| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Year | 2023 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 2 |
| Type | Centre of mass of composite lamina |
| Difficulty | Standard +0.3 This is a standard composite centre of mass problem requiring calculation of individual centres of mass and application of the formula x̄ = Σ(mx)/Σm. The geometry is straightforward (disc tangent to rectangle sides), and it's a multiple-choice question with 2-3 marks typical for this type. While it requires careful coordinate setup and arithmetic, it follows a well-practiced procedure with no novel insight needed. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.8\times10\times2(1-\cos30°) = \frac{1}{2}\times0.8\times v^2\) | M1 | Forms energy equation with GPE and KE terms |
| \(v = \sqrt{40(1-\cos30°)}\) | A1 | Correct equation |
| \(v = 2.31 = 2\ \text{ms}^{-1}\) (to 1 sf) | A1 | Condone missing units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| String is inextensible | B1 | States string is inextensible or inelastic or light |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(mgr(\cos\theta - \cos30°) = \frac{1}{2}mv^2\) | M1 | Forms energy equation involving \(\cos\theta\) |
| \(v^2 = 2gr(\cos\theta - \cos30°) = 40\cos\theta - 20\sqrt{3}\) | M1 | Forms three-term equation of motion by resolving radially |
| \(T - mg\cos\theta = \frac{mv^2}{r}\) leading to \(T = 3mg\cos\theta - 2mg\cos30°\) | A1 | Correct equation in terms of \(T\), \(v\) and \(\theta\) |
| \(T = 24\cos\theta - 8\sqrt{3}\) | M1 | Combines equations to eliminate \(v\) |
| Maximum \(T\) requires \(\cos\theta = 1\), so \(\theta = 0\) (particle directly below \(O\)) | B1 | Deduces angle for maximum tension |
| Reasoned argument to obtain result | R1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T = 24 - 8\sqrt{3} = 10.14...\text{ N}\) | B1F | Follow through their expression for tension |
| Air resistance would reduce speed and hence tension when string is vertical, so result is reasonable as it is less than predicted tension; or \(g\) will be less than 10 giving lower tension value | E1 | Only awarded if value greater than 9.5 is obtained and explained reasonably |
## Question 6(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.8\times10\times2(1-\cos30°) = \frac{1}{2}\times0.8\times v^2$ | M1 | Forms energy equation with GPE and KE terms |
| $v = \sqrt{40(1-\cos30°)}$ | A1 | Correct equation |
| $v = 2.31 = 2\ \text{ms}^{-1}$ (to 1 sf) | A1 | Condone missing units |
---
## Question 6(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| String is inextensible | B1 | States string is inextensible or inelastic or light |
---
## Question 6(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $mgr(\cos\theta - \cos30°) = \frac{1}{2}mv^2$ | M1 | Forms energy equation involving $\cos\theta$ |
| $v^2 = 2gr(\cos\theta - \cos30°) = 40\cos\theta - 20\sqrt{3}$ | M1 | Forms three-term equation of motion by resolving radially |
| $T - mg\cos\theta = \frac{mv^2}{r}$ leading to $T = 3mg\cos\theta - 2mg\cos30°$ | A1 | Correct equation in terms of $T$, $v$ and $\theta$ |
| $T = 24\cos\theta - 8\sqrt{3}$ | M1 | Combines equations to eliminate $v$ |
| Maximum $T$ requires $\cos\theta = 1$, so $\theta = 0$ (particle directly below $O$) | B1 | Deduces angle for maximum tension |
| Reasoned argument to obtain result | R1 | |
---
## Question 6(d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T = 24 - 8\sqrt{3} = 10.14...\text{ N}$ | B1F | Follow through their expression for tension |
| Air resistance would reduce speed and hence tension when string is vertical, so result is reasonable as it is less than predicted tension; or $g$ will be less than 10 giving lower tension value | E1 | Only awarded if value greater than 9.5 is obtained and explained reasonably |
---6 Nm\\
8 Nm\\
10 Nm\\
14 Nm
3 A uniform disc has mass 6 kg and diameter 8 cm
A uniform rectangular lamina, $A B C D$, has mass 4 kg , width 8 cm and length 20 cm\\
The disc is fixed to the lamina to form a composite body as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{cd0d239b-ab92-4d17-9cb8-45722e2894cb-03_448_881_587_577}
The sides $A B , A D$ and $C D$ are tangents to the disc.\\
Calculate the distance of the centre of mass of the composite body from $A D$\\
Circle your answer.\\
4 cm\\
5.6 cm\\
6.4 cm\\
8.8 cm
4 A car of mass 1400 kg drives around a horizontal circular bend of radius 60 metres.\\
The car has a constant speed of $12 \mathrm {~ms} ^ { - 1 }$ on the bend.\\
Calculate the magnitude of the resultant force acting on the car.\\[0pt]
[2 marks]\\
$5 \quad$ A region bounded by the curve with equation $y = 4 - x ^ { 2 }$, the $x$-axis and the $y$-axis is shown below.\\
\includegraphics[max width=\textwidth, alt={}, center]{cd0d239b-ab92-4d17-9cb8-45722e2894cb-04_641_380_408_831}
The region is rotated through $360 ^ { \circ }$ around the $x$-axis to create a uniform solid.\\
5
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of the solid from the circular face is $\frac { 5 } { 8 }$\\[0pt]
[5 marks]\\
5
\item The solid is suspended in equilibrium from a point on the edge of the circular face.\\
Find the angle between the circular face and the horizontal, giving your answer to the nearest degree.\\
6 In this question use $g = 10 \mathrm {~m} \mathrm {~s} ^ { - 2 }$
A sphere of mass 0.8 kg is attached to one end of a string of length 2 metres.\\
The other end of the string is attached to a fixed point $O$\\
The sphere is released from rest with the string taut and at an angle of $30 ^ { \circ }$ to the vertical, as shown in the diagram below.\\
\includegraphics[max width=\textwidth, alt={}, center]{cd0d239b-ab92-4d17-9cb8-45722e2894cb-06_464_218_676_909}
6 (a) Find the speed of the sphere when it is directly below $O$\\
6 (b) State one assumption that you made about the string.\\
6
\item As the sphere moves, the string makes an angle $\theta$ with the downward vertical.
By finding an expression for the tension in the string in terms of $\theta$, show that the tension is a maximum when the sphere is directly below $O$
6
\item A physics student conducts an experiment and uses a device to measure the tension in the string when the sphere is directly below $O$
They find that the tension is 9.5 newtons.\\
Explain why this result is reasonable, showing any calculations that you make.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2023 Q6 [12]}}