| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2022 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Horizontal elastic string on smooth surface |
| Difficulty | Standard +0.3 This is a standard energy conservation problem with elastic strings. Part (a) requires applying conservation of energy between two positions with given speeds and extensions, leading to a straightforward algebraic solution. Part (b) uses Hooke's law to find tension and then applies F=ma. Both parts are routine applications of standard Further Mechanics techniques with no novel insight required, making it slightly easier than average for Further Maths. |
| Spec | 3.03u Static equilibrium: on rough surfaces6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Loss in \(KE =\) Gain in EPE | B1 | EPE terms correct |
| \(\frac{1}{2}mv^2 - \frac{1}{2}m\left(\frac{v}{2}\right)^2 = \frac{1}{2} \times \frac{4mg}{a}\left(\left(\frac{1}{2}a\right)^2 - \left(\frac{1}{4}a\right)^2\right)\) | M1 | All 4 terms and no extras |
| \(\frac{3}{4}mv^2 = \frac{4mg}{a} \times \frac{3}{16}a^2\) | M1 | Simplify |
| \(v^2 = ag, \quad v = \sqrt{ag}\) | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Hooke's law: tension \(= \frac{4mg}{a} \times \frac{1}{2}a \ (= 2mg)\) | M1 | |
| Acceleration \(= \frac{2mg}{m} = 2g\) | A1 | Accept \(-2g\) |
| Total: 2 |
**Question 2(a):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Loss in $KE =$ Gain in EPE | B1 | EPE terms correct |
| $\frac{1}{2}mv^2 - \frac{1}{2}m\left(\frac{v}{2}\right)^2 = \frac{1}{2} \times \frac{4mg}{a}\left(\left(\frac{1}{2}a\right)^2 - \left(\frac{1}{4}a\right)^2\right)$ | M1 | All 4 terms and no extras |
| $\frac{3}{4}mv^2 = \frac{4mg}{a} \times \frac{3}{16}a^2$ | M1 | Simplify |
| $v^2 = ag, \quad v = \sqrt{ag}$ | A1 | |
| **Total: 4** | | |
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**Question 2(b):**
| Answer | Marks | Guidance |
|--------|-------|----------|
| Hooke's law: tension $= \frac{4mg}{a} \times \frac{1}{2}a \ (= 2mg)$ | M1 | |
| Acceleration $= \frac{2mg}{m} = 2g$ | A1 | Accept $-2g$ |
| **Total: 2** | | |2 A light elastic string has natural length $a$ and modulus of elasticity 4 mg . One end of the string is fixed to a point $O$ on a smooth horizontal surface. A particle $P$ of mass $m$ is attached to the other end of the string. The particle $P$ is projected along the surface in the direction $O P$. When the length of the string is $\frac { 5 } { 4 } a$, the speed of $P$ is $v$. When the length of the string is $\frac { 3 } { 2 } a$, the speed of $P$ is $\frac { 1 } { 2 } v$.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for $v$ in terms of $a$ and $g$.
\item Find, in terms of $g$, the acceleration of $P$ when the stretched length of the string is $\frac { 3 } { 2 } a$.\\
\includegraphics[max width=\textwidth, alt={}, center]{5e95e0c9-d47d-4f2b-89da-ab949b9661f4-04_552_1059_264_502}
A smooth cylinder is fixed to a rough horizontal surface with its axis of symmetry horizontal. A uniform rod $A B$, of length $4 a$ and weight $W$, rests against the surface of the cylinder. The end $A$ of the rod is in contact with the horizontal surface. The vertical plane containing the rod $A B$ is perpendicular to the axis of the cylinder. The point of contact between the rod and the cylinder is $C$, where $A C = 3 a$. The angle between the rod and the horizontal surface is $\theta$ where $\tan \theta = \frac { 3 } { 4 }$ (see diagram). The coefficient of friction between the rod and the horizontal surface is $\frac { 6 } { 7 }$.
A particle of weight $k W$ is attached to the rod at $B$. The rod is about to slip. The normal reaction between the rod and the cylinder is $N$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q2 [6]}}