| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2020 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of lamina by integration |
| Difficulty | Standard +0.8 This is a Further Maths FM2 question requiring integration to find the y-coordinate of centre of mass using the standard formula ȳ = (1/A)∫½y² dx, followed by applying equilibrium conditions to a suspended lamina. While the integration itself is straightforward (exponential function), students must recall the correct centre of mass formula, execute the integration accurately, and then apply geometric reasoning for part (b). The multi-step nature and Further Maths context place it moderately above average difficulty. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2.4\bar{y} = \frac{1}{2}\int y^2\,dx = \frac{1}{2}\int\{64e^{-2x}\}\,dx\) | M1 | |
| \(= -16\left[e^{-2x}\right]_{\ln2}^{\ln5}\) | A1 | |
| Complete strategy to find \(\bar{y}\) | M1 | |
| \(2.4\bar{y} = -16e^{-\ln 25} + 16e^{-\ln 4} = \frac{16}{4} - \frac{16}{25} = \frac{84}{25}\), \(\bar{y} = \frac{84}{25}\times\frac{10}{24} \left(= \frac{7}{5}\right) = 1.4\) | A1* |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2.4\bar{x} = \int(8xe^{-x})\,dx\) | M1 | |
| \(= \left[-8xe^{-x} - 8e^{-x}\right]_{\ln2}^{\ln5}\) | ||
| \(= -\frac{8}{5}(\ln5+1) + \frac{8}{2}(\ln2+1)\ (= 2.5974\ldots)\) | M1 | |
| \(\bar{x} = 1.08\) | A1 | |
| Complete strategy to find \(\theta\) | M1 | |
| \(\tan\theta° = \frac{\ln5 - \bar{x}}{8e^{-\ln5} - 1.4}\ (= 2.63\ldots)\) | A1ft | |
| \(\theta = 69\) | A1 |
## Question 2(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2.4\bar{y} = \frac{1}{2}\int y^2\,dx = \frac{1}{2}\int\{64e^{-2x}\}\,dx$ | M1 | |
| $= -16\left[e^{-2x}\right]_{\ln2}^{\ln5}$ | A1 | |
| Complete strategy to find $\bar{y}$ | M1 | |
| $2.4\bar{y} = -16e^{-\ln 25} + 16e^{-\ln 4} = \frac{16}{4} - \frac{16}{25} = \frac{84}{25}$, $\bar{y} = \frac{84}{25}\times\frac{10}{24} \left(= \frac{7}{5}\right) = 1.4$ | A1* | |
**(4 marks)**
---
## Question 2(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2.4\bar{x} = \int(8xe^{-x})\,dx$ | M1 | |
| $= \left[-8xe^{-x} - 8e^{-x}\right]_{\ln2}^{\ln5}$ | | |
| $= -\frac{8}{5}(\ln5+1) + \frac{8}{2}(\ln2+1)\ (= 2.5974\ldots)$ | M1 | |
| $\bar{x} = 1.08$ | A1 | |
| Complete strategy to find $\theta$ | M1 | |
| $\tan\theta° = \frac{\ln5 - \bar{x}}{8e^{-\ln5} - 1.4}\ (= 2.63\ldots)$ | A1ft | |
| $\theta = 69$ | A1 | |
**(6 marks)**2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{962c2b40-3c45-4eed-a0af-a59068bda0e1-04_506_590_255_429}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{962c2b40-3c45-4eed-a0af-a59068bda0e1-04_296_327_456_1311}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A uniform plane figure $R$, shown shaded in Figure 1, is bounded by the $x$-axis, the line with equation $x = \ln 5$, the curve with equation $y = 8 \mathrm { e } ^ { - x }$ and the line with equation $x = \ln 2$. The unit of length on each axis is one metre.
The area of $R$ is $2.4 \mathrm {~m} ^ { 2 }$\\
The centre of mass of $R$ is at the point with coordinates $( \bar { x } , \bar { y } )$.
\begin{enumerate}[label=(\alph*)]
\item Use algebraic integration to show that $\bar { y } = 1.4$
Figure 2 shows a uniform lamina $A B C D$, which is the same size and shape as $R$. The lamina is freely suspended from $C$ and hangs in equilibrium with $C B$ at an angle $\theta ^ { \circ }$ to the downward vertical.
\item Find the value of $\theta$
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 2020 Q2 [10]}}