| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2018 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Lamina with particle attached |
| Difficulty | Challenging +1.2 This is a standard M3 centre of mass question with three parts: (a) proving a given result using integration (routine calculus with spherical symmetry), (b) applying the result to find combined centre of mass (straightforward formula application), and (c) using equilibrium condition (toppling criterion). While it requires multiple techniques and careful algebra, all steps follow standard M3 procedures without requiring novel insight. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \((\pi)\int_0^r y^2 x\,dx = (\pi)\int_0^r (r^2-x^2)x\,dx\) | M1A1 | Using \((\pi)\int_0^r y^2 x\,dx\) with or without \(\pi\); integrand must be of form \((a^2-x^2)x\); limits not needed |
| \(= (\pi)\left[\frac{1}{2}r^2x^2 - \frac{1}{4}x^4\right]_0^r = \frac{1}{2}\pi r^4 - \frac{1}{4}(\pi)r^4 \left(=\frac{1}{4}(\pi)r^4\right)\) | dM1, A1 | Attempt integration with correct limits \(0,r\) or \(0,a\); award for \(\frac{1}{2}\pi r^4 - \frac{1}{4}(\pi)r^4\) or \(\frac{1}{4}(\pi)r^4\) |
| \(\bar{x} = \dfrac{\int_0^r \pi y^2 x\,dx}{\frac{2}{3}\pi r^3} = \dfrac{\frac{1}{4}\pi r^4}{\frac{2}{3}\pi r^3} = \frac{3}{8}r\) ✱ | M1A1cso (6) | Using \(\bar{x} = \frac{\int_0^r \pi y^2 x\,dx}{\frac{2}{3}\pi r^3}\); \(\pi\) in numerator and denominator or in neither; correct given result with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Mass ratio: \(\frac{2}{3}\pi a^3\) : \(\frac{2}{3}\pi\left(\frac{1}{2}a\right)^3 k\) : \(\frac{2}{3}\pi a^3\left(1+\frac{1}{8}k\right)\), simplified to \(1\) : \(\frac{1}{8}k\) : \(\left(1+\frac{1}{8}k\right)\) | B1 | Correct mass ratio - any equivalent; NB no penalty if \(\frac{4}{3}\pi r^3\) used instead of \(\frac{2}{3}\pi r^3\) |
| Distance from \(O\): \(\frac{3}{8}a\) : \((-)\frac{3}{16}a\) : \(\bar{x}\) | B1 | Correct distances from \(O\) or any other point |
| \(\frac{3}{8}a - \frac{3}{8\times16}ak = \left(1+\frac{1}{8}k\right)\bar{x}\) | M1A1ft | Attempting dimensionally consistent moments equation; correct equation following through mass ratio and distances; signs correct |
| \(\bar{x} = \dfrac{ | (48-3k)a | }{16(8+k)}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| \(\bar{x}=0\), \(\therefore k=16\), valid only if \(k>0\) | M1, A1ft (2) [13] | Set \(\bar{x}=0\) and solve for \(k\); correct \(k\) following through their \(\bar{x}\); award even if no modulus signs in (b) |
# Question 5:
## Part (a):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $(\pi)\int_0^r y^2 x\,dx = (\pi)\int_0^r (r^2-x^2)x\,dx$ | M1A1 | Using $(\pi)\int_0^r y^2 x\,dx$ with or without $\pi$; integrand must be of form $(a^2-x^2)x$; limits not needed |
| $= (\pi)\left[\frac{1}{2}r^2x^2 - \frac{1}{4}x^4\right]_0^r = \frac{1}{2}\pi r^4 - \frac{1}{4}(\pi)r^4 \left(=\frac{1}{4}(\pi)r^4\right)$ | dM1, A1 | Attempt integration with correct limits $0,r$ or $0,a$; award for $\frac{1}{2}\pi r^4 - \frac{1}{4}(\pi)r^4$ or $\frac{1}{4}(\pi)r^4$ |
| $\bar{x} = \dfrac{\int_0^r \pi y^2 x\,dx}{\frac{2}{3}\pi r^3} = \dfrac{\frac{1}{4}\pi r^4}{\frac{2}{3}\pi r^3} = \frac{3}{8}r$ ✱ | M1A1cso (6) | Using $\bar{x} = \frac{\int_0^r \pi y^2 x\,dx}{\frac{2}{3}\pi r^3}$; $\pi$ in numerator and denominator or in neither; correct given result with no errors seen |
## Part (b):
| Working/Answer | Marks | Guidance |
|---|---|---|
| Mass ratio: $\frac{2}{3}\pi a^3$ : $\frac{2}{3}\pi\left(\frac{1}{2}a\right)^3 k$ : $\frac{2}{3}\pi a^3\left(1+\frac{1}{8}k\right)$, simplified to $1$ : $\frac{1}{8}k$ : $\left(1+\frac{1}{8}k\right)$ | B1 | Correct mass ratio - any equivalent; NB no penalty if $\frac{4}{3}\pi r^3$ used instead of $\frac{2}{3}\pi r^3$ |
| Distance from $O$: $\frac{3}{8}a$ : $(-)\frac{3}{16}a$ : $\bar{x}$ | B1 | Correct distances from $O$ or any other point |
| $\frac{3}{8}a - \frac{3}{8\times16}ak = \left(1+\frac{1}{8}k\right)\bar{x}$ | M1A1ft | Attempting dimensionally consistent moments equation; correct equation following through mass ratio and distances; signs correct |
| $\bar{x} = \dfrac{|(48-3k)a|}{16(8+k)}$ | A1 (5) | Must have modulus signs (sign of $48-3k$ unknown); numerator can be $|(3k-48)a|$; no fractions within fractions |
## Part (c):
| Working/Answer | Marks | Guidance |
|---|---|---|
| $\bar{x}=0$, $\therefore k=16$, valid only if $k>0$ | M1, A1ft (2) [13] | Set $\bar{x}=0$ and solve for $k$; correct $k$ following through their $\bar{x}$; award even if no modulus signs in (b) |
---5. A uniform solid hemisphere has radius $r$. The centre of the plane face of the hemisphere is $O$.
\begin{enumerate}[label=(\alph*)]
\item Use algebraic integration to show that the distance from $O$ to the centre of mass of the hemisphere is $\frac { 3 } { 8 } r$.\\[0pt]
[You may assume that the volume of a sphere of radius $r$ is $\frac { 4 } { 3 } \pi r ^ { 3 }$ ]
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2cf74ba3-857a-4ce9-ab5b-e6203b279161-14_378_602_740_671}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A solid $S$ is formed by joining a uniform solid hemisphere of radius $a$ to a uniform solid hemisphere of radius $\frac { 1 } { 2 } a$. The plane faces of the hemispheres are joined together so that their centres coincide at $O$, as shown in Figure 1. The mass per unit volume of the smaller hemisphere is $k$ times the mass per unit volume of the larger hemisphere.
\item Find the distance from $O$ to the centre of mass of $S$.
When $S$ is placed on a horizontal plane with any point on the curved surface of the larger hemisphere in contact with the plane, $S$ remains in equilibrium.
\item Find the value of $k$.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2018 Q5 [13]}}