| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle in hemispherical bowl |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring energy conservation across two circular arcs, circular motion dynamics at the point of losing contact (N=0 condition), and algebraic manipulation to derive a given result. The multi-stage setup and need to track height changes across different arc centers makes it significantly harder than standard A-level mechanics, though the individual techniques are familiar to FM students. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\dfrac{1}{2}mv^2 = mg\!\left(\dfrac{r}{4} + (r - r\cos\theta)\right)\) | M1 | Uses conservation of energy |
| \(\dfrac{1}{2}v^2 = gr\!\left(\dfrac{1}{4} + 1 - \cos\theta\right)\) | A1 | Correct equation |
| \(v^2 = 2gr\!\left(\dfrac{5}{4} - \cos\theta\right) = \dfrac{gr}{2}(5 - 4\cos\theta)\) | R1 | Completes argument to find \(v\); fully correct, clear solution required |
| \(v = \sqrt{\dfrac{gr}{2}(5 - 4\cos\theta)}\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(mg\cos\theta = \dfrac{m}{r}\cdot\dfrac{gr}{2}(5 - 4\cos\theta)\) | M1 | Resolves radially to form equation for \(\theta\) |
| \(2\cos\theta = 5 - 4\cos\theta\) | ||
| \(\cos\theta = \dfrac{5}{6},\quad \theta = 34°\) | A1F | Correct \(\theta\) from their equation; FT if both M1 marks awarded |
| Answer | Marks | Guidance |
|---|---|---|
| Air resistance would decrease the speed and therefore increase \(\theta\) | E1 | Explains air resistance decreases speed |
| \(\theta\) will increase | R1 | Correct deduction |
| Answer | Marks | Guidance |
|---|---|---|
| Because \(F = \mu N\) and \(N\) varies as child moves down the slope | E1 | Normal reaction changes, friction proportional to normal reaction |
## Question 8:
### Part (a):
| $\dfrac{1}{2}mv^2 = mg\!\left(\dfrac{r}{4} + (r - r\cos\theta)\right)$ | M1 | Uses conservation of energy |
| $\dfrac{1}{2}v^2 = gr\!\left(\dfrac{1}{4} + 1 - \cos\theta\right)$ | A1 | Correct equation |
| $v^2 = 2gr\!\left(\dfrac{5}{4} - \cos\theta\right) = \dfrac{gr}{2}(5 - 4\cos\theta)$ | R1 | Completes argument to find $v$; fully correct, clear solution required |
| $v = \sqrt{\dfrac{gr}{2}(5 - 4\cos\theta)}$ | | |
### Part (b):
| $mg\cos\theta = \dfrac{m}{r}\cdot\dfrac{gr}{2}(5 - 4\cos\theta)$ | M1 | Resolves radially to form equation for $\theta$ |
| $2\cos\theta = 5 - 4\cos\theta$ | | |
| $\cos\theta = \dfrac{5}{6},\quad \theta = 34°$ | A1F | Correct $\theta$ from their equation; FT if both M1 marks awarded |
### Part (c):
| Air resistance would decrease the speed and therefore increase $\theta$ | E1 | Explains air resistance decreases speed |
| $\theta$ will increase | R1 | Correct deduction |
### Part (d):
| Because $F = \mu N$ and $N$ varies as child moves down the slope | E1 | Normal reaction changes, friction proportional to normal reaction |
---8 The diagram shows part of a water park slide, $A B C$.\\
The slide is in the shape of two circular arcs, $A B$ and $B C$, each of radius $r$.\\
The point $A$ is at a height of $\frac { r } { 4 }$ above $B$.\\
The circular $\operatorname { arc } B C$ has centre $O$ and $B$ is vertically above $O$.\\
These points are joined as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{4fdb2637-6368-422c-99da-85b80efe31c5-12_590_1173_756_443}
A child starts from rest at $A$, moves along the slide past the point $B$ and then loses contact with the slide at a point $D$.
The angle between the vertical, $O B$, and $O D$ is $\theta$\\
Assume that the slide is smooth.
8
\begin{enumerate}[label=(\alph*)]
\item Show that the speed $v$ of the child at $D$ is given by $v = \sqrt { \frac { g r } { 2 } ( 5 - 4 \cos \theta ) }$, where $g$ is the acceleration due to gravity.
8
\item Find $\theta$, giving your answer to the nearest degree.\\
8
\item A refined model takes into account air resistance. Explain how taking air resistance into account would affect your answer to part (b).\\[0pt]
[2 marks]\\
8
\item In reality the slide is not smooth. It has a surface with the same coefficient of friction between the slide and the child for its entire length.
Explain why the frictional force experienced by the child is not constant.\\[0pt]
[1 mark]
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics Q8 [8]}}