Adam: I think all the modulus stuff is misleading and have never met it before. In any particular answer you wouldn't need it and indeed it would be confusing* or even wrong. I think the mark scheme writer was trying to use a short-hand to show what answers were acceptable. The trap, if any, is muddling the signs on δm. It might seem "natural" to draw the diagram with "BEFORE" showing "m" and "AFTER" showing "m-δm" (or equivalently, "BEFORE" showing "m" and "AFTER" showing "m-δm", with the implication that δm is a positive quantity. However, you would then have to remember that dm/dt is the limit not of δm/δt but rather -δm/δt: dm/dt is negative however you approach the sign convention. That's why I prefer the convention in my diagram. It's not confusing, IMHO, provided you have at the back of your mind that δm is a negative number, so that m+δm is slightly smaller than m, -δm is the mass of exhaust gas, and δm/δt tends to dm/dt.
Small point: I was brought up to show forces with simple arrows, velocities with filled triangular arrowheads (doubled for acceleration), and impulses (and momentums) with open triangular arrowheads. Then to construct a careful "BEFORE" and "AFTER" diagram showing a SYSTEM before and after an IMPULSE. So BEFORE+IMPULSE=AFTER. It's easy to get muddled or baffled by rocket questions. The idea of a system helps, because Newton's Law applies to a collection of particles, even if they change shape (as here, into two parts: burnt fuel and unused fuel). (It's not useful to consider, say, the rocket casing as a system, as you know nothing about the innards and the pressure etc.)The subtle thing with rocket algebra that this is all instantaneous: in the next instant the system is slightly different (less mass). In GCSE mechanics (e.g. colliding snooker balls) this doesn't happen.
"So small that they can be ignored?" Sort of, though a pure mathematician looking at this would cut through all this δm, δt, δv stuff as really it's unnecessarily going back to first principle of differentiation and limits, such as #lim x to 0 x+δx=x# and suchlike, and wouldn't dirty their hands with "approximations" and physcial interpretaion In fact, you can practically write down E1 at sight and much more cleanly, if you start with notion that OVERALL force on a SYSTEM is equal to its RATE OF CHANGE of momentum (just Newton's Law, slightly extended). In this case "the system" *instantaneously" consists of the (A) mass of the rocket plus *unused* fuel. The first term on the left is just the rate of change of momentum of everything except the fuel instantaneously being burnt, and the second term on the left is the rate of change of momentum of the fuel being expelled, and the term on the right is the instantaneous weight of the entire system. MEI4 rockets is an extension of GCSE(? MEI1??) exploding trolleys where a spring suddenly blasts a small mass out of a larger mass (usually horizontally). If you imagine a trolley contraption blasting out one marble a second out of the back in order to go forwards, and do the maths, you get the same sort of equations. If you then blast half a marble every half second, and see how the equations change (if at all) you find that "in the limit" (which is like blasting out an infinitessimally massed marble every infinitessimal fraction of a second, but with constant marblemass/interval) you get equations like E1. This "quantising" is similar to replacing a jet of water with discrete lumps of ice! Were you to do the same "quantizing" for the given example (I don't suggest you do!) you would get cluttered up with all manner of second order terms which disappear as δt goes to zero.
Can somebody please explain to me the signs used for ∂m in this solution, with regard to the 'u'.
I do not understand why '-|∂m|' is used, and then how it is later changed to a '+∂m'?
Thanks in advance!