| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | January |
| 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 straightforward M2 circular motion question requiring standard differentiation of position vectors and application of F=ma. Parts (a) and (b) involve routine substitution and differentiation, while part (c) requires finding acceleration by differentiating twice and computing force magnitude—all standard techniques with no novel problem-solving required. Slightly easier than average due to the mechanical nature of the calculations. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02f Non-uniform acceleration: using differentiation and integration3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) \(t = 0, r = 2\mathbf{i} + 10\mathbf{k}\) | B1 | |
| (ii) \(t = 2\pi, r = 2\mathbf{i} + 7.49\mathbf{k}\) | B1 | Or \(r = 2\mathbf{i} + (10 - 0.8\pi)\mathbf{k}\), accept 7.5k |
| (iii) \(t = 2\pi, \quad t = 4\pi\) | B1, B1 | |
| (b) \(\mathbf{v} = -2\sin t\mathbf{i} + 2\cos t\mathbf{j} - 0.4\mathbf{k}\) | M1, A1, A1 | Differentiation; Trig; k |
| Answer | Marks | Guidance |
|---|---|---|
| \( | \mathbf{F} | = \sqrt{50^2\cos^2 t + 50^2\sin^2 t}\) giving \( |
**(a)(i)** $t = 0, r = 2\mathbf{i} + 10\mathbf{k}$ | B1 | | **Total: 1 mark**
**(ii)** $t = 2\pi, r = 2\mathbf{i} + 7.49\mathbf{k}$ | B1 | Or $r = 2\mathbf{i} + (10 - 0.8\pi)\mathbf{k}$, accept 7.5k | **Total: 1 mark**
**(iii)** $t = 2\pi, \quad t = 4\pi$ | B1, B1 | | **Total: 2 marks**
**(b)** $\mathbf{v} = -2\sin t\mathbf{i} + 2\cos t\mathbf{j} - 0.4\mathbf{k}$ | M1, A1, A1 | Differentiation; Trig; k | **Total: 3 marks**
**(c)** $\mathbf{a} = -2\cos t\mathbf{i} - 2\sin t\mathbf{j}$
$\mathbf{F} = -50\cos t\mathbf{i} - 50\sin t\mathbf{j}$
$|\mathbf{F}| = \sqrt{50^2\cos^2 t + 50^2\sin^2 t}$ giving $|\mathbf{F}| = 50 \text{ (N)}$ | M1A1, M1, M1, A1 | No unit vectors | **Total: 5 marks**
**Question 5 Total: 12 marks**
---5 Tom is on a fairground ride.\\
Tom's position vector, $\mathbf { r }$ metres, at time $t$ seconds is given by
$$\mathbf { r } = 2 \cos t \mathbf { i } + 2 \sin t \mathbf { j } + ( 10 - 0.4 t ) \mathbf { k }$$
The perpendicular unit vectors $\mathbf { i }$ and $\mathbf { j }$ are in the horizontal plane and the unit vector $\mathbf { k }$ is directed vertically upwards.
\begin{enumerate}[label=(\alph*)]
\item \begin{enumerate}[label=(\roman*)]
\item Find Tom's position vector when $t = 0$.
\item Find Tom's position vector when $t = 2 \pi$.
\item Write down the first two values of $t$ for which Tom is directly below his starting point.
\end{enumerate}\item Find an expression for Tom's velocity at time $t$.
\item Tom has mass 25 kg .
Show that the resultant force acting on Tom during the motion has constant magnitude. State the magnitude of the resultant force.\\
(5 marks)
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2007 Q5 [12]}}