| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2022 |
| Session | October |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particles at coordinate positions |
| Difficulty | Moderate -0.8 This is a straightforward application of the centre of mass formula for particles in 2D. Part (a) requires substituting given values into the standard formula (routine calculation), and part (b) involves solving a simple linear equation by substituting the centre of mass coordinates into the line equation. No problem-solving insight or novel techniques required—purely mechanical application of a standard M2 formula. |
| Spec | 6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(M(x \text{ axis})\) | M1 | Need all terms. Dimensionally consistent. Condone if \(m\) missing throughout. Accept as part of a vector equation |
| \(2m \times (-2) + 3m \times 2 + 4m \times 3k = 9m \times \bar{x}\) leading to \(\bar{x} = \dfrac{2+12k}{9}\) | A1* | Obtain given result |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(M(y \text{ axis})\) | M1 | Need all terms. Dimensionally consistent. Might be seen as part of a vector equation in (a). Does not score marks until referred to in part (b). Condone if \(m\) missing throughout |
| \(2m \times 5 + 3m \times (-3) + 4m \times k = 9m \times \bar{y}\) leading to \(\bar{y} = \dfrac{1+4k}{9}\) | A1 | Correct unsimplified equation. Allow if \(m\) missing throughout |
| Form and solve equation in \(k\): \((2+12k+2+8k=27)\) | DM1 | Use their \(\bar{y}\) and \(\bar{x} + 2\bar{y} = 3\). Dependent on the two preceding M marks |
| \(k = \dfrac{23}{20}\) \((1.15)\) | A1 | Correct answer only |
| Total: 4 | ||
| Question Total: (6) |
# Question 1:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M(x \text{ axis})$ | M1 | Need all terms. Dimensionally consistent. Condone if $m$ missing throughout. Accept as part of a vector equation |
| $2m \times (-2) + 3m \times 2 + 4m \times 3k = 9m \times \bar{x}$ leading to $\bar{x} = \dfrac{2+12k}{9}$ | A1* | Obtain given result |
| **Total: 2** | | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M(y \text{ axis})$ | M1 | Need all terms. Dimensionally consistent. Might be seen as part of a vector equation in (a). Does not score marks until referred to in part (b). Condone if $m$ missing throughout |
| $2m \times 5 + 3m \times (-3) + 4m \times k = 9m \times \bar{y}$ leading to $\bar{y} = \dfrac{1+4k}{9}$ | A1 | Correct unsimplified equation. Allow if $m$ missing throughout |
| Form and solve equation in $k$: $(2+12k+2+8k=27)$ | DM1 | Use their $\bar{y}$ and $\bar{x} + 2\bar{y} = 3$. Dependent on the two preceding M marks |
| $k = \dfrac{23}{20}$ $(1.15)$ | A1 | Correct answer only |
| **Total: 4** | | |
| **Question Total: (6)** | | |
---\begin{enumerate}
\item Three particles of masses $2 m , 3 m$ and $4 m$ are placed at the points with coordinates $( - 2,5 ) , ( 2 , - 3 )$ and $( 3 k , k )$ respectively, where $k$ is a constant. The centre of mass of the three particles is at the point $( \bar { x } , \bar { y } )$.\\
(a) Show that $\bar { x } = \frac { 2 + 12 k } { 9 }$
\end{enumerate}
The centre of mass of the three particles lies at a point on the straight line with equation $x + 2 y = 3$\\
(b) Find the value of $k$.
\hfill \mbox{\textit{Edexcel M2 2022 Q1 [6]}}