| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2023 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle at midpoint of string between two horizontal fixed points: horizontal surface motion |
| Difficulty | Challenging +1.2 This is a standard Further Mechanics elastic energy problem requiring energy conservation and Hooke's law force resolution. Part (a) uses energy methods (EPE to KE conversion) with straightforward algebra, while part (b) applies force resolution with symmetry. The setup is clear, methods are routine for Further Maths students, and the algebra is manageable, making it moderately above average difficulty but not requiring novel insight. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks |
|---|---|
| 4(a) | Energy equation: gain in KE = loss in EPE |
| Answer | Marks |
|---|---|
| 2 2 8a | B1 |
| Answer | Marks |
|---|---|
| A1 | At least one correct EPE. |
| Answer | Marks | Guidance |
|---|---|---|
| 4L2 −8aL−20a2 =0 | M1 | Simplify to quadratic in L. |
| L=10a | A1 | Correct single answer. |
| Answer | Marks | Guidance |
|---|---|---|
| 4(b) | At P: 2Tcos=macceleration | M1 |
| Answer | Marks |
|---|---|
| 8a 2 | M1 |
| Solve, acceleration = 12g | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 4:
--- 4(a) ---
4(a) | Energy equation: gain in KE = loss in EPE
1 1 5mg( )
m80ag= (2L−8a)2 −(4a)2
2 2 8a | B1
M1
A1 | At least one correct EPE.
3-term energy equation.
All correct.
4L2 −8aL−20a2 =0 | M1 | Simplify to quadratic in L.
L=10a | A1 | Correct single answer.
5
--- 4(b) ---
4(b) | At P: 2Tcos=macceleration | M1
5mg 15
Hooke’s law: T = 12a= mg
8a 2 | M1
Solve, acceleration = 12g | A1
3
Question | Answer | Marks | Guidance\includegraphics{figure_4}
A light elastic string has natural length $8a$ and modulus of elasticity $5mg$. A particle $P$ of mass $m$ is attached to the midpoint of the string. The ends of the string are attached to points $A$ and $B$ which are a distance $12a$ apart on a smooth horizontal table. The particle $P$ is held on the table so that $AP = BP = L$ (see diagram). The particle $P$ is released from rest. When $P$ is at the midpoint of $AB$ it has speed $\sqrt{80ag}$.
\begin{enumerate}[label=(\alph*)]
\item Find $L$ in terms of $a$. [5]
\item Find the initial acceleration of $P$ in terms of $g$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q4 [8]}}