| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Centre of mass of composite shapes |
| Difficulty | Challenging +1.2 This is a standard M3 non-uniform body equilibrium problem requiring knowledge that the center of mass of a cone is at h/4 from the base, taking moments about the suspension point, and solving a trigonometric equation. While it involves 3D geometry and requires careful angle work, it follows a well-established method taught in M3 with no novel insight needed—moderately above average difficulty due to the geometric setup and calculation complexity. |
| Spec | 6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\text{Vol} = \pi \int_0^1 4e^{2x}\,dx\) | M1 | Using \(\pi \int y^2\,dx\); no limits needed |
| \(= \pi \left[2e^{2x}\right]_0^1\) | DM1A1 | Integrating their expression for volume; correct integration inc limits |
| \(= 2\pi(e^2 - 1)\) ✱ | A1cso (4) | Substituting limits to obtain GIVEN answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\pi \int_0^1 4xe^{2x}\,dx\) | M1 | Using \((\pi)\int xy^2\,dx\); no limits needed; \(\pi\) can be omitted |
| \(= 4\pi \left\{ \left[x \times \frac{1}{2}e^{2x}\right]_0^1 - \int_0^1 \frac{1}{2}e^{2x}\,dx \right\}\) | DM1 | Attempting integration by parts; allow \(\pm\) between two parts; no limits needed |
| \(= 4\pi \left[\frac{1}{2}e^2 - 0\right] - 4\pi \left[\frac{1}{4}e^{2x}\right]_0^1\) | A1 | Correct integration including limits |
| \(= \pi(e^2 + 1)\) | A1 | Correct after limits substituted |
| \(x\text{ coord} = \frac{\pi(e^2+1)}{2\pi(e^2-1)} = \frac{e^2+1}{2(e^2-1)}\) | M1A1 (6) [10] | Use of \(\frac{\pi\int xy^2\,dx}{\pi\int y^2\,dx}\); \(\pi\) must appear in both or neither; A1cao correct answer |
# Question 2:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\text{Vol} = \pi \int_0^1 4e^{2x}\,dx$ | M1 | Using $\pi \int y^2\,dx$; no limits needed |
| $= \pi \left[2e^{2x}\right]_0^1$ | DM1A1 | Integrating their expression for volume; correct integration inc limits |
| $= 2\pi(e^2 - 1)$ ✱ | A1cso (4) | Substituting limits to obtain GIVEN answer |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\pi \int_0^1 4xe^{2x}\,dx$ | M1 | Using $(\pi)\int xy^2\,dx$; no limits needed; $\pi$ can be omitted |
| $= 4\pi \left\{ \left[x \times \frac{1}{2}e^{2x}\right]_0^1 - \int_0^1 \frac{1}{2}e^{2x}\,dx \right\}$ | DM1 | Attempting integration by parts; allow $\pm$ between two parts; no limits needed |
| $= 4\pi \left[\frac{1}{2}e^2 - 0\right] - 4\pi \left[\frac{1}{4}e^{2x}\right]_0^1$ | A1 | Correct integration including limits |
| $= \pi(e^2 + 1)$ | A1 | Correct after limits substituted |
| $x\text{ coord} = \frac{\pi(e^2+1)}{2\pi(e^2-1)} = \frac{e^2+1}{2(e^2-1)}$ | M1A1 (6) [10] | Use of $\frac{\pi\int xy^2\,dx}{\pi\int y^2\,dx}$; $\pi$ must appear in both or neither; A1cao correct answer |
---2.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\includegraphics[alt={},max width=\textwidth]{ab85ec29-b1fc-45a9-9343-09feb33ab6c5-004_513_399_303_785}
\end{center}
\end{figure}
A uniform solid right circular cone has base radius $a$ and semi-vertical angle $\alpha$, where $\tan \alpha = \frac { 1 } { 3 }$. The cone is freely suspended by a string attached at a point $A$ on the rim of its base, and hangs in equilibrium with its axis of symmetry making an angle of $\theta ^ { \circ }$ with the upward vertical, as shown in Figure 1.
Find, to one decimal place, the value of $\theta$.\\
\hfill \mbox{\textit{Edexcel M3 Q2}}