| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2018 |
| 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 | Challenging +1.8 This is a challenging mechanics problem requiring energy conservation with elastic strings, involving multiple phases (free fall then extension). Students must identify when the string becomes taut, set up the correct energy equation with elastic potential energy, and solve a quadratic. The second part requires physical insight about energy loss. This is substantially harder than typical A-level mechanics but standard for Further Maths FM3, requiring careful setup and multi-step reasoning rather than novel mathematical insight. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| 3(a) | Let \(x\) be the extension in the string when \(P\) is instantaneously at rest for the 1st time. | |
| Loss in PE \(= mgh = 30 \times 9.8(0.9+x)\) | M1 | attempted, \(h\) a distance |
| A1 | ||
| Gain in EE \(= \frac{1}{2} \times \lambda \times \frac{x^2}{l} = \frac{1}{2} \times 490 \times \frac{x^2}{1.5}\) | M1 | attempted |
| A1 | ||
| Conservation of energy | M1 | |
| \(\frac{1}{2} \times 490 \times \frac{x^2}{1.5} = 30 \times 9.8(0.9+x)\) | A1 | |
| \(x^2 - 1.8x - 1.62 = 0\) | m1 | attempt to solve quadratic |
| \(x = \frac{1.8 \pm \sqrt{1.8^2 + 4 \times 1 \times 1.62}}{2}\) | ||
| \(x = 2.4588\) | ||
| \(AP = 3.96\) (m) | A1 | cao |
| 3(b) | When \(P\) is instantaneously at rest for the 2nd time \(AP = 0.6\) (m) | B1 |
| External resistance to motion have been assumed to be negligible. | B1 |
3(a) | Let $x$ be the extension in the string when $P$ is instantaneously at rest for the 1st time. | | |
| Loss in PE $= mgh = 30 \times 9.8(0.9+x)$ | M1 | attempted, $h$ a distance |
| | A1 | |
| Gain in EE $= \frac{1}{2} \times \lambda \times \frac{x^2}{l} = \frac{1}{2} \times 490 \times \frac{x^2}{1.5}$ | M1 | attempted |
| | A1 | |
| Conservation of energy | M1 | |
| $\frac{1}{2} \times 490 \times \frac{x^2}{1.5} = 30 \times 9.8(0.9+x)$ | A1 | |
| $x^2 - 1.8x - 1.62 = 0$ | m1 | attempt to solve quadratic |
| $x = \frac{1.8 \pm \sqrt{1.8^2 + 4 \times 1 \times 1.62}}{2}$ | | |
| $x = 2.4588$ | | |
| $AP = 3.96$ (m) | A1 | cao |
3(b) | When $P$ is instantaneously at rest for the 2nd time $AP = 0.6$ (m) | B1 | |
| External resistance to motion have been assumed to be negligible. | B1 | |A light elastic string of natural length $1.5$ m and modulus of elasticity $490$ N has one end attached to a fixed point $A$ and the other end attached to a particle $P$ of mass $30$ kg. Initially, $P$ is held at rest vertically below $A$ such that the distance $AP$ is $0.6$ m. It is then allowed to fall vertically.
\begin{enumerate}[label=(\alph*)]
\item Calculate the distance $AP$ when $P$ is instantaneously at rest for the first time, giving your answer correct to 2 decimal places. [8]
\item Estimate the distance $AP$ when $P$ is instantaneously at rest for the second time and clearly state one assumption that you have made in making your estimate. [2]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2018 Q3 [10]}}