| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Session | Specimen |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particles at coordinate positions |
| Difficulty | Standard +0.3 This is a straightforward centre of mass problem requiring standard formulas: taking moments about a point to find the centre of mass position, then using equilibrium conditions (sum of forces and moments) to find support reactions. All steps are routine applications of well-practiced techniques with no conceptual challenges or novel problem-solving required. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{x} = \frac{3.7 \times 20 + 2 \times 22.5}{42.5} = 2.8\) | M1 | Forms equation to find the centre of mass |
| \(\bar{x} = 2.8\) | A1 | Obtains correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moments about \(C\) | M1 | Forms an equation using moments |
| \(42.5 \times 9.8 \times 2.3 = 2.5R_D\) | A1F | Correct moment equation; ft 'their' centre of mass provided both M1 marks awarded |
| \(R_D = \frac{957.95}{2.5} = 383.18\) N \(= 380\) N (2sf) | A1F | One correct reaction force; allow more than 2sf; FT provided both M1 marks awarded |
| \(R_C + R_D = 42.5 \times 9.8\) → \(R_C = 416.5 - 383.18 = 33.32\) N \(= 33\) N (2sf) | A1F | Both reaction forces correctly to 2sf; ft 'their' reaction force, provided both M1 marks awarded |
# Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{x} = \frac{3.7 \times 20 + 2 \times 22.5}{42.5} = 2.8$ | M1 | Forms equation to find the centre of mass |
| $\bar{x} = 2.8$ | A1 | Obtains correct answer |
**Total for 3(a): 2 marks**
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# Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $C$ | M1 | Forms an equation using moments |
| $42.5 \times 9.8 \times 2.3 = 2.5R_D$ | A1F | Correct moment equation; ft 'their' centre of mass provided both M1 marks awarded |
| $R_D = \frac{957.95}{2.5} = 383.18$ N $= 380$ N (2sf) | A1F | One correct reaction force; allow more than 2sf; FT provided both M1 marks awarded |
| $R_C + R_D = 42.5 \times 9.8$ → $R_C = 416.5 - 383.18 = 33.32$ N $= 33$ N (2sf) | A1F | Both reaction forces correctly to 2sf; ft 'their' reaction force, provided both M1 marks awarded |
**Total for 3(b): 4 marks**
---3 In this question use $g = 9.8 \mathrm {~m} \mathrm {~s} ^ { - 2 }$.\\
A composite body consists of a uniform rod, $A B$, and a particle.\\
The rod has length 4 metres and mass 22.5 kilograms.\\
The particle, $P$, has mass 20 kilograms and is placed on the rod at a distance of 0.3 metres from $B$, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{4fdb2637-6368-422c-99da-85b80efe31c5-04_163_1323_767_402}
3
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of the body from $A$.
3
\item The body rests in equilibrium in a horizontal position on two supports, $C$ and $D$.\\
The support $C$ is 0.5 metres from $A$ and the support $D$ is 1 metre from $B$.
Find the magnitudes of the forces exerted on the body by the supports.\\[0pt]
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics Q3 [6]}}