| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Session | Specimen |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: tension at specific point |
| Difficulty | Standard +0.8 This is a standard Further Maths mechanics problem involving circular motion with energy conservation and tension calculations. While it requires multiple techniques (geometry, energy methods, circular motion dynamics), the approach is methodical and follows well-established patterns for this topic. The geometry setup and multi-part structure add some complexity, but it remains a typical Further Maths exercise rather than requiring novel insight. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| (a) For P falling freely: \(v^2 = u^2 + 2as\), \(u = 0\), \(s = 2\), \(a = g\) gives \(v^2 = 2g \times 2 = 4g\), so \(v = 2\sqrt{g}\) | M1 A1 | AO3 AO1 |
| When string tightens, P has vertical speed \(2\sqrt{g}\). The component of this along the string is destroyed and P begins to move in a vertical circle with initial speed \(2\sqrt{g}\cos 30° = 2\sqrt{g} \times \frac{\sqrt{3}}{2} = \sqrt{3g}\) | m1 A1 | AO3 AO1 |
| (b)(i) Conservation of energy: \(\frac{1}{2}mv^2 = \frac{1}{2}m \times 3g + mg \times 4(\cos 45° - \sin 30°)\) | M1 A1 | AO3 AO1 |
| \(v^2 = 3g - 4g + 8g\cos 45° = 3g - 4g + 8g\cos 45°\) | A1 | AO1 |
| \(v^2 = 45.63717(165)\) | A1 | AO1 |
| \(v = 6.76 \text{ ms}^{-1}\) \((6.7555289...)\) | A1 | AO1 |
| (b)(ii) N2L towards centre: \(T - mg\cos 45° = \frac{mv^2}{4}\) | M1 A1 | AO3 AO2 |
| \(T = 3g\left(\frac{1}{\sqrt{2}}\right) + \frac{3}{4}(45.63717)\) | m1 | AO1 |
| \(T = 55.02\) \((55.0168...)\) | A1 | AO1 |
**(a)** For P falling freely: $v^2 = u^2 + 2as$, $u = 0$, $s = 2$, $a = g$ gives $v^2 = 2g \times 2 = 4g$, so $v = 2\sqrt{g}$ | M1 A1 | AO3 AO1
When string tightens, P has vertical speed $2\sqrt{g}$. The component of this along the string is destroyed and P begins to move in a vertical circle with initial speed $2\sqrt{g}\cos 30° = 2\sqrt{g} \times \frac{\sqrt{3}}{2} = \sqrt{3g}$ | m1 A1 | AO3 AO1
**(b)(i)** Conservation of energy: $\frac{1}{2}mv^2 = \frac{1}{2}m \times 3g + mg \times 4(\cos 45° - \sin 30°)$ | M1 A1 | AO3 AO1 | KE and PE in dimensionally correct equation
$v^2 = 3g - 4g + 8g\cos 45° = 3g - 4g + 8g\cos 45°$ | A1 | AO1 | KE
$v^2 = 45.63717(165)$ | A1 | AO1 | PE
$v = 6.76 \text{ ms}^{-1}$ $(6.7555289...)$ | A1 | AO1
**(b)(ii)** N2L towards centre: $T - mg\cos 45° = \frac{mv^2}{4}$ | M1 A1 | AO3 AO2 | dimensionally correct equation
$T = 3g\left(\frac{1}{\sqrt{2}}\right) + \frac{3}{4}(45.63717)$ | m1 | AO1 | substitute for $v^2$
$T = 55.02$ $(55.0168...)$ | A1 | AO1
**Total: [12]**
---A particle $P$, of mass 3 kg, is attached to a fixed point $O$ by a light inextensible string of length 4 m. Initially, particle $P$ is held at rest at a point which is $2\sqrt{3}$ m horizontally from $O$. It is then released and allowed to fall under gravity.
\begin{enumerate}[label=(\alph*)]
\item Show that the speed of $P$ when it first begins to move in a circle is $\sqrt{3g}$. [4]
\item In the subsequent motion, when the string first makes an angle of 45° with the downwards vertical,
\begin{enumerate}[label=(\roman*)]
\item calculate the speed $v$ of $P$,
\item determine the tension in the string. [8]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 Q2 [12]}}