| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Uniform beam on two supports |
| Difficulty | Moderate -0.3 This is a standard M2 moments question requiring force diagrams, taking moments about a point, and resolving forces vertically. Part (b) is routine calculation, part (c) extends to a modified scenario but uses identical methods, and part (d) tests conceptual understanding. Slightly easier than average A-level due to straightforward setup and standard techniques, though multi-part structure provides some substance. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| (a) [Correct force diagram with labels and arrows] | B1 | 1 mark |
| (b) \(2T = 0.5 \times 98\) and \(T = 24.5 \text{ N}\) | M1, A1, A1 | 3 marks |
| (c)(i) \(2 \times 2 \times 24.5 = 3 \times 9.8 \times m + 0.5 \times 98\) and \(m = \frac{98 - 49}{3 \times 9.8} = \frac{5}{3} = 1.67 \text{ kg (to 3 sf)}\) | B1, M1, A1, A1 | 4 marks |
| Or: \(2 \times 2.45 = 3 \times 9.8m\) and \(m = \frac{49}{29.4} = \frac{5}{3} = 1.67 \text{ kg}\) | (M1A1), (M1A1) | — |
| (ii) \(R = 24.5 \times 2 + 98 + \frac{5}{3} \times 9.8 = 163 \text{ N}\) | M1, A1, A1 | 3 marks |
| (d) This allows the centre of mass to be placed at the centre of the rod for the moment calculations. | B1 | 1 mark |
**(a)** [Correct force diagram with labels and arrows] | B1 | 1 mark | Correct force diagram, with labels and arrows
**(b)** $2T = 0.5 \times 98$ and $T = 24.5 \text{ N}$ | M1, A1, A1 | 3 marks | Moment equation; correct equation; correct positive value for the tension from correct working. (AG)
**(c)(i)** $2 \times 2 \times 24.5 = 3 \times 9.8 \times m + 0.5 \times 98$ and $m = \frac{98 - 49}{3 \times 9.8} = \frac{5}{3} = 1.67 \text{ kg (to 3 sf)}$ | B1, M1, A1, A1 | 4 marks | Tension doubled; moment equation; correct equation; correct mass
Or: $2 \times 2.45 = 3 \times 9.8m$ and $m = \frac{49}{29.4} = \frac{5}{3} = 1.67 \text{ kg}$ | (M1A1), (M1A1) | — | For equation; for finding $m$
**(ii)** $R = 24.5 \times 2 + 98 + \frac{5}{3} \times 9.8 = 163 \text{ N}$ | M1, A1, A1 | 3 marks | Considering vertical equilibrium with 3 terms; correct equation; correct reaction. Must be consistent with 3(c)(i) if awarding accuracy marks
**(d)** This allows the centre of mass to be placed at the centre of the rod for the moment calculations. | B1 | 1 mark | Correct explanation
**Total for Question 3: 12 marks**
---3 The diagram shows a uniform rod, $A B$, of mass 10 kg and length 5 metres. The rod is held in equilibrium in a horizontal position, by a support at $C$ and a light vertical rope attached to $A$, where $A C$ is 2 metres.\\
\includegraphics[max width=\textwidth, alt={}, center]{c02cf013-365b-44e2-8c16-aa8209cbe250-3_237_680_479_648}
\begin{enumerate}[label=(\alph*)]
\item Draw and label a diagram to show the forces acting on the rod.
\item Show that the tension in the rope is 24.5 N .
\item A package of mass $m \mathrm {~kg}$ is suspended from $B$. The tension in the rope has to be doubled to maintain equilibrium.
\begin{enumerate}[label=(\roman*)]
\item Find $m$.
\item Find the magnitude of the force exerted on the rod by the support.
\end{enumerate}\item Explain how you have used the fact that the rod is uniform in your solution.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2006 Q3 [12]}}