| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2011 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Particle motion: 2D constant acceleration |
| Difficulty | Moderate -0.8 This is a straightforward application of SUVAT equations in 2D vector form. Part (a) uses v = u + at directly, part (b) substitutes a value and finds direction from components, and part (c) requires solving |v| = 5 which leads to a quadratic equation. All steps are routine M1 techniques with no conceptual challenges or novel problem-solving required. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{v} = (4\mathbf{i} + 0.5\mathbf{j}) + (-0.4\mathbf{i} + 0.2\mathbf{j})t\) | M1 | Using \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) |
| \(\mathbf{v} = (4 - 0.4t)\mathbf{i} + (0.5 + 0.2t)\mathbf{j}\) | A1 | Accept equivalent forms |
| Answer | Marks | Guidance |
|---|---|---|
| When \(t = 22.5\): \(\mathbf{v} = (4 - 0.4(22.5))\mathbf{i} + (0.5 + 0.2(22.5))\mathbf{j}\) | M1 | Substituting \(t = 22.5\) |
| \(\mathbf{v} = -5\mathbf{i} + 5\mathbf{j}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| North-west (or equivalent, e.g. bearing of 315°) | B1 | Must be consistent with their velocity vector |
| Answer | Marks | Guidance |
|---|---|---|
| Speed = 5, so \((4 - 0.4t)^2 + (0.5 + 0.2t)^2 = 25\) | M1 | Setting up speed equation |
| \(16 - 3.2t + 0.16t^2 + 0.25 + 0.2t + 0.04t^2 = 25\) | M1 | Expanding brackets |
| \(0.2t^2 - 3t + 16.25 = 25\) | A1 | Correct simplification |
| \(0.2t^2 - 3t - 8.75 = 0\) | ||
| \(2t^2 - 30t - 87.5 = 0\) or \(4t^2 - 60t - 175 = 0\) | A1 | Correct equation |
| \((2t - 35)(2t + 5) = 0\) or using quadratic formula | DM1 | Solving quadratic |
| \(t = 17.5\) | A1 | (reject \(t = -2.5\)) |
# Question 5:
## Part (a):
| $\mathbf{v} = (4\mathbf{i} + 0.5\mathbf{j}) + (-0.4\mathbf{i} + 0.2\mathbf{j})t$ | M1 | Using $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ |
|---|---|---|
| $\mathbf{v} = (4 - 0.4t)\mathbf{i} + (0.5 + 0.2t)\mathbf{j}$ | A1 | Accept equivalent forms |
## Part (b)(i):
| When $t = 22.5$: $\mathbf{v} = (4 - 0.4(22.5))\mathbf{i} + (0.5 + 0.2(22.5))\mathbf{j}$ | M1 | Substituting $t = 22.5$ |
|---|---|---|
| $\mathbf{v} = -5\mathbf{i} + 5\mathbf{j}$ | A1 | |
## Part (b)(ii):
| North-west (or equivalent, e.g. bearing of 315°) | B1 | Must be consistent with their velocity vector |
|---|---|---|
## Part (c):
| Speed = 5, so $(4 - 0.4t)^2 + (0.5 + 0.2t)^2 = 25$ | M1 | Setting up speed equation |
|---|---|---|
| $16 - 3.2t + 0.16t^2 + 0.25 + 0.2t + 0.04t^2 = 25$ | M1 | Expanding brackets |
| $0.2t^2 - 3t + 16.25 = 25$ | A1 | Correct simplification |
| $0.2t^2 - 3t - 8.75 = 0$ | | |
| $2t^2 - 30t - 87.5 = 0$ or $4t^2 - 60t - 175 = 0$ | A1 | Correct equation |
| $(2t - 35)(2t + 5) = 0$ or using quadratic formula | DM1 | Solving quadratic |
| $t = 17.5$ | A1 | (reject $t = -2.5$) |
---5 A particle moves with constant acceleration $( - 0.4 \mathbf { i } + 0.2 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }$. Initially, it has velocity $( 4 \mathbf { i } + 0.5 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 1 }$. The unit vectors $\mathbf { i }$ and $\mathbf { j }$ are directed east and north respectively.
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the velocity of the particle at time $t$ seconds.
\item \begin{enumerate}[label=(\roman*)]
\item Find the velocity of the particle when $t = 22.5$.
\item State the direction in which the particle is travelling at this time.
\end{enumerate}\item Find the time when the speed of the particle is $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2011 Q5 [11]}}