| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2024 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string with compression (spring) |
| Difficulty | Standard +0.3 This is a straightforward energy conservation problem with elastic strings. Part (a) is a simple 'show that' calculation using EPE = λx²/(2l). Part (b) requires conservation of energy with one equation to solve, though students must correctly account for the change in gravitational PE and identify the spring length from the given elastic energy. Standard Further Maths mechanics with no novel insight required. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
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Question 2:
2
For Examiner’s use only
Maximum
estion
Mark | Mark
Awarded
1 14
2 10
3 5
4 7
5 9
6 10
7 15
otal 70
Question
number | Additional page, if required.
Write the question number(s) in the left-hand margin.
Question
number | Additional page, if required.
Write the question number(s) in the left-hand margin.\begin{enumerate}
\item The diagram below shows a light spring of natural length 1.2 m and modulus of elasticity 84 N . One end of the spring $A$ is fixed and the other end is attached to an object $P$ of mass 4 kg .\\
\includegraphics[max width=\textwidth, alt={}, center]{ae23a093-1419-4be4-8285-951650dc5a35-06_542_451_466_808}
\end{enumerate}
Initially, $P$ is held at rest with the spring stretched to a total length of 2.2 m and $A P$ vertical.\\
(a) Show that the elastic energy stored in the spring is 35 J .\\
(b) The object $P$ is then released. Find the speed of $P$ at the instant when the elastic energy in the spring is reduced to $5 \cdot 6 \mathrm {~J}$.\\
\hfill \mbox{\textit{WJEC Further Unit 3 2024 Q2 [10]}}