| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2008 |
| Session | January |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Constant acceleration vector problems |
| Difficulty | Standard +0.3 This is a straightforward mechanics question requiring standard application of constant acceleration formulae in vector form. Part (a) is a 'show that' using v = u + at, part (b) uses s = ut + ½at², and part (c) requires setting velocity components equal for southeast direction. All steps are routine M1 techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10e Position vectors: and displacement3.02e Two-dimensional constant acceleration: with vectors3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| \(4\mathbf{i} = 5\mathbf{j} + 40\mathbf{a}\) → \(\mathbf{a} = \frac{4\mathbf{i} - 5\mathbf{j}}{40} = 0.1\mathbf{i} - 0.125\mathbf{j}\) | M1, A1, dM1, A1 (Total: 4) | Forming a vector equation based on constant acceleration. Correct equation. Solving for \(\mathbf{a}\). Correct \(\mathbf{a}\) from correct working. For \(\frac{4\mathbf{i} - 5\mathbf{j}}{40}\) on its own give M0. Allow verification. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{r} = 5\mathbf{j} \times 40 + \frac{1}{2}(0.1\mathbf{i} - 0.125\mathbf{j}) \times 40^2 = 80\mathbf{i} + 100\mathbf{j}\) | M1, A1, A1 (Total: 3) | Finding position vector. Correct expression. Correct simplified result. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{v} = 5\mathbf{j} + (0.1\mathbf{i} - 0.125\mathbf{j})t = 0.1t\mathbf{i} + (5 - 0.125t)\mathbf{j}\) → \(5 - 0.125t = -0.1t\) → \(5 = 0.025t\) → \(t = \frac{5}{0.025} = 200\) | M1, A1, dM1, A1, A1 (Total: 5) | Expression for \(\mathbf{v}\). Correct expression for \(\mathbf{v}\) seen or implied. Equating components, with or without a minus sign. Correct equation. Correct time. |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{v} = 0.1 \times 200\mathbf{i} + (5 - 0.125 \times 200)\mathbf{j} = 20\mathbf{i} - 20\mathbf{j}\) | M1, AIF (Total: 2) | Finding velocity using their time. Correct velocity for their time. |
### Part (a)
$4\mathbf{i} = 5\mathbf{j} + 40\mathbf{a}$ → $\mathbf{a} = \frac{4\mathbf{i} - 5\mathbf{j}}{40} = 0.1\mathbf{i} - 0.125\mathbf{j}$ | M1, A1, dM1, A1 (Total: 4) | Forming a vector equation based on constant acceleration. Correct equation. Solving for $\mathbf{a}$. Correct $\mathbf{a}$ from correct working. For $\frac{4\mathbf{i} - 5\mathbf{j}}{40}$ on its own give M0. Allow verification.
### Part (b)
$\mathbf{r} = 5\mathbf{j} \times 40 + \frac{1}{2}(0.1\mathbf{i} - 0.125\mathbf{j}) \times 40^2 = 80\mathbf{i} + 100\mathbf{j}$ | M1, A1, A1 (Total: 3) | Finding position vector. Correct expression. Correct simplified result.
### Part (c)(i)
$\mathbf{v} = 5\mathbf{j} + (0.1\mathbf{i} - 0.125\mathbf{j})t = 0.1t\mathbf{i} + (5 - 0.125t)\mathbf{j}$ → $5 - 0.125t = -0.1t$ → $5 = 0.025t$ → $t = \frac{5}{0.025} = 200$ | M1, A1, dM1, A1, A1 (Total: 5) | Expression for $\mathbf{v}$. Correct expression for $\mathbf{v}$ seen or implied. Equating components, with or without a minus sign. Correct equation. Correct time.
### Part (c)(ii)
$\mathbf{v} = 0.1 \times 200\mathbf{i} + (5 - 0.125 \times 200)\mathbf{j} = 20\mathbf{i} - 20\mathbf{j}$ | M1, AIF (Total: 2) | Finding velocity using their time. Correct velocity for their time.
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**TOTAL: 75**
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**Note for question 8:** Consistent use of $u = 4\mathbf{i}$ or $5\mathbf{i}$ or $a = 0.1\mathbf{i} + 0.125\mathbf{j}$ award method marks only.8 A Jet Ski is at the origin and is travelling due north at $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ when it begins to accelerate uniformly. After accelerating for 40 seconds, it is travelling due east at $4 \mathrm {~ms} ^ { - 1 }$. The unit vectors $\mathbf { i }$ and $\mathbf { j }$ are directed east and north respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that the acceleration of the Jet Ski is $( 0.1 \mathbf { i } - 0.125 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }$.
\item Find the position vector of the Jet Ski at the end of the 40 second period.
\item The Jet Ski is travelling southeast $t$ seconds after it leaves the origin.
\begin{enumerate}[label=(\roman*)]
\item Find $t$.
\item Find the velocity of the Jet Ski at this time.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M1 2008 Q8 [14]}}