| Exam Board | AQA |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2023 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Uniform beam on two supports |
| Difficulty | Standard +0.3 This is a standard A-level mechanics question on moments and equilibrium requiring routine application of ΣF=0 and taking moments about a point. Part (a)(i) is a 'show that' with the answer given, (a)(ii) uses vertical equilibrium, and (b) tests conceptual understanding of symmetry. Slightly easier than average due to straightforward setup and clear methodology. |
| Spec | 3.04b Equilibrium: zero resultant moment and force3.04c Use moments: beams, ladders, static problems |
| Answer | Marks | Guidance |
|---|---|---|
| Taking moments about \(Y\): \(1.5mg = (3.6)(4g)\) | M1 | Forms moments equation about \(Y\) with one term correct; must use force \(\times\) distance for every term |
| \(m = \frac{(3.6)(4g)}{1.5g} \Rightarrow m = 9.6\) kilograms | R1 | Completes reasoned argument with at least one intermediate step; AG; condone 9.6 |
| Answer | Marks | Guidance |
|---|---|---|
| Resolving vertical forces: \(4g + R = 9.6g\) | M1 | Resolves vertically to find \(R\), or forms moments equation about any point other than \(Y\) |
| \(R = 5.6g\) N | A1 | Obtains \(5.6g\) N; condone missing units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Obtains \(4.8g\), or states the two supports are equidistant from the centre/ends of the plank, or refers to symmetry | B1 | AO 3.1b |
| States one of: the reaction force at \(Y\) changes; the reaction force at \(Y\) decreases; \(4.8g \neq 5.6g\); \(4.8g < 5.6g\) and concludes the claim is incorrect | E1 | AO 2.4 |
## Question 17(a)(i):
Taking moments about $Y$: $1.5mg = (3.6)(4g)$ | M1 | Forms moments equation about $Y$ with one term correct; must use force $\times$ distance for every term
$m = \frac{(3.6)(4g)}{1.5g} \Rightarrow m = 9.6$ kilograms | R1 | Completes reasoned argument with at least one intermediate step; AG; condone 9.6
---
## Question 17(a)(ii):
Resolving vertical forces: $4g + R = 9.6g$ | M1 | Resolves vertically to find $R$, or forms moments equation about any point other than $Y$
$R = 5.6g$ N | A1 | Obtains $5.6g$ N; condone missing units
## Question 17(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Obtains $4.8g$, or states the two supports are equidistant from the centre/ends of the plank, or refers to symmetry | B1 | AO 3.1b |
| States one of: the reaction force at $Y$ changes; the reaction force at $Y$ decreases; $4.8g \neq 5.6g$; $4.8g < 5.6g$ **and** concludes the claim is incorrect | E1 | AO 2.4 |
**Subtotal: 2 marks**
---17 A uniform plank $P Q$, of length 7 metres, lies horizontally at rest, in equilibrium, on two fixed supports at points $X$ and $Y$
The distance $P X$ is 1.4 metres and the distance $Q Y$ is 2 metres as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{de8a7d38-a665-4feb-854e-ac83f413d133-26_56_689_534_762}\\
\includegraphics[max width=\textwidth, alt={}, center]{de8a7d38-a665-4feb-854e-ac83f413d133-26_225_830_607_694}
17
\begin{enumerate}[label=(\alph*)]
\item The reaction force on the plank at $X$ is $4 g$ newtons.\\
17 (a) (i) Show that the mass of the plank is 9.6 kilograms.\\
17 (a) (ii) Find the reaction force, in terms of $g$, on the plank at $Y$\\
17
\item The support at $Y$ is moved so that the distance $Q Y = 1.4$ metres.
The plank remains horizontally at rest in equilibrium.\\
It is claimed that the reaction force at $Y$ remains unchanged.\\
Explain, with a reason, whether this claim is correct.
\end{enumerate}
\hfill \mbox{\textit{AQA Paper 2 2023 Q17 [6]}}