| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2004 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | River crossing: reach point directly opposite (find angle and/or time) |
| Difficulty | Standard +0.3 This is a standard relative velocity problem involving vector components. Part (a) requires resolving velocities to find an angle using basic trigonometry, (b) is straightforward time calculation, while (c-d) extend to a two-region river requiring tracking position but using the same fundamental techniques. The problem is slightly easier than average as it's methodical application of standard M4 relative velocity methods without requiring novel insight or complex problem-solving. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Vector ! or \(\leftarrow\) | M1 | |
| \(\cos\theta = \frac{0.4}{0.85}\) | M1 | |
| \(\theta \approx 61.9°\) | A1 (awrt 62°) | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(u = \sqrt{(0.85^2 - 0.4^2)}\) or \(u = 0.85\sin\theta\) | M1 | |
| \(t = \frac{60}{u} = \frac{60}{0.75} = 80\) (s) | M1, A1 (cao) | (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v_{N \text{ rel } W} = -0.4i + (0.75i)\) | M1 | |
| Allow for \(\pm0.4i\) | ||
| \(v_N = v_{N \text{ rel } W} + 0.5i = 0.1i + (0.75i)\) | A1 | |
| \(t = \frac{40}{0.75} = \frac{160}{3}\) | M1 | |
| \(\delta = 0.1 \times \frac{160}{3} = \frac{16}{3}\) (awrt 5.3) | M1, A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| As in (c) \(v_N = -0.2i + 0.75i\) | M1 (±0.2i) | |
| \(t = \frac{20}{0.75} = \frac{80}{3}\) | M1 | |
| \(\delta = 0.2 \times \frac{80}{3} = \frac{16}{3}\) | M1 | |
| Hence \(N\) lands at \(D\) | A1 (cso) | (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(v^2 = 0.5^2 + 0.85^2 - 2 \times 0.5 \times 0.85 \times \cos\theta\) | M1 | |
| \(= 0.5725\) (\(v = \frac{\sqrt{229}}{20} \approx 82.4°\)) | ||
| \(\frac{\sin\varphi}{\sin\theta} = \frac{0.85}{v}\) | M1 | |
| \(\sin\varphi = \frac{15}{\sqrt{229}}\) (\(\approx 0.9912; \varphi \approx 82.4°\)) | A1 | |
| \(\frac{\delta}{40} = \cot\varphi\); \(\delta = 40 \times \frac{2}{5} = \frac{16}{3}\) (awrt 5.3) | M1, A1 | (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(w^2 = 0.2^2 + 0.85^2 - 2 \times 0.2 \times 0.85 \times \cos\theta = 0.6025\) | M1 | |
| (\(w = \frac{\sqrt{241}}{20} \approx 0.7762\ldots\)) | ||
| \(\frac{\sin\psi}{\sin\theta} = \frac{0.85}{w}\) | M1 | |
| \(\sin\psi = \frac{15}{\sqrt{241}}\) (\(\approx 0.9662; \psi \approx 104.9°\)) | M1 | |
| \(\psi = 75.1°\) gains M1 | ||
| \(\frac{\varepsilon}{20} = \cot(180° - \psi) = \frac{4}{15}\) | ||
| \(\varepsilon = \frac{16}{3} = \delta\) | M1 | |
| Hence \(N\) lands at \(D\) | A1 | (4 marks) |
## Part (a)
Vector ! or $\leftarrow$ | M1 |
$\cos\theta = \frac{0.4}{0.85}$ | M1 |
$\theta \approx 61.9°$ | A1 (awrt 62°) | (3 marks)
## Part (b)
$u = \sqrt{(0.85^2 - 0.4^2)}$ or $u = 0.85\sin\theta$ | M1 |
$t = \frac{60}{u} = \frac{60}{0.75} = 80$ (s) | M1, A1 (cao) | (3 marks)
## Part (c)
$v_{N \text{ rel } W} = -0.4i + (0.75i)$ | M1 |
Allow for $\pm0.4i$ | |
$v_N = v_{N \text{ rel } W} + 0.5i = 0.1i + (0.75i)$ | A1 |
$t = \frac{40}{0.75} = \frac{160}{3}$ | M1 |
$\delta = 0.1 \times \frac{160}{3} = \frac{16}{3}$ (awrt 5.3) | M1, A1 | (5 marks)
## Part (d)
As in (c) $v_N = -0.2i + 0.75i$ | M1 (±0.2i) |
$t = \frac{20}{0.75} = \frac{80}{3}$ | M1 |
$\delta = 0.2 \times \frac{80}{3} = \frac{16}{3}$ | M1 |
Hence $N$ lands at $D$ | A1 (cso) | (4 marks)
