| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2021 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Toppling and sliding of solids |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics problem on composite bodies and toppling. Part (a) requires routine application of center of mass formulas for cone and cylinder (standard results), then combining them—straightforward but multi-step. Part (b) uses the toppling condition (vertical line through COM passes through pivot point) with given trigonometry, leading to an algebraic equation to solve for k. While it requires careful setup and algebraic manipulation, the techniques are all standard for Further Maths mechanics with no novel insight required. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| 4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw). |
| Answer | Marks |
|---|---|
| 4(a) | Centre of mass |
| Answer | Marks |
|---|---|
| 3 | M1 |
| Answer | Marks |
|---|---|
| 3 3 4 2 | A1 |
| A1 | 2 terms correct. |
| Answer | Marks | Guidance |
|---|---|---|
| 4 ( 3+ k ) | A1 | AG. Shown convincingly. |
| Answer | Marks |
|---|---|
| 4(b) | r |
| Answer | Marks |
|---|---|
| x | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| 2k2 −k −15=0 | M1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| k=3 | A1 | CAO. No other solutions. |
| Answer | Marks |
|---|---|
| Volume | Centre of mass |
| Answer | Marks |
|---|---|
| Cone | 1 |
| Answer | Marks |
|---|---|
| 3 | kh |
| Answer | Marks | Guidance |
|---|---|---|
| Cylinder | πr2h | h |
| Answer | Marks | Guidance |
|---|---|---|
| Combined | πr2h 1 k +1 | |
| 3 | x | |
| Question | Answer | Marks |
Question 4:
4 | Unless otherwise indicated, marks once gained cannot subsequently be lost, e.g. wrong working following a correct form of answer is ignored (isw).
--- 4(a) ---
4(a) | Centre of mass
Volume from base of
cylinder
1 kh
Cone πr2kh h+
3 4
h
Cylinder πr2h
2
Combined πr2h 1 k +1 x
3 | M1
Take moments about base:
πr2h 1 k +1 x = 1 πr2kh h+ kh + πr2h h
3 3 4 2 | A1
A1 | 2 terms correct.
All terms correct.
( )
h k2 + 4k +6
x =
4 ( 3+ k ) | A1 | AG. Shown convincingly.
4
--- 4(b) ---
4(b) | r
tanθ=
x | M1
4 6h ( k +3 )
=
( )
3 h k2 +4k +6
2k2 −k −15=0 | M1 | 4
Equate to and simplify to quadratic.
3
k=3 | A1 | CAO. No other solutions.
3
Volume | Centre of mass
from base of
cylinder
Cone | 1
πr2kh
3 | kh
h+
4
Cylinder | πr2h | h
2
Combined | πr2h 1 k +1
3 | x
Question | Answer | Marks | Guidance\includegraphics{figure_4}
A uniform solid circular cone has vertical height $kh$ and radius $r$. A uniform solid cylinder has height $h$ and radius $r$. The base of the cone is joined to one of the circular faces of the cylinder so that the axes of symmetry of the two solids coincide (see diagram, which shows a cross-section). The cone and the cylinder are made of the same material.
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of the combined solid from the base of the cylinder is $\frac{h(k^2 + 4k + 6)}{4(3 + k)}$. [4]
\end{enumerate}
The solid is placed on a plane that is inclined to the horizontal at an angle $\theta$. The base of the cylinder is in contact with the plane. The plane is sufficiently rough to prevent sliding. It is given that $3h = 2r$ and that the solid is on the point of toppling when $\tan \theta = \frac{4}{3}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Find the value of $k$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q4 [7]}}