AQA Paper 2 2022 June — Question 14 4 marks

Exam BoardAQA
ModulePaper 2 (Paper 2)
Year2022
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicMoments
TypeUniform beam on two supports
DifficultyStandard +0.3 This is a straightforward moments problem requiring taking moments about the support point with the coin's center of mass 14mm from one end and the load at the other end. The setup is clearly defined, requiring only one equilibrium equation and basic arithmetic. Slightly easier than average due to its routine application of moments principles with no complicating factors.
Spec3.04b Equilibrium: zero resultant moment and force

14 A \(\pounds 2\) coin has a diameter of 28 mm and a mass of 12 grams. A uniform rod \(A B\) of length 160 mm and a fixed load of mass \(m\) grams are used to check that a \(\pounds 2\) coin has the correct mass. The rod rests with its midpoint on a support.
A \(\pounds 2\) coin is placed face down on the rod with part of its curved edge directly above \(A\). The fixed load is hung by a light inextensible string from a point directly below the other end of the rod at \(B\), as shown in the diagram. \includegraphics[max width=\textwidth, alt={}, center]{ad6590e8-6673-45ca-bef3-a14716978827-22_195_766_854_639} 14
  1. Given that the rod is horizontal and rests in equilibrium, find \(m\).
    14
  2. State an assumption you have made about the \(\pounds 2\) coin to answer part (a).

Question 14(a):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
\(66 \times 12g = 80mg\)B1 (3.3) Forms dimensionally correct non-zero moment e.g. \(0.012g \times 0.066\), \(12g \times 66\), \(160mg\)
\(m = 9.9\)M1 (1.1a) Obtains correct single equation where only unknown is the mass; PI by 9.9
\(m = 9.9\)A1 (1.1b) Condone \(m = 0.0099\text{ kg}\) OE; provided units included if \(m\) not in grams
Question 14(b):
AnswerMarks Guidance
Answer/WorkingMarks Guidance
The coin is uniformE1 (3.5a) States valid assumption about £2 coin with no contradictions; e.g. uniform, weight acts through centre, centre of mass at centre. Do not accept 'mass acts at' or 'centre of mass acts at the centre' or 'centre of weight is at' or 'weight is at'
## Question 14(a):

| Answer/Working | Marks | Guidance |
|---|---|---|
| $66 \times 12g = 80mg$ | B1 (3.3) | Forms dimensionally correct non-zero moment e.g. $0.012g \times 0.066$, $12g \times 66$, $160mg$ |
| $m = 9.9$ | M1 (1.1a) | Obtains correct single equation where only unknown is the mass; PI by 9.9 |
| $m = 9.9$ | A1 (1.1b) | Condone $m = 0.0099\text{ kg}$ OE; provided units included if $m$ not in grams |

## Question 14(b):

| Answer/Working | Marks | Guidance |
|---|---|---|
| The coin is uniform | E1 (3.5a) | States valid assumption about £2 coin with no contradictions; e.g. uniform, weight acts through centre, centre of mass at centre. Do not accept 'mass acts at' or 'centre of mass acts at the centre' or 'centre of weight is at' or 'weight is at' |
14 A $\pounds 2$ coin has a diameter of 28 mm and a mass of 12 grams.

A uniform rod $A B$ of length 160 mm and a fixed load of mass $m$ grams are used to check that a $\pounds 2$ coin has the correct mass.

The rod rests with its midpoint on a support.\\
A $\pounds 2$ coin is placed face down on the rod with part of its curved edge directly above $A$.

The fixed load is hung by a light inextensible string from a point directly below the other end of the rod at $B$, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{ad6590e8-6673-45ca-bef3-a14716978827-22_195_766_854_639}

14
\begin{enumerate}[label=(\alph*)]
\item Given that the rod is horizontal and rests in equilibrium, find $m$.\\

14
\item State an assumption you have made about the $\pounds 2$ coin to answer part (a).
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 2 2022 Q14 [4]}}