| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2023 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: released from rest, string starts taut |
| Difficulty | Standard +0.3 Part (a) is a straightforward application of Hooke's law at equilibrium (T = mg), requiring simple algebraic manipulation. Part (b) is a standard energy conservation problem with elastic potential energy and gravitational potential energy—a common Further Maths mechanics exercise with clear methodology. The multi-step nature and 8 marks elevate it slightly above average, but it follows a well-practiced template without requiring novel insight. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Use of Hooke's Law: \(T = \frac{\lambda x}{l}\) | M1 | Used, \(T = \frac{\lambda x}{l}\) |
| \(\lambda = 5g = 49\) and \(4g = 39 \cdot 2\) | A1 | Convincing |
| \(x = 0 \cdot 16\) (m) | A1 | Correct expression |
| Before (at lowest point): Using expression for \(EE = \frac{\lambda x^2}{2l}\) | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \end{cases}\] | A1 | A correct expression |
| Using expression for \(PE = mgh\): | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| \end{cases}\] | A1 | A correct expression |
| \(KE = \frac{1}{2}(4)v^2\) | B1 | |
| Conservation of Energy (before = after): KE, EE and PE all present | M1 | |
| \[\frac{5g(0 \cdot 28)^2}{2(0 \cdot 2)} = \frac{1}{2}(4)v^2 + \frac{5g(0 \cdot 16)^2}{2(0 \cdot 2)} + 4g(0 \cdot 12)\] | A1 | All correct, oe |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = 0 \cdot 9(391 \ldots)\) (ms\(^{-1}\)) | A1 | Note \(v = \frac{21\sqrt{5}}{50}\) or \(v = 0 \cdot 3\sqrt{g}\) |
Use of Hooke's Law: $T = \frac{\lambda x}{l}$ | M1 | Used, $T = \frac{\lambda x}{l}$
$\lambda = 5g = 49$ and $4g = 39 \cdot 2$ | A1 | Convincing
$x = 0 \cdot 16$ (m) | A1 | Correct expression
Before (at lowest point): Using expression for $EE = \frac{\lambda x^2}{2l}$ | M1 |
$$EE = \begin{cases}
\frac{5g(0.28^2 - 0.16^2)}{2(0.2)} = 6 \cdot 468 = 0 \cdot 66g \\
\frac{5g(0.28)^2}{2(0.2)} = 9 \cdot 604 = 0 \cdot 98g \\
\frac{5g(0.16)^2}{2(0.2)} = 3 \cdot 136 = 0 \cdot 32g
\end{cases}$$ | A1 | A correct expression
Using expression for $PE = mgh$: | M1 |
$$PE = \begin{cases}
4g(0 \cdot 28 - 0 \cdot 16) = 4 \cdot 704 = 0 \cdot 48g \\
4g(0 \cdot 28) = 10 \cdot 976 = 1 \cdot 12g \\
4g(0 \cdot 16) = 6 \cdot 272 = 0 \cdot 64g \\
4g(0 \cdot 48) = 18 \cdot 816 = 1 \cdot 92g \\
4g(0 \cdot 36) = 14 \cdot 112 = 1 \cdot 44g
\end{cases}$$ | A1 | A correct expression
$KE = \frac{1}{2}(4)v^2$ | B1 |
Conservation of Energy (before = after): KE, EE and PE all present | M1 |
$$\frac{5g(0 \cdot 28)^2}{2(0 \cdot 2)} = \frac{1}{2}(4)v^2 + \frac{5g(0 \cdot 16)^2}{2(0 \cdot 2)} + 4g(0 \cdot 12)$$ | A1 | All correct, oe
$9 \cdot 604 = 2v^2 + 3 \cdot 136 + 4 \cdot 704$
$v^2 = 0 \cdot 882$ ($= 0 \cdot 09g$)
$v = 0 \cdot 9(391 \ldots)$ (ms$^{-1}$) | A1 | Note $v = \frac{21\sqrt{5}}{50}$ or $v = 0 \cdot 3\sqrt{g}$
---One end of a light elastic string, of natural length $0.2$ m and modulus of elasticity $5g$ N, is attached to a fixed point $O$. The other end is attached to a particle of mass $4$ kg. The particle hangs in equilibrium vertically below $O$.
\begin{enumerate}[label=(\alph*)]
\item Show that the extension of the string is $0.16$ m. [2]
\item The particle is pulled down vertically and held at rest so that the extension of the string is $0.28$ m. The particle is then released. Determine the speed of the particle as it passes through the equilibrium position. [8]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2023 Q1 [10]}}