d)  a = - v’/m  dm/dt
e)  y = v’/C  [(m- m ln ( m)]
d)
Explanation:
a) To model this problem we can use the law of conservation of the moment where the system is formed by the rocket plus the ejected fuel that creates the propulsion
b) in the attached we can see a diagram of the free body the cohetees the aja grade and the fuel the small box
 The ejected mass is m ’= Δm
c) we write the initial moment
         p₀ = (m + Δm) v
   Final moment
        pf = m (v + Δv) + Δm (v- v ’)
Where v ’is the gas velocity relative to the rocket
       p₀ = pf
       (m + Δm) v = m (v + Δv) + Δm (v- v ’)
       Δm v ’= m Δv
The amount Δm is negative because it is a decrease in rocket masses
         Δv = -v’ Δm / m
         dv = -v' dm/m
We integrate
         v = v’ ln m
We evaluate between the lower limit v = 0 m = m0 and the upper limit v = v for m = m
       v = v’ ln (m₀ / m)
d) acceleration is defined
       a = dv / dt
       a = dv / dm dm / dt
       a = - v’/m  dm/dt
The amount dm/dt in the amount of fuel burned per unit of time, in general is given for a specific engine
e) the rocket speed is found in part c
The position, we can find it from the define of the speed
       v = dy / dt = dy /dm  dm/dt
Dm / dt = C
        dy = v dm / C
        dy = v’/C  ln (mo / m) dm
We integrate
        y = v’/C  [(m- m ln ( m)]
f) Â dimensional analysis
  a = - v’/m  dm/dt
 [ m/s²] =[m/s]/[km] [km/s] = [ m/s²]
  y = v’/C  [(m- m ln ( m)]
 [m] = [m/s]/[kg/s] [kg] = [m]