| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2023 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Conical or hemispherical shell composite |
| Difficulty | Challenging +1.2 This is a standard M3 centre of mass question requiring calculation of composite body COM using standard formulas (square with circular hole, conical shell), then applying toppling condition. While multi-step with some algebraic manipulation, it follows a well-established template with given formulas and clear structure. The toppling part is routine once COM is found. Slightly above average due to length and algebraic complexity, but no novel insight required. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Slant height \(l=\sqrt{\left(\frac{7a}{4}\right)^2+(6a)^2}=\frac{25a}{4}\) | M1 | Use of Pythagoras (unsimplified). May be seen on diagram |
| Square mass: \(16a^2\) | B1 | Mass/area of square |
| Circle mass: \(\pi\left(\frac{7a}{4}\right)^2\) | B1 | Mass/area of circle |
| Conical shell mass: \(\pi\times\frac{7a}{4}\times\frac{25a}{4}\) | B1ft | Mass/area of conical shell and total. Common error to use \(6a\) as slant height, only ft on calculated slant height |
| Total: \(\left[16a^2-\pi\left(\frac{7a}{4}\right)^2+\pi\times\frac{7a}{4}\times\frac{25a}{4}\right]\) | B1ft | All distances correct |
| Distances: Square \(0\), Circle \(0\), Conical shell \(2a\), Total \(\bar{x}\) | B1 | All distances correct |
| \(\pi\times\frac{7a}{4}\times\frac{25a}{4}\times 2a=\left[16a^2-\pi\left(\frac{7a}{4}\right)^2+\pi\times\frac{7a}{4}\times\frac{25a}{4}\right]\bar{x}\) | M1 A1 | Dimensionally correct moments equation, correct number of terms including attempt to subtract circle. Condone slip with \(a\) in one term |
| \(\bar{x}=\frac{175\pi a}{(63\pi+128)}\) | A1* | Given answer correctly obtained. Condone missing brackets from denominator |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\tan\alpha=\frac{2a}{\left(\frac{175\pi a}{63\pi+128}\right)}\) | M1 | Allow reciprocal. Must use \(2a\) and given \(\bar{x}\) |
| \(\tan\alpha=\frac{126\pi+256}{175\pi}\) or \(\frac{2(63\pi+128)}{175\pi}\) | A1 | Cao. Exact fraction required |
# Question 3:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Slant height $l=\sqrt{\left(\frac{7a}{4}\right)^2+(6a)^2}=\frac{25a}{4}$ | M1 | Use of Pythagoras (unsimplified). May be seen on diagram |
| Square mass: $16a^2$ | B1 | Mass/area of square |
| Circle mass: $\pi\left(\frac{7a}{4}\right)^2$ | B1 | Mass/area of circle |
| Conical shell mass: $\pi\times\frac{7a}{4}\times\frac{25a}{4}$ | B1ft | Mass/area of conical shell and total. Common error to use $6a$ as slant height, only ft on calculated slant height |
| Total: $\left[16a^2-\pi\left(\frac{7a}{4}\right)^2+\pi\times\frac{7a}{4}\times\frac{25a}{4}\right]$ | B1ft | All distances correct |
| Distances: Square $0$, Circle $0$, Conical shell $2a$, Total $\bar{x}$ | B1 | All distances correct |
| $\pi\times\frac{7a}{4}\times\frac{25a}{4}\times 2a=\left[16a^2-\pi\left(\frac{7a}{4}\right)^2+\pi\times\frac{7a}{4}\times\frac{25a}{4}\right]\bar{x}$ | M1 A1 | Dimensionally correct moments equation, correct number of terms including attempt to subtract circle. Condone slip with $a$ in one term |
| $\bar{x}=\frac{175\pi a}{(63\pi+128)}$ | A1* | Given answer correctly obtained. Condone missing brackets from denominator |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\tan\alpha=\frac{2a}{\left(\frac{175\pi a}{63\pi+128}\right)}$ | M1 | Allow reciprocal. Must use $2a$ and given $\bar{x}$ |
| $\tan\alpha=\frac{126\pi+256}{175\pi}$ or $\frac{2(63\pi+128)}{175\pi}$ | A1 | Cao. Exact fraction required |
---3.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{631b78c4-2763-4a1e-9d30-2f301fe3af2e-06_908_1367_269_349}
\end{center}
A square ABCD of side 4a is made from thin uniform cardboard. The centre of the square is 0 . A circle with centre 0 and radius $\frac { 7 a } { 4 }$ is then removed from the square to form a template T, shown shaded in Figure 3.\\
A right conical shell, with no base, has radius $\frac { 7 a } { 4 }$ and perpendicular height $6 a$.\\
The shell is made of the same thin uniform cardboard as T.\\
The shell is attached to T so that the circumference of the end of the shell coincides with the circumference of the circle centre 0 , to form the hat H , shown in Figure 4.\\[0pt]
[The surface area of a right conical shell of radius r and slant height I is $\pi r l$.]
\begin{enumerate}[label=(\alph*)]
\item Show that the exact distance of the centre of mass of H from O is
$$\frac { 175 \pi a } { ( 63 \pi + 128 ) }$$
A fixed rough plane is inclined to the horizontal at an angle $\alpha$. The hat H is placed on the plane, with ABCD in contact with the plane, and AB parallel to a line of greatest slope of the plane. The plane is sufficiently rough to prevent the hat from sliding down the plane.
Given that the hat is on the point of toppling,
\item find the exact value of $\tan \alpha$, giving your answer in simplest form.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2023 Q3 [10]}}