| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string on rough inclined plane |
| Difficulty | Challenging +1.2 This is a multi-step Further Maths mechanics problem involving elastic strings, friction, and energy methods on an inclined plane. While it requires careful bookkeeping of energy terms (elastic PE, gravitational PE, work against friction) and understanding of equilibrium conditions, the solution follows a standard work-energy framework with clearly defined start and end states. Part (b) requires force analysis at rest to determine subsequent motion, which is routine for Further Maths students. The 'show that' format and 10 total marks indicate moderate difficulty within Further Maths, placing it above average overall but not exceptionally challenging. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| \(\text{EPE} = \dfrac{65 \times 0.7^2}{2 \times 1.3} = 12.25\text{ J}\) | B1 | Correct EPE |
| Friction \(= 2 \times 9.8 \times \cos(30°) \times 0.6 = 10.184\text{ N}\) | B1 | Correct friction value |
| \(\text{EPE} = F_r(2-d) + mg(2-d)\sin(30°) + \dfrac{65(d-1.3)^2}{2\times 1.3}\) | M1 | Work-energy equation with at least two correct terms |
| \(12.25 = 10.184(2-d) + 9.8(2-d) + 25(d-1.3)^2\) | A1 | All terms correct with correct signs |
| \(25d^2 - 84.984d + 69.968 = 0\) | M1 | Solves to obtain two solutions |
| \(d = 1.39936\text{ or }2\) | A1 | |
| \(d = 2\) rejected as it is the starting position | A1 | Rejects \(d=2\) with reason |
| \(d = 1.4\text{ m}\) | R1 | Rigorous argument; fully correct solution required |
| Answer | Marks | Guidance |
|---|---|---|
| Component of weight down plane \(= 2 \times 9.8\sin(30°) = 9.6\text{ N}\) | M1 | Calculates component of weight parallel to plane |
| Limiting friction \(= 10.184 > 9.6\) (and there is tension too) | R1 | Compares friction (and tension) with weight component |
| \(\therefore\) Particle does not slide back down the plane | R1 | Correct determination |
## Question 9:
### Part (a):
| $\text{EPE} = \dfrac{65 \times 0.7^2}{2 \times 1.3} = 12.25\text{ J}$ | B1 | Correct EPE |
| Friction $= 2 \times 9.8 \times \cos(30°) \times 0.6 = 10.184\text{ N}$ | B1 | Correct friction value |
| $\text{EPE} = F_r(2-d) + mg(2-d)\sin(30°) + \dfrac{65(d-1.3)^2}{2\times 1.3}$ | M1 | Work-energy equation with at least two correct terms |
| $12.25 = 10.184(2-d) + 9.8(2-d) + 25(d-1.3)^2$ | A1 | All terms correct with correct signs |
| $25d^2 - 84.984d + 69.968 = 0$ | M1 | Solves to obtain two solutions |
| $d = 1.39936\text{ or }2$ | A1 | |
| $d = 2$ rejected as it is the starting position | A1 | Rejects $d=2$ with reason |
| $d = 1.4\text{ m}$ | R1 | Rigorous argument; fully correct solution required |
### Part (b):
| Component of weight down plane $= 2 \times 9.8\sin(30°) = 9.6\text{ N}$ | M1 | Calculates component of weight parallel to plane |
| Limiting friction $= 10.184 > 9.6$ (and there is tension too) | R1 | Compares friction (and tension) with weight component |
| $\therefore$ Particle does not slide back down the plane | R1 | Correct determination |9 In this question use $g = 9.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
A light elastic string has one end attached to a fixed point, $A$, on a rough plane inclined at $30 ^ { \circ }$ to the horizontal.
The other end of the string is attached to a particle, $P$, of mass 2 kg .\\
The elastic string has natural length 1.3 metres and modulus of elasticity 65 N .\\
The particle is pulled down the plane in the direction of the line of greatest slope through $A$.\\
The particle is released from rest when it is 2 metres from $A$, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{4fdb2637-6368-422c-99da-85b80efe31c5-14_549_744_861_785}
The coefficient of friction between the particle and the plane is 0.6\\
After the particle is released it moves up the plane.\\
The particle comes to rest at a point $B$, which is a distance, $d$ metres, from $A$.
9
\begin{enumerate}[label=(\alph*)]
\item Show that the value of $d$ is 1.4.\\[0pt]
[7 marks]
9
\item Determine what happens after $P$ reaches the point $B$.
Fully justify your answer.\\[0pt]
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics Q9 [10]}}