| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2022 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: released from rest at natural length or above (string initially slack) |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics problem involving elastic strings and energy methods. Part (a) requires routine application of energy conservation (6 marks suggests showing working clearly), part (b) is a straightforward calculation setting v=0, and part (c) requires recognizing that maximum speed occurs when acceleration=0 (when weight equals tension). While it involves multiple steps and careful algebra, the techniques are well-practiced in Further Maths syllabi with no novel insight required—moderately above average difficulty due to the multi-part nature and algebraic manipulation needed. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Using expression for PE = \(mgh\) or EE = \(\frac{\lambda x^2}{2l}\) | M1 | |
| Loss in PE = \(2g(2 \cdot 5 + x)\) (= \(5g + 2gx\)) | A1 | |
| Gain in EE = \(\frac{\lambda x^2}{2(2 \cdot 5)} - \frac{30gx^2}{2(2 \cdot 5)}\) (= \(6gx^2\)) | A1 | |
| Gain in KE = \(\frac{1}{2}(2)v^2\) (= \(v^2\)) | B1 | |
| Conservation of energy | M1 | Used with PE,KE and EE; All terms, allow sign errors |
| Answer | Marks | Guidance |
|---|---|---|
| \(v^2 = g(5 + 2x - 6x^2)\) | A1[6] | Convincing |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = \frac{2\pm\sqrt{124}}{12}\) | m1 | \(x = \frac{1\pm\sqrt{31}}{6}\) from calculator |
| \(x = 1 \cdot 09(4627 \ldots)\) (or \(x = -0 \cdot 76(1294 \ldots)\)) | A1 | cao; \(x = -0 \cdot 76 \ldots\) clearly discarded |
| Answer | Marks | Guidance |
|---|---|---|
| When \(P\) attains its maximum speed, \(a = 0\) so that Tension in OP = \(2g\) | M1 | Hooke's Law used with \(T = 2g\) |
| Answer | Marks |
|---|---|
| \(x = \frac{1}{6}\) (m) | A1, A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = \frac{1}{6}\) | (M1), (A1), (A1) | Condone the following incorrect notation \(\frac{dv}{dx} = g(2-12x)\); oe |
| Answer | Marks | Guidance |
|---|---|---|
| Maximum speed is \(7 \cdot 11(57103 \ldots)\) (ms\(^{-1}\)) | M1, A1[5] | FT their \(x \geq 0\); \(v = \sqrt{\frac{319}{6}} = \sqrt{\frac{1519}{30}}\); FT their \(x \neq 0\) for \(v^2 > 0\) |
## Part (a)
Using expression for PE = $mgh$ or EE = $\frac{\lambda x^2}{2l}$ | M1 |
Loss in PE = $2g(2 \cdot 5 + x)$ (= $5g + 2gx$) | A1 |
Gain in EE = $\frac{\lambda x^2}{2(2 \cdot 5)} - \frac{30gx^2}{2(2 \cdot 5)}$ (= $6gx^2$) | A1 |
Gain in KE = $\frac{1}{2}(2)v^2$ (= $v^2$) | B1 |
Conservation of energy | M1 | Used with PE,KE and EE; All terms, allow sign errors
$v^2 + 6gx^2 = 5g + 2gx$
$v^2 = g(5 + 2x - 6x^2)$ | A1[6] | Convincing
## Part (b)
At maximum extension, $v = 0$
$0 = g(5 + 2x - 6x^2)$
$6x^2 - 2x - 5 = 0$
Attempting to solve
$x = \frac{2\pm\sqrt{124}}{12}$ | m1 | $x = \frac{1\pm\sqrt{31}}{6}$ from calculator
$x = 1 \cdot 09(4627 \ldots)$ (or $x = -0 \cdot 76(1294 \ldots)$) | A1 | cao; $x = -0 \cdot 76 \ldots$ clearly discarded
**[3]**
## Part (c)
### (i)
When $P$ attains its maximum speed, $a = 0$ so that Tension in OP = $2g$ | M1 | Hooke's Law used with $T = 2g$
$\frac{30gx}{2 \cdot 5} = 2g$
$x = \frac{1}{6}$ (m) | A1, A1 |
### Alternative Solution to (i)
(i) Differentiating to find for maximum $v^2$(or $v$)
$\frac{d(v^2)}{dx} = 0$
$g(2 - 12x) = 0$
$x = \frac{1}{6}$ | (M1), (A1), (A1) | Condone the following incorrect notation $\frac{dv}{dx} = g(2-12x)$; oe
### (ii)
Sub. $x = \frac{1}{6}$ into $v^2 = g(5 + 2x - 6x^2)$
Maximum speed is $7 \cdot 11(57103 \ldots)$ (ms$^{-1}$) | M1, A1[5] | FT their $x \geq 0$; $v = \sqrt{\frac{319}{6}} = \sqrt{\frac{1519}{30}}$; FT their $x \neq 0$ for $v^2 > 0$
**Total for Question 5: 14**
---One end of a light elastic string, of natural length 2.5 m and modulus of elasticity $30g$ N, is fixed to a point O. A particle $P$, of mass 2 kg, is attached to the other end of the string. Initially, $P$ is held at rest at the point O. It is then released and allowed to fall under gravity.
\begin{enumerate}[label=(\alph*)]
\item Show that, while the string is taut,
$$v^2 = g(5 + 2x - 6x^2),$$
where $v\text{ ms}^{-1}$ denotes the velocity of the particle when the extension in the string is $x$ m. [6]
\item Calculate the maximum extension of the string. [3]
\item \begin{enumerate}[label=(\roman*)]
\item Find the extension of the string when $P$ attains its maximum speed.
\item Hence determine the maximum speed of $P$. [5]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2022 Q5 [14]}}