Question 2 - A-Level Maths

WJEC Further Unit 6 Specimen — Question 2 15 marks

Exam Board	WJEC
Module	Further Unit 6 (Further Unit 6)
Session	Specimen
Marks	15
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Centre of Mass 1
Type	Solid with removed cone from cone or cylinder
Difficulty	Challenging +1.2 This is a structured multi-part centre of mass question requiring standard techniques: proving a standard result, applying the composite body formula with removed volume, and using equilibrium conditions for a suspended body. While it involves several steps and careful bookkeeping of coordinates/volumes, each part follows well-established methods taught in Further Maths mechanics with no novel insight required.
Spec	6.04c Composite bodies: centre of mass 6.04d Integration: for centre of mass of laminas/solids 6.04e Rigid body equilibrium: coplanar forces

2. (a) Prove that the centre of mass of a uniform solid cone of height \(h\) and base radius \(b\) is at a height of \(\frac { 1 } { 4 } h\) above its base.
(b) A uniform solid cone \(C _ { 1 }\) has height 3 m and base radius 2 m . A smaller cone \(C _ { 2 }\) of height 2 m and base radius 1 m is contained symmetrically inside \(C _ { 1 }\). The bases of \(C _ { 1 }\) and \(C _ { 2 }\) have a common centre and the axis of \(C _ { 2 }\) is part of the axis of \(C _ { 1 }\). If \(C _ { 2 }\) is removed from \(C _ { 1 }\), show that the centre of mass of the remaining solid is at a distance of \(\frac { 11 } { 5 } \mathrm {~m}\) from the vertex of \(C _ { 1 }\).
(c) The remaining solid is suspended from a string which is attached to a point on the outer curved surface at a distance of \(\frac { 1 } { 3 } \sqrt { 13 } \mathrm {~m}\) from the vertex of \(C _ { 1 }\). Given that the axis of symmetry is inclined at an angle of \(\alpha\) to the vertical, find \(\tan \alpha\).

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Question 2:

Part (a)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Let \(\rho\) be mass per unit volume. By symmetry, c of m lies on \(Ox\). Divide cone into slices parallel to base. Consider slice \(PQ\), distance \(x\) from \(O\), thickness \(\delta x\).	M1
By similar triangles, radius of slice is \(\frac{bx}{h}\)
Mass of slice \(= \frac{\pi b^2 x^2}{h^2}\rho\,\delta x\) acting \(x\) from \(O\)
Mass of cone \(= \frac{\pi b^2 h}{3}\rho\) acting at \(\bar{x}\) from \(O\)
Take moments about \(y\) axis	m1
\(\frac{\pi b^2 h}{3}\rho\bar{x} = \int_0^h \frac{\pi b^2 x^2}{h^2}\times x\rho\,dx\)	A1
\(\frac{1}{3}h\bar{x} = \frac{1}{h^2}\left[\frac{1}{4}x^4\right]_0^h\)
\(\bar{x} = \frac{3}{h^3}\cdot\frac{h^4}{4} = \frac{3h}{4}\)	A1

Part (b)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
\(C_1\): mass \(\frac{\pi}{3}(2)^2\times 3\rho\), distance \(\frac{3}{4}\times 3\)	B1
\(C_2\): mass \(\frac{\pi}{3}\times 1^2\times 2\rho\), distance \(1+\frac{3}{4}\times 2\)	B1
Remainder: mass \(\frac{\pi}{3}\rho(12-2)\), distance \(\bar{h}\)	B1
Take moments about \(y\) axis: \(\frac{\pi}{3}\rho\times 10\times\bar{h} = \frac{\pi}{3}\times 12\rho\times\frac{9}{4} - \frac{\pi}{3}\times 2\rho\times\frac{5}{2}\)	M1, A1
\(\bar{h} = \frac{11}{5}\)	A1

Part (c)

Answer	Marks	Guidance
Answer/Working	Mark	Guidance
Draw \(HK\) perpendicular to \(OG\); \(OH = \frac{\sqrt{13}}{3}\), \(OG = \frac{11}{5}\)
Angle \(HOK = \theta\), \(\tan\theta = \frac{2}{3}\)	B1
\(\sin\theta = \frac{2}{\sqrt{13}}\), \(\cos\theta = \frac{3}{\sqrt{13}}\)	B1
\(HK = OH\sin\theta = \frac{\sqrt{13}}{3}\times\frac{2}{\sqrt{13}} = \frac{2}{3}\)	B1
\(KG = \frac{11}{5} - OH\cos\theta = \frac{11}{5} - \frac{\sqrt{13}}{3}\times\frac{3}{\sqrt{13}} = \frac{6}{5}\)	B1
\(\tan\alpha = \frac{2}{3}\div\frac{6}{5} = \frac{2}{3}\times\frac{5}{6} = \frac{5}{9}\)	B1

