| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Vector motion with components |
| Difficulty | Moderate -0.8 This is a straightforward M1 kinematics question using vectors with constant acceleration. Part (a) is trivial vector notation, (b) is direct integration/application of v=u+at, (c) requires setting the i-component to zero, and (d) needs position vector calculation and basic trigonometry. All steps are routine applications of standard formulae with no problem-solving insight required. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors3.02e Two-dimensional constant acceleration: with vectors3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{u} = 5\mathbf{i}\) or \(\begin{bmatrix} 5 \\ 0 \end{bmatrix}\) | B1 | Correct velocity |
| Total: 1 mark |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{v} = 5\mathbf{i} + (-0.2\mathbf{i} + 0.25\mathbf{j})t\) | M1 | Use of constant acceleration equation, with \(\mathbf{u}\) and \(\mathbf{a}\) not zero |
| A1 | Correct velocity. M1A0 for using 5j or just 5 | |
| OR | ||
| \(\mathbf{v} = \begin{bmatrix} 5 - 0.2t \\ 0.25t \end{bmatrix}\) | ||
| Total: 2 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(5 - 0.2t = 0\) | M1 | Easterly component zero |
| \(t = \frac{5}{0.2} = 25 \text{ seconds}\) | A1 | Correct equation |
| A1 | Correct \(t\) | |
| Total: 3 marks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{r} = 5\mathbf{i} \times 25 + \frac{1}{2}(-0.2\mathbf{i} + 0.25\mathbf{j}) \times 25^2\) | M1 | Use of constant acceleration equation with \(t\) from part (c) |
| \(= 62.5 + 78.125\mathbf{j}\) | A1F | Correct expression based on \(t\) from part (c) |
| \(\theta = \tan^{-1}\left(\frac{62.5}{78.125}\right)\) | dM1 | Using tan to find the angle |
| \(= 038.7°\) | A1F | Correct expression based on \(t\) from part (c), with correct two values (either way) |
| A1 | Correct angle. Accept \(38.6°\) or \(039°\) | |
| OR | ||
| \(\mathbf{r} = \frac{1}{2}(5\mathbf{i} + 6.25\mathbf{j}) \times 25\) | (M1) (A1F) (A1) (dM1) (A1F) (A1) | |
| \(\theta = \tan^{-1}\left(\frac{6.25}{6.25}\right) = 038.7°\) | ||
| Total: 12 marks |
**8(a)**
| $\mathbf{u} = 5\mathbf{i}$ or $\begin{bmatrix} 5 \\ 0 \end{bmatrix}$ | B1 | Correct velocity |
| **Total: 1 mark** | | |
**8(b)**
| $\mathbf{v} = 5\mathbf{i} + (-0.2\mathbf{i} + 0.25\mathbf{j})t$ | M1 | Use of constant acceleration equation, with $\mathbf{u}$ and $\mathbf{a}$ not zero |
| | A1 | Correct velocity. M1A0 for using 5j or just 5 |
| **OR** | | |
| $\mathbf{v} = \begin{bmatrix} 5 - 0.2t \\ 0.25t \end{bmatrix}$ | | |
| **Total: 2 marks** | | |
**8(c)**
| $5 - 0.2t = 0$ | M1 | Easterly component zero |
| $t = \frac{5}{0.2} = 25 \text{ seconds}$ | A1 | Correct equation |
| | A1 | Correct $t$ |
| **Total: 3 marks** | | |
**8(d)**
| $\mathbf{r} = 5\mathbf{i} \times 25 + \frac{1}{2}(-0.2\mathbf{i} + 0.25\mathbf{j}) \times 25^2$ | M1 | Use of constant acceleration equation with $t$ from part (c) |
| $= 62.5 + 78.125\mathbf{j}$ | A1F | Correct expression based on $t$ from part (c) |
| $\theta = \tan^{-1}\left(\frac{62.5}{78.125}\right)$ | dM1 | Using tan to find the angle |
| $= 038.7°$ | A1F | Correct expression based on $t$ from part (c), with correct two values (either way) |
| | A1 | Correct angle. Accept $38.6°$ or $039°$ |
| **OR** | | |
| $\mathbf{r} = \frac{1}{2}(5\mathbf{i} + 6.25\mathbf{j}) \times 25$ | (M1) (A1F) (A1) (dM1) (A1F) (A1) | |
| $\theta = \tan^{-1}\left(\frac{6.25}{6.25}\right) = 038.7°$ | | |
| **Total: 12 marks** | | |
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**TOTAL FOR PAPER: 75 marks**8 A boat is initially at the origin, heading due east at $5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. It then experiences a constant acceleration of $( - 0.2 \mathbf { i } + 0.25 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }$. The unit vectors $\mathbf { i }$ and $\mathbf { j }$ are directed east and north respectively.
\begin{enumerate}[label=(\alph*)]
\item State the initial velocity of the boat as a vector.
\item Find an expression for the velocity of the boat $t$ seconds after it has started to accelerate.
\item Find the value of $t$ when the boat is travelling due north.
\item Find the bearing of the boat from the origin when the boat is travelling due north.
\end{enumerate}
\hfill \mbox{\textit{AQA M1 2007 Q8 [12]}}