| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 3 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: kinematics extension |
| Difficulty | Moderate -0.8 This is a straightforward application of Newton's second law in vector form with routine algebraic manipulation. Part (a) requires simple vector addition, part (b) uses F=ma to set up two linear equations, and part (c) applies standard SUVAT kinematics. All steps are standard M1 procedures with no problem-solving insight required, making it easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation3.02e Two-dimensional constant acceleration: with vectors3.03b Newton's first law: equilibrium3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \((4\mathbf{i} + 7\mathbf{j}) + (p\mathbf{i} + 5\mathbf{j}) + (-8\mathbf{i} + q\mathbf{j}) = (p - 4)\mathbf{i} + (12 + q)\mathbf{j}\) | B1 | Any correct version of the resultant force. (Does not need to be simplified, can be awarded for the first line.) Condone missing brackets. |
| Answer | Marks | Guidance |
|---|---|---|
| \((p - 4)\mathbf{i} + (12 + q)\mathbf{j} = 5(2\mathbf{i} - \mathbf{j})\), \(p - 4 = 5 \times 2\), \(p = 14\); \(12 + q = 5 \times (-1)\), \(q = -17\) | M1, A1; M1, A1 | Their \(\mathbf{i}\) component equated to \(5 \times 2\). Correct value for \(p\). Condone 14i. Their \(\mathbf{j}\) component equated to \(5 \times (-1)\). Correct value for \(q\). Condone -17j. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{r} = \frac{1}{2}(2\mathbf{i} - \mathbf{j}) \times 4^2 = 16\mathbf{i} - 8\mathbf{j}\) OR \(\mathbf{v} = (2\mathbf{i} - \mathbf{j}) \times 4 = 8\mathbf{i} - 4\mathbf{j}\), \(\mathbf{r} = \frac{((0\mathbf{i} + 0\mathbf{j}) + (8\mathbf{i} - 4\mathbf{j}))}{2} \times 4 = 16\mathbf{i} - 8\mathbf{j}\) | M1A1, A1; (M1A1), (A1) | M1: Using \(\mathbf{r} = \mathbf{u}t + \frac{1}{2}a t^2\) (or \(\mathbf{v} = \mathbf{u} + a t\) with \(\mathbf{u} = 0\mathbf{i} + 0\mathbf{j}\)). A1: Correct expression or components. A1: Correct displacement. Do not penalise candidates who go on to find the magnitude of this. (ISW). |
**2. (a)**
$(4\mathbf{i} + 7\mathbf{j}) + (p\mathbf{i} + 5\mathbf{j}) + (-8\mathbf{i} + q\mathbf{j}) = (p - 4)\mathbf{i} + (12 + q)\mathbf{j}$ | B1 | Any correct version of the resultant force. (Does not need to be simplified, can be awarded for the first line.) Condone missing brackets. | 1 mark
**2. (b)**
$(p - 4)\mathbf{i} + (12 + q)\mathbf{j} = 5(2\mathbf{i} - \mathbf{j})$, $p - 4 = 5 \times 2$, $p = 14$; $12 + q = 5 \times (-1)$, $q = -17$ | M1, A1; M1, A1 | Their $\mathbf{i}$ component equated to $5 \times 2$. Correct value for $p$. Condone 14i. Their $\mathbf{j}$ component equated to $5 \times (-1)$. Correct value for $q$. Condone -17j. | 4 marks
**2. (c)**
$\mathbf{r} = \frac{1}{2}(2\mathbf{i} - \mathbf{j}) \times 4^2 = 16\mathbf{i} - 8\mathbf{j}$ OR $\mathbf{v} = (2\mathbf{i} - \mathbf{j}) \times 4 = 8\mathbf{i} - 4\mathbf{j}$, $\mathbf{r} = \frac{((0\mathbf{i} + 0\mathbf{j}) + (8\mathbf{i} - 4\mathbf{j}))}{2} \times 4 = 16\mathbf{i} - 8\mathbf{j}$ | M1A1, A1; (M1A1), (A1) | M1: Using $\mathbf{r} = \mathbf{u}t + \frac{1}{2}a t^2$ (or $\mathbf{v} = \mathbf{u} + a t$ with $\mathbf{u} = 0\mathbf{i} + 0\mathbf{j}$). A1: Correct expression or components. A1: Correct displacement. Do not penalise candidates who go on to find the magnitude of this. (ISW). | 3 marks
**Total for Question 2: 8 marks**
---2 Three forces $( 4 \mathbf { i } + 7 \mathbf { j } ) \mathrm { N } , ( p \mathbf { i } + 5 \mathbf { j } ) \mathrm { N }$ and $( - 8 \mathbf { i } + q \mathbf { j } ) \mathrm { N }$ act on a particle of mass 5 kg to produce an acceleration of $( 2 \mathbf { i } - \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }$. No other forces act on the particle.
\begin{enumerate}[label=(\alph*)]
\item Find the resultant force acting on the particle in terms of $p$ and $q$.
\item $\quad$ Find $p$ and $q$.
\item Given that the particle is initially at rest, find the displacement of the particle from its initial position when these forces have been acting for 4 seconds.\\[0pt]
[3 marks]
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2016 Q2 [3]}}