| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2019 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Energy considerations in circular motion |
| Difficulty | Standard +0.8 This is a multi-part Further Maths mechanics problem requiring volume of revolution (shown), center of mass calculation for a solid of revolution, composite body center of mass with different densities, and toppling equilibrium analysis. While the techniques are standard for FM2, the combination of several concepts and the need to carefully track the geometry and density ratios makes it moderately challenging but still within typical Further Maths scope. |
| Spec | 6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \pi x^2 y\, dy = \pi\int \frac{y^5}{4}dy = \pi\left[\frac{y^6}{24}\right]_0^2\) | M1 | Use model to find moment about base; usual rules for integration; correct limits |
| \(= \frac{64\pi}{24}\left(=\frac{8\pi}{3}\right)\) | A1 | Any equivalent form |
| \(\Rightarrow \bar{y} = \left(\text{their } \frac{8\pi}{3}\right)\times\frac{5}{8\pi}\left(=\frac{5}{3}\right)\) | M1 | Use model and volume to find \(\bar{y}\); their moments and \(\frac{8\pi}{5}\) used correctly |
| Distance from plane face \(= 2 - \frac{5}{3} = \frac{1}{3}\) * | A1* | Obtain given answer from correct working |
| (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(S\): mass ratio \(3\times\frac{8\pi}{5}\), c of m from base \(\frac{13}{3}\); Cylinder: \(16\pi\), \(2\); \(M\): \((16+4.8)\pi\), \(d\) | B1 | Correct mass ratios seen or implied |
| Moments about diameter of base | M1 | Dimensionally correct; allow for their mass ratios |
| \(\frac{24}{5}\times\frac{13}{3} + 32 = \frac{104}{5}d\left(=\frac{264}{5}\right)\) | A1 | Correct unsimplified equation |
| \(d = \frac{264}{104}\left(=\frac{33}{13}\right)\) cm | A1 | Any equivalent form (2.54 cm) |
| Complete strategy for \(\theta\) | M1 | Complete strategy to find position of centre of mass of composite body and use trig to find \(\theta\) |
| About to topple: \(\tan\theta = \frac{2}{d}\left(=\frac{26}{33}\right)\) | A1ft | Trig ratio for a relevant angle; follow their \(d\) |
| \(\theta = 38.2°\) (or 38) | A1 | Correct angle (38 or better) (38.2338...) |
| (7) |
## Question 5(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \pi x^2 y\, dy = \pi\int \frac{y^5}{4}dy = \pi\left[\frac{y^6}{24}\right]_0^2$ | M1 | Use model to find moment about base; usual rules for integration; correct limits |
| $= \frac{64\pi}{24}\left(=\frac{8\pi}{3}\right)$ | A1 | Any equivalent form |
| $\Rightarrow \bar{y} = \left(\text{their } \frac{8\pi}{3}\right)\times\frac{5}{8\pi}\left(=\frac{5}{3}\right)$ | M1 | Use model and volume to find $\bar{y}$; their moments and $\frac{8\pi}{5}$ used correctly |
| Distance from plane face $= 2 - \frac{5}{3} = \frac{1}{3}$ * | A1* | Obtain given answer from correct working |
| **(4)** | | |
## Question 5(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $S$: mass ratio $3\times\frac{8\pi}{5}$, c of m from base $\frac{13}{3}$; Cylinder: $16\pi$, $2$; $M$: $(16+4.8)\pi$, $d$ | B1 | Correct mass ratios seen or implied |
| Moments about diameter of base | M1 | Dimensionally correct; allow for their mass ratios |
| $\frac{24}{5}\times\frac{13}{3} + 32 = \frac{104}{5}d\left(=\frac{264}{5}\right)$ | A1 | Correct unsimplified equation |
| $d = \frac{264}{104}\left(=\frac{33}{13}\right)$ cm | A1 | Any equivalent form (2.54 cm) |
| Complete strategy for $\theta$ | M1 | Complete strategy to find position of centre of mass of composite body and use trig to find $\theta$ |
| About to topple: $\tan\theta = \frac{2}{d}\left(=\frac{26}{33}\right)$ | A1ft | Trig ratio for a relevant angle; follow their $d$ |
| $\theta = 38.2°$ (or 38) | A1 | Correct angle (38 or better) (38.2338...) |
| **(7)** | | |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9b995178-a4be-4d5a-95f8-6c2978ff01b3-16_560_560_283_749}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
The region $R$, shown shaded in Figure 4, is bounded by part of the curve with equation $y ^ { 2 } = 2 x$, the line with equation $y = 2$ and the $y$-axis. The unit of length on both axes is one centimetre. A uniform solid, $S$, is formed by rotating $R$ through $360 ^ { \circ }$ about the $y$-axis.\\
Given that the volume of $S$ is $\frac { 8 } { 5 } \pi \mathrm {~cm} ^ { 3 }$,
\begin{enumerate}[label=(\alph*)]
\item show that the centre of mass of $S$ is $\frac { 1 } { 3 } \mathrm {~cm}$ from its plane face.
A uniform solid cylinder, $C$, has base radius 2 cm and height 4 cm . The cylinder $C$ is attached to $S$ so that the plane face of $S$ coincides with a plane face of $C$, to form the paperweight $P$, shown in Figure 5. The density of the material used to make $S$ is three times the density of the material used to make $C$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{9b995178-a4be-4d5a-95f8-6c2978ff01b3-16_572_456_1617_758}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
The plane face of $P$ rests in equilibrium on a desk lid that is inclined at an angle $\theta ^ { \circ }$ to the horizontal. The lid is sufficiently rough to prevent $P$ from slipping. Given that $P$ is on the point of toppling,
\item find the value of $\theta$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 2019 Q5 [11]}}