| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2018 |
| Session | Specimen |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of solid of revolution |
| Difficulty | Standard +0.8 This is a multi-part M3 question requiring volume and centre of mass calculations for solids of revolution, then combining two solids with different densities, and finally applying equilibrium conditions. While the individual integrations are standard, the question requires careful bookkeeping across multiple parts, understanding of composite centre of mass with different densities, and applying suspension equilibrium—making it moderately challenging but within typical M3 scope. |
| Spec | 4.08d Volumes of revolution: about x and y axes6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
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| Q6 | |
| VIIIV SIHI NI JAIIM ION OC | VIIIV SIHI NI JIHMM ION OO | VI4V SIHI NI JIIYM IONOO |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\text{Vol} = \pi\int_0^2 (x^2+3)^2\, dx\) | M1 | Using \(\pi\int y^2\,dx\) with the equation of the curve, no limits needed |
| \(= \pi\int_0^2 (x^4 + 6x^2 + 9)\,dx\) | ||
| \(= \pi\left[\frac{1}{5}x^5 + 2x^3 + 9x\right]_0^2\) | dM1, A1 | Integrating their expression; correct integration including limits |
| \(= \frac{202}{5}\pi\) cm\(^3\) | A1 | Substituting limits to obtain the given answer |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\pi\int_0^2 x(x^2+3)^2\,dx = \pi\int_0^2(x^5+6x^3+9x)\,dx\) | M1 | Using \((\pi)\int xy^2\,dx\) with the equation of the curve; \(\pi\) can be omitted |
| \(= \pi\left[\frac{1}{6}x^6 + \frac{3}{2}x^4 + \frac{9}{2}x^2\right]_0^2\) | A1 | Correct integration including limits |
| \(= \frac{158}{3}\pi\) | A1 | Correct substitution of limits |
| \(\text{C of m} = \frac{158}{3} \times \frac{5}{202} = 1.3036... = 1.30\) cm | M1, A1 | Use of \(\frac{\pi\int xy^2\,dx}{\pi\int y^2\,dx}\); cso correct answer, must be 1.30 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | 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 \(V\): \(6.7 \quad 4.5 \quad \bar{x}\) | B1 | Correct distances from \(V\) or any other point, provided consistent |
| \(\frac{404}{5}\times 6.7 + 98\times 4.5 = \left(\frac{404}{5}+98\right)\bar{x}\) | M1, A1ft | Attempting a moments equation; 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 (accept 5.49 or better) | A1 | Correct distance from \(V\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\tan\theta = \frac{6-\bar{x}}{7} = \frac{0.5058...}{7}\) | M1 | Attempting the tan of an appropriate angle, numbers either way up |
| \(\alpha = \tan^{-1}\!\left(\frac{6}{7}\right) - \tan^{-1}\!\left(\frac{0.5058...}{7}\right) = 36.468...° = 36°\) or better | M1, A1 | Attempting to obtain the required angle; correct final answer 2sf or more |
## Question 6:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\text{Vol} = \pi\int_0^2 (x^2+3)^2\, dx$ | M1 | Using $\pi\int y^2\,dx$ with the equation of the curve, no limits needed |
| $= \pi\int_0^2 (x^4 + 6x^2 + 9)\,dx$ | | |
| $= \pi\left[\frac{1}{5}x^5 + 2x^3 + 9x\right]_0^2$ | dM1, A1 | Integrating their expression; correct integration including limits |
| $= \frac{202}{5}\pi$ cm$^3$ | A1 | Substituting limits to obtain the given answer |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\pi\int_0^2 x(x^2+3)^2\,dx = \pi\int_0^2(x^5+6x^3+9x)\,dx$ | M1 | Using $(\pi)\int xy^2\,dx$ with the equation of the curve; $\pi$ can be omitted |
| $= \pi\left[\frac{1}{6}x^6 + \frac{3}{2}x^4 + \frac{9}{2}x^2\right]_0^2$ | A1 | Correct integration including limits |
| $= \frac{158}{3}\pi$ | A1 | Correct substitution of limits |
| $\text{C of m} = \frac{158}{3} \times \frac{5}{202} = 1.3036... = 1.30$ cm | M1, A1 | Use of $\frac{\pi\int xy^2\,dx}{\pi\int y^2\,dx}$; **cso** correct answer, must be 1.30 |
### Part (c):
| Working | Marks | 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 $V$: $6.7 \quad 4.5 \quad \bar{x}$ | B1 | Correct distances from $V$ or any other point, provided consistent |
| $\frac{404}{5}\times 6.7 + 98\times 4.5 = \left(\frac{404}{5}+98\right)\bar{x}$ | M1, A1ft | Attempting a moments equation; 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 (accept 5.49 or better) | A1 | Correct distance from $V$ |
### Part (d):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\tan\theta = \frac{6-\bar{x}}{7} = \frac{0.5058...}{7}$ | M1 | Attempting the tan of an appropriate angle, numbers either way up |
| $\alpha = \tan^{-1}\!\left(\frac{6}{7}\right) - \tan^{-1}\!\left(\frac{0.5058...}{7}\right) = 36.468...° = 36°$ or better | M1, A1 | Attempting to obtain the required angle; correct final answer 2sf or more |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bb73b211-7629-4ed7-9b71-91841c29bb85-20_442_723_237_605}
\captionsetup{labelformat=empty}
\caption{Figure 4}
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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]{bb73b211-7629-4ed7-9b71-91841c29bb85-20_483_469_1402_767}
\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.
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VIIIV SIHI NI JAIIM ION OC & VIIIV SIHI NI JIHMM ION OO & VI4V SIHI NI JIIYM IONOO \\
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\hfill \mbox{\textit{Edexcel M3 2018 Q6 [17]}}