AQA M2 2010 January — Question 2 6 marks

Exam BoardAQA
ModuleM2 (Mechanics 2)
Year2010
SessionJanuary
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeParticles at coordinate positions
DifficultyModerate -0.8 This is a straightforward application of the centre of mass formula for a system of particles. Students need to use the formula $\bar{x} = \frac{\sum m_i x_i}{\sum m_i}$ (and similarly for $\bar{y}$), treating the lamina as a single particle at its given centre of mass. The calculation involves basic arithmetic with given coordinates and masses—no problem-solving insight or geometric reasoning required, making it easier than average.
Spec6.04b Find centre of mass: using symmetry

2 A piece of modern art is modelled as a uniform lamina and three particles. The diagram shows the lamina, the three particles \(A , B\) and \(C\), and the \(x\) - and \(y\)-axes. \includegraphics[max width=\textwidth, alt={}, center]{06b431ca-d3a8-46d6-b9f8-bac08d3fd51e-2_875_1004_1414_502} The lamina, which is fixed in the \(x - y\) plane, has mass 10 kg and its centre of mass is at the point (12, 9). The three particles are attached to the lamina.
Particle \(A\) has mass 3 kg and is at the point (15, 6).
Particle \(B\) has mass 1 kg and is at the point ( 7,14 ).
Particle \(C\) has mass 6 kg and is at the point ( 8,7 ).
Find the coordinates of the centre of mass of the piece of modern art.

Question 2:
AnswerMarks Guidance
WorkingMarks Guidance
\(\bar{X} = \frac{3\times15 + 1\times7 + 6\times8 + 10\times12}{3+1+6+10}\)M1A1 M1 for at least 3 multiplication & addition
\(= \frac{220}{20}\) or \(11\)A1
\(\bar{Y} = \frac{3\times6 + 1\times14 + 6\times7 + 10\times9}{20}\)M1A1
\(= \frac{164}{20}\) or \(8.2\)A1 SC 4 (10, 7.4) [omit lamina] ie: B2, B2
\(\therefore\) Centre of mass is at \((11, 8.2)\) Total: 6
## Question 2:

| Working | Marks | Guidance |
|---------|-------|----------|
| $\bar{X} = \frac{3\times15 + 1\times7 + 6\times8 + 10\times12}{3+1+6+10}$ | M1A1 | M1 for at least 3 multiplication & addition |
| $= \frac{220}{20}$ or $11$ | A1 | |
| $\bar{Y} = \frac{3\times6 + 1\times14 + 6\times7 + 10\times9}{20}$ | M1A1 | |
| $= \frac{164}{20}$ or $8.2$ | A1 | SC 4 (10, 7.4) [omit lamina] ie: B2, B2 |
| $\therefore$ Centre of mass is at $(11, 8.2)$ | | **Total: 6** |

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2 A piece of modern art is modelled as a uniform lamina and three particles. The diagram shows the lamina, the three particles $A , B$ and $C$, and the $x$ - and $y$-axes.\\
\includegraphics[max width=\textwidth, alt={}, center]{06b431ca-d3a8-46d6-b9f8-bac08d3fd51e-2_875_1004_1414_502}

The lamina, which is fixed in the $x - y$ plane, has mass 10 kg and its centre of mass is at the point (12, 9).

The three particles are attached to the lamina.\\
Particle $A$ has mass 3 kg and is at the point (15, 6).\\
Particle $B$ has mass 1 kg and is at the point ( 7,14 ).\\
Particle $C$ has mass 6 kg and is at the point ( 8,7 ).\\
Find the coordinates of the centre of mass of the piece of modern art.

\hfill \mbox{\textit{AQA M2 2010 Q2 [6]}}