| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2019 |
| Session | June |
| Marks | 8 |
| 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 two-part mechanics question requiring Hooke's law (F = λx/l) to find the modulus, then energy conservation (EPE = KE) to find the final speed. Both are standard applications of well-defined formulas with clear setup and minimal problem-solving 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 |
| Answer | Marks | Guidance |
|---|---|---|
| (a) Use of Hooke's Law: \(21 = \frac{\lambda x}{0.15}\) | M1 | |
| \(\lambda = 35\) (N) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Energy at start: \(EE = \frac{4x^2}{2(0.15)} - \frac{35(0.09)^2}{2(0.15)} = 0.945\) | M1, A1 | FT \(\lambda\) and \(x\) from (a) |
| Energy at end: \(KE = \frac{1}{2}(0 \cdot 1)v^2 = (0 \cdot 05v^2)\) | A1 | |
| Conservation of energy: \(0 \cdot 05v^2 = 0.945\) | M1 | Used with EE and KE |
| \(v = 4 \cdot 3\) (ms\(^{-1}\)) (2 sig. figs) | A1 | cao |
**(a)** Use of Hooke's Law: $21 = \frac{\lambda x}{0.15}$ | M1 |
$\lambda = 35$ (N) | A1 | cao
**Subtotal: [3]**
**(b)** Using expression for EE or KE:
Energy at start: $EE = \frac{4x^2}{2(0.15)} - \frac{35(0.09)^2}{2(0.15)} = 0.945$ | M1, A1 | FT $\lambda$ and $x$ from (a)
Energy at end: $KE = \frac{1}{2}(0 \cdot 1)v^2 = (0 \cdot 05v^2)$ | A1 |
Conservation of energy: $0 \cdot 05v^2 = 0.945$ | M1 | Used with EE and KE
$v = 4 \cdot 3$ (ms$^{-1}$) (2 sig. figs) | A1 | cao
**Subtotal: [5]**
**Total for Question 1: 8**
---\begin{enumerate}
\item The diagram shows a spring of natural length 0.15 m enclosed in a smooth horizontal tube. One end of the spring $A$ is fixed and the other end $B$ is compressed against a ball of mass $0 \cdot 1 \mathrm {~kg}$.\\
\includegraphics[max width=\textwidth, alt={}, center]{b430aa50-27e3-46f7-afef-7b8e75d46e1f-2_241_714_639_632}
\end{enumerate}
Initially, the ball is held in equilibrium by a force of 21 N so that the compressed length of the spring is $\frac { 2 } { 5 }$ of its natural length.\\
(a) Calculate the modulus of elasticity of the spring.\\
(b) The ball is released by removing the force. Determine the speed of the ball when the spring reaches its natural length. Give your answer correct to two significant figures.\\
\hfill \mbox{\textit{WJEC Further Unit 3 2019 Q1 [8]}}