| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Session | Specimen |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Solid on inclined plane - toppling |
| Difficulty | Challenging +1.8 This is a Further Maths mechanics question requiring volume/centre of mass integration for a solid of revolution, then applying toppling conditions on an inclined plane. While the integration is standard (using Pappus or direct methods), the multi-step nature, Further Maths content, and application of toppling criteria (requiring geometric insight about the critical angle) make this significantly harder than average A-level questions. |
| Spec | 3.03t Coefficient of friction: F <= mu*R model3.03u Static equilibrium: on rough surfaces6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| \(M = \rho\displaystyle\int_0^2 \dfrac{\pi x^4}{4}\,dx\) | M1 | Forms integral to find mass |
| \(= \pi\rho\left[\dfrac{x^5}{20}\right]_0^2 = \dfrac{8}{5}\pi\rho\) | A1 | Correct mass |
| \(\bar{x} \times \dfrac{8}{5}\pi\rho = \displaystyle\int_0^2 \dfrac{\pi\rho x^5}{4}\,dx\) | M1 | Forms integral of form \(\int xy^2\,dx\) |
| \(= \pi\rho\left[\dfrac{x^6}{24}\right]_0^2 = \dfrac{8}{3}\pi\rho\) | A1 | Correct value of \(\int xy^2\,dx\) |
| \(\bar{x} = \dfrac{5}{3}\) | A1F | Correct centre of mass; FT if both M1 marks awarded |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\alpha = \dfrac{2}{2 - \dfrac{5}{3}}\) | M1 | Uses trigonometry with their centre of mass to form equation |
| \(\alpha = 80.5°\) | A1F | Correct angle; FT if all M1 marks awarded |
## Question 6:
### Part (a):
| $M = \rho\displaystyle\int_0^2 \dfrac{\pi x^4}{4}\,dx$ | M1 | Forms integral to find mass |
| $= \pi\rho\left[\dfrac{x^5}{20}\right]_0^2 = \dfrac{8}{5}\pi\rho$ | A1 | Correct mass |
| $\bar{x} \times \dfrac{8}{5}\pi\rho = \displaystyle\int_0^2 \dfrac{\pi\rho x^5}{4}\,dx$ | M1 | Forms integral of form $\int xy^2\,dx$ |
| $= \pi\rho\left[\dfrac{x^6}{24}\right]_0^2 = \dfrac{8}{3}\pi\rho$ | A1 | Correct value of $\int xy^2\,dx$ |
| $\bar{x} = \dfrac{5}{3}$ | A1F | Correct centre of mass; FT if both M1 marks awarded |
### Part (b):
| $\tan\alpha = \dfrac{2}{2 - \dfrac{5}{3}}$ | M1 | Uses trigonometry with their centre of mass to form equation |
| $\alpha = 80.5°$ | A1F | Correct angle; FT if all M1 marks awarded |
---6 A uniform solid is formed by rotating the region enclosed by the positive $x$-axis, the line $x = 2$ and the curve $y = \frac { 1 } { 2 } x ^ { 2 }$ through $360 ^ { \circ }$ around the $x$-axis.
6
\begin{enumerate}[label=(\alph*)]
\item Find the centre of mass of this solid.\\
6
\item The solid is placed with its plane face on a rough inclined plane and does not slide. The angle between the inclined plane and the horizontal is gradually increased.
When the angle between the inclined plane and the horizontal is $\alpha$, the solid is on the point of toppling.
Find $\alpha$, giving your answer to the nearest $0.1 ^ { \circ }$
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics Q6 [7]}}