| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2023 |
| Session | June |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: general symbolic/proof questions |
| Difficulty | Standard +0.8 This is a two-part Further Maths mechanics question requiring energy conservation with elastic strings (including careful handling of slack/taut transitions) and oblique collision analysis with restitution coefficient. Part (a) needs energy methods accounting for elastic PE and gravitational PE across different string states; part (b) combines geometry, restitution, and energy loss conditions. While systematic, it requires confident handling of multiple mechanics principles and algebraic manipulation beyond standard A-level. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings6.03i Coefficient of restitution: e6.03j Perfectly elastic/inelastic: collisions6.03k Newton's experimental law: direct impact6.03l Newton's law: oblique impacts |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{3mg}{2a}(2a)^2\) | B1 | Correct EPE term seen |
| \(\frac{1}{2}mv^2 + mg \times \left(3a - \frac{3}{4}a\right) = \frac{3mg}{2a}(2a)^2\) | M1 | Dimensionally correct energy equation. Must have one KE, one EPE term and at least one GPE. Allow sign errors. |
| \(v = \sqrt{\frac{15}{2}ag}\) \(\left[2.74\sqrt{ag}\right]\) | A1 | AEF |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T - mg = mA\) and \(T = \frac{3mg}{a} \times 2a\) | M1 | N2L and Hooke's law |
| Acceleration \(= 5g\) [upwards] | A1 | Allow \(\pm 50\) or \(\pm 5g\) |
| 2 |
## Question 1:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{3mg}{2a}(2a)^2$ | **B1** | Correct EPE term seen |
| $\frac{1}{2}mv^2 + mg \times \left(3a - \frac{3}{4}a\right) = \frac{3mg}{2a}(2a)^2$ | **M1** | Dimensionally correct energy equation. Must have one KE, one EPE term and at least one GPE. Allow sign errors. |
| $v = \sqrt{\frac{15}{2}ag}$ $\left[2.74\sqrt{ag}\right]$ | **A1** | AEF |
| | **3** | |
---
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T - mg = mA$ and $T = \frac{3mg}{a} \times 2a$ | **M1** | N2L and Hooke's law |
| Acceleration $= 5g$ [upwards] | **A1** | Allow $\pm 50$ or $\pm 5g$ |
| | **2** | |1 One end of a light elastic string, of natural length $a$ and modulus of elasticity $3 m g$, is attached to a fixed point $O$. The other end of the string is attached to a particle $P$ of mass $m$. The string hangs with $P$ vertically below $O$. The particle $P$ is pulled vertically downwards so that the extension of the string is $2 a$. The particle $P$ is then released from rest.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $P$ when it is at a distance $\frac { 3 } { 4 } a$ below $O$.
\item Find the initial acceleration of $P$ when it is released from rest.\\
\includegraphics[max width=\textwidth, alt={}, center]{454be64a-204f-4fa4-a5fc-72fd88e1289f-03_741_473_269_836}
A particle $P$ of mass $m$ is moving with speed $u$ on a fixed smooth horizontal surface. It collides at an angle $\alpha$ with a fixed smooth vertical barrier. After the collision, $P$ moves at an angle $\theta$ with the barrier, where $\tan \theta = \frac { 1 } { 2 }$ (see diagram). The coefficient of restitution between $P$ and the barrier is $e$. The particle $P$ loses 20\% of its kinetic energy as a result of the collision.
Find the value of $e$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2023 Q1 [5]}}