| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2008 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Position vector circular motion |
| Difficulty | Moderate -0.3 This is a structured multi-part question on circular motion that systematically guides students through standard differentiation and verification steps. While it requires multiple techniques (differentiation of trig functions, magnitude calculations, proving circular motion), each part is routine for M2 level with clear signposting, making it slightly easier than average overall. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration6.05a Angular velocity: definitions6.05b Circular motion: v=r*omega and a=v^2/r |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = \frac{dr}{dt}\), \(v = -2\sin\frac{1}{4}ti - 2\cos\frac{1}{4}tj\) | M1 A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Speed is \(\{(-2\sin\frac{1}{4}t)^2 + (-2\cos\frac{1}{4}t)^2\}^{\frac{1}{2}} = 2(\sin^2\frac{1}{4}t + \cos^2\frac{1}{4}t)^{\frac{1}{2}} = 2\) (constant) | M1 m1 A1 | 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Magnitude of \(r\) is \(\{(8\cos\frac{1}{4}t)^2 + (8\sin\frac{1}{4}t)^2\}^{\frac{1}{2}} = 8\) (constant). Particle is moving in a circle | M1 A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Using \(v = a\omega\), Angular speed is 0.25 | M1 A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(a = -\frac{1}{2}\cos\frac{1}{4}ti + \frac{1}{2}\sin\frac{1}{4}tj\) | M1 A1 | 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| Magnitude of acceleration is \(\frac{1}{2}\) | B1 | 1 mark |
**Part (a)**
$v = \frac{dr}{dt}$, $v = -2\sin\frac{1}{4}ti - 2\cos\frac{1}{4}tj$ | M1 A1 | 2 marks | No i, j: no marks
**Part (b)**
Speed is $\{(-2\sin\frac{1}{4}t)^2 + (-2\cos\frac{1}{4}t)^2\}^{\frac{1}{2}} = 2(\sin^2\frac{1}{4}t + \cos^2\frac{1}{4}t)^{\frac{1}{2}} = 2$ (constant) | M1 m1 A1 | 3 marks | Clear use of $\sin^2\theta + \cos^2\theta = 1$. Use of 2 values SC1
**Part (c)**
Magnitude of $r$ is $\{(8\cos\frac{1}{4}t)^2 + (8\sin\frac{1}{4}t)^2\}^{\frac{1}{2}} = 8$ (constant). Particle is moving in a circle | M1 A1 | 2 marks | $a = -kr \Rightarrow$ circle SC2
**Part (d)**
Using $v = a\omega$, Angular speed is 0.25 | M1 A1 | 2 marks | M1 for their $\frac{b}{c}$ if both found
**Part (e)**
$a = -\frac{1}{2}\cos\frac{1}{4}ti + \frac{1}{2}\sin\frac{1}{4}tj$ | M1 A1 | 2 marks
**Part (f)**
Magnitude of acceleration is $\frac{1}{2}$ | B1 | 1 mark
**Total: 12 marks**
---5 A particle moves on a horizontal plane in which the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are directed east and north respectively.
At time $t$ seconds, the particle's position vector, $\mathbf { r }$ metres, is given by
$$\mathbf { r } = 8 \left( \cos \frac { 1 } { 4 } t \right) \mathbf { i } - 8 \left( \sin \frac { 1 } { 4 } t \right) \mathbf { j }$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the velocity of the particle at time $t$.
\item Show that the speed of the particle is a constant.
\item Prove that the particle is moving in a circle.
\item Find the angular speed of the particle.
\item Find an expression for the acceleration of the particle at time $t$.
\item State the magnitude of the acceleration of the particle.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2008 Q5 [12]}}