| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2007 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors 3D & Lines |
| Type | Kinematics with position vectors |
| Difficulty | Standard +0.3 This is a standard M3 relative velocity question requiring routine application of formulas: finding relative velocity by subtraction, writing position as a function of time, then minimizing distance using calculus (or recognizing perpendicularity). The multi-step nature and 3D vectors add slight complexity, but the techniques are well-practiced and algorithmic with no novel insight required. |
| 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 vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(_B\mathbf{v}_A = \mathbf{v}_A - \mathbf{v}_B = (20\mathbf{i} - 10\mathbf{j} + 20\mathbf{k}) - (30\mathbf{i} + 10\mathbf{j} + 10\mathbf{k})\) | M1A1 | Simplification not necessary; Total: 2 marks |
| \(= -10\mathbf{i} - 20\mathbf{j} + 10\mathbf{k}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(_B\mathbf{r}_{0A} = (8000\mathbf{i} + 1500\mathbf{j} + 3000\mathbf{k}) - (2000\mathbf{i} + 500\mathbf{j} + 1000\mathbf{k})\) | M1 | |
| \(= 6000\mathbf{i} + 1000\mathbf{j} + 2000\mathbf{k}\) | ||
| \(_B\mathbf{r}_A = (6000\mathbf{i} + 1000\mathbf{j} + 2000\mathbf{k}) + (-10\mathbf{i} - 20\mathbf{j} + 10\mathbf{k})t\) | M1, A1F | Simplification not necessary; Total: 3 marks |
| \(_B\mathbf{r}_A = (6000 - 10t)\mathbf{i} + (1000 - 20t)\mathbf{j} + (2000 + 10t)\mathbf{k}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\ | _B\mathbf{r}_A\ | ^2 = (6000-10t)^2 + (1000-20t)^2 + (2000+10t)^2\) |
| Minimise \(y = (6000-10t)^2 + (1000-20t)^2 + (2000+10t)^2\) | ||
| \(\frac{dy}{dt} = 2(-10)(6000-10t) + 2(-20)(1000-20t) + 2(10)(2000+10t) = 0\) | m1, A1F | |
| \(t = 100\) | A1F | Total: 5 marks |
| Alternative: \(\begin{pmatrix} 6000-10t \\ 1000-20t \\ 2000+10t \end{pmatrix} \cdot \begin{pmatrix} -10 \\ -20 \\ 10 \end{pmatrix} = 0\) | (M1)(A1F) | |
| \(-60000 + 100t - 20000 + 400t + 20000 + 100t = 0\) | (m1)(A1F) | |
| \(600t = 60000\), \(t = 100\) | (A1F) | (5) |
## Question 2:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $_B\mathbf{v}_A = \mathbf{v}_A - \mathbf{v}_B = (20\mathbf{i} - 10\mathbf{j} + 20\mathbf{k}) - (30\mathbf{i} + 10\mathbf{j} + 10\mathbf{k})$ | M1A1 | Simplification not necessary; Total: 2 marks |
| $= -10\mathbf{i} - 20\mathbf{j} + 10\mathbf{k}$ | | |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $_B\mathbf{r}_{0A} = (8000\mathbf{i} + 1500\mathbf{j} + 3000\mathbf{k}) - (2000\mathbf{i} + 500\mathbf{j} + 1000\mathbf{k})$ | M1 | |
| $= 6000\mathbf{i} + 1000\mathbf{j} + 2000\mathbf{k}$ | | |
| $_B\mathbf{r}_A = (6000\mathbf{i} + 1000\mathbf{j} + 2000\mathbf{k}) + (-10\mathbf{i} - 20\mathbf{j} + 10\mathbf{k})t$ | M1, A1F | Simplification not necessary; Total: 3 marks |
| $_B\mathbf{r}_A = (6000 - 10t)\mathbf{i} + (1000 - 20t)\mathbf{j} + (2000 + 10t)\mathbf{k}$ | | |
### Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\|_B\mathbf{r}_A\|^2 = (6000-10t)^2 + (1000-20t)^2 + (2000+10t)^2$ | M1, A1F | |
| Minimise $y = (6000-10t)^2 + (1000-20t)^2 + (2000+10t)^2$ | | |
| $\frac{dy}{dt} = 2(-10)(6000-10t) + 2(-20)(1000-20t) + 2(10)(2000+10t) = 0$ | m1, A1F | |
| $t = 100$ | A1F | Total: 5 marks |
| **Alternative:** $\begin{pmatrix} 6000-10t \\ 1000-20t \\ 2000+10t \end{pmatrix} \cdot \begin{pmatrix} -10 \\ -20 \\ 10 \end{pmatrix} = 0$ | (M1)(A1F) | |
| $-60000 + 100t - 20000 + 400t + 20000 + 100t = 0$ | (m1)(A1F) | |
| $600t = 60000$, $t = 100$ | (A1F) | (5) |
---2 The unit vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ are directed due east, due north and vertically upwards respectively.
Two helicopters, $A$ and $B$, are flying with constant velocities of $( 20 \mathbf { i } - 10 \mathbf { j } + 20 \mathbf { k } ) \mathrm { ms } ^ { - 1 }$ and $( 30 \mathbf { i } + 10 \mathbf { j } + 10 \mathbf { k } ) \mathrm { m } \mathrm { s } ^ { - 1 }$ respectively. At noon, the position vectors of $A$ and $B$ relative to a fixed origin, $O$, are $( 8000 \mathbf { i } + 1500 \mathbf { j } + 3000 \mathbf { k } ) \mathrm { m }$ and $( 2000 \mathbf { i } + 500 \mathbf { j } + 1000 \mathbf { k } ) \mathrm { m }$ respectively.
\begin{enumerate}[label=(\alph*)]
\item Write down the velocity of $A$ relative to $B$.
\item Find the position vector of $A$ relative to $B$ at time $t$ seconds after noon.
\item Find the value of $t$ when $A$ and $B$ are closest together.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2007 Q2 [10]}}