| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2021 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: general symbolic/proof questions |
| Difficulty | Challenging +1.2 This is a standard energy conservation problem for elastic strings requiring setup of equilibrium condition (to find natural extension), then applying conservation of energy with elastic PE, gravitational PE, and KE. The algebra is moderately involved but the method is routine for Further Mechanics students. Slightly above average difficulty due to algebraic manipulation required. |
| Spec | 6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Loss in EPE \(= \frac{1}{2}\times\frac{12mge^2}{a} - \frac{1}{2}\times\frac{12mg}{a}\times\left(e-\frac{a}{3}\right)^2 \left(=\frac{2mg}{3}(6e-a)\right)\) | B1 | Either term correct |
| Gain in KE \(= \frac{1}{2}mv^2\) and Gain in GPE \(= \frac{mga}{3}\) | B1 | |
| Gain in KE + Gain in GPE = Loss in EPE | M1 | KE, GPE and at least one EPE term |
| \(\frac{1}{2}mv^2 + \frac{mga}{3} = \frac{2mg}{3}(6e-a)\) | A1 | All terms correct |
| Simplify to a linear equation in \(e\) | M1 | |
| \(e = \frac{1}{2}a\) | A1 | |
| Total | 6 |
## Question 3:
| Answer | Mark | Guidance |
|--------|------|----------|
| Loss in EPE $= \frac{1}{2}\times\frac{12mge^2}{a} - \frac{1}{2}\times\frac{12mg}{a}\times\left(e-\frac{a}{3}\right)^2 \left(=\frac{2mg}{3}(6e-a)\right)$ | B1 | Either term correct |
| Gain in KE $= \frac{1}{2}mv^2$ and Gain in GPE $= \frac{mga}{3}$ | B1 | |
| Gain in KE + Gain in GPE = Loss in EPE | M1 | KE, GPE and at least one EPE term |
| $\frac{1}{2}mv^2 + \frac{mga}{3} = \frac{2mg}{3}(6e-a)$ | A1 | All terms correct |
| Simplify to a linear equation in $e$ | M1 | |
| $e = \frac{1}{2}a$ | A1 | |
| **Total** | **6** | |
---3 A light elastic string has natural length $a$ and modulus of elasticity 12 mg . One end of the string is attached to a fixed point $O$. The other end of the string is attached to a particle of mass $m$. The particle hangs in equilibrium vertically below $O$. The particle is pulled vertically down and released from rest with the extension of the string equal to $e$, where $\mathrm { e } > \frac { 1 } { 3 } \mathrm { a }$. In the subsequent motion the particle has speed $\sqrt { 2 \mathrm { ga } }$ when it has ascended a distance $\frac { 1 } { 3 } a$.
Find $e$ in terms of $a$.\\
\includegraphics[max width=\textwidth, alt={}, center]{b10c65ef-dafd-4746-be5b-789130b7d030-06_488_496_269_781}
A uniform lamina $A E C F$ is formed by removing two identical triangles $B C E$ and $C D F$ from a square lamina $A B C D$. The square has side $3 a$ and $E B = D F = h$ (see diagram).
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the lamina $A E C F$ from $A D$ and from $A B$, giving your answers in terms of $a$ and $h$.\\
The lamina $A E C F$ is placed vertically on its edge $A E$ on a horizontal plane.
\item Find, in terms of $a$, the set of values of $h$ for which the lamina remains in equilibrium.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q3 [6]}}