| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2021 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Solid with removed cone from cone or cylinder |
| Difficulty | Standard +0.8 This is a multi-step M3 centre of mass problem requiring calculation of the centre of mass of a composite solid (cone with two cones removed), followed by an equilibrium analysis with suspension. It requires understanding of similar solids, volume/mass ratios, and geometric reasoning, but follows standard M3 techniques without requiring novel insight. The 5-mark part (a) and subsequent equilibrium part (b) make it moderately challenging but within typical M3 scope. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mass ratio: Top cone \((-1)\), inside cone \((-1)\), \(C\): \(8\), \(S\): \(6\) | B1 | Correct mass ratio for 3 cones and \(S\); allow consistent \((-)\) |
| \(y\) distance: Top cone \(5a\), inside cone \(3a\), \(C\): \(2a\), \(S\): \(\bar{y}\) | B1 | Correct distances for the 3 cones; allow distances from vertex \((3a, 5a, 6a)\) or small plane face \((-a, a, 2a)\); condone missing \((-)\) |
| \(8\times 2a - 1\times 5a - 1\times 3a = 6\bar{y}\) | M1A1ft | Dimensionally correct moments equation about any parallel axis, must include 4 terms; correct moments equation following through their distances |
| \(6\bar{y} = 8a \rightarrow \bar{y} = \frac{4}{3}a\) | A1 | \(\bar{y} = \frac{4}{3}a\) o.e. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan\alpha = \frac{3}{8}\) \((\alpha = 20.556...\ \text{or}\ 69.44...)\) | B1 | Correct expression for \(\tan\alpha\) or \(\alpha\) seen (either way round) |
| \(\tan\beta = \dfrac{\frac{3a}{2}}{4a - \frac{4a}{3}} = \frac{9}{16}\) \((\beta = 29.357...\ \text{or}\ 60.642...)\) | M1A1ft | Correct attempt to use their \(\bar{y}\) to find \(\tan\beta\); correct expression for \(\tan\beta\) or \(\beta\), ft their \(\bar{y}\) |
| \(\alpha + \beta = \theta = 50°\) (or better \(49.91379...\)) | A1 | \(50°\) or better (\(0.87\) rad or better \(0.87116...\)) |
| or \(180 - 69.44... - 60.64... = 50°\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos\theta = \dfrac{AB^2 + BG^2 - AG^2}{2\times AB\times BG}\) | B1 | Correct expression for \(\cos\theta\) in terms of \(AB\), \(AG\) and \(BG\) |
| \(BG^2 = (1.5a)^2 + (4a-\bar{y})^2\ \left(=\frac{337a^2}{36}\right)\) | M1 | Correct attempt to use their \(\bar{y}\) to find \(AG\) and \(BG\) |
| \(AG^2 = (3a)^2 + (\bar{y})^2\ \left(=\frac{97a^2}{9}\right)\) | ||
| \(\{AB^2 = (1.5a)^2 + (4a)^2\ \left(=\frac{73a^2}{4}\right)\}\) | ||
| \(\cos\theta = \ldots = 0.6439\ldots\) | A1ft | Correct expression for \(\cos\theta\), ft their \(\bar{y}\) |
| \(\theta = 50°\) (or better \(49.91379...\)) | A1 | \(50°\) or better |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass ratio: Top cone $(-1)$, inside cone $(-1)$, $C$: $8$, $S$: $6$ | B1 | Correct mass ratio for 3 cones and $S$; allow consistent $(-)$ |
| $y$ distance: Top cone $5a$, inside cone $3a$, $C$: $2a$, $S$: $\bar{y}$ | B1 | Correct distances for the 3 cones; allow distances from vertex $(3a, 5a, 6a)$ or small plane face $(-a, a, 2a)$; condone missing $(-)$ |
| $8\times 2a - 1\times 5a - 1\times 3a = 6\bar{y}$ | M1A1ft | Dimensionally correct moments equation about any parallel axis, must include 4 terms; correct moments equation following through their distances |
| $6\bar{y} = 8a \rightarrow \bar{y} = \frac{4}{3}a$ | A1 | $\bar{y} = \frac{4}{3}a$ o.e. |
*SC: if $a$'s are missing, B1B0M1A1ftA0 is maximum available*
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\alpha = \frac{3}{8}$ $(\alpha = 20.556...\ \text{or}\ 69.44...)$ | B1 | Correct expression for $\tan\alpha$ or $\alpha$ seen (either way round) |
| $\tan\beta = \dfrac{\frac{3a}{2}}{4a - \frac{4a}{3}} = \frac{9}{16}$ $(\beta = 29.357...\ \text{or}\ 60.642...)$ | M1A1ft | Correct attempt to use their $\bar{y}$ to find $\tan\beta$; correct expression for $\tan\beta$ or $\beta$, ft their $\bar{y}$ |
| $\alpha + \beta = \theta = 50°$ (or better $49.91379...$) | A1 | $50°$ or better ($0.87$ rad or better $0.87116...$) |
| or $180 - 69.44... - 60.64... = 50°$ | | |
### Part (b) ALT:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos\theta = \dfrac{AB^2 + BG^2 - AG^2}{2\times AB\times BG}$ | B1 | Correct expression for $\cos\theta$ in terms of $AB$, $AG$ and $BG$ |
| $BG^2 = (1.5a)^2 + (4a-\bar{y})^2\ \left(=\frac{337a^2}{36}\right)$ | M1 | Correct attempt to use their $\bar{y}$ to find $AG$ and $BG$ |
| $AG^2 = (3a)^2 + (\bar{y})^2\ \left(=\frac{97a^2}{9}\right)$ | | |
| $\{AB^2 = (1.5a)^2 + (4a)^2\ \left(=\frac{73a^2}{4}\right)\}$ | | |
| $\cos\theta = \ldots = 0.6439\ldots$ | A1ft | Correct expression for $\cos\theta$, ft their $\bar{y}$ |
| $\theta = 50°$ (or better $49.91379...$) | A1 | $50°$ or better |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8a687d17-ec7e-463f-84dd-605f5c230db1-12_442_506_251_721}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
A uniform right solid cone $C$ has diameter $6 a$ and height $8 a$, as shown in Figure 3.\\
The solid $S$ is formed by removing a cone of height $4 a$ from the top of $C$ and then removing an identical, inverted cone. The vertex of the removed cone is at the point $O$ in the centre of the base of $C$, as shown in Figure 4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8a687d17-ec7e-463f-84dd-605f5c230db1-12_236_502_1126_721}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of $S$ from $O$.\\
(5)
The point $A$ lies on the circumference of the base of $S$ and the point $B$ lies on the circumference of the top of $S$. The points $O$, $A$ and $B$ all lie in the same vertical plane, as shown in Figure 5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8a687d17-ec7e-463f-84dd-605f5c230db1-12_248_449_1845_749}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
The solid $S$ is freely suspended from the point $B$ and hangs in equilibrium.
\item Find the size of the angle that $A B$ makes with the downward vertical.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2021 Q4 [9]}}