| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Position vector at time t (constant velocity) |
| Difficulty | Moderate -0.3 This is a standard M1 mechanics question on relative motion and interception. It involves routine vector operations (resolving velocities, position vectors, displacement vectors) with clear scaffolding through parts (a)-(e). The only slightly challenging element is part (e) requiring interpretation of bearing conditions, but overall this is a typical textbook exercise requiring methodical application of standard techniques rather than novel insight. |
| Spec | 1.05a Sine, cosine, tangent: definitions for all arguments1.10a 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 displacement3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{v}_B = (20\sin\alpha)\mathbf{i} + (20\cos\alpha)\mathbf{j}\) | M1 | Condone sign errors and sin/cos confusion but both components must be resolved. Allow \(\cos(\frac{3}{5})\) or similar. If \(12\mathbf{i}+16\mathbf{j}\) appears without working, award M1A0 |
| \(= 16\mathbf{i} + 12\mathbf{j}\) (km h\(^{-1}\)) | A1 (2) | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{s} = (10\mathbf{i}+5\mathbf{j}) + t(16\mathbf{i}+12\mathbf{j})\) or \((10+16t)\mathbf{i}+(5+12t)\mathbf{j}\) | M1, A1ft (2) | M1: Correct structure, condone slips. A1ft: ft on answer to (a) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{AB} = \mathbf{s} - \mathbf{r} = (10\mathbf{i}+5\mathbf{j}) + t(16\mathbf{i}+12\mathbf{j}) - [20\mathbf{j}+40t\mathbf{i}]\) | M1 | Allow \(\mathbf{r}-\mathbf{s}\); \(\mathbf{r}\) and \(\mathbf{s}\) must be substituted |
| \(\overrightarrow{AB} = [(10-24t)\mathbf{i} + (12t-15)\mathbf{j}]\) km | A1* (2) | Correct given answer, correctly obtained. Need to see \(\overrightarrow{AB}\) at start or finish. Answer must be exactly as printed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(10-24t = 0\) and \(12t-15=0\) OR \(40t=10+16t\) and \(20=5+12t\) | M1 | They may use \(\mathbf{r}=\mathbf{s}\) with both \(\mathbf{i}\) and \(\mathbf{j}\) components equated |
| \(t = \frac{5}{12}\) and \(\frac{5}{4}\), or one correct \(t\) value used in other equation correctly to show not true | A1 | Need both \(t\) values. Accept \(0.42\) or better and \(1.25\) |
| Different \(t\) values, so never collide | A1* (3) | Correct conclusion. ALT 1: \((10-24t)^2+(12t-15)^2=0\); discriminant \((-840)^2-4\times720\times325 < 0\); no real roots so never collide. ALT 2: Minimum of \(720t^2-840t+325\) is \(80\) (or \(\sqrt{80}\)), so never collide |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(10-24t = 12t-15\) | M1 | Correct method |
| \(t = \frac{25}{36}\) or \(0.69\) or better | A1 | cao |
| \(\overrightarrow{AB} = \left[(10-24\times\frac{25}{36})\mathbf{i}+(12\times\frac{25}{36}-15)\mathbf{j}\right]\) km | M1 | Sub calculated \(t\) into \(\overrightarrow{AB}\) or \(\overrightarrow{BA}\); independent M mark |
| \(AB = 20\dfrac{\sqrt{2}}{3}\), \(9.4\) or better (km) | A1 (4) | cao |
# Question 7:
## Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{v}_B = (20\sin\alpha)\mathbf{i} + (20\cos\alpha)\mathbf{j}$ | M1 | Condone sign errors and sin/cos confusion but both components must be resolved. Allow $\cos(\frac{3}{5})$ or similar. If $12\mathbf{i}+16\mathbf{j}$ appears without working, award M1A0 |
| $= 16\mathbf{i} + 12\mathbf{j}$ (km h$^{-1}$) | A1 (2) | cao |
## Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{s} = (10\mathbf{i}+5\mathbf{j}) + t(16\mathbf{i}+12\mathbf{j})$ **or** $(10+16t)\mathbf{i}+(5+12t)\mathbf{j}$ | M1, A1ft (2) | M1: Correct structure, condone slips. A1ft: ft on answer to (a) |
## Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{AB} = \mathbf{s} - \mathbf{r} = (10\mathbf{i}+5\mathbf{j}) + t(16\mathbf{i}+12\mathbf{j}) - [20\mathbf{j}+40t\mathbf{i}]$ | M1 | Allow $\mathbf{r}-\mathbf{s}$; $\mathbf{r}$ and $\mathbf{s}$ must be substituted |
| $\overrightarrow{AB} = [(10-24t)\mathbf{i} + (12t-15)\mathbf{j}]$ km | A1* (2) | Correct given answer, correctly obtained. Need to see $\overrightarrow{AB}$ at start or finish. Answer must be exactly as printed |
## Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $10-24t = 0$ and $12t-15=0$ **OR** $40t=10+16t$ and $20=5+12t$ | M1 | They may use $\mathbf{r}=\mathbf{s}$ with both $\mathbf{i}$ and $\mathbf{j}$ components equated |
| $t = \frac{5}{12}$ and $\frac{5}{4}$, or one correct $t$ value used in other equation correctly to show not true | A1 | Need both $t$ values. Accept $0.42$ or better and $1.25$ |
| Different $t$ values, so never collide | A1* (3) | Correct conclusion. **ALT 1:** $(10-24t)^2+(12t-15)^2=0$; discriminant $(-840)^2-4\times720\times325 < 0$; no real roots so never collide. **ALT 2:** Minimum of $720t^2-840t+325$ is $80$ (or $\sqrt{80}$), so never collide |
## Part (e):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $10-24t = 12t-15$ | M1 | Correct method |
| $t = \frac{25}{36}$ or $0.69$ or better | A1 | cao |
| $\overrightarrow{AB} = \left[(10-24\times\frac{25}{36})\mathbf{i}+(12\times\frac{25}{36}-15)\mathbf{j}\right]$ km | M1 | Sub calculated $t$ into $\overrightarrow{AB}$ or $\overrightarrow{BA}$; independent M mark |
| $AB = 20\dfrac{\sqrt{2}}{3}$, $9.4$ or better (km) | A1 (4) | cao |
---\begin{enumerate}
\item \hspace{0pt} [In this question, the horizontal unit vectors $\mathbf { i }$ and $\mathbf { j }$ are directed due east and due north respectively and position vectors are given relative to a fixed origin $O$.]
\end{enumerate}
Two speedboats, $A$ and $B$, are each moving with constant velocity.
\begin{itemize}
\item the velocity of $A$ is $40 \mathrm { kmh } ^ { - 1 }$ due east
\item the velocity of $B$ is $20 \mathrm { kmh } ^ { - 1 }$ on a bearing of angle $\alpha \left( 0 ^ { \circ } < \alpha < 90 ^ { \circ } \right)$, where $\tan \alpha = \frac { 4 } { 3 }$ The boats are modelled as particles.\\
(a) Find, in terms of $\mathbf { i }$ and $\mathbf { j }$, the velocity of $B$ in $\mathrm { km } \mathrm { h } ^ { - 1 }$
\end{itemize}
At noon
\begin{itemize}
\item the position vector of $A$ is $20 \mathbf { j } \mathrm {~km}$
\item the position vector of $B$ is $( 10 \mathbf { i } + 5 \mathbf { j } ) \mathrm { km }$
\end{itemize}
At time $t$ hours after noon
\begin{itemize}
\item the position vector of $A$ is $\mathbf { r k m }$, where $\mathbf { r } = 20 \mathbf { j } + 40 t \mathbf { i }$
\item the position vector of $B$ is $\mathbf { s }$ km\\
(b) Find an expression for $\mathbf { s }$ in terms of $t , \mathbf { i }$ and $\mathbf { j }$.\\
(c) Show that at time $t$ hours after noon,
\end{itemize}
$$\overrightarrow { A B } = [ ( 10 - 24 t ) \mathbf { i } + ( 12 t - 15 ) \mathbf { j } ] \mathrm { km }$$
(d) Show that the boats will never collide.\\
(e) Find the distance between the boats when the bearing of $B$ from $A$ is $225 ^ { \circ }$
\hfill \mbox{\textit{Edexcel M1 2024 Q7 [13]}}