| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Session | Specimen |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string equilibrium and statics |
| Difficulty | Challenging +1.2 This is a static equilibrium problem with elastic strings requiring geometry setup, Hooke's law application, and force resolution. While it involves multiple components (string tension, elastic energy, rod thrust) and requires careful geometric analysis to find extension, the solution follows standard mechanics procedures without requiring novel insight. The multi-part structure and geometric complexity place it above average difficulty, but it remains a recognizable Further Maths mechanics exercise. |
| Spec | 3.03n Equilibrium in 2D: particle under forces6.02h Elastic PE: 1/2 k x^2 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(AB = 2 \times 2\cos 50°\) | B1 | AO3 |
| Hooke's Law: \(T_{AB} = \frac{\lambda}{2}(4\cos 50° - 2) = \lambda(2\cos 50° - 1)\) | M1 A1 | AO2 AO1 |
| \(T_{AB} = 0.286\lambda\) (N) | A1 | AO1 |
| (b) \(EE = \frac{1}{2} \times \frac{\lambda(4\cos 50° - 2)^2}{2}\) | M1 | AO1 |
| \(EE = 0.0816\lambda\) (J) | A1 | AO1 |
| (c) For vertical equilibrium: \(T_{AB}\cos 50° + T_{BC}\cos 80° = mg\) | M1 A1 | AO3 AO2 |
| \(T_{BC}\cos 80° = 49 - 20.286\cos 50°\) | m1 | AO1 |
| \(T_{BC} = 282.180 - 1.057\lambda\) (N) | A1 | AO1 |
| OR For horizontal equilibrium: \(T_{AB}\sin 50° = T_{BC}\sin 80°\) | (M1) | (AO3) |
| \(T_{BC} = \lambda(2\cos 50° - 1) \times \frac{\sin 50°}{\sin 80°}\) | (A1) | (AO2) |
| \(T_{BC} = 0.222\lambda\) (N) | (A1) | (AO1) |
**(a)** $AB = 2 \times 2\cos 50°$ | B1 | AO3
Hooke's Law: $T_{AB} = \frac{\lambda}{2}(4\cos 50° - 2) = \lambda(2\cos 50° - 1)$ | M1 A1 | AO2 AO1
$T_{AB} = 0.286\lambda$ (N) | A1 | AO1
**(b)** $EE = \frac{1}{2} \times \frac{\lambda(4\cos 50° - 2)^2}{2}$ | M1 | AO1
$EE = 0.0816\lambda$ (J) | A1 | AO1
**(c)** For vertical equilibrium: $T_{AB}\cos 50° + T_{BC}\cos 80° = mg$ | M1 A1 | AO3 AO2 | resolve vertically
$T_{BC}\cos 80° = 49 - 20.286\cos 50°$ | m1 | AO1
$T_{BC} = 282.180 - 1.057\lambda$ (N) | A1 | AO1
OR For horizontal equilibrium: $T_{AB}\sin 50° = T_{BC}\sin 80°$ | (M1) | (AO3) | Resolve horizontally
$T_{BC} = \lambda(2\cos 50° - 1) \times \frac{\sin 50°}{\sin 80°}$ | (A1) | (AO2)
$T_{BC} = 0.222\lambda$ (N) | (A1) | (AO1)
**Total: [9]**
---\includegraphics{figure_6}
A particle of mass 5 kg is attached to a string $AB$ and a rod $BC$ at the point $B$. The string $AB$ is light and elastic with modulus $\lambda$ N and natural length 2 m. The rod $BC$ is light and of length 2 m. The end $A$ of the string is attached to a fixed point and the end $C$ of the rod is attached to another fixed point such that $A$ is vertically above $C$ with $AC = 2$ m. When the particle rests in equilibrium, $AB$ makes an angle of 50° with the downward vertical.
\begin{enumerate}[label=(\alph*)]
\item Determine, in terms of $\lambda$, the tension in the string $AB$. [3]
\item Calculate, in terms of $\lambda$, the energy stored in the string $AB$. [2]
\item Find, in terms of $\lambda$, the thrust in the rod $BC$. [4]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 Q6 [9]}}