| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2024 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Solid on inclined plane - toppling |
| Difficulty | Challenging +1.2 This is a standard M3 centre of mass question requiring routine application of formulas for solids of revolution (∫x·πy²dx / ∫πy²dx) followed by a straightforward toppling condition (tan α = horizontal distance to COM / vertical distance). The algebraic integration involves expanding (3-x)² and integrating polynomials, which is mechanical. While it's a multi-step problem worth several marks, it follows a well-practiced template with no novel insight required, making it moderately above average difficulty for A-level but typical for M3. |
| Spec | 4.08d Volumes of revolution: about x and y axes6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moment of \(S\) about the \(y\)-axis | M1 | Use of formula \((\pi)(\rho)\int xy^2\,\mathrm{d}x\). No need to see correct limits. The curve equation must be substituted correctly and an attempt to integrate seen (at least one term must have a power of \(x\) raised by 1). Note the correct expression for integrating is \(x\!\left(\tfrac{1}{4}x(3-x)\right)^2 = \tfrac{1}{16}x^3(3-x)^2 = \tfrac{1}{16}(9x^3 - 6x^4 + x^5)\) |
| \(= (\pi)(\rho)\dfrac{1}{16}\left[\dfrac{9}{4}x^4 - \dfrac{6}{5}x^5 + \dfrac{1}{6}x^6\right]\) | A1 | Correct integrated expression. |
| \(= \dfrac{31}{60}(\pi)(\rho)\) | A1 | Correct use of correct limits (0 and 2). Must see \(\dfrac{31}{60}\) or equivalent numerical evaluation of integral. |
| \(\bar{x} = \dfrac{\dfrac{31}{60}(\pi)(\rho)}{\dfrac{2}{5}(\pi)(\rho)}\) | M1 | Complete method to find the distance. Formula must be \(\bar{x} = \dfrac{(\pi)(\rho)\int xy^2\,\mathrm{d}x}{M}\). Must have consistent use of \(\pi\) and of \(\rho\). |
| \(= \dfrac{31}{24}\) * | A1* | Obtain given answer from correct working |
| [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Correct use of trig; \(\tan\alpha° = \dfrac{1}{2} \div \dfrac{17}{24}\left(= \dfrac{12}{17}\right)\) | M1, A1 | Correct trig ratio to find a relevant angle, \(\alpha°\) or \((90-\alpha)°\). Must use curve equation with \(x = 2\) and \(\left(2 - \dfrac{31}{24}\right)\). Or equivalent. Condone reciprocal. |
| \(\alpha = 35\) | A1 | 2 sf or better (35.2175…). A0 for use of radians. |
| [3] | ||
| (8) total |
# Question 3:
## Part 3a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moment of $S$ about the $y$-axis | M1 | Use of formula $(\pi)(\rho)\int xy^2\,\mathrm{d}x$. No need to see correct limits. The curve equation must be substituted correctly and an attempt to integrate seen (at least one term must have a power of $x$ raised by 1). Note the correct expression for integrating is $x\!\left(\tfrac{1}{4}x(3-x)\right)^2 = \tfrac{1}{16}x^3(3-x)^2 = \tfrac{1}{16}(9x^3 - 6x^4 + x^5)$ |
| $= (\pi)(\rho)\dfrac{1}{16}\left[\dfrac{9}{4}x^4 - \dfrac{6}{5}x^5 + \dfrac{1}{6}x^6\right]$ | A1 | Correct integrated expression. |
| $= \dfrac{31}{60}(\pi)(\rho)$ | A1 | Correct use of correct limits (0 and 2). Must see $\dfrac{31}{60}$ or equivalent numerical evaluation of integral. |
| $\bar{x} = \dfrac{\dfrac{31}{60}(\pi)(\rho)}{\dfrac{2}{5}(\pi)(\rho)}$ | M1 | Complete method to find the distance. Formula must be $\bar{x} = \dfrac{(\pi)(\rho)\int xy^2\,\mathrm{d}x}{M}$. Must have consistent use of $\pi$ and of $\rho$. |
| $= \dfrac{31}{24}$ * | A1* | Obtain given answer from correct working |
| | [5] | |
## Part 3b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Correct use of trig; $\tan\alpha° = \dfrac{1}{2} \div \dfrac{17}{24}\left(= \dfrac{12}{17}\right)$ | M1, A1 | Correct trig ratio to find a relevant angle, $\alpha°$ or $(90-\alpha)°$. Must use curve equation with $x = 2$ and $\left(2 - \dfrac{31}{24}\right)$. Or equivalent. Condone reciprocal. |
| $\alpha = 35$ | A1 | 2 sf or better (35.2175…). A0 for use of radians. |
| | [3] | |
| | (8) total | |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{92131234-bfc1-4e0e-87d4-db9335fbf343-08_246_734_296_667}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
The shaded region in Figure 2 is bounded by the $x$-axis, the line with equation $x = 2$ and the curve with equation $y = \frac { 1 } { 4 } x ( 3 - x )$.\\
This region is rotated through $2 \pi$ radians about the $x$-axis, to form a solid of revolution which is used to model a uniform solid $S$.
The volume of $S$ is $\frac { 2 } { 5 } \pi$
\begin{enumerate}[label=(\alph*)]
\item Use the model and algebraic integration to show that the $x$ coordinate of the centre of mass of $S$ is $\frac { 31 } { 24 }$
The solid $S$ is placed with its circular face on a rough plane which is inclined at $\alpha ^ { \circ }$ to the horizontal. The plane is sufficiently rough to prevent $S$ from sliding.
The solid $S$ is on the point of toppling.
\item Find the value of $\alpha$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2024 Q3 [8]}}