| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2023 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Centre of mass of composite shapes |
| Difficulty | Challenging +1.2 This is a multi-step M3 mechanics problem requiring center of mass calculation for composite bodies and equilibrium with moments. Part (a) involves standard cone center of mass formulas and algebraic manipulation (shown result guides the work). Part (b) requires setting up moment and force equations for a suspended body. While requiring careful bookkeeping and multiple techniques, the methods are standard M3 fare with no novel geometric insight needed. Slightly above average due to the algebraic complexity and multi-stage reasoning, but well within expected M3 difficulty. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Mass: \(\frac{1}{3}\pi r^2 H\), \(\frac{1}{3}\pi r^2 h\), \(\frac{1}{3}\pi r^2 H - \frac{1}{3}\pi r^2 h\) | B1 | Three correct mass ratios: \(H\), \(h\), \(H-h\) |
| Distance from \(V\): \(\frac{3H}{4}\), \(H - \frac{1}{4}h\), \(\bar{x}\) | B1 | Three correct distances (allow if measured from some other axis) |
| \(\frac{1}{3}\pi r^2 H \times \frac{3H}{4} - \frac{1}{3}\pi r^2 h\!\left(H - \frac{1}{4}h\right) = \left(\frac{1}{3}\pi r^2 H - \frac{1}{3}\pi r^2 h\right)\bar{x}\) | M1A1 | Moments equation with correct no. of terms, dimensionally correct; condone addition treated as sign error; must work with solids not conical shell; correct unsimplified equation |
| \(\bar{x} = \dfrac{(3H-h)(H-h)}{4(H-h)} = \dfrac{1}{4}(3H-h)\) * | A1* (5) | Given answer correctly obtained including cancelling \((H-h)\); factorised expression must be seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\text{M}(G),\ T_1\bar{x} = T_2(H - \bar{x})\) | M1 | Complete method for equation in \(H\), \(h\), \(T_1\), \(T_2\) only; dimensionally correct; may take moments about \(G\) and substitute for \(\bar{x}\); or use two equations and eliminate weight |
| \(\dfrac{T_1}{T_2} = \dfrac{H - \frac{1}{4}(3H-h)}{\frac{1}{4}(3H-h)}\) | A1 | Correct equation in \(T_1\), \(T_2\), \(H\) and \(h\) only |
| \(\dfrac{T_1}{T_2} = \dfrac{H+h}{3H-h}\) | A1 (3) | Correct answer; simplest form required |
# Question 3(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Mass: $\frac{1}{3}\pi r^2 H$, $\frac{1}{3}\pi r^2 h$, $\frac{1}{3}\pi r^2 H - \frac{1}{3}\pi r^2 h$ | B1 | Three correct mass ratios: $H$, $h$, $H-h$ |
| Distance from $V$: $\frac{3H}{4}$, $H - \frac{1}{4}h$, $\bar{x}$ | B1 | Three correct distances (allow if measured from some other axis) |
| $\frac{1}{3}\pi r^2 H \times \frac{3H}{4} - \frac{1}{3}\pi r^2 h\!\left(H - \frac{1}{4}h\right) = \left(\frac{1}{3}\pi r^2 H - \frac{1}{3}\pi r^2 h\right)\bar{x}$ | M1A1 | Moments equation with correct no. of terms, dimensionally correct; condone addition treated as sign error; must work with solids not conical shell; correct unsimplified equation |
| $\bar{x} = \dfrac{(3H-h)(H-h)}{4(H-h)} = \dfrac{1}{4}(3H-h)$ * | A1* (5) | Given answer correctly obtained including cancelling $(H-h)$; factorised expression must be seen |
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# Question 3(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\text{M}(G),\ T_1\bar{x} = T_2(H - \bar{x})$ | M1 | Complete method for equation in $H$, $h$, $T_1$, $T_2$ only; dimensionally correct; may take moments about $G$ and substitute for $\bar{x}$; or use two equations and eliminate weight |
| $\dfrac{T_1}{T_2} = \dfrac{H - \frac{1}{4}(3H-h)}{\frac{1}{4}(3H-h)}$ | A1 | Correct equation in $T_1$, $T_2$, $H$ and $h$ only |
| $\dfrac{T_1}{T_2} = \dfrac{H+h}{3H-h}$ | A1 (3) | Correct answer; simplest form required |
---\begin{enumerate}
\item A uniform solid right circular cone $C$ has base radius $r$, height $H$ and vertex $V$. A uniform solid $S$, shown in Figure 3, is formed by removing from $C$ a uniform solid right circular cone of height $h ( h < H )$ that has the same base and axis of symmetry as $C$.
\end{enumerate}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{dceb2432-117c-40fe-bf3d-782beeb42e41-08_725_1152_422_461}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}
(a) Show that the distance of the centre of mass of $S$ from $V$ is
$$\frac { 1 } { 4 } ( 3 H - h )$$
The solid $S$ is suspended by two vertical light strings. The first string is attached to $S$ at $V$ and the second string is attached to $S$ at a point on the circumference of the circular base of $S$.\\
The solid $S$ hangs freely in equilibrium with its axis of symmetry horizontal.\\
The tension in the first string is $T _ { 1 }$ and the tension in the second string is $T _ { 2 }$\\
(b) Find $\frac { T _ { 1 } } { T _ { 2 } }$, giving your answer in terms of $H$ and $h$, in its simplest form.
\hfill \mbox{\textit{Edexcel M3 2023 Q3 [8]}}