| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Position vector from velocity integration |
| Difficulty | Moderate -0.3 This is a straightforward application of Newton's second law followed by standard integration with initial conditions. Part (a) requires F=ma, part (b) is routine integration of acceleration to verify given velocity (with clear initial conditions), and part (c) is another standard integration. The vector notation and time-dependent force add minimal complexity beyond basic M1 content. Slightly easier than average due to the 'show that' format in part (b) which guides students to the answer. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02a Kinematics language: position, displacement, velocity, acceleration3.02f Non-uniform acceleration: using differentiation and integration3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Part (a): Using \(F = ma\): \(2400i - 4800j = 800a\); \(a = 3i - 6j\) | M1, A1 | |
| Part (b): \(v = \int a \, dt = 3ti - 3t^2j + \mathbf{c}\); When \(t = 0\), \(v = 6i + 30j\); \(\therefore \mathbf{c} = 6i + 30j\); \(\therefore v = (3t + 6)i + (30 - 3t^2)j\) | M1, A1, M1, A1 | Condone no '+' c'; Needs '+' c' above; AG |
| Part (c): \(\mathbf{r} = \int v \, dt = (\frac{3}{2}t^2 + 6t)i + (30t - t^3)j + \mathbf{d}\); When \(t = 0\), \(\mathbf{r} = 2i + 5j\); \(\therefore \mathbf{d} = 2i + 5j\); \(\therefore \mathbf{r} = (\frac{3}{2}t^2 + 6t + 2)i + (30t - t^3 + 5)j\) | M1, A1,A1, M1, A1 | A1 i term, A1 j term; condone no '+' d'; |
| Total: 11 |
| Answer/Working | Marks | Guidance |
|---|---|---|
| **Part (a):** Using $F = ma$: $2400i - 4800j = 800a$; $a = 3i - 6j$ | M1, A1 | |
| **Part (b):** $v = \int a \, dt = 3ti - 3t^2j + \mathbf{c}$; When $t = 0$, $v = 6i + 30j$; $\therefore \mathbf{c} = 6i + 30j$; $\therefore v = (3t + 6)i + (30 - 3t^2)j$ | M1, A1, M1, A1 | Condone no '+' c'; Needs '+' c' above; AG |
| **Part (c):** $\mathbf{r} = \int v \, dt = (\frac{3}{2}t^2 + 6t)i + (30t - t^3)j + \mathbf{d}$; When $t = 0$, $\mathbf{r} = 2i + 5j$; $\therefore \mathbf{d} = 2i + 5j$; $\therefore \mathbf{r} = (\frac{3}{2}t^2 + 6t + 2)i + (30t - t^3 + 5)j$ | M1, A1,A1, M1, A1 | A1 i term, A1 j term; condone no '+' d'; |
| | | **Total: 11** |3 A particle has mass 800 kg . A single force of $( 2400 \mathbf { i } - 4800 t \mathbf { j } )$ newtons acts on the particle at time $t$ seconds. No other forces act on the particle.
\begin{enumerate}[label=(\alph*)]
\item Find the acceleration of the particle at time $t$.
\item At time $t = 0$, the velocity of the particle is $( 6 \mathbf { i } + 30 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$. The velocity of the particle at time $t$ is $\mathbf { v } \mathrm { m } \mathrm { s } ^ { - 1 }$.
Show that
$$\mathbf { v } = ( 6 + 3 t ) \mathbf { i } + \left( 30 - 3 t ^ { 2 } \right) \mathbf { j }$$
\item Initially, the particle is at the point with position vector $( 2 \mathbf { i } + 5 \mathbf { j } ) \mathrm { m }$.
Find the position vector, $\mathbf { r }$ metres, of the particle at time $t$.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2007 Q3 [11]}}