| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 13 |
| 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 composite body calculations and toppling analysis. Part (a) involves routine application of centre of mass formulas for standard shapes (cone and cylinder) with algebraic manipulation to reach the given result. Part (b) requires understanding that toppling occurs when the vertical line through the centre of mass passes outside the base, leading to a straightforward tan α calculation. While it requires multiple steps and careful geometry, the techniques are well-practiced M3 material with no novel insight needed. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| portion | mass | \(y\) |
| cone | \(\rho \times \frac{\pi r^2(2r)}{3} = \frac{2}{3}\rho\pi r^3\) | \(h + \frac{1}{4}(2r) = h + \frac{1}{2}r\) |
| cylinder | \(\rho\pi r^2 h\) | \(\frac{1}{2}h\) |
| firework | \(\rho\pi r^2(h + \frac{2}{3}r)\) | \(\overline{y}\) |
| M2 A4 | ||
| \(\rho\pi r^2(h + \frac{2}{3}r) \times y = \rho\pi r^2(\frac{1}{2}h^2 + \frac{2}{3}rh + \frac{1}{4}r^2)\) | M1 | |
| \(\therefore 2(3h + 2r) \times y = 3h^2 + 4rh + 2r^2\) | M1 | |
| giving \(\overline{y} = \frac{3h^2 + 4rh + 2r^2}{2(3h + 2r)}\) | A1 | |
| (b) \(h = 4r \therefore \overline{y} = \frac{13}{14}r\) | M1 | |
| \(\tan \alpha = r + (\frac{13}{14}r) = \frac{14}{33}\) | M1 A1 | |
| \(\therefore \alpha = 23°\) (nearest degree) | A1 | (13) |
(a)
| portion | mass | $y$ | $my$ |
|---------|------|-----|------|
| cone | $\rho \times \frac{\pi r^2(2r)}{3} = \frac{2}{3}\rho\pi r^3$ | $h + \frac{1}{4}(2r) = h + \frac{1}{2}r$ | $\frac{2}{3}\rho\pi r^3(h + \frac{1}{2}r)$ |
| cylinder | $\rho\pi r^2 h$ | $\frac{1}{2}h$ | $\frac{1}{2}\rho\pi r^2 h^2$ |
| firework | $\rho\pi r^2(h + \frac{2}{3}r)$ | $\overline{y}$ | $\rho\pi r^2(\frac{1}{2}h^2 + \frac{2}{3}rh + \frac{1}{4}r^2)$ |
| M2 A4
$\rho\pi r^2(h + \frac{2}{3}r) \times y = \rho\pi r^2(\frac{1}{2}h^2 + \frac{2}{3}rh + \frac{1}{4}r^2)$ | M1
$\therefore 2(3h + 2r) \times y = 3h^2 + 4rh + 2r^2$ | M1
giving $\overline{y} = \frac{3h^2 + 4rh + 2r^2}{2(3h + 2r)}$ | A1
(b) $h = 4r \therefore \overline{y} = \frac{13}{14}r$ | M1
$\tan \alpha = r + (\frac{13}{14}r) = \frac{14}{33}$ | M1 A1
$\therefore \alpha = 23°$ (nearest degree) | A1 | (13)
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{8b85b908-bb74-4532-a1b4-3826946bd43b-3_588_291_1126_662}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
A firework is modelled as a uniform solid formed by joining the plane surface of a right circular cone of height $2 r$ and base radius $r$, to one of the plane surfaces of a cylinder of height $h$ and base radius $r$ as shown in Figure 2.
Using this model,
\begin{enumerate}[label=(\alph*)]
\item show that the distance of the centre of mass of the firework from its plane base is
$$\frac { 3 h ^ { 2 } + 4 h r + 2 r ^ { 2 } } { 2 ( 3 h + 2 r ) }$$
The firework is to be launched from rough ground inclined at an angle $\alpha$ to the horizontal. Given that the firework does not slip or topple and that $h = 4 r$,
\item Find, correct to the nearest degree, the maximum value of $\alpha$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 Q5 [13]}}