| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2021 |
| Session | October |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Solid with removed cone from cone or cylinder |
| Difficulty | Challenging +1.2 This is a structured multi-part centre of mass question with guided steps. Part (a) is a standard textbook result, part (b) requires composite body calculation with similar cones (straightforward ratio work), and part (c) applies toppling equilibrium condition. While it requires careful algebra and understanding of equilibrium, the question provides clear guidance and uses standard M3 techniques without requiring novel insight. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{x} = \dfrac{\displaystyle\int_0^h x\!\left(\frac{rx}{h}\right)^2 dx}{\displaystyle\int_0^h \left(\frac{rx}{h}\right)^2 dx}\) | M1DM1 | M1 for use of \(\int xy^2\,dx\); DM1 for use of full \(\bar{x}\) formula (depends on M); allow cone formula quoted with \(\pi\) in numerator |
| \(= \dfrac{\left[\dfrac{x^4}{4}\right]_0^h}{\left[\dfrac{x^3}{3}\right]_0^h}\) | A1 | |
| \(= \frac{3h}{4}\) * | A1* (4) | Given answer correctly obtained; upper limits must be substituted |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Distances: \(F\): \(\bar{y} = \frac{1}{4}h\); \(C\): \(\frac{2h}{3}\); \(C'\): \(\left(\frac{1}{4}\times\frac{h}{3}\right) = \frac{3h}{4}\) | B1 | B1 for distances from larger plane face or any parallel axis |
| Mass ratios: \(F\): 26, \(C\): 27, \(C'\): 1 | B1 | B1 for mass (volume) ratios |
| \(26\bar{y} = \frac{1}{4}h\times 27 - \left[\frac{2h}{3}+\left(\frac{1}{4}\times\frac{h}{3}\right)\right]\times 1\) | M1A1ft | M1 for moments about larger plane face; A1ft for correct equation following through distances and masses |
| \(\bar{y} = \frac{3}{13}h\) * | A1* (5) | Given answer correctly obtained; at least one working step from equation must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| For equilibrium, \(\bar{y} = \frac{3}{13}h \leq CB\) | M1 | M1 for overall method using suitable inequality (comparing lengths or angles); if limiting case used, reason for direction of inequality must be seen (e.g. \(\bar{y}\), \(CB\)) |
| \(\frac{AB}{AN} = \frac{AN}{AV} \Rightarrow AB = \frac{1}{3}r\times\frac{r}{h} = \frac{r^2}{3h}\) | M1A1 | M1 for finding appropriate length; A1 for correct relevant distance in terms of \(r\) and \(h\) |
| So for equilibrium: \(\frac{3}{13}h \leq \frac{2}{3}h - \frac{r^2}{3h}\) | M1 | M1 for producing an inequality in \(r\) and \(h\); must be right way round |
| \(13r^2 \leq 17h^2\) * | A1* (5) | A1* for correctly showing given inequality; at least one working step from previous inequality must be seen |
## Question 7:
### Part 7(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x} = \dfrac{\displaystyle\int_0^h x\!\left(\frac{rx}{h}\right)^2 dx}{\displaystyle\int_0^h \left(\frac{rx}{h}\right)^2 dx}$ | M1DM1 | M1 for use of $\int xy^2\,dx$; DM1 for use of full $\bar{x}$ formula (depends on M); allow cone formula quoted with $\pi$ in numerator |
| $= \dfrac{\left[\dfrac{x^4}{4}\right]_0^h}{\left[\dfrac{x^3}{3}\right]_0^h}$ | A1 | |
| $= \frac{3h}{4}$ * | A1* (4) | Given answer correctly obtained; upper limits must be substituted |
### Part 7(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Distances: $F$: $\bar{y} = \frac{1}{4}h$; $C$: $\frac{2h}{3}$; $C'$: $\left(\frac{1}{4}\times\frac{h}{3}\right) = \frac{3h}{4}$ | B1 | B1 for distances from larger plane face or any parallel axis |
| Mass ratios: $F$: 26, $C$: 27, $C'$: 1 | B1 | B1 for mass (volume) ratios |
| $26\bar{y} = \frac{1}{4}h\times 27 - \left[\frac{2h}{3}+\left(\frac{1}{4}\times\frac{h}{3}\right)\right]\times 1$ | M1A1ft | M1 for moments about larger plane face; A1ft for correct equation following through distances and masses |
| $\bar{y} = \frac{3}{13}h$ * | A1* (5) | Given answer correctly obtained; at least one working step from equation must be seen |
### Part 7(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| For equilibrium, $\bar{y} = \frac{3}{13}h \leq CB$ | M1 | M1 for overall method using suitable inequality (comparing lengths or angles); if limiting case used, reason for direction of inequality must be seen (e.g. $\bar{y}$, $CB$) |
| $\frac{AB}{AN} = \frac{AN}{AV} \Rightarrow AB = \frac{1}{3}r\times\frac{r}{h} = \frac{r^2}{3h}$ | M1A1 | M1 for finding appropriate length; A1 for correct relevant distance in terms of $r$ and $h$ |
| So for equilibrium: $\frac{3}{13}h \leq \frac{2}{3}h - \frac{r^2}{3h}$ | M1 | M1 for producing an inequality in $r$ and $h$; must be right way round |
| $13r^2 \leq 17h^2$ * | A1* (5) | A1* for correctly showing given inequality; at least one working step from previous inequality must be seen |\begin{enumerate}
\item \hspace{0pt} [You may assume that the volume of a cone of height $h$ and base radius $r$ is $\frac { 1 } { 3 } \pi r ^ { 2 } h$.] A uniform solid right circular cone $C$, with vertex $V$, has base radius $r$ and height $h$.\\
(a) Show that the centre of mass of $C$ is $\frac { 3 } { 4 } h$ from $V$
\end{enumerate}
A solid $F$, shown below in Figure 4, is formed by removing the solid right circular cone $C ^ { \prime }$ from $C$, where cone $C ^ { \prime }$ has height $\frac { 1 } { 3 } h$ and vertex $V$
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9777abb8-a564-40d5-8d96-d5649913737b-24_666_670_854_639}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
(b) Show that the distance of the centre of mass of $F$ from its larger plane face is $\frac { 3 } { 13 } h$
The solid $F$ rests in equilibrium with its curved surface in contact with a horizontal plane.\\
(c) Show that $13 r ^ { 2 } \leqslant 17 h ^ { 2 }$\\
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{9777abb8-a564-40d5-8d96-d5649913737b-28_2642_1844_116_114}
\end{center}
\hfill \mbox{\textit{Edexcel M3 2021 Q7 [14]}}