| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Conical pendulum (horizontal circle) |
| Difficulty | Challenging +1.2 This is a standard conical pendulum mechanics problem requiring resolution of forces and circular motion equations. Part (a) involves routine force resolution (normal reaction, weight, centripetal force) with given speed, while part (b) requires geometric relationships in the cone. The problem is structured with clear guidance ('show that') and uses standard Further Maths mechanics techniques, though it requires careful multi-step working across both dynamics and geometry. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Resolve vertically: \(R\sin\theta = Mg\) | M1 | Dimensionally correct |
| \[\sin\theta = \frac{5}{13}\] | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \[R = \frac{13Mg}{5}\] | A1 | Convincing |
| B1 | si | |
| (b) N2L towards centre: \(R\cos\theta = Ma\) | M1 | Dimensionally correct |
| \[\frac{13Mg}{5} \times \frac{12}{13} = M\frac{(3g)^2}{r}\] | m1 | \(a = \frac{v^2}{r}\) |
| \[CP = r = \frac{15}{4}\] (\(= 3 \cdot 75\) m) | A1 | |
| \[\frac{r}{CY} = \frac{5}{12}\] (\(= \tan\theta\)) | M1 | oe, similar triangles |
| \[CY = 9\] (m) | A1 | cao |
(a) Resolve vertically: $R\sin\theta = Mg$ | M1 | Dimensionally correct
$$\sin\theta = \frac{5}{13}$$ | A1 |
$$R = Mg \times \frac{13}{5}$$
$$R = \frac{13Mg}{5}$$ | A1 | Convincing
| B1 | si
(b) N2L towards centre: $R\cos\theta = Ma$ | M1 | Dimensionally correct
$$\frac{13Mg}{5} \times \frac{12}{13} = M\frac{(3g)^2}{r}$$ | m1 | $a = \frac{v^2}{r}$
$$CP = r = \frac{15}{4}$$ ($= 3 \cdot 75$ m) | A1 |
$$\frac{r}{CY} = \frac{5}{12}$$ ($= \tan\theta$) | M1 | oe, similar triangles
$$CY = 9$$ (m) | A1 | cao
---The diagram below shows a hollow cone, of base radius $5$ m and height $12$ m, which is fixed with its axis vertical and vertex $V$ downwards. A particle $P$, of mass $M$ kg, moves in a horizontal circle with centre $C$ on the smooth inner surface of the cone with constant speed $v = 3\sqrt{g}$ ms$^{-1}$.
\includegraphics{figure_3}
\begin{enumerate}[label=(\alph*)]
\item Show that the normal reaction of the surface of the cone on the particle is $\frac{13Mg}{5}$ N. [4]
\item Calculate the length of $CP$ and hence determine the height of $C$ above $V$. [6]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2023 Q3 [10]}}