| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particles at coordinate positions |
| Difficulty | Moderate -0.3 This is a straightforward application of centre of mass formulas for particles in 2D. Part (a) is algebraic verification, parts (b-c) involve standard substitution and solving, and part (d) extends the method predictably. While multi-part, each step follows directly from the formula with no conceptual challenges beyond routine algebraic manipulation. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moments about \(y\) axis: \(km \times -1 + 4m \times -3 + 2m \times 6 = (k+6)m \times \bar{x}\) | M1 | Moments about \(y\) axis (or parallel axis). All terms required. Dimensionally consistent. Condone sign errors. |
| Correct unsimplified equation | A1 | Correct unsimplified equation |
| \(\Rightarrow \bar{x} = \dfrac{-k}{k+6}\) * | A1* | Obtain given answer from full and correct working. Accept with \(6+k\) |
| (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moments about \(x\) axis: \(km \times 5 + 4m \times -1 + 2m \times 1 = (k+6)m \times \bar{y}\) | M1 | Moments about \(x\) axis (or parallel axis). All terms required. Dimensionally consistent. Condone sign errors. |
| \(\Rightarrow \bar{y} = \dfrac{5k-2}{k+6}\) | A1 | Correct unsimplified expression for \(\bar{y}\). Any equivalent simplified form. The first 5 marks are available for a combined equation in vector form. |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\bar{y} = 2\bar{x}+3 \Rightarrow \dfrac{5k-2}{k+6} = \dfrac{-2k}{k+6}+3 \quad (5k-2 = -2k+(3k+18))\) | M1 | Correct use of their \(\bar{y}\) and given \(\bar{x}\) to find \(k\). |
| \(\Rightarrow k = 5\) | A1 | Correct only |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4 = 2\lambda + 3 \Rightarrow\) | M1 | |
| \(\lambda = \dfrac{1}{2}\) | A1 | |
| (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use the model to obtain \(\lambda = \ldots\) or \(x = \ldots\) | M1 | If working from the beginning and the new particle has mass \(M\) then \(\frac{23+4M}{11+M} = 2\left(\frac{-5+\lambda M}{11+M}\right)+3\) |
| Accept \(x = \frac{1}{2}\) | A1 |
## Question 1:
### Part 1a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $y$ axis: $km \times -1 + 4m \times -3 + 2m \times 6 = (k+6)m \times \bar{x}$ | M1 | Moments about $y$ axis (or parallel axis). All terms required. Dimensionally consistent. Condone sign errors. |
| Correct unsimplified equation | A1 | Correct unsimplified equation |
| $\Rightarrow \bar{x} = \dfrac{-k}{k+6}$ * | A1* | Obtain given answer from full and correct working. Accept with $6+k$ |
| **(3 marks)** | | |
### Part 1b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $x$ axis: $km \times 5 + 4m \times -1 + 2m \times 1 = (k+6)m \times \bar{y}$ | M1 | Moments about $x$ axis (or parallel axis). All terms required. Dimensionally consistent. Condone sign errors. |
| $\Rightarrow \bar{y} = \dfrac{5k-2}{k+6}$ | A1 | Correct unsimplified expression for $\bar{y}$. Any equivalent simplified form. The first 5 marks are available for a combined equation in vector form. |
| **(2 marks)** | | |
### Part 1c:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\bar{y} = 2\bar{x}+3 \Rightarrow \dfrac{5k-2}{k+6} = \dfrac{-2k}{k+6}+3 \quad (5k-2 = -2k+(3k+18))$ | M1 | Correct use of their $\bar{y}$ and given $\bar{x}$ to find $k$. |
| $\Rightarrow k = 5$ | A1 | Correct only |
| **(2 marks)** | | |
### Part 1d:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4 = 2\lambda + 3 \Rightarrow$ | M1 | |
| $\lambda = \dfrac{1}{2}$ | A1 | |
| **(2 marks)** | | |
**Total: 9 Marks**
## Question 1d:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use the model to obtain $\lambda = \ldots$ or $x = \ldots$ | M1 | If working from the beginning and the new particle has mass $M$ then $\frac{23+4M}{11+M} = 2\left(\frac{-5+\lambda M}{11+M}\right)+3$ |
| Accept $x = \frac{1}{2}$ | A1 | |
---\begin{enumerate}
\item Three particles of masses $4 m , 2 m$ and $k m$ are placed at the points with coordinates $( - 3 , - 1 ) , ( 6,1 )$ and $( - 1,5 )$ respectively.
\end{enumerate}
Given that the centre of mass of the three particles is at the point with coordinates $( \bar { x } , \bar { y } )$\\
(a) show that $\bar { x } = \frac { - k } { k + 6 }$\\
(b) find $\bar { y }$ in terms of $k$.
Given that the centre of mass of the three particles lies on the line with equation $y = 2 x + 3$\\
(c) find the value of $k$.
A fourth particle is placed at the point with coordinates $( \lambda , 4 )$.\\
Given that the centre of mass of the four particles also lies on the line with equation $y = 2 x + 3$\\
(d) find the value of $\lambda$.
\hfill \mbox{\textit{Edexcel FM2 AS 2023 Q1 [9]}}