| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 15 |
| 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 techniques: converting direction vectors to velocities using speed/magnitude, finding relative velocity, expressing relative position as a function of time, and minimizing distance using calculus. All steps are textbook procedures with no novel insight required, making it slightly easier than average. |
| 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 |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Direction of A: \(-2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}\), magnitude \(= \sqrt{4+9+36} = 7\) | M1 | Finding magnitude of direction vector for A |
| Unit vector for A: \(\frac{1}{7}(-2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k})\) | A1 | Correct unit vector |
| Velocity of A \(= \frac{140}{7}(-2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}) = -40\mathbf{i} + 60\mathbf{j} + 120\mathbf{k}\) | A1 | Correct velocity of A |
| Direction of B: \(2\mathbf{i} - \mathbf{j} + 2\mathbf{k}\), magnitude \(= \sqrt{4+1+4} = 3\); Velocity of B \(= \frac{60}{3}(2\mathbf{i} - \mathbf{j} + 2\mathbf{k}) = 40\mathbf{i} - 20\mathbf{j} + 40\mathbf{k}\) | M1 | Finding velocity of B |
| \(\mathbf{v}_{A} - \mathbf{v}_{B} = (-40-40)\mathbf{i} + (60+20)\mathbf{j} + (120-40)\mathbf{k} = -80\mathbf{i} + 80\mathbf{j} + 80\mathbf{k}\) | A1 | Correct relative velocity |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Initial relative position: \((4-(-3))\mathbf{i} + (-2-6)\mathbf{j} + (3-3)\mathbf{k} = 7\mathbf{i} - 8\mathbf{j} + 0\mathbf{k}\) | M1 | Subtracting initial position vectors |
| Position of A relative to B \(= (7-80t)\mathbf{i} + (-8+80t)\mathbf{j} + 80t\mathbf{k}\) | A1 | Correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(d^2 = (7-80t)^2 + (-8+80t)^2 + (80t)^2\) | M1 | Setting up expression for \(d^2\) |
| \(= 49 - 1120t + 6400t^2 + 64 - 1280t + 6400t^2 + 6400t^2\) | M1 | Expanding |
| \(= 19200t^2 - 2400t + 113\) | A1 | Correct simplified expression |
| \(\frac{d(d^2)}{dt} = 38400t - 2400 = 0\) | M1 | Differentiating and setting equal to zero |
| \(t = \frac{2400}{38400} = \frac{1}{16}\) | A1 | Correct value of \(t\) |
| \(d^2 = 19200\left(\frac{1}{16}\right)^2 - 2400\left(\frac{1}{16}\right) + 113\) | M1 | Substituting back |
| \(d^2 = 75 - 150 + 113 = 38\) | A1 | Correct value of \(d^2\) |
| \(d = \sqrt{38} \approx 6.16\) km | A1 | Correct distance |
# Question 4:
## Part (a): Find the velocity of A relative to B (5 marks)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Direction of A: $-2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}$, magnitude $= \sqrt{4+9+36} = 7$ | M1 | Finding magnitude of direction vector for A |
| Unit vector for A: $\frac{1}{7}(-2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k})$ | A1 | Correct unit vector |
| Velocity of A $= \frac{140}{7}(-2\mathbf{i} + 3\mathbf{j} + 6\mathbf{k}) = -40\mathbf{i} + 60\mathbf{j} + 120\mathbf{k}$ | A1 | Correct velocity of A |
| Direction of B: $2\mathbf{i} - \mathbf{j} + 2\mathbf{k}$, magnitude $= \sqrt{4+1+4} = 3$; Velocity of B $= \frac{60}{3}(2\mathbf{i} - \mathbf{j} + 2\mathbf{k}) = 40\mathbf{i} - 20\mathbf{j} + 40\mathbf{k}$ | M1 | Finding velocity of B |
| $\mathbf{v}_{A} - \mathbf{v}_{B} = (-40-40)\mathbf{i} + (60+20)\mathbf{j} + (120-40)\mathbf{k} = -80\mathbf{i} + 80\mathbf{j} + 80\mathbf{k}$ | A1 | Correct relative velocity |
## Part (b): Position vector of A relative to B (2 marks)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Initial relative position: $(4-(-3))\mathbf{i} + (-2-6)\mathbf{j} + (3-3)\mathbf{k} = 7\mathbf{i} - 8\mathbf{j} + 0\mathbf{k}$ | M1 | Subtracting initial position vectors |
| Position of A relative to B $= (7-80t)\mathbf{i} + (-8+80t)\mathbf{j} + 80t\mathbf{k}$ | A1 | Correct expression |
## Part (c): Distance between A and B when closest (8 marks)
| Working/Answer | Mark | Guidance |
|---|---|---|
| $d^2 = (7-80t)^2 + (-8+80t)^2 + (80t)^2$ | M1 | Setting up expression for $d^2$ |
| $= 49 - 1120t + 6400t^2 + 64 - 1280t + 6400t^2 + 6400t^2$ | M1 | Expanding |
| $= 19200t^2 - 2400t + 113$ | A1 | Correct simplified expression |
| $\frac{d(d^2)}{dt} = 38400t - 2400 = 0$ | M1 | Differentiating and setting equal to zero |
| $t = \frac{2400}{38400} = \frac{1}{16}$ | A1 | Correct value of $t$ |
| $d^2 = 19200\left(\frac{1}{16}\right)^2 - 2400\left(\frac{1}{16}\right) + 113$ | M1 | Substituting back |
| $d^2 = 75 - 150 + 113 = 38$ | A1 | Correct value of $d^2$ |
| $d = \sqrt{38} \approx 6.16$ km | A1 | Correct distance |4 The unit vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ are directed due east, due north and vertically upwards respectively.
A helicopter, $A$, is travelling in the direction of the vector $- 2 \mathbf { i } + 3 \mathbf { j } + 6 \mathbf { k }$ with constant speed $140 \mathrm {~km} \mathrm {~h} ^ { - 1 }$. Another helicopter, $B$, is travelling in the direction of the vector $2 \mathbf { i } - \mathbf { j } + 2 \mathbf { k }$ with constant speed $60 \mathrm {~km} \mathrm {~h} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the velocity of $A$ relative to $B$.
\item Initially, the position vectors of $A$ and $B$ are $( 4 \mathbf { i } - 2 \mathbf { j } + 3 \mathbf { k } ) \mathrm { km }$ and $( - 3 \mathbf { i } + 6 \mathbf { j } + 3 \mathbf { k } ) \mathrm { km }$ respectively, relative to a fixed origin.
Write down the position vector of $A$ relative to $B , t$ hours after they leave their initial positions.
\item Find the distance between $A$ and $B$ when they are closest together.\\
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{0590950d-145c-4ae2-bc3c-f61a9433d158-10_2486_1714_221_153}
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\begin{center}
\includegraphics[max width=\textwidth, alt={}]{0590950d-145c-4ae2-bc3c-f61a9433d158-11_2486_1714_221_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2011 Q4 [15]}}