Edexcel M3 2006 June — Question 4 11 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2006
SessionJune
Marks11
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 1
TypeParticle on cone surface – no string (normal reaction only)
DifficultyStandard +0.3 This is a standard circular motion problem on a conical surface requiring resolution of forces (normal reaction, weight) perpendicular and parallel to the surface, use of F=mrω², and geometric relationships. While it involves multiple steps and careful geometry (finding the cone's semi-vertical angle from dimensions, relating radius to height), it follows a well-established template for M3 conical pendulum problems with no novel insight required. The 11 marks reflect length rather than conceptual difficulty.
Spec3.03e Resolve forces: two dimensions6.05c Horizontal circles: conical pendulum, banked tracks

\includegraphics{figure_1} A hollow cone, of base radius \(3a\) and height \(4a\), is fixed with its axis vertical and vertex \(V\) downwards, as shown in Figure 1. A particle moves in a horizontal circle with centre \(C\), on the smooth inner surface of the cone with constant angular speed \(\sqrt{\frac{8g}{9a}}\). Find the height of \(C\) above \(V\). [11]

AnswerMarks
\(\tan \alpha = \frac{3}{4}\) or equivalentB1
\(\tan \alpha = \frac{r}{h}\) or \(\frac{r}{h} = \frac{3a}{4a}\)B1
\(R(\uparrow)\) \(R\sin\alpha = mg\)M1 A1
\(R(\leftarrow)\) \(R\cos\alpha = mr\omega^2\)M1 A1
\(= mr \times \frac{8g}{9a}\) \(\left(R = \frac{10mrg}{9a}\right)\)A1
\(\tan\alpha = \frac{9a}{8r}\left(\frac{5}{3}mg = \frac{10mrg}{9a}\right)\)M1 A1
\(\left(\frac{3}{4} = \frac{9a}{8r} \Rightarrow r = \frac{3}{2}a\right)\)M1 A1
\(h = \frac{r}{\tan\alpha} = \frac{3a}{2} \times \frac{4}{3} = 2a\)M1 A1
Total: [11]
Note: Eliminating \(R\)
| $\tan \alpha = \frac{3}{4}$ or equivalent | B1 |
| $\tan \alpha = \frac{r}{h}$ or $\frac{r}{h} = \frac{3a}{4a}$ | B1 |
| $R(\uparrow)$ $R\sin\alpha = mg$ | M1 A1 |
| $R(\leftarrow)$ $R\cos\alpha = mr\omega^2$ | M1 A1 |
| $= mr \times \frac{8g}{9a}$ $\left(R = \frac{10mrg}{9a}\right)$ | A1 |
| $\tan\alpha = \frac{9a}{8r}\left(\frac{5}{3}mg = \frac{10mrg}{9a}\right)$ | M1 A1 |
| $\left(\frac{3}{4} = \frac{9a}{8r} \Rightarrow r = \frac{3}{2}a\right)$ | M1 A1 |
| $h = \frac{r}{\tan\alpha} = \frac{3a}{2} \times \frac{4}{3} = 2a$ | M1 A1 |
| **Total:** [11] |

**Note:** Eliminating $R$

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\includegraphics{figure_1}

A hollow cone, of base radius $3a$ and height $4a$, is fixed with its axis vertical and vertex $V$ downwards, as shown in Figure 1. A particle moves in a horizontal circle with centre $C$, on the smooth inner surface of the cone with constant angular speed $\sqrt{\frac{8g}{9a}}$.

Find the height of $C$ above $V$.
[11]

\hfill \mbox{\textit{Edexcel M3 2006 Q4 [11]}}