| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2012 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Interception: find bearing/direction to intercept (exact intercept) |
| Difficulty | Challenging +1.2 This is a standard M4 relative velocity problem requiring vector resolution and optimization. While it involves multiple parts and careful handling of current effects, the techniques (resolving velocities, minimizing time using calculus or geometry) are well-practiced at this level. The problem is more computational than conceptually challenging, placing it moderately above average difficulty. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| \(\frac{\sin\theta}{\sin 45°} = \frac{5}{20}\) | M1 A1 | Use a vector triangle to find \(\theta\). Condone the 5 ms⁻¹ in the wrong direction. Correct equation for \(\theta\) |
| \(\theta = 10.182...\) Bearing is \(45° - \theta = 34.8 = 35°\) (nearest degree) | M1 A1 | Use their angle correctly in their triangle to find the bearing. Accept alternative forms e.g. N 35 E 45° rt angle triangle t substitution leading to correct equation in \(t\), use of \(R\cos(\theta + \alpha)\) o.e. |
| Answer | Marks |
|---|---|
| \(SW \to (20\sin\theta)T = (5 + 20\cos\theta)T\) | M1 A1 |
| \(3t^2 + 8t - 5 = 0, t = \frac{-8 + \sqrt{124}}{6} = 0.5225...\) | A1 |
| \(\theta = 55.18...\) Bearing is \(90° - \theta = 34.8°\) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(v^2 = 5^2 + 20^2 - 2 \times 5 \times 20\cos 124.818...\) | M1 | Complete method to find \(v\) |
| \(OR\ v = \frac{20}{\sin 45°} \times \sin 124.8\) | ||
| \(OR\ v = 5\cos 45° + 20\cos\theta\) | ||
| \(v = 23.22\) | A1 | Or better \(\left(\frac{5\sqrt{2} + 5\sqrt{62}}{2}\right)\) |
| \(t = \frac{15}{23.22} = 0.646\) h = 39 min (nearest min) | M1 A1 | \(\frac{15}{\text{their }v}\) The Q specifies "nearest minute" |
| Answer | Marks | Guidance |
|---|---|---|
| Due N. (since current affects both equally) | B1 | (1) cao eso |
| Answer | Marks |
|---|---|
| \(t = \frac{4}{20} = 0.2\) h = 12 min | B1 |
**Part (a)**
| $\frac{\sin\theta}{\sin 45°} = \frac{5}{20}$ | M1 A1 | Use a vector triangle to find $\theta$. Condone the 5 ms⁻¹ in the wrong direction. Correct equation for $\theta$ |
| $\theta = 10.182...$ Bearing is $45° - \theta = 34.8 = 35°$ (nearest degree) | M1 A1 | Use their angle correctly in their triangle to find the bearing. Accept alternative forms e.g. N 35 E 45° rt angle triangle t substitution leading to correct equation in $t$, use of $R\cos(\theta + \alpha)$ o.e. |
**OR**
| $SW \to (20\sin\theta)T = (5 + 20\cos\theta)T$ | M1 A1 | |
| $3t^2 + 8t - 5 = 0, t = \frac{-8 + \sqrt{124}}{6} = 0.5225...$ | A1 | |
| $\theta = 55.18...$ Bearing is $90° - \theta = 34.8°$ | M1A1 | |
**(4) marks**(13)
**Part (b)**
| $v^2 = 5^2 + 20^2 - 2 \times 5 \times 20\cos 124.818...$ | M1 | Complete method to find $v$ |
| $OR\ v = \frac{20}{\sin 45°} \times \sin 124.8$ | | |
| $OR\ v = 5\cos 45° + 20\cos\theta$ | | |
| $v = 23.22$ | A1 | Or better $\left(\frac{5\sqrt{2} + 5\sqrt{62}}{2}\right)$ |
| $t = \frac{15}{23.22} = 0.646$ h = 39 min (nearest min) | M1 A1 | $\frac{15}{\text{their }v}$ The Q specifies "nearest minute" |
**(4) marks**(13)
**Part (c)**
| Due N. (since current affects both equally) | B1 | (1) cao eso |
**(1)10 marks total for Question 4**
**Part (d)**
| $t = \frac{4}{20} = 0.2$ h = 12 min | B1 | |
**(1)10 marks**
---4. A rescue boat, whose maximum speed is $20 \mathrm {~km} \mathrm {~h} ^ { - 1 }$, receives a signal which indicates that a yacht is in distress near a fixed point $P$. The rescue boat is 15 km south-west of $P$. There is a constant current of $5 \mathrm {~km} \mathrm {~h} ^ { - 1 }$ flowing uniformly from west to east. The rescue boat sets the course needed to get to $P$ as quickly as possible. Find
\begin{enumerate}[label=(\alph*)]
\item the course the rescue boat sets,
\item the time, to the nearest minute, to get to $P$.
When the rescue boat arrives at $P$, the yacht is just visible 4 km due north of $P$ and is drifting with the current. Find
\item the course that the rescue boat should set to get to the yacht as quickly as possible,
\item the time taken by the rescue boat to reach the yacht from $P$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M4 2012 Q4 [10]}}