# Question 2:

## Part (a)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Let $\rho$ be mass per unit volume. By symmetry, c of m lies on $Ox$. Divide cone into slices parallel to base. Consider slice $PQ$, distance $x$ from $O$, thickness $\delta x$. | M1 | |
| By similar triangles, radius of slice is $\frac{bx}{h}$ | | |
| Mass of slice $= \frac{\pi b^2 x^2}{h^2}\rho\,\delta x$ acting $x$ from $O$ | | |
| Mass of cone $= \frac{\pi b^2 h}{3}\rho$ acting at $\bar{x}$ from $O$ | | |
| Take moments about $y$ axis | m1 | |
| $\frac{\pi b^2 h}{3}\rho\bar{x} = \int_0^h \frac{\pi b^2 x^2}{h^2}\times x\rho\,dx$ | A1 | |
| $\frac{1}{3}h\bar{x} = \frac{1}{h^2}\left[\frac{1}{4}x^4\right]_0^h$ | | |
| $\bar{x} = \frac{3}{h^3}\cdot\frac{h^4}{4} = \frac{3h}{4}$ | A1 | |

## Part (b)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $C_1$: mass $\frac{\pi}{3}(2)^2\times 3\rho$, distance $\frac{3}{4}\times 3$ | B1 | |
| $C_2$: mass $\frac{\pi}{3}\times 1^2\times 2\rho$, distance $1+\frac{3}{4}\times 2$ | B1 | |
| Remainder: mass $\frac{\pi}{3}\rho(12-2)$, distance $\bar{h}$ | B1 | |
| Take moments about $y$ axis: $\frac{\pi}{3}\rho\times 10\times\bar{h} = \frac{\pi}{3}\times 12\rho\times\frac{9}{4} - \frac{\pi}{3}\times 2\rho\times\frac{5}{2}$ | M1, A1 | |
| $\bar{h} = \frac{11}{5}$ | A1 | |

## Part (c)
| Answer/Working | Mark | Guidance |
|---|---|---|
| Draw $HK$ perpendicular to $OG$; $OH = \frac{\sqrt{13}}{3}$, $OG = \frac{11}{5}$ | | |
| Angle $HOK = \theta$, $\tan\theta = \frac{2}{3}$ | B1 | |
| $\sin\theta = \frac{2}{\sqrt{13}}$, $\cos\theta = \frac{3}{\sqrt{13}}$ | B1 | |
| $HK = OH\sin\theta = \frac{\sqrt{13}}{3}\times\frac{2}{\sqrt{13}} = \frac{2}{3}$ | B1 | |
| $KG = \frac{11}{5} - OH\cos\theta = \frac{11}{5} - \frac{\sqrt{13}}{3}\times\frac{3}{\sqrt{13}} = \frac{6}{5}$ | B1 | |
| $\tan\alpha = \frac{2}{3}\div\frac{6}{5} = \frac{2}{3}\times\frac{5}{6} = \frac{5}{9}$ | B1 | |

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2. (a) Prove that the centre of mass of a uniform solid cone of height $h$ and base radius $b$ is at a height of $\frac { 1 } { 4 } h$ above its base.\\
(b) A uniform solid cone $C _ { 1 }$ has height 3 m and base radius 2 m . A smaller cone $C _ { 2 }$ of height 2 m and base radius 1 m is contained symmetrically inside $C _ { 1 }$. The bases of $C _ { 1 }$ and $C _ { 2 }$ have a common centre and the axis of $C _ { 2 }$ is part of the axis of $C _ { 1 }$. If $C _ { 2 }$ is removed from $C _ { 1 }$, show that the centre of mass of the remaining solid is at a distance of $\frac { 11 } { 5 } \mathrm {~m}$ from the vertex of $C _ { 1 }$.\\
(c) The remaining solid is suspended from a string which is attached to a point on the outer curved surface at a distance of $\frac { 1 } { 3 } \sqrt { 13 } \mathrm {~m}$ from the vertex of $C _ { 1 }$. Given that the axis of symmetry is inclined at an angle of $\alpha$ to the vertical, find $\tan \alpha$.\\

\hfill \mbox{\textit{WJEC Further Unit 6  Q2 [15]}}

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Q1 14 Q2 15 Q3 10 Q4 11 Q5 13 Q6 17