| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Conical pendulum – horizontal circle in free space (no surface) |
| Difficulty | Moderate -0.5 This is a straightforward conical pendulum problem requiring resolution of forces (vertical equilibrium gives T cos θ = mg, leading to the given angle) and application of circular motion (horizontal component provides centripetal force). The 'show that' part guides students to the answer, and the calculations are routine for Further Maths students who have learned this standard setup. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Resolve vertically: \(490\sqrt{3} \cos \theta = 75g\) | M1, A1 | |
| \(\cos \theta = \frac{\sqrt{3}}{2}\) | ||
| \(\theta = 30°\) | A1 | Convincing |
| Answer | Marks | Guidance |
|---|---|---|
| (b) N2L towards centre: \(490\sqrt{3} \sin \theta = 75a\) | M1, A1 | \(a = \omega^2 r\) |
| \(490\sqrt{3} \sin \theta = 75(1.4)^2 r\) | m1 | |
| length of chain \(= l\) | ||
| \(l \sin \theta = r\) | m1 | |
| \(490\sqrt{3} \sin \theta = 75(1.4)^2 l \sin \theta\) | ||
| \(l = \frac{490\sqrt{3}}{75(1.4)^2}\) | ||
| \(l = 5 \cdot 77(3502 \ldots)\) (m) | A1 | cao Accept \(\frac{10\sqrt{3}}{3}\) |
**(a)** Resolve vertically: $490\sqrt{3} \cos \theta = 75g$ | M1, A1 |
$\cos \theta = \frac{\sqrt{3}}{2}$ | |
$\theta = 30°$ | A1 | Convincing
**Subtotal: [3]**
**(b)** N2L towards centre: $490\sqrt{3} \sin \theta = 75a$ | M1, A1 | $a = \omega^2 r$
$490\sqrt{3} \sin \theta = 75(1.4)^2 r$ | m1 |
length of chain $= l$ | |
$l \sin \theta = r$ | m1 |
$490\sqrt{3} \sin \theta = 75(1.4)^2 l \sin \theta$ | |
$l = \frac{490\sqrt{3}}{75(1.4)^2}$ | |
$l = 5 \cdot 77(3502 \ldots)$ (m) | A1 | cao Accept $\frac{10\sqrt{3}}{3}$
**Subtotal: [5]**
**Total for Question 5: 8**
---5. The diagram shows a fairground ride that consists of a number of seats suspended by chains that swing out as the centre rotates.\\
\includegraphics[max width=\textwidth, alt={}, center]{b430aa50-27e3-46f7-afef-7b8e75d46e1f-4_711_718_466_678}
When the ride rotates at a constant angular speed of $\omega = 1.4 \mathrm { rads } ^ { - 1 }$, the seats move in a horizontal circle with each chain making an angle $\theta$ with the vertical. Each of the seats and the chains may be modelled as light. Assume that all chains have the same length and are inextensible.
When a man of mass 75 kg occupies a seat, the tension in the chain is $490 \sqrt { 3 } \mathrm {~N}$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\theta = 30 ^ { \circ }$.
\item Calculate the length of each chain.
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2019 Q5 [8]}}