| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Vector motion with components |
| Difficulty | Standard +0.3 This is a straightforward M2 mechanics question requiring standard differentiation and integration of vector functions. Part (a) involves differentiating velocity to find acceleration, (b)(i) applies F=ma, (b)(ii) substitutes and finds magnitude, (c) sets the i-component to zero, and (d) integrates velocity with initial conditions. All techniques are routine for M2 students with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02g Two-dimensional variable acceleration3.03d Newton's second law: 2D vectors |
A particle moves in a horizontal plane under the action of a single force, $\mathbf{F}$ newtons.
The unit vectors $\mathbf{i}$ and $\mathbf{j}$ are directed east and north respectively.
At time $t$ seconds, the velocity of the particle, $\mathbf{v} \text{ m s}^{-1}$, is given by
$$\mathbf{v} = (8t - t^4)\mathbf{i} + 6e^{-3t}\mathbf{j}$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the acceleration of the particle at time $t$.
[2 marks]
\item The mass of the particle is $2$ kg.
\begin{enumerate}[label=(\roman*)]
\item Find an expression for the force $\mathbf{F}$ acting on the particle at time $t$.
[2 marks]
\item Find the magnitude of $\mathbf{F}$ when $t = 1$.
[3 marks]
\end{enumerate}
\item Find the value of $t$ when $\mathbf{F}$ acts due south.
[2 marks]
\item When $t = 0$, the particle is at the point with position vector $(3\mathbf{i} - 5\mathbf{j})$ metres.
Find an expression for the position vector, $\mathbf{r}$ metres, of the particle at time $t$.
[4 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2016 Q2 [13]}}