Question 5 - A-Level Maths

Edexcel M5 — Question 5 11 marks

Exam Board	Edexcel
Module	M5 (Mechanics 5)
Marks	11
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Variable mass problems
Type	Find ejected or remaining mass
Difficulty	Challenging +1.8 This is a variable mass/rocket equation problem requiring application of the rocket equation in an unusual configuration (fuel ejected forwards, not backwards). It demands careful setup of momentum conservation with relative velocities, integration or use of the Tsiolkovsky equation adapted to deceleration, and algebraic manipulation. The 11 marks indicate substantial working. While M5 students should know the rocket equation, applying it correctly when fuel ejects forwards (to slow down) requires solid conceptual understanding and is more challenging than standard textbook examples.
Spec	6.03b Conservation of momentum: 1D two particles 6.03f Impulse-momentum: relation

5. A spaceship is moving in deep space with no external forces acting on it. Initially it has total mass \(M\) and is moving with speed \(V\). The spaceship reduces its speed to \(\frac { 2 } { 3 } V\) by ejecting fuel from its front end with a speed of \(c\) relative to itself and in the same direction as its own motion. Find the mass of fuel ejected.
(11 marks)

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Question 5:

Answer	Marks	Guidance
Answer/Working	Marks	Guidance
Diagram: mass \(m\) velocity \(v\); after time \(\delta t\): mass \(m+\delta m\) velocity \(v+\delta v\); ejected mass \(-\delta m\) velocity \(v+c\)
\((m+\delta m)(v+\delta v) + (-\delta m)(v+c) - mv = 0\)	M1 A2, 1, 0
\(\Rightarrow m\,\mathrm{d}v - c\,\mathrm{d}m = 0\)	M1	reducing equation
\(\int_{V}^{\frac{2}{3}V} \mathrm{d}v = c\int_{M}^{m} \frac{\mathrm{d}m}{m}\)	M1	separating variables
\(\left[v\right]_{V}^{\frac{2}{3}V} = c\left[\ln m\right]_{M}^{m}\)	M1 A1	integration
\(-\frac{1}{3}v = c\ln\left(\frac{m}{M}\right)\)	M1	applying limits
\(\Rightarrow m = Me^{-\frac{v}{3c}}\)	M1	eliminating ln
\(\therefore\) Fuel used \(= M - m = M(1 - e^{-\frac{v}{3c}})\)	M1 A1	(11)

# Question 5:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Diagram: mass $m$ velocity $v$; after time $\delta t$: mass $m+\delta m$ velocity $v+\delta v$; ejected mass $-\delta m$ velocity $v+c$ | | |
| $(m+\delta m)(v+\delta v) + (-\delta m)(v+c) - mv = 0$ | M1 A2, 1, 0 | |
| $\Rightarrow m\,\mathrm{d}v - c\,\mathrm{d}m = 0$ | M1 | reducing equation |
| $\int_{V}^{\frac{2}{3}V} \mathrm{d}v = c\int_{M}^{m} \frac{\mathrm{d}m}{m}$ | M1 | separating variables |
| $\left[v\right]_{V}^{\frac{2}{3}V} = c\left[\ln m\right]_{M}^{m}$ | M1 A1 | integration |
| $-\frac{1}{3}v = c\ln\left(\frac{m}{M}\right)$ | M1 | applying limits |
| $\Rightarrow m = Me^{-\frac{v}{3c}}$ | M1 | eliminating ln |
| $\therefore$ Fuel used $= M - m = M(1 - e^{-\frac{v}{3c}})$ | M1 A1 | **(11)** |

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5. A spaceship is moving in deep space with no external forces acting on it. Initially it has total mass $M$ and is moving with speed $V$. The spaceship reduces its speed to $\frac { 2 } { 3 } V$ by ejecting fuel from its front end with a speed of $c$ relative to itself and in the same direction as its own motion.

Find the mass of fuel ejected.\\
(11 marks)\\

\hfill \mbox{\textit{Edexcel M5  Q5 [11]}}

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Q1 7 Q2 7 Q3 10 Q4 11 Q5 11 Q6 12 Q7 17