| Exam Board | AQA |
|---|---|
| Module | Further AS Paper 2 Mechanics (Further AS Paper 2 Mechanics) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Maximum/minimum speed in elastic motion |
| Difficulty | Challenging +1.2 This is a Further Maths mechanics problem requiring energy conservation with elastic potential energy in vertical motion. While it involves multiple energy forms (gravitational PE, elastic PE, kinetic energy) and requires careful identification of the equilibrium position, it follows a standard template for this topic with straightforward application of energy principles once set up correctly. The 6 marks reflect moderate length rather than exceptional difficulty. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0.5 \times 2 \times 10 = \frac{\lambda}{2 \times 0.2} \times 0.4^2\) | M1 | Forms an energy equation between PE and EPE |
| \(\lambda = \frac{10}{0.4} = 25 \text{ N}\) | A1 | Obtains correct value for \(\lambda\) or \(k\) |
| \(2 \times 10 = \frac{25}{0.2}e\) | M1 | Using Hooke's Law to find extension at equilibrium |
| \(e = \frac{2 \times 10 \times 0.2}{25} = 0.16\) | A1F | Obtains correct extension at equilibrium using 'their' \(\lambda\) or \(k\) |
| \(2 \times 10 \times 0.26 = \frac{1}{2} \times 2v^2 + \frac{25 \times 0.16^2}{2 \times 0.2}\) | M1 | Forms equation using conservation of energy |
| \(v^2 = 3.6\), \(v = 1.897... = 2 \text{ m s}^{-1}\) (to 1 sf) | A1 | Obtains the correct speed. Only accept \(V = 2 \text{ m s}^{-1}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Could include air resistance. | E1 | States appropriate refinement |
## Question 8:
### Part 8(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0.5 \times 2 \times 10 = \frac{\lambda}{2 \times 0.2} \times 0.4^2$ | M1 | Forms an energy equation between PE and EPE |
| $\lambda = \frac{10}{0.4} = 25 \text{ N}$ | A1 | Obtains correct value for $\lambda$ or $k$ |
| $2 \times 10 = \frac{25}{0.2}e$ | M1 | Using Hooke's Law to find extension at equilibrium |
| $e = \frac{2 \times 10 \times 0.2}{25} = 0.16$ | A1F | Obtains correct extension at equilibrium using 'their' $\lambda$ or $k$ |
| $2 \times 10 \times 0.26 = \frac{1}{2} \times 2v^2 + \frac{25 \times 0.16^2}{2 \times 0.2}$ | M1 | Forms equation using conservation of energy |
| $v^2 = 3.6$, $v = 1.897... = 2 \text{ m s}^{-1}$ (to 1 sf) | A1 | Obtains the correct speed. Only accept $V = 2 \text{ m s}^{-1}$ |
### Part 8(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Could include air resistance. | E1 | States appropriate refinement |8 In this question use $g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
A particle, of mass 2 kg , is attached to one end of a light elastic string of natural length 0.2 metres.
The other end of the string is attached to a fixed point $O$.\\
The particle is pulled down and released from rest at a point 0.6 metres directly below $O$.\\
The particle then moves vertically and next comes to rest when it is 0.1 metres below $O$.\\
Assume that no air resistance acts on the particle.\\
8
\begin{enumerate}[label=(\alph*)]
\item Find the maximum speed of the particle.\\[0pt]
[6 marks]\\
8
\item Describe one way in which the model you have used could be refined.
\end{enumerate}
\hfill \mbox{\textit{AQA Further AS Paper 2 Mechanics Q8 [6]}}