| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Year | 2020 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Position vector circular motion |
| Difficulty | Standard +0.3 This is a standard circular motion question requiring routine application of formulas (ω = 2π/T, position vector in component form, centripetal acceleration). Part (b) requires careful setup of the parametric equations given the geometry, but all parts follow predictable patterns with no novel problem-solving. Slightly above average difficulty due to the vector notation and multiple parts, but well within typical Further Maths mechanics scope. |
| Spec | 6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(4\omega = 2\pi\), so \(\omega = \frac{\pi}{2}\) | M1 (AO 1.1a) | Forms equation to obtain angular speed |
| Angular speed \(= \frac{\pi}{2} \text{ rad s}^{-1}\) | A1 (AO 1.1b) | Obtains correct angular speed; condone missing units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{r} = 2\sin(\omega t)\,\mathbf{i} + (2\sin/\cos(\omega t) + c)\,\mathbf{j}\) with their \(\omega\); allow \(c = 0\) | M1 (AO 3.3) | States position vector of this form |
| \(\mathbf{r} = 2\sin\!\left(\frac{\pi t}{2}\right)\mathbf{i} + \left(2 - 2\cos\!\left(\frac{\pi t}{2}\right)\right)\mathbf{j}\) metres | A1F (AO 1.1b) | Obtains correct position vector; FT their \(\omega\); condone missing units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Differentiates position vector to find velocity | M1 (AO 1.1a) | — |
| \(\mathbf{v} = \pi\cos\!\left(\frac{\pi t}{2}\right)\mathbf{i} + \pi\sin\!\left(\frac{\pi t}{2}\right)\mathbf{j}\) | M1 (AO 1.1a) | Differentiates velocity to find acceleration |
| \(\mathbf{a} = -\frac{\pi^2}{2}\sin\!\left(\frac{\pi t}{2}\right)\mathbf{i} + \frac{\pi^2}{2}\cos\!\left(\frac{\pi t}{2}\right)\mathbf{j} \text{ ms}^{-2}\) | A1F (AO 2.1) | Obtains correct acceleration from fully correct working; condone missing units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{\pi^2}{2} \text{ ms}^{-2}\) | B1F (AO 1.1b) | States correct magnitude of acceleration; condone missing units |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2\) seconds | B1 (AO 2.2a) | Deduces correctly that acceleration is directed towards origin at 2 seconds; condone missing units |
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $4\omega = 2\pi$, so $\omega = \frac{\pi}{2}$ | M1 (AO 1.1a) | Forms equation to obtain angular speed |
| Angular speed $= \frac{\pi}{2} \text{ rad s}^{-1}$ | A1 (AO 1.1b) | Obtains correct angular speed; condone missing units |
---
## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{r} = 2\sin(\omega t)\,\mathbf{i} + (2\sin/\cos(\omega t) + c)\,\mathbf{j}$ with their $\omega$; allow $c = 0$ | M1 (AO 3.3) | States position vector of this form |
| $\mathbf{r} = 2\sin\!\left(\frac{\pi t}{2}\right)\mathbf{i} + \left(2 - 2\cos\!\left(\frac{\pi t}{2}\right)\right)\mathbf{j}$ metres | A1F (AO 1.1b) | Obtains correct position vector; FT their $\omega$; condone missing units |
---
## Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Differentiates position vector to find velocity | M1 (AO 1.1a) | — |
| $\mathbf{v} = \pi\cos\!\left(\frac{\pi t}{2}\right)\mathbf{i} + \pi\sin\!\left(\frac{\pi t}{2}\right)\mathbf{j}$ | M1 (AO 1.1a) | Differentiates velocity to find acceleration |
| $\mathbf{a} = -\frac{\pi^2}{2}\sin\!\left(\frac{\pi t}{2}\right)\mathbf{i} + \frac{\pi^2}{2}\cos\!\left(\frac{\pi t}{2}\right)\mathbf{j} \text{ ms}^{-2}$ | A1F (AO 2.1) | Obtains correct acceleration from fully correct working; condone missing units |
---
## Question 6(d):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{\pi^2}{2} \text{ ms}^{-2}$ | B1F (AO 1.1b) | States correct magnitude of acceleration; condone missing units |
---
## Question 6(e):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2$ seconds | B1 (AO 2.2a) | Deduces correctly that acceleration is directed towards origin at 2 seconds; condone missing units |
---6 A particle moves with constant speed on a circular path of radius 2 metres.
The centre of the circle has position vector $2 \mathbf { j }$ metres.\\
At time $t = 0$, the particle is at the origin and is moving in the positive $\mathbf { i }$ direction.\\
The particle returns to the origin every 4 seconds.\\
The unit vectors $\mathbf { i }$ and $\mathbf { j }$ are perpendicular.\\
6
\begin{enumerate}[label=(\alph*)]
\item Calculate the angular speed of the particle.\\
6
\item Write down an expression for the position vector of the particle at time $t$ seconds.\\
6
\item Find an expression for the acceleration of the particle at time $t$ seconds.\\
6
\item State the magnitude of the acceleration of the particle.\\
6
\item State the time when the acceleration is first directed towards the origin.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2020 Q6 [9]}}