M5 June 2014 Q4

EdexcelOld spec17 marksImpulse & Momentum

4. A spacecraft is travelling in a straight line in deep space where all external forces can be assumed to be negligible. The spacecraft decelerates by ejecting fuel at a constant speed \(k\) relative to the spacecraft, in the direction of motion of the spacecraft. At time \(t\), the spacecraft has speed \(v\) and mass \(m\).

(a) Show, from first principles, that while the spacecraft is ejecting fuel,\[\frac{\mathrm{d}v}{\mathrm{d}m} - \frac{k}{m} = 0\] (5)

At time \(t = 0\), the spacecraft has speed \(U\) and mass \(M\).

(b) Find the mass of the spacecraft when it comes to rest. (6)

Given that \(m = M\mathrm{e}^{-\alpha t^2}\), where \(\alpha\) is a positive constant, and that the spacecraft comes to rest at time \(t = T\),

(c) find, in terms of \(U\) and \(T\) only, the distance travelled by the spacecraft in decelerating from speed \(U\) to rest. (6)

Mark scheme (a)

Scheme	Marks
\((m + \delta m)(v + \delta v) + (-\delta m)(v + k) = mv\)	M1 A2
\(mv + v\delta m + m\delta v - v\delta m - k\delta m = mv\)
\(m\delta v - k\delta m = 0\)
\(\dfrac{\mathrm{d}v}{\mathrm{d}m} - \dfrac{k}{m} = 0\)	DM1 A1
	(5)

Notes

First M1 for momentum equation (correct number of terms, excluding any \(\delta m\delta v\) terms)

A2 for a correct equation −1 e.e.

Second M1, dependent on first M1, for simplifying and dividing by \(m\delta m\) and taking limits

Third A1 for PRINTED ANSWER

Mark scheme (b)

Scheme	Marks
\(\displaystyle\int \mathrm{d}v = k\int \frac{\mathrm{d}m}{m}\)	M1
\(v = k\ln m + C\)	A1
\(v = U,\ m = M \Rightarrow C = U - k\ln M\)	M1
\(v = k\ln m + U - k\ln M\)
\(v = U + k\ln\left(\dfrac{m}{M}\right)\)	A1
\(v = 0 \Rightarrow m = M\mathrm{e}^{-\frac{U}{k}}\)	M1 A1
	(6)

Notes

First M1 for separating and integrating

First A1 correct expression (without \(C\))

Second M1 for using limits

Second A1 for a correct \(v\) (seen or implied)

Third M1 for putting \(v = 0\) and solving for \(m\)

Third A1 for correct answer

Mark scheme (c)

Scheme	Marks
\(m = M\mathrm{e}^{-\alpha t^2} \Rightarrow v = U - k\alpha t^2\)	M1
\(s = Ut - \tfrac{1}{3}k\alpha t^3\ (+D)\)	M1 A1
At \(t = T,\ \ s = UT - \tfrac{1}{3}k\alpha T^3\)
At \(t = T,\ v = 0 \Rightarrow k\alpha T^2 = U\)	M1
\(s = UT - \tfrac{1}{3}UT = \tfrac{2}{3}UT\)	M1 A1
	(6)
	(17 marks)

Notes

First M1 for obtaining \(v = U - k\alpha t^2\) (method)

Second M1 for integrating wrt time

First A1 for a correct expression for \(s\) (without \(D\))

Third M1 for using \(v = 0\) at \(t = T\) to obtain \(U = k\alpha T^2\) (method)

Fourth M1 for obtaining \(s\) in terms of \(U\) and \(T\)

Second A1 for correct answer