| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2020 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Solid with removed cone from cone or cylinder |
| Difficulty | Standard +0.8 This is a two-part Further Maths mechanics question combining conical pendulum dynamics with center of mass of composite solids. Part (a) requires resolving forces and applying Hooke's law with circular motion equations—standard but multi-step. Part (b) involves finding the center of mass of a truncated cone using integration or formula application, then applying equilibrium conditions for suspension. While systematic, it requires confident handling of 3D geometry, composite bodies, and multiple mechanics principles, placing it moderately above average difficulty. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T = 4mg \cdot \frac{ka}{a}\) | B1 | Use Hooke's law |
| \(T\sin\theta = \left(\frac{mrg}{a}\right) = m(k+1)a\sin\theta \cdot \frac{g}{a}\) | M1 | N2L horizontally. Must see \(T\) and \(k\) |
| \(T = mg(k+1)\) | A1 | |
| Equate: \(k = \frac{1}{3}\) | A1 | |
| Total | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\uparrow T\cos\theta = mg\) | M1 | |
| \(\left(T = \frac{4}{3}mg\right) \quad \cos\theta = \frac{mg}{\frac{4}{3}mg} = \frac{3}{4}\) | A1 | |
| Total | 2 |
## Question 3:
### Part 3(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T = 4mg \cdot \frac{ka}{a}$ | B1 | Use Hooke's law |
| $T\sin\theta = \left(\frac{mrg}{a}\right) = m(k+1)a\sin\theta \cdot \frac{g}{a}$ | M1 | N2L horizontally. Must see $T$ and $k$ |
| $T = mg(k+1)$ | A1 | |
| Equate: $k = \frac{1}{3}$ | A1 | |
| **Total** | **4** | |
### Part 3(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\uparrow T\cos\theta = mg$ | M1 | |
| $\left(T = \frac{4}{3}mg\right) \quad \cos\theta = \frac{mg}{\frac{4}{3}mg} = \frac{3}{4}$ | A1 | |
| **Total** | **2** | |
---3 One end of a light elastic string, of natural length $a$ and modulus of elasticity $4 m g$, is attached to a fixed point $O$. The other end of the string is attached to a particle of mass $m$. The particle moves in a horizontal circle with a constant angular speed $\sqrt { \frac { \mathrm { g } } { \mathrm { a } } }$ with the string inclined at an angle $\theta$ to the downward vertical through $O$. The length of the string during this motion is $( \mathrm { k } + 1 ) \mathrm { a }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$.
\item Find the value of $\cos \theta$.\\
\includegraphics[max width=\textwidth, alt={}, center]{1c53c407-25ea-43fc-a571-74ba1fffea8f-06_584_695_264_667}
The diagram shows the cross-section $A B C D$ of a uniform solid object which is formed by removing a cone with cross-section $D C E$ from the top of a larger cone with cross-section $A B E$. The perpendicular distance between $A B$ and $D C$ is $h$, the diameter $A B$ is $6 r$ and the diameter $D C$ is $2 r$.\\
(a) Find an expression, in terms of $h$, for the distance of the centre of mass of the solid object from $A B$.\\
The object is freely suspended from the point $B$ and hangs in equilibrium. The angle between $A B$ and the downward vertical through $B$ is $\theta$.\\
(b) Given that $h = \frac { 13 } { 4 } r$, find the value of $\tan \theta$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q3 [6]}}