| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Year | 2022 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Centre of mass of composite shapes |
| Difficulty | Challenging +1.2 This is a multi-part mechanics question involving standard centre of mass calculations for composite bodies and equilibrium with friction. Part (a) is routine moments calculation, part (b)(i) is trivial force identification, part (b)(ii) is standard moments about a point (though marked 'show that'), and part (b)(iii) requires checking both sliding and toppling conditions. While it has multiple steps and requires careful application of friction laws and stability criteria, all techniques are standard A-level Further Maths mechanics with no novel insight required. The 'show that' and toppling check elevate it slightly above average difficulty. |
| Spec | 3.03u Static equilibrium: on rough surfaces3.04b Equilibrium: zero resultant moment and force6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} = \dfrac{9 \times 10 + 1 \times 18}{10} = 10.8\) | M1 | Forms equation to find distance of centre of mass from \(AF\). Condone one error in distances. |
| \(\bar{x} = 10.8\) | A1 | Correct distance |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R = (12 + m)g\) | B1 | Obtains correct vertical reaction force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Take moments about \(B\): \(P\sin 60° = 12g \times 0.5\cos 60° + mg \times 0.75\cos 60°\) | M1 | Takes moments about any point on the rod, with at least two correct terms |
| \(S = \dfrac{g(6 + 0.75m)}{\tan 60°} = \dfrac{g(6 + 0.75m)}{\sqrt{3}}\) | A1 | Correct moment equation |
| \(P = S\), therefore horizontal reaction force at \(B\) obtained | R1 | Completes reasoned argument to obtain correct expression for horizontal reaction force at \(B\). AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(10g \times 9.2 + (12+m)g \times 4 = 20 \times \dfrac{g(6+0.75m)}{\sqrt{3}}\) | M1 | Forms equation/inequality with at least three terms to find \(m\) when on point of toppling, taking moments about \(E\) with moment of horizontal force correct |
| \(140\sqrt{3} + 4m\sqrt{3} = 120 + 15m\) | A1 | Correct equation/inequality |
| \(m = \dfrac{140\sqrt{3} - 120}{15 - 4\sqrt{3}} = 15.2\), AWRT 15 | A1 | Correct \(m\) for point of toppling |
| \(\dfrac{g(6+0.75m)}{\sqrt{3}} = 0.3 \times (22+m)g\); \(6 + 0.75m = 6.6\sqrt{3} + 0.3\sqrt{3}m\) | M1 | Forms equation/inequality to find \(m\) when on point of sliding using \(F = \mu R\). Condone omission of mass of block. |
| \(m = \dfrac{6 - 6.6\sqrt{3}}{0.3\sqrt{3} - 0.75} = 23.6\), AWRT 24 | A1 | Correct \(m\) for point of sliding |
| Maximum value of \(m\) is 15.2 | R1 | Deduces value of \(m\) is AWRT 15 from correct working |
## Question 9(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = \dfrac{9 \times 10 + 1 \times 18}{10} = 10.8$ | M1 | Forms equation to find distance of centre of mass from $AF$. Condone one error in distances. |
| $\bar{x} = 10.8$ | A1 | Correct distance |
**Total: 2 marks**
---
## Question 9(b)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R = (12 + m)g$ | B1 | Obtains correct vertical reaction force |
**Total: 1 mark**
---
## Question 9(b)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Take moments about $B$: $P\sin 60° = 12g \times 0.5\cos 60° + mg \times 0.75\cos 60°$ | M1 | Takes moments about any point on the rod, with at least two correct terms |
| $S = \dfrac{g(6 + 0.75m)}{\tan 60°} = \dfrac{g(6 + 0.75m)}{\sqrt{3}}$ | A1 | Correct moment equation |
| $P = S$, therefore horizontal reaction force at $B$ obtained | R1 | Completes reasoned argument to obtain correct expression for horizontal reaction force at $B$. AG |
**Total: 3 marks**
---
## Question 9(b)(iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $10g \times 9.2 + (12+m)g \times 4 = 20 \times \dfrac{g(6+0.75m)}{\sqrt{3}}$ | M1 | Forms equation/inequality with at least three terms to find $m$ when on point of toppling, taking moments about $E$ with moment of horizontal force correct |
| $140\sqrt{3} + 4m\sqrt{3} = 120 + 15m$ | A1 | Correct equation/inequality |
| $m = \dfrac{140\sqrt{3} - 120}{15 - 4\sqrt{3}} = 15.2$, AWRT 15 | A1 | Correct $m$ for point of toppling |
| $\dfrac{g(6+0.75m)}{\sqrt{3}} = 0.3 \times (22+m)g$; $6 + 0.75m = 6.6\sqrt{3} + 0.3\sqrt{3}m$ | M1 | Forms equation/inequality to find $m$ when on point of sliding using $F = \mu R$. Condone omission of mass of block. |
| $m = \dfrac{6 - 6.6\sqrt{3}}{0.3\sqrt{3} - 0.75} = 23.6$, AWRT 24 | A1 | Correct $m$ for point of sliding |
| Maximum value of $m$ is 15.2 | R1 | Deduces value of $m$ is AWRT 15 from correct working |
**Total: 6 marks**9 Two blocks have square cross sections.
One block has mass 9 kg and its cross section has sides of length 20 cm\\
The other block has mass 1 kg and its cross section has sides of length 4 cm\\
The blocks are fixed together to form the composite body shown in Figure 1.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{0afe3ff2-0af5-4aeb-98c5-1346fa803388-13_570_492_717_776}
\end{center}
\end{figure}
9
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the composite body from $A F$\\[0pt]
[2 marks]\\
Question 9 continues on the next page
9
\item A uniform rod has mass 12 kg and length 1 metre.
One end of the rod rests against a smooth vertical wall.\\
The other end of the rod rests on the composite body at point $B$\\
The composite body is on a horizontal surface.\\
The coefficient of friction between the composite body and the horizontal surface is 0.3
The angle between the rod and $A B$ is $60 ^ { \circ }$\\
A particle of mass $m \mathrm {~kg}$ is fixed to the rod at a distance of 75 cm from $B$\\
The rod, particle and composite body are shown in Figure 2.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{0afe3ff2-0af5-4aeb-98c5-1346fa803388-14_939_1020_1133_511}
\end{center}
\end{figure}
9 (b) (i) Write down the magnitude of the vertical reaction force acting on the rod at $B$ in terms of $m$ and $g$\\[0pt]
[1 mark]
9 (b) (ii) Show that the magnitude of the horizontal reaction force acting on the rod at $B$ is
$$\frac { g ( 6 + 0.75 m ) } { \sqrt { 3 } }$$
9 (b) (iii) Find the maximum value of $m$ for which the composite body does not slide or topple.
Fully justify your answer.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2022 Q9 [14]}}