M5 June 2018 Q5

EdexcelOld spec14 marksImpulse & Momentum

5. At time \(t = 0\) a rocket is launched. The rocket has initial mass \(M\), of which mass \(\lambda M\), \(0 \lt \lambda \lt 1\), is fuel. The rocket is launched vertically upwards, from rest, from the surface of the Earth. The rocket burns fuel and the burnt fuel is ejected vertically downwards with constant speed \(U\) relative to the rocket. At time \(t\), the rocket has mass \(m\) and velocity \(v\). Ignoring air resistance and any variation in \(g\),

(a) show, from first principles, that until all the fuel is used, \[m\frac{\mathrm{d}v}{\mathrm{d}t} + U\frac{\mathrm{d}m}{\mathrm{d}t} = -mg\] (4)

The rocket accelerates vertically upwards with constant acceleration \(g\).

(b) Show that \(m = M\mathrm{e}^{\frac{-2gt}{U}}\) (4)

(c) Find, in terms of \(M\), \(U\) and \(\lambda\), an expression for the kinetic energy of the rocket at the instant when all of the fuel has been used. (6)

Mark scheme (a)

Scheme	Marks
\((m + \delta m)(v + \delta v) + (-\delta m)(v - U) - mv = -mg\,\delta t\)	M1 A2
\(mv + v\,\delta m + m\,\delta v + U\,\delta m - v\,\delta m - mv = -mg\,\delta t\)
\(m\dfrac{\mathrm{d}v}{\mathrm{d}t} + U\dfrac{\mathrm{d}m}{\mathrm{d}t} = -mg\) GIVEN ANSWER	A1
	(4)

Notes

First M1 for use of Impulse-Momentum principle, dimensionally correct, with correct no. of terms, with usual rules.

First and second A1 for a correct equation

Third A1 for the correct given answer correctly obtained.

Mark scheme (b)

Scheme	Marks
\(mg + U\dfrac{\mathrm{d}m}{\mathrm{d}t} = -mg\)
\(U\dfrac{\mathrm{d}m}{\mathrm{d}t} = -2mg\)	M1
\(\displaystyle\int_M^m \dfrac{\mathrm{d}m}{m} = \dfrac{-2g}{U}\int_0^t \mathrm{d}t\)	M1
\(\left[\ln m\right]_M^m = \dfrac{-2gt}{U}\)	A1
\(m = M\mathrm{e}^{\frac{-2gt}{U}}\) GIVEN ANSWER	A1
	(4)

Notes

First M1 for putting \(\dfrac{\mathrm{d}v}{\mathrm{d}t} = g\) into the DE and collecting terms

Second M1 for separating the variables and attempting to integrate both sides

First A1 for a correct integral on both sides with correct limits oe

Second A1 for the GIVEN ANSWER correctly obtained.

Mark scheme (c)

Scheme	Marks
\((1 - \lambda)M = M\mathrm{e}^{\frac{-2gT}{U}}\)	M1
\(T = \dfrac{U}{2g}\ln\left(\dfrac{1}{1 - \lambda}\right)\)	A1
\(v = gT = \dfrac{U}{2}\ln\left(\dfrac{1}{1 - \lambda}\right)\)	M1 A1
\(\text{KE} = \dfrac{1}{2}(1 - \lambda)M\left[\dfrac{U}{2}\ln\left(\dfrac{1}{1 - \lambda}\right)\right]^2 = \dfrac{1}{8}MU^2(1 - \lambda)\left[\ln\left(\dfrac{1}{1 - \lambda}\right)\right]^2\)	M1 A1
	(6)
	(14 marks)

Notes

First M1 for putting \(m = (1 - \lambda)M\) and solving for \(T\)

First A1 for a correct expression for \(T\)

Second M1 for using \(v = u + at\) oe with \(u = 0\), \(a = g\) and \(t\) = their \(T\) to find \(v\)

Second A1 for a correct \(v\)

Third M1 for use of KE formula (M0 if they use \(M\) for mass)

Third A1 for a correct expression in any form.