| Exam Board | AQA |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2010 |
| 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 kinematics question with position vectors requiring routine application of formulas (r = r₀ + vt, relative position, closest approach via differentiation or dot product). Parts (a)-(c) are straightforward bookwork, while part (d) requires standard minimization technique. The multi-step nature and 3D vectors add slight complexity, but this is typical textbook material 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 notation1.10d 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 |
|---|---|---|
| \(\mathbf{r}_A = (-60 + 250t)\mathbf{i} + 50t\mathbf{j} + (30 - 100t)\mathbf{k}\) | B1 | Accept column vector form |
| \(\mathbf{r}_B = (-40 + 200t)\mathbf{i} + (10 + 25t)\mathbf{j} + (-10 + 50t)\mathbf{k}\) | B1 | |
| Both correct | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{r}_A - \mathbf{r}_B = [(-60+250t)-(-40+200t)]\mathbf{i} + [(50t)-(10+25t)]\mathbf{j} + [(30-100t)-(-10+50t)]\mathbf{k}\) | M1 | Subtract position vectors |
| \(= (-20+50t)\mathbf{i} + (-10+25t)\mathbf{j} + (40-150t)\mathbf{k}\) | A1 | Shown (given result) |
| Answer | Marks | Guidance |
|---|---|---|
| For collision need all three components zero simultaneously | M1 | |
| \(-20+50t = 0 \Rightarrow t = 0.4\) | A1 | |
| \(-10+25t = 0 \Rightarrow t = 0.4\) | A1 | |
| \(40-150t = 0 \Rightarrow t = \frac{4}{15}\) | A1 | Since \(t\) values differ, no collision |
| Answer | Marks | Guidance |
|---|---|---|
| \(d^2 = (-20+50t)^2 + (-10+25t)^2 + (40-150t)^2\) | M1 | |
| \(= (2500+625+22500)t^2 + (-2000-500-12000)t + (400+100+1600)\) | M1 | Expand |
| \(= 25625t^2 - 14500t + 2100\) | A1 | |
| \(\frac{d(d^2)}{dt} = 51250t - 14500 = 0\) | M1 | Differentiate and set to zero |
| \(t = \frac{14500}{51250} = \frac{58}{205} \approx 0.283\) hours | A1 A1 |
# Question 4:
## Part (a):
| $\mathbf{r}_A = (-60 + 250t)\mathbf{i} + 50t\mathbf{j} + (30 - 100t)\mathbf{k}$ | B1 | Accept column vector form |
| $\mathbf{r}_B = (-40 + 200t)\mathbf{i} + (10 + 25t)\mathbf{j} + (-10 + 50t)\mathbf{k}$ | B1 | |
| Both correct | B1 | |
## Part (b):
| $\mathbf{r}_A - \mathbf{r}_B = [(-60+250t)-(-40+200t)]\mathbf{i} + [(50t)-(10+25t)]\mathbf{j} + [(30-100t)-(-10+50t)]\mathbf{k}$ | M1 | Subtract position vectors |
| $= (-20+50t)\mathbf{i} + (-10+25t)\mathbf{j} + (40-150t)\mathbf{k}$ | A1 | Shown (given result) |
## Part (c):
| For collision need all three components zero simultaneously | M1 | |
| $-20+50t = 0 \Rightarrow t = 0.4$ | A1 | |
| $-10+25t = 0 \Rightarrow t = 0.4$ | A1 | |
| $40-150t = 0 \Rightarrow t = \frac{4}{15}$ | A1 | Since $t$ values differ, no collision |
## Part (d):
| $d^2 = (-20+50t)^2 + (-10+25t)^2 + (40-150t)^2$ | M1 | |
| $= (2500+625+22500)t^2 + (-2000-500-12000)t + (400+100+1600)$ | M1 | Expand |
| $= 25625t^2 - 14500t + 2100$ | A1 | |
| $\frac{d(d^2)}{dt} = 51250t - 14500 = 0$ | M1 | Differentiate and set to zero |
| $t = \frac{14500}{51250} = \frac{58}{205} \approx 0.283$ hours | A1 A1 | |
---4 The unit vectors $\mathbf { i } , \mathbf { j }$ and $\mathbf { k }$ are directed east, north and vertically upwards respectively.
At time $t = 0$, the position vectors of two small aeroplanes, $A$ and $B$, relative to a fixed origin $O$ are $( - 60 \mathbf { i } + 30 \mathbf { k } ) \mathrm { km }$ and $( - 40 \mathbf { i } + 10 \mathbf { j } - 10 \mathbf { k } ) \mathrm { km }$ respectively.
The aeroplane $A$ is flying with constant velocity $( 250 \mathbf { i } + 50 \mathbf { j } - 100 \mathbf { k } ) \mathrm { km } \mathrm { h } ^ { - 1 }$ and the aeroplane $B$ is flying with constant velocity $( 200 \mathbf { i } + 25 \mathbf { j } + 50 \mathbf { k } ) \mathrm { km } \mathrm { h } ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Write down the position vectors of $A$ and $B$ at time $t$ hours.
\item Show that the position vector of $A$ relative to $B$ at time $t$ hours is $( ( - 20 + 50 t ) \mathbf { i } + ( - 10 + 25 t ) \mathbf { j } + ( 40 - 150 t ) \mathbf { k } ) \mathrm { km }$.
\item Show that $A$ and $B$ do not collide.
\item Find the value of $t$ when $A$ and $B$ are closest together.
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{01071eb0-2c48-4028-8cd3-6021ce86d7e5-13_2484_1709_223_153}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{AQA M3 2010 Q4 [15]}}