| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2015 |
| Session | June |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of solid of revolution |
| Difficulty | Challenging +1.8 This is a substantial M3 mechanics question requiring multiple integrations for volume and centre of mass of a solid of revolution, followed by composite body calculations and equilibrium analysis. While the techniques are standard for Further Maths M3, the multi-part nature, algebraic complexity (particularly the x̄ integral involving x⁵ terms), and the suspension equilibrium in part (d) requiring geometric reasoning make this significantly harder than average A-level questions but still within standard Further Maths scope. |
| Spec | 4.08d Volumes of revolution: about x and y axes6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\text{Vol} = \pi\int_0^2(x^2+3)^2\,dx\) | M1 | Using \(\pi\int y^2\,dx\) with equation of curve, no limits needed |
| \(= \pi\int_0^2(x^4+6x^2+9)\,dx\) | Expanding | |
| \(= \pi\left[\frac{1}{5}x^5+2x^3+9x\right]_0^2\) | DM1, A1 | DM1: Integrating their expression; A1: Correct integration inc limits |
| \(= \frac{202}{5}\pi\) cm³ | A1 (4) | Substituting limits to obtain the given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\pi\int_0^2 x(x^2+3)^2\,dx = \pi\int_0^2(x^5+6x^3+9x)\,dx\) | M1 | Using \((\pi)\int xy^2\,dx\), \(\pi\) can be omitted, no limits needed |
| \(= \pi\left[\frac{1}{6}x^6+\frac{3}{2}x^4+\frac{9}{2}x^2\right]_0^2\) | A1 | Correct integration including limits |
| \(= \frac{158}{3}\pi\) | A1 | Correct substitution of limits |
| \(\bar{x} = \frac{\pi\int xy^2\,dx}{\pi\int y^2\,dx} = \frac{158}{3}\times\frac{5}{202} = 1.3036... = 1.30\) cm | M1, A1 (5) | M1: Use of correct \(\bar{x}\) formula with \(\pi\) in both or neither; A1cao: Must be 1.30 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Mass ratio: \(2\times\frac{202}{5}\pi \quad \frac{1}{3}\pi\times7^2\times6 \quad \left(\frac{404}{5}+98\right)\pi\) | B1 | Correct mass ratio |
| Dist from \(V\): \(\quad 6.7 \quad\quad\quad 4.5 \quad\quad\quad \bar{x}\) | B1 | Correct distances from \(V\) or any other point, provided consistent |
| \(\frac{404}{5}\times6.7+98\times4.5=\left(\frac{404}{5}+98\right)\bar{x}\) | M1, A1ft | M1: Attempting moments equation; A1ft: Correct equation following through distances and mass ratio |
| \(\bar{x} = \dfrac{\frac{404}{5}\times6.7+98\times4.5}{\left(\frac{404}{5}+98\right)} = 5.494... = 5.5\) cm | A1 (5) | Accept 5.49 or better |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(\tan\theta = \frac{6-\bar{x}}{7} = \frac{0.5058...}{7}\) | M1 | Attempting tan of an appropriate angle, numbers either way up |
| \(\alpha = \tan^{-1}\left(\frac{6}{7}\right) - \tan^{-1}\left(\frac{0.5058...}{7}\right) = 36.468...° = 36°\) | M1, A1 (3) [17] | M1: Attempting to obtain required angle; A1: Correct final answer 2sf or more |
# Question 6:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| $\text{Vol} = \pi\int_0^2(x^2+3)^2\,dx$ | M1 | Using $\pi\int y^2\,dx$ with equation of curve, no limits needed |
| $= \pi\int_0^2(x^4+6x^2+9)\,dx$ | | Expanding |
| $= \pi\left[\frac{1}{5}x^5+2x^3+9x\right]_0^2$ | DM1, A1 | DM1: Integrating their expression; A1: Correct integration inc limits |
| $= \frac{202}{5}\pi$ cm³ | A1 (4) | Substituting limits to obtain the given answer |
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $\pi\int_0^2 x(x^2+3)^2\,dx = \pi\int_0^2(x^5+6x^3+9x)\,dx$ | M1 | Using $(\pi)\int xy^2\,dx$, $\pi$ can be omitted, no limits needed |
| $= \pi\left[\frac{1}{6}x^6+\frac{3}{2}x^4+\frac{9}{2}x^2\right]_0^2$ | A1 | Correct integration including limits |
| $= \frac{158}{3}\pi$ | A1 | Correct substitution of limits |
| $\bar{x} = \frac{\pi\int xy^2\,dx}{\pi\int y^2\,dx} = \frac{158}{3}\times\frac{5}{202} = 1.3036... = 1.30$ cm | M1, A1 (5) | M1: Use of correct $\bar{x}$ formula with $\pi$ in both or neither; A1cao: Must be 1.30 |
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| Mass ratio: $2\times\frac{202}{5}\pi \quad \frac{1}{3}\pi\times7^2\times6 \quad \left(\frac{404}{5}+98\right)\pi$ | B1 | Correct mass ratio |
| Dist from $V$: $\quad 6.7 \quad\quad\quad 4.5 \quad\quad\quad \bar{x}$ | B1 | Correct distances from $V$ or any other point, provided consistent |
| $\frac{404}{5}\times6.7+98\times4.5=\left(\frac{404}{5}+98\right)\bar{x}$ | M1, A1ft | M1: Attempting moments equation; A1ft: Correct equation following through distances and mass ratio |
| $\bar{x} = \dfrac{\frac{404}{5}\times6.7+98\times4.5}{\left(\frac{404}{5}+98\right)} = 5.494... = 5.5$ cm | A1 (5) | Accept 5.49 or better |
## Part (d):
| Working | Mark | Guidance |
|---------|------|----------|
| $\tan\theta = \frac{6-\bar{x}}{7} = \frac{0.5058...}{7}$ | M1 | Attempting tan of an appropriate angle, numbers either way up |
| $\alpha = \tan^{-1}\left(\frac{6}{7}\right) - \tan^{-1}\left(\frac{0.5058...}{7}\right) = 36.468...° = 36°$ | M1, A1 (3) [17] | M1: Attempting to obtain required angle; A1: Correct final answer 2sf or more |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b7cfcf0a-8f54-4350-8e07-a3b51d94d0f2-11_442_727_237_603}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
The shaded region $R$ is bounded by part of the curve with equation $y = x ^ { 2 } + 3$, the $x$-axis, the $y$-axis and the line with equation $x = 2$, as shown in Figure 4. The unit of length on each axis is one centimetre. The region $R$ is rotated through $2 \pi$ radians about the $x$-axis to form a uniform solid $S$.\\
Using algebraic integration,
\begin{enumerate}[label=(\alph*)]
\item show that the volume of $S$ is $\frac { 202 } { 5 } \pi \mathrm {~cm} ^ { 3 }$,
\item show that, to 2 decimal places, the centre of mass of $S$ is 1.30 cm from $O$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b7cfcf0a-8f54-4350-8e07-a3b51d94d0f2-11_478_472_1407_762}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
A uniform right circular solid cone, of base radius 7 cm and height 6 cm , is joined to $S$ to form a solid $T$. The base of the cone coincides with the larger plane face of $S$, as shown in Figure 5. The vertex of the cone is $V$.\\
The mass per unit volume of $S$ is twice the mass per unit volume of the cone.
\item Find the distance from $V$ to the centre of mass of $T$.
The point $A$ lies on the circumference of the base of the cone. The solid $T$ is suspended from $A$ and hangs freely in equilibrium.
\item Find the size of the angle between $V A$ and the vertical.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2015 Q6 [17]}}