| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Composite solid with cone and cylinder |
| Difficulty | Standard +0.8 This is a multi-part centre of mass problem requiring: (a) calculation using composite volumes and standard cone COM formula, (b) geometric reasoning about limiting equilibrium with the vertical through contact points, and (c) applying equilibrium conditions for a suspended body. While M3 level, it demands careful geometric visualization, multiple COM calculations, and understanding of stability conditions—significantly above average difficulty but uses standard techniques throughout. |
| Spec | 6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) Mass ratios: Small cone \(= 4\), Large cone \(= k\), \(S = 4+k\) | Volumes or ratio of volumes used | |
| Disp from \(O\): Small cone \(= -r\), Large cone \(= \frac{kr}{4}\), \(S = \bar{x}\) | ||
| \(-4r + \frac{k^2r}{4} = (4+k)\bar{x}\) | M1A1A1 | M1 moments equation about any suitable point; A1 LHS correct; A1 RHS correct |
| \(\bar{x} = \frac{(k^2-16)r}{4(4+k)} = \frac{1}{4}(k-4)r\) | A1 (4) | Correct distance from \(O\), inc use of \(k>4\); single fraction only in expression |
| (b) \(k\) greatest when \(\frac{\bar{x}}{r} = \frac{r}{4r}\) | M1 | Using vertical through c of m passes through \(A\) to obtain connection between \(\bar{x}\) and \(r\) or numerical value for \(\bar{x}\) or any other complete valid method |
| \(\frac{1}{4}(k-4) = \frac{1}{4}\) | A1ft | Correct equation for \(k\) with their \(\bar{x}\) |
| Greatest \(k = 5\) | A1 (3) | cao \(k=5\) (inequality gets A0) |
| (c) \(\tan 12° = \frac{\bar{x}}{r} = \frac{1}{4}(k-4)\) | M1A1ft | M1 \(\tan 12° = \frac{\bar{x}}{r}\) either way up; A1ft substitute for \(\bar{x}\) correct way up |
| \(k = 4.85\) or \(4.9\) \((4.8502...)\) | A1 (3) [10] | Final answer 4.9, 4.85 or better |
# Question 5:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(a)** Mass ratios: Small cone $= 4$, Large cone $= k$, $S = 4+k$ | | Volumes or ratio of volumes used |
| Disp from $O$: Small cone $= -r$, Large cone $= \frac{kr}{4}$, $S = \bar{x}$ | | |
| $-4r + \frac{k^2r}{4} = (4+k)\bar{x}$ | M1A1A1 | M1 moments equation about any suitable point; A1 LHS correct; A1 RHS correct |
| $\bar{x} = \frac{(k^2-16)r}{4(4+k)} = \frac{1}{4}(k-4)r$ | A1 (4) | Correct distance from $O$, inc use of $k>4$; single fraction only in expression |
| **(b)** $k$ greatest when $\frac{\bar{x}}{r} = \frac{r}{4r}$ | M1 | Using vertical through c of m passes through $A$ to obtain connection between $\bar{x}$ and $r$ or numerical value for $\bar{x}$ or any other complete valid method |
| $\frac{1}{4}(k-4) = \frac{1}{4}$ | A1ft | Correct equation for $k$ with their $\bar{x}$ |
| Greatest $k = 5$ | A1 (3) | cao $k=5$ (inequality gets A0) |
| **(c)** $\tan 12° = \frac{\bar{x}}{r} = \frac{1}{4}(k-4)$ | M1A1ft | M1 $\tan 12° = \frac{\bar{x}}{r}$ either way up; A1ft substitute for $\bar{x}$ correct way up |
| $k = 4.85$ or $4.9$ $(4.8502...)$ | A1 (3) [10] | Final answer 4.9, 4.85 or better |5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3706a02d-95c6-4e7a-bf38-88b338d77892-09_270_919_267_557}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
Figure 3 shows a uniform solid $S$ formed by joining the plane faces of two solid right circular cones, of base radius $r$, so that the centres of their bases coincide at $O$. One cone, with vertex $V$, has height $4 r$ and the other cone has height $k r$, where $k > 4$
\begin{enumerate}[label=(\alph*)]
\item Find the distance of the centre of mass of $S$ from $O$.\\
(4)
The point $A$ lies on the circumference of the common base of the cones. The solid is placed on a horizontal surface with VA in contact with the surface. Given that $S$ rests in equilibrium,
\item find the greatest possible value of $k$.
When $S$ is suspended from $A$ and hangs freely in equilibrium, $O A$ makes an angle of $12 ^ { \circ }$ with the downward vertical.
\item Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2015 Q5 [10]}}