| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Distance between two moving objects |
| Difficulty | Standard +0.3 This is a standard M1 kinematics question using vectors. Parts (a) and (b) are routine (finding speed from velocity vector and writing position vector equation). Part (c) requires finding minimum distance by differentiating |PS|² or using the perpendicular distance formula, which is a well-practiced technique. The calculations are straightforward with no conceptual surprises, making it slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sqrt{12^2+16^2}=20\ (\text{km h}^{-1})\) | M1 A1 | Use of Pythagoras; since 3,4,5 triangle correct answer may appear without working. cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((19\mathbf{i}+22\mathbf{j})+t(12\mathbf{i}-16\mathbf{j})\) | M1 A1 | Correct structure; correct answer with \(\mathbf{i}\)'s and \(\mathbf{j}\)'s |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\overrightarrow{LS}=(19+12t-26)\mathbf{i}+(22-16t-15)\mathbf{j}\) or \(\overrightarrow{SL}=(26-19-12t)\mathbf{i}+(15-22+16t)\mathbf{j}\) | M1 | Subtraction used to find displacement vector \(\overrightarrow{LS}\) or \(\overrightarrow{SL}\) |
| \(\overrightarrow{LS}=(12t-7)\mathbf{i}+(7-16t)\mathbf{j}\) or \(\overrightarrow{SL}=(7-12t)\mathbf{i}+(16t-7)\mathbf{j}\) | A1 | Correct components with \(\mathbf{i}\)'s and \(\mathbf{j}\)'s collected |
| \( | \overrightarrow{LS} | =\sqrt{(12t-7)^2+(7-16t)^2}\) |
| \(400t^2-392t+98\) | A1 | Correct 3TQ, seen or implied, e.g. allow \(400t^2-392t+96.31\) |
| Min occurs when \(t=0.49\) | A1 | cao; may appear without working via quadratic solver. Other methods include completing the square or differentiation |
| Alternative: \([(12t-7)\mathbf{i}+(7-16t)\mathbf{j}].(12\mathbf{i}-16\mathbf{j})=0\) | M1 | Closest when relative velocity is perpendicular to relative displacement |
| \(400t-196=0\) | A1 | |
| Min occurs when \(t=0.49\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\sqrt{1.96}=1.4\ \text{(km)}\) | M1 | Use of their \(t\) value from minimising distance \(LS\) |
| \(1.4>1.3\) so it is safe for \(S\) to continue its course | A1 cso | Correct conclusion comparing 1.4 and 1.3. Accept e.g. '1.4 therefore it is safe' |
# Question 7:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{12^2+16^2}=20\ (\text{km h}^{-1})$ | M1 A1 | Use of Pythagoras; since 3,4,5 triangle correct answer may appear without working. cao |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(19\mathbf{i}+22\mathbf{j})+t(12\mathbf{i}-16\mathbf{j})$ | M1 A1 | Correct structure; correct answer with $\mathbf{i}$'s and $\mathbf{j}$'s |
## Part (c)(i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\overrightarrow{LS}=(19+12t-26)\mathbf{i}+(22-16t-15)\mathbf{j}$ or $\overrightarrow{SL}=(26-19-12t)\mathbf{i}+(15-22+16t)\mathbf{j}$ | M1 | Subtraction used to find displacement vector $\overrightarrow{LS}$ or $\overrightarrow{SL}$ |
| $\overrightarrow{LS}=(12t-7)\mathbf{i}+(7-16t)\mathbf{j}$ or $\overrightarrow{SL}=(7-12t)\mathbf{i}+(16t-7)\mathbf{j}$ | A1 | Correct components with $\mathbf{i}$'s and $\mathbf{j}$'s collected |
| $|\overrightarrow{LS}|=\sqrt{(12t-7)^2+(7-16t)^2}$ | M1 | Use of Pythagoras with components from attempt at subtracting positions to form 3TQ for distance or distance squared |
| $400t^2-392t+98$ | A1 | Correct 3TQ, seen or implied, e.g. allow $400t^2-392t+96.31$ |
| Min occurs when $t=0.49$ | A1 | cao; may appear without working via quadratic solver. Other methods include completing the square or differentiation |
| **Alternative:** $[(12t-7)\mathbf{i}+(7-16t)\mathbf{j}].(12\mathbf{i}-16\mathbf{j})=0$ | M1 | Closest when relative velocity is perpendicular to relative displacement |
| $400t-196=0$ | A1 | |
| Min occurs when $t=0.49$ | A1 | |
## Part (c)(ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\sqrt{1.96}=1.4\ \text{(km)}$ | M1 | Use of their $t$ value from minimising distance $LS$ |
| $1.4>1.3$ so it is safe for $S$ to continue its course | A1 cso | Correct conclusion comparing 1.4 and 1.3. Accept e.g. '1.4 therefore it is safe' |
---\begin{enumerate}
\item \hspace{0pt} [In this question $\mathbf { i }$ and $\mathbf { j }$ are horizontal unit vectors directed due east and due north respectively and position vectors are given relative to a fixed origin $O$.]
\end{enumerate}
At midnight, a ship $S$ is at the point with position vector ( $19 \mathbf { i } + 22 \mathbf { j }$ )km\\
The ship travels with constant velocity $( 12 \mathbf { i } - 16 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }$\\
(a) Find the speed of $S$.
At time $t$ hours after midnight, the position vector of $S$ is $\mathbf { s } \mathrm { km }$.\\
(b) Find an expression for $\mathbf { s }$ in terms of $\mathbf { i } , \mathbf { j }$ and $t$.
A lighthouse stands on a small rocky island. The lighthouse is modelled as being at the point with position vector $( 26 \mathbf { i } + 15 \mathbf { j } ) \mathrm { km }$.
It is not safe for ships to be within 1.3 km of the lighthouse.\\
(c) (i) Find the value of $t$ when $S$ is closest to the lighthouse.\\
(ii) Hence determine whether it is safe for $S$ to continue its course.
\hfill \mbox{\textit{Edexcel M1 2024 Q7 [11]}}