| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach of two objects |
| Difficulty | Moderate -0.3 This is a standard M3 relative velocity question with straightforward vector subtraction, position vector formation, and closest approach using calculus or dot product. All steps follow routine procedures with no novel problem-solving required, making it slightly easier than average for A-level. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| \(_Av_B = (12i - 8j) - (6i + 12j)\) | M1 | |
| \(= 6i - 20j\) | A1 | Needs to be in terms of \(i\) and \(j\) |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| \(_Ar_B = r_0 + v_Bt\) | M1A1 | Attempted use |
| \(_Ar_B = (18i + 5j) - (5i - j) + (6i - 20j)t\) | A1F | |
| \(_Ar_B = (13 + 6t)i + (6 - 20t)j\) | A1 | AG (not penalised if not in terms of \(i\) and \(j\)) |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(s^2 = (13 + 6t)^2 + (6 - 20t)^2\) | M1A1F | Attempt for squaring and tidying up |
| \(A \text{ and } B \text{ are closest when } \frac{ds}{dt} = 0\) or \(\frac{ds^2}{dt} = 0\) | M1 | |
| \(2x\frac{ds}{dt} = 2(13 + 6t)6 - 2(6 - 20t)20 = 0\) | M1A1 | Accuracy of differentiation |
| \(t = 0.0963\) (or \(0.096\) or \(\frac{21}{218}\)) | A1F | |
| 6 |
| Answer | Marks | Guidance |
|---|---|---|
| \(s = \sqrt{(13 + 6 \times 0.0963)^2 + (6 - 20 \times 0.0963)^2}\) | m1 | Dependent on M1s in part (c) |
| \(s = 14.2 \text{ km}\) | A1F | AWRT |
| 2 | ||
| 14 | Total |
### Part (a)
$_Av_B = (12i - 8j) - (6i + 12j)$ | M1 |
$= 6i - 20j$ | A1 | Needs to be in terms of $i$ and $j$
| | **2** |
### Part (b)
$_Ar_B = r_0 + v_Bt$ | M1A1 | Attempted use
$_Ar_B = (18i + 5j) - (5i - j) + (6i - 20j)t$ | A1F |
$_Ar_B = (13 + 6t)i + (6 - 20t)j$ | A1 | AG (not penalised if not in terms of $i$ and $j$)
| | **4** |
### Part (c)
$s^2 = (13 + 6t)^2 + (6 - 20t)^2$ | M1A1F | Attempt for squaring and tidying up
$A \text{ and } B \text{ are closest when } \frac{ds}{dt} = 0$ or $\frac{ds^2}{dt} = 0$ | M1 |
$2x\frac{ds}{dt} = 2(13 + 6t)6 - 2(6 - 20t)20 = 0$ | M1A1 | Accuracy of differentiation
$t = 0.0963$ (or $0.096$ or $\frac{21}{218}$) | A1F |
| | **6** |
### Part (d)
$s = \sqrt{(13 + 6 \times 0.0963)^2 + (6 - 20 \times 0.0963)^2}$ | m1 | Dependent on M1s in part (c)
$s = 14.2 \text{ km}$ | A1F | AWRT
| | **2** |
| | **14** | **Total** |4 The unit vectors $\mathbf { i }$ and $\mathbf { j }$ are directed due east and due north respectively.\\
Two cyclists, Aazar and Ben, are cycling on straight horizontal roads with constant velocities of $( 6 \mathbf { i } + 12 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }$ and $( 12 \mathbf { i } - 8 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }$ respectively. Initially, Aazar and Ben have position vectors $( 5 \mathbf { i } - \mathbf { j } ) \mathrm { km }$ and $( 18 \mathbf { i } + 5 \mathbf { j } ) \mathrm { km }$ respectively, relative to a fixed origin.
\begin{enumerate}[label=(\alph*)]
\item Find, as a vector in terms of $\mathbf { i }$ and $\mathbf { j }$, the velocity of Ben relative to Aazar.
\item The position vector of Ben relative to Aazar at time $t$ hours after they start is $\mathbf { r } \mathrm { km }$.
Show that
$$\mathbf { r } = ( 13 + 6 t ) \mathbf { i } + ( 6 - 20 t ) \mathbf { j }$$
\item Find the value of $t$ when Aazar and Ben are closest together.
\item Find the closest distance between Aazar and Ben.
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2006 Q4 [10]}}