| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Position vector circular motion |
| Difficulty | Moderate -0.3 This is a standard M2 circular motion question requiring routine differentiation and verification. Parts (a)-(c) involve straightforward calculus on trigonometric functions, part (d) is simple comparison, and part (e) requires recognizing that acceleration points toward the center. While multi-part, each step follows directly from standard techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration6.05b Circular motion: v=r*omega and a=v^2/r |
4 A particle moves on a horizontal plane, in which the unit vectors $\mathbf { i }$ and $\mathbf { j }$ are perpendicular.
At time $t$, the particle's position vector, $\mathbf { r }$, is given by
$$\mathbf { r } = 4 \cos 3 t \mathbf { i } - 4 \sin 3 t \mathbf { j }$$
\begin{enumerate}[label=(\alph*)]
\item Prove that the particle is moving on a circle, which has its centre at the origin.
\item Find an expression for the velocity of the particle at time $t$.
\item Find an expression for the acceleration of the particle at time $t$.
\item The acceleration of the particle can be written as
$$\mathbf { a } = k \mathbf { r }$$
where $k$ is a constant.
Find the value of $k$.
\item State the direction of the acceleration of the particle.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2012 Q4 [9]}}