| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 14 |
| 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 problem requiring systematic application of known formulas (hemispherical shell CM at r/2, hemisphere volume at 3r/8) and algebraic manipulation across multiple parts. While it involves several components and extended calculation, the approach is methodical with no novel insight required—typical of higher-tier mechanics questions but not exceptionally challenging. |
| Spec | 6.04c Composite bodies: centre of mass |
| Answer | Marks |
|---|---|
| (a) Cup: S.A. \(= 2\pi r^2\), mass \(M\) | B1 |
| Base: Area \(= \pi r^2\), so mass \(= \frac{1}{2}M\) | |
| (b) \(M(base): M\frac{3r}{2} + m\frac{r}{2} = \left(M + \frac{M}{2} + m\right)\frac{12r}{14}\) | M1 M1 A1 A1 |
| \(42M + 14m = 39M + 26m\) | |
| \(3M = 12m\), \(M = 4m\) | M1 A1 |
| (c) \(M(base): \frac{1M \cdot 13r}{14} + 2M\frac{13r}{14} = \left(2M + \frac{1M}{4}\right)y_1\) | M1 A1 M1 A1 A1 |
| \(y_1 = \frac{13r}{10}\) | |
| C. of M. rises by \(\frac{13r}{10} - \frac{13r}{14} = \frac{13r}{35}\) | A1 |
| (d) Assumed liquid is a solid hemisphere | B1 |
**(a)** Cup: S.A. $= 2\pi r^2$, mass $M$ | B1 |
Base: Area $= \pi r^2$, so mass $= \frac{1}{2}M$ | |
**(b)** $M(base): M\frac{3r}{2} + m\frac{r}{2} = \left(M + \frac{M}{2} + m\right)\frac{12r}{14}$ | M1 M1 A1 A1 |
$42M + 14m = 39M + 26m$ | |
$3M = 12m$, $M = 4m$ | M1 A1 |
**(c)** $M(base): \frac{1M \cdot 13r}{14} + 2M\frac{13r}{14} = \left(2M + \frac{1M}{4}\right)y_1$ | M1 A1 M1 A1 A1 |
$y_1 = \frac{13r}{10}$ | |
C. of M. rises by $\frac{13r}{10} - \frac{13r}{14} = \frac{13r}{35}$ | A1 |
**(d)** Assumed liquid is a solid hemisphere | B1 |
**Total: 14 marks**The figure show a wine glass consisting of a hemispherical cup of radius $r$, a cylindrical solid stem of height $r$ and a circular base of radius $r$. The cup has mass $M$ and the stem has mass $m$. Modelling the cup as a thin, uniform hemispherical shell, the base as a uniform lamina made of the same thin material as the cup, and the stem as a uniform solid cylinder,
\begin{enumerate}[label=(\alph*)]
\item show that the mass of the circular base is $\frac{1}{2}M$. [1 mark]
\end{enumerate}
Given that the centre of mass of the glass is at a distance $\frac{13r}{14}$ from the base along the vertical axis of symmetry,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item express $M$ in terms of $m$. [6 marks]
\end{enumerate}
If the cup is now filled with liquid whose mass is $2M$,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item show that the position of the centre of mass rises through a distance $\frac{13r}{35}$. [6 marks]
\item State an assumption that you have made about the liquid. [1 mark]
\end{enumerate}
\includegraphics{figure_6}
\hfill \mbox{\textit{Edexcel M3 Q6 [14]}}