| Exam Board | AQA |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2007 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | River crossing: perpendicular heading or minimum time (find drift and/or time) |
| Difficulty | Moderate -0.8 This is a straightforward relative velocity problem requiring vector addition using Pythagoras and trigonometry. All steps are routine M1 mechanics: find resultant magnitude (√(0.3² + 0.1²)), find angle (tan⁻¹(0.1/0.3)), calculate time (15/0.3), and distance (time × resultant speed). No problem-solving insight needed, just direct application of standard formulas. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors3.02d Constant acceleration: SUVAT formulae |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(v = \sqrt{0.3^2 + 0.1^2} = \sqrt{0.1} = 0.316 \text{ ms}^{-1}\) | M1 A1 | Total: 2 — Use of Pythagoras to find \(v\). Correct \(v\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\alpha = \tan^{-1}\left(\frac{0.3}{0.1}\right) = 71.6°\) | M1 A1 | Use of trigonometry with reasonable choice of sides to find \(\alpha\). Correct expression |
| A1 | Total: 3 — Correct angle CAO |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(t = \frac{15}{0.3} = 50\text{s}\) | M1 | Use of \(s/v\) to find \(t\) with \(s\) and \(t\) consistent |
| A1 | Total: 2 — Correct \(t\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(s = 50 \times \sqrt{0.1} = 15.8\text{m}\) | M1 A1 | Total: 2 — Use of their \(t\) in \(t \times v\) to find \(s\) or the use of trigonometry. Correct distance CAO |
## Question 5:
### Part (a)
| Working | Marks | Guidance |
|---------|-------|----------|
| $v = \sqrt{0.3^2 + 0.1^2} = \sqrt{0.1} = 0.316 \text{ ms}^{-1}$ | M1 A1 | **Total: 2** — Use of Pythagoras to find $v$. Correct $v$ |
### Part (b)
| Working | Marks | Guidance |
|---------|-------|----------|
| $\alpha = \tan^{-1}\left(\frac{0.3}{0.1}\right) = 71.6°$ | M1 A1 | Use of trigonometry with reasonable choice of sides to find $\alpha$. Correct expression |
| | A1 | **Total: 3** — Correct angle **CAO** |
### Part (i)
| Working | Marks | Guidance |
|---------|-------|----------|
| $t = \frac{15}{0.3} = 50\text{s}$ | M1 | Use of $s/v$ to find $t$ with $s$ and $t$ consistent |
| | A1 | **Total: 2** — Correct $t$ |
### Part (ii)
| Working | Marks | Guidance |
|---------|-------|----------|
| $s = 50 \times \sqrt{0.1} = 15.8\text{m}$ | M1 A1 | **Total: 2** — Use of their $t$ in $t \times v$ to find $s$ or the use of trigonometry. Correct distance **CAO** |
---5 A girl in a boat is rowing across a river, in which the water is flowing at $0.1 \mathrm {~ms} ^ { - 1 }$. The velocity of the boat relative to the water is $0.3 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and is perpendicular to the bank, as shown in the diagram.\\
\includegraphics[max width=\textwidth, alt={}, center]{965a176a-848c-478d-a748-80fc9dfe4399-4_314_1152_468_450}
\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the resultant velocity of the boat.
\item Find the acute angle between the resultant velocity and the bank.
\item The width of the river is 15 metres.
\begin{enumerate}[label=(\roman*)]
\item Find the time that it takes the boat to cross the river.
\item Find the total distance travelled by the boat as it crosses the river.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{AQA M1 2007 Q5 [9]}}