Initially a total mass of a rocket is M, of which kM is the mass of the fuel. Starting from rest, the rocket gives itself a constant vertical acceleration of magnitude g by ejecting fuel with constant speed u relative to itself.
a) If m denotes its remaining mass at time t, show that the rate of decrease of m with respect to t is 2mg/u and deduce that m=Me^(-2gt/u).
b) Show that the kinetic energy of the rocket when the fuel is exhausted is (1/8)Mu^2(1-k)[ln(1-k)]^2
c) Show the value of k for which energy is a maximum is 1-e^-2.
See the attached file.
Following is the text part of the solution. Please see the attached file for complete solution. Equations, diagrams, graphs and special characters will not appear correctly here.
Above diagram shows the rocket at time t = 0, t = t and t = t + dt. Let the mass of the rocket be m when t = t.
In small time dt, rocket ejects fuel of mass dm. So in time t + dt, rocket's mass is m - dm.
Further, velocity of the fuel ejected is given ...
This is the solution to a rocket propulsion question. As the rocket moves upward, its mass gradually decreases. This causes the kinetic energy to increase to a certain maximum value. This problem requires careful set up of equations using basic physics principles and solving them. I have done just that in this 4-page word document.