| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2020 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Particles at coordinate positions |
| Difficulty | Standard +0.3 This is a straightforward optimization problem combining center of mass formula with minimizing distance from origin. It requires setting up the center of mass coordinates as functions of p, then minimizing the distance using calculus (differentiation). The algebra is routine and the conceptual steps are standard for FM2, making it slightly easier than average. |
| Spec | 6.04b Find centre of mass: using symmetry |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct method to find equation in \(\bar{x}\) | M1 | Take moments about axis parallel to \(x = 0\). Need all terms and dimensionally correct. |
| \(-3\times2 + 4\times3 + 2\times p = 9\bar{x}\) giving \((6 + 2p = 9\bar{x})\) | A1 | Correct unsimplified equation in \(\bar{x}\). Seen or implied. |
| Correct method to find equation in \(\bar{y}\) | M1 | Take moments about axis parallel to \(y = 0\). Need all terms and dimensionally correct. |
| \(3\times2 + 4\times1 + 2\times p = 9\bar{y}\) giving \(10 + 2p = 9\bar{y}\) | A1 | Correct unsimplified equation in \(\bar{y}\). Seen or implied. |
| \((9\bar{x})^2 + (9\bar{y})^2 = (6+2p)^2 + (10+2p)^2\) \(\left(= 136 + 64p + 8p^2\right)\) | M1 | Use of Pythagoras to find distance (or square of distance) from origin. |
| \(= 8\left[(p+4)^2 + 17 - 16\right]\) | M1 | Correct strategy to find value of \(p\) to minimise the distance e.g. use of calculus or complete the square. |
| \(\Rightarrow p = -4\) | A1 | Correct answer only. |
## Question 1:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct method to find equation in $\bar{x}$ | M1 | Take moments about axis parallel to $x = 0$. Need all terms and dimensionally correct. |
| $-3\times2 + 4\times3 + 2\times p = 9\bar{x}$ giving $(6 + 2p = 9\bar{x})$ | A1 | Correct unsimplified equation in $\bar{x}$. Seen or implied. |
| Correct method to find equation in $\bar{y}$ | M1 | Take moments about axis parallel to $y = 0$. Need all terms and dimensionally correct. |
| $3\times2 + 4\times1 + 2\times p = 9\bar{y}$ giving $10 + 2p = 9\bar{y}$ | A1 | Correct unsimplified equation in $\bar{y}$. Seen or implied. |
| $(9\bar{x})^2 + (9\bar{y})^2 = (6+2p)^2 + (10+2p)^2$ $\left(= 136 + 64p + 8p^2\right)$ | M1 | Use of Pythagoras to find distance (or square of distance) from origin. |
| $= 8\left[(p+4)^2 + 17 - 16\right]$ | M1 | Correct strategy to find value of $p$ to minimise the distance e.g. use of calculus or complete the square. |
| $\Rightarrow p = -4$ | A1 | Correct answer only. |
**(7 marks)**
---\begin{enumerate}
\item Three particles of masses $3 m$, $4 m$ and $2 m$ are placed at the points $( - 2,2 ) , ( 3,1 )$ and ( $p , p$ ) respectively.
\end{enumerate}
The value of $p$ is such that the distance of the centre of mass of the three particles from the point ( 0,0 ) is as small as possible.
Find the value of $p$.
\hfill \mbox{\textit{Edexcel FM2 2020 Q1 [7]}}