| Exam Board | AQA |
|---|---|
| Module | Further Paper 3 Mechanics (Further Paper 3 Mechanics) |
| Year | 2022 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Position vector circular motion |
| Difficulty | Standard +0.3 This is a standard Further Maths circular motion question requiring differentiation of position vectors and verification of perpendicularity (dot product = 0) and constant force magnitude. The calculations are straightforward with no novel insight needed, making it slightly easier than average for A-level but typical for Further Maths mechanics. |
| Spec | 1.10c Magnitude and direction: of vectors1.10h Vectors in kinematics: uniform acceleration in vector form3.02g Two-dimensional variable acceleration6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{v} = 6\cos 3t\,\mathbf{i} - 6\sin 3t\,\mathbf{j}\) | M1, A1 | Differentiates to obtain velocity with at least one component correct; then obtains correct velocity |
| \(\mathbf{v} \cdot \mathbf{r} = 12\sin 3t\cos 3t - 12\cos 3t\sin 3t = 0\) | M1 | Calculates scalar product of velocity and position vector |
| As the scalar product is zero the vectors are perpendicular | R1 | Shows scalar product is zero and explains the two vectors are perpendicular |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{a} = -18\sin 3t\,\mathbf{i} - 18\cos 3t\,\mathbf{j}\), \(\mathbf{F} = -90\sin 3t\,\mathbf{i} - 90\cos 3t\,\mathbf{j}\) | B1 | Obtains correct acceleration |
| \(F = \sqrt{(-90\sin 3t)^2 + (-90\cos 3t)^2} = \sqrt{90^2(\sin^2 3t + \cos^2 3t)} = 90\text{ N}\) | M1 | Finds magnitude of acceleration or applies Newton's second law |
| So the magnitude of the resultant force is constant | R1 | Applies Newton's second law and simplifies using trigonometric identity; must show use of identity; condone missing units; conclusion does not have to be stated |
## Question 6(a):
$\mathbf{v} = 6\cos 3t\,\mathbf{i} - 6\sin 3t\,\mathbf{j}$ | M1, A1 | Differentiates to obtain velocity with at least one component correct; then obtains correct velocity
$\mathbf{v} \cdot \mathbf{r} = 12\sin 3t\cos 3t - 12\cos 3t\sin 3t = 0$ | M1 | Calculates scalar product of velocity and position vector
As the scalar product is zero the vectors are perpendicular | R1 | Shows scalar product is zero and explains the two vectors are perpendicular
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## Question 6(b):
$\mathbf{a} = -18\sin 3t\,\mathbf{i} - 18\cos 3t\,\mathbf{j}$, $\mathbf{F} = -90\sin 3t\,\mathbf{i} - 90\cos 3t\,\mathbf{j}$ | B1 | Obtains correct acceleration
$F = \sqrt{(-90\sin 3t)^2 + (-90\cos 3t)^2} = \sqrt{90^2(\sin^2 3t + \cos^2 3t)} = 90\text{ N}$ | M1 | Finds magnitude of acceleration or applies Newton's second law
So the magnitude of the resultant force is constant | R1 | Applies Newton's second law and simplifies using trigonometric identity; must show use of identity; condone missing units; conclusion does not have to be stated
---6 A particle, of mass 5 kg , moves on a circular path so that at time $t$ seconds it has position vector $\mathbf { r }$ metres, where
$$\mathbf { r } = ( 2 \sin 3 t ) \mathbf { i } + ( 2 \cos 3 t ) \mathbf { j }$$
6
\begin{enumerate}[label=(\alph*)]
\item Prove that the velocity of the particle is perpendicular to its position vector.\\
6
\item Prove that the magnitude of the resultant force on the particle is constant.
\end{enumerate}
\hfill \mbox{\textit{AQA Further Paper 3 Mechanics 2022 Q6 [7]}}