| Exam Board | Edexcel |
|---|---|
| Module | M4 (Mechanics 4) |
| Year | 2012 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | When is one object due north/east/west/south of another |
| Difficulty | Standard +0.8 This M4 mechanics problem requires setting up position vectors for two moving ships, using bearing conversions, solving simultaneous equations to find when a specific geometric configuration occurs, then determining a future configuration. It demands careful coordinate geometry, vector manipulation, and multi-step reasoning beyond standard textbook exercises, though it follows a recognizable interception problem structure. |
| 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 |
|---|---|---|
| \(\mathbf{r}_A = (6\sin 30i + 6\cos 30j)\) | M1 A1 | Express the original relative positions in component (vector) form – one term correct. Both terms correct (substitution of trig values not required). |
| \(= (3i + 3\sqrt{3}j)\) | ||
| \(\mathbf{r}_B = v ti\) or \(v_B = vi\) | B1 | Position of \(B\) at time \(t\) (seen or implied) |
| \((v-4)i + (4\sqrt{3})j\) | M1 A1 | Express the relative velocity in component form – one term correct. Both terms correct |
| or \((v-8\sin 30)i + (8\cos 30)j\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(-3\sqrt{3} + 4\sqrt{3}t = -2\sqrt{3} \Rightarrow t = \frac{1}{4}\) | M1 A1 | Compare j displacement with \(±2\sqrt{3}\) and solve for \(t\) cao |
| \(vt - 3 - 4t = 0 \Rightarrow v = 16\) | M1 A1 | Equate i displacement to zero and substitute their value of \(t\). cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(-3\sqrt{3} + 4\sqrt{3}t = 0 \Rightarrow t = \frac{3}{4}\) i.e. at 12.45 pm | M1 A1 | Equate j displacement to zero and solve for \(t\). Any equivalent form for the time. |
| then distance \(AB = 16x\frac{3}{4} - 3 - 4x\frac{3}{4} = 6\) km. | M1 A1 | Substitute their v & t in the i displacement and evaluate cao. Must be a scalar. |
| Answer | Marks | Guidance |
|---|---|---|
| \(BC^2 = 36 + 12 - 2 \times 6 \times 2\sqrt{3}\cos 30\) | M1A1 | The given information provides us with two triangles - velocities in bold. Fix \(A\) and \(B\) follows the path \(BP\). \(C\) is the point when \(B\) is due South of \(A\), and \(P\) when it is due East. |
| Solve for \(BC\), \(\angle ABC\) | M1A1 | |
| \(BC = 2\sqrt{3},\) \(\to\) triangle is isosceles | ||
| \(\angle B\) in velocity triangle is \(30°\) | B1 | |
| Trig in rt \(\triangle\) gives relative velocity \(= 8 \times \tan 60° = 8\sqrt{3}\) | M1A1 | |
| \(\angle APB = 30°\) (angles of a triangle) so triangle is isosceles and distance \(AP = 6\)km | M1A1 | |
| Using cosine rule or symmetry of isosceles triangle, distance \(BP = 6\sqrt{3}\) | M1A1 | |
| Time taken \(= \frac{6\sqrt{3}}{8\sqrt{3}} = \frac{3}{4}\) hr, time is now 12.45 | M1A1 |
With $B$ as origin.
| $\mathbf{r}_A = (6\sin 30i + 6\cos 30j)$ | M1 A1 | Express the original relative positions in component (vector) form – one term correct. Both terms correct (substitution of trig values not required). |
| $= (3i + 3\sqrt{3}j)$ | | |
| $\mathbf{r}_B = v ti$ or $v_B = vi$ | B1 | Position of $B$ at time $t$ (seen or implied) |
| $(v-4)i + (4\sqrt{3})j$ | M1 A1 | Express the relative velocity in component form – one term correct. Both terms correct |
| or $(v-8\sin 30)i + (8\cos 30)j$ | | |
When $B$ is $2\sqrt{3}$ km south of $A$,
| $-3\sqrt{3} + 4\sqrt{3}t = -2\sqrt{3} \Rightarrow t = \frac{1}{4}$ | M1 A1 | Compare j displacement with $±2\sqrt{3}$ and solve for $t$ cao |
| $vt - 3 - 4t = 0 \Rightarrow v = 16$ | M1 A1 | Equate i displacement to zero and substitute their value of $t$. cao |
When $B$ is due east of $A$,
| $-3\sqrt{3} + 4\sqrt{3}t = 0 \Rightarrow t = \frac{3}{4}$ i.e. at 12.45 pm | M1 A1 | Equate j displacement to zero and solve for $t$. Any equivalent form for the time. |
| then distance $AB = 16x\frac{3}{4} - 3 - 4x\frac{3}{4} = 6$ km. | M1 A1 | Substitute their v & t in the i displacement and evaluate cao. Must be a scalar. |
**OR (Geometrical alternative)**
| $BC^2 = 36 + 12 - 2 \times 6 \times 2\sqrt{3}\cos 30$ | M1A1 | The given information provides us with two triangles - velocities in bold. Fix $A$ and $B$ follows the path $BP$. $C$ is the point when $B$ is due South of $A$, and $P$ when it is due East. |
| Solve for $BC$, $\angle ABC$ | M1A1 | |
| $BC = 2\sqrt{3},$ $\to$ triangle is isosceles | | |
| $\angle B$ in velocity triangle is $30°$ | B1 | |
| Trig in rt $\triangle$ gives relative velocity $= 8 \times \tan 60° = 8\sqrt{3}$ | M1A1 | |
| $\angle APB = 30°$ (angles of a triangle) so triangle is isosceles and distance $AP = 6$km | M1A1 | |
| Using cosine rule or symmetry of isosceles triangle, distance $BP = 6\sqrt{3}$ | M1A1 | |
| Time taken $= \frac{6\sqrt{3}}{8\sqrt{3}} = \frac{3}{4}$ hr, time is now 12.45 | M1A1 | |
**(13) Total for Question 2: 13 marks**
---\begin{enumerate}
\item A $\operatorname { ship } A$ is moving at a constant speed of $8 \mathrm {~km} \mathrm {~h} \mathrm {~h} ^ { - 1 }$ on a bearing of $150 ^ { \circ }$. At noon a second ship $B$ is 6 km from $A$, on a bearing of $210 ^ { \circ }$. Ship $B$ is moving due east at a constant speed. At a later time, $B$ is $2 \sqrt { 3 } \mathrm {~km}$ due south of $A$.
\end{enumerate}
Find\\
(i) the time at which $B$ will be due east of $A$,\\
(ii) the distance between the ships at that time.
\hfill \mbox{\textit{Edexcel M4 2012 Q2 [13]}}