| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2021 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Conical or hemispherical shell composite |
| Difficulty | Challenging +1.2 This is a standard Further Mechanics 2 centre of mass question requiring composite body techniques (subtracting hemispheres) and equilibrium on an inclined plane. Part (a) uses the standard formula for hemisphere COM with straightforward algebra. Part (b) requires geometric reasoning about the equilibrium position but follows a predictable pattern. While it's a multi-step problem requiring several techniques, these are well-practiced in FM2 and don't require novel insight—slightly above average difficulty due to the composite geometry and multi-part nature. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mass ratios: \(\frac{2}{3}\pi(2a)^3\), \(\frac{2}{3}\pi a^3\), \(\frac{2}{3}\pi(2a)^3 - \frac{2}{3}\pi a^3\) giving \((8, 1, 7)\) | B1 | Correct unsimplified \((8, 1, 7)\) |
| Distances: \(\frac{3}{8}(2a)\), \(\frac{3}{8}a\), \(\bar{x}\) | B1 | Correct unsimplified but distances could be measured from a parallel axis |
| \(\left(\frac{2}{3}\pi(2a)^3 - \frac{2}{3}\pi a^3\right)\bar{x} = \frac{2}{3}\pi(2a)^3 \times \frac{3}{8}(2a) - \frac{2}{3}\pi a^3 \times \frac{3}{8}a\) | M1 | All terms needed and must be dimensionally correct |
| \(\bar{x} = \frac{45a}{56}\) | A1* | Correct answer correctly derived |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Use of appropriate trig ratio e.g. \(\sin\alpha = \dfrac{\frac{45a}{56}}{2a}\) | M1 | Must be using an appropriate trig ratio |
| \(\alpha = 23.7°\) (3 sf) | A1 | cao to 3SF |
## Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratios: $\frac{2}{3}\pi(2a)^3$, $\frac{2}{3}\pi a^3$, $\frac{2}{3}\pi(2a)^3 - \frac{2}{3}\pi a^3$ giving $(8, 1, 7)$ | B1 | Correct unsimplified $(8, 1, 7)$ |
| Distances: $\frac{3}{8}(2a)$, $\frac{3}{8}a$, $\bar{x}$ | B1 | Correct unsimplified but distances could be measured from a parallel axis |
| $\left(\frac{2}{3}\pi(2a)^3 - \frac{2}{3}\pi a^3\right)\bar{x} = \frac{2}{3}\pi(2a)^3 \times \frac{3}{8}(2a) - \frac{2}{3}\pi a^3 \times \frac{3}{8}a$ | M1 | All terms needed and must be dimensionally correct |
| $\bar{x} = \frac{45a}{56}$ | A1* | Correct answer correctly derived |
## Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Use of appropriate trig ratio e.g. $\sin\alpha = \dfrac{\frac{45a}{56}}{2a}$ | M1 | Must be using an appropriate trig ratio |
| $\alpha = 23.7°$ (3 sf) | A1 | cao to 3SF |3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d4fc2ea6-3ffc-42f2-b462-9694adfe2ec1-10_552_807_246_630}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A uniform solid hemisphere $H$ has radius $2 a$. A solid hemisphere of radius $a$ is removed from the hemisphere $H$ to form a bowl. The plane faces of the hemispheres coincide and the centres of the two hemispheres coincide at the point $O$, as shown in Figure 2.
The centre of mass of the bowl is at the point $G$.
\begin{enumerate}[label=(\alph*)]
\item Show that $O G = \frac { 45 a } { 56 }$
Figure 3 below shows a cross-section of the bowl which is resting in equilibrium with a point $P$ on its curved surface in contact with a rough plane. The plane is inclined to the horizontal at an angle $\alpha$ and is sufficiently rough to prevent the bowl from slipping. The line $O G$ is horizontal and the points $O , G$ and $P$ lie in a vertical plane which passes through a line of greatest slope of the inclined plane.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d4fc2ea6-3ffc-42f2-b462-9694adfe2ec1-10_812_1086_1667_493}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
\item Find the size of $\alpha$, giving your answer in degrees to 3 significant figures.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 2021 Q3 [6]}}