**Total: 15 marks**
**Notes:**
1. In (c) and (d), candidates can take components without using vectors. Mark as vector method.
2. After the first line in (d), the result is clear by proportion. Allow as long as some explanation given.
3. $\cos\theta = \frac{8}{17} = 0.4705\ldots$, $\sin\theta = \frac{15}{17} = 0.8823\ldots$
4. Alternatives to (c) and (d) using vector triangles are given on next page.
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# Question 6 - Alternatives to (c) and (d):
## Part (c) - Vector Triangle Method:
$v^2 = 0.5^2 + 0.85^2 - 2 \times 0.5 \times 0.85 \times \cos\theta$ | M1 |
$= 0.5725$ ($v = \frac{\sqrt{229}}{20} \approx 82.4°$) | |
$\frac{\sin\varphi}{\sin\theta} = \frac{0.85}{v}$ | M1 |
$\sin\varphi = \frac{15}{\sqrt{229}}$ ($\approx 0.9912; \varphi \approx 82.4°$) | A1 |
$\frac{\delta}{40} = \cot\varphi$; $\delta = 40 \times \frac{2}{5} = \frac{16}{3}$ (awrt 5.3) | M1, A1 | (5 marks)
## Part (d) - Vector Triangle Method:
$w^2 = 0.2^2 + 0.85^2 - 2 \times 0.2 \times 0.85 \times \cos\theta = 0.6025$ | M1 |
($w = \frac{\sqrt{241}}{20} \approx 0.7762\ldots$) | |
$\frac{\sin\psi}{\sin\theta} = \frac{0.85}{w}$ | M1 |
$\sin\psi = \frac{15}{\sqrt{241}}$ ($\approx 0.9662; \psi \approx 104.9°$) | M1 |
$\psi = 75.1°$ gains M1 | |
$\frac{\varepsilon}{20} = \cot(180° - \psi) = \frac{4}{15}$ | |
$\varepsilon = \frac{16}{3} = \delta$ | M1 |
Hence $N$ lands at $D$ | A1 | (4 marks)
**Note:** Exact working is needed for final A1 but all previous marks in (c) and (d) may be gained by approximate working.\includegraphics{figure_4}
Mary swims in still water at 0.85 m s$^{-1}$. She swims across a straight river which is 60 m wide and flowing at 0.4 m s$^{-1}$. She sets off from a point $A$ on the near bank and lands at a point $B$, which is directly opposite $A$ on the far bank, as shown in Fig. 4.
Find
\begin{enumerate}[label=(\alph*)]
\item the angle between the near bank and the direction in which Mary swims,
[3]
\item the time she takes to cross the river.
[3]
\end{enumerate}
\includegraphics{figure_5}
A little further downstream a large tree has fallen from the far bank into the river. The river is modelled as flowing at 0.5 m s$^{-1}$ for a width of 40 m from the near bank, and 0.2 m s$^{-1}$ for the 20 m beyond this. Nassim swims at 0.85 m s$^{-1}$ in still water. He swims across the river from a point $C$ on the near bank. The point $D$ on the far bank is directly opposite $C$, as shown in Fig. 5. Nassim swims at the same angle to the near bank as Mary.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Find the maximum distance, downstream from $CD$, of Nassim during the crossing.
[5]
\item Show that he will land at the point $D$.
[4]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2004 Q6 [15]}}