| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| 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 position vectors with constant velocity. Parts (a)-(c) involve routine application of r = râ‚€ + vt formula and vector subtraction. Part (d) requires minimizing distance using calculus, which is a common textbook exercise. The multi-part structure and 2D vectors make it slightly easier than average A-level difficulty. |
| Spec | 1.10e Position vectors: and displacement1.10f Distance between points: using position vectors1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{v} = \dfrac{(9\mathbf{i}+23\mathbf{j})-(-2\mathbf{i}+\mathbf{j})}{11}\) | M1 | Use of displacement/time to find velocity. Allow the difference either way round. |
| Expression for \(\mathbf{r}\) with correct structure | M1 | Expression for \(\mathbf{r}\) with correct structure using *their* \(\mathbf{v}\) and the correct initial position vector. |
| \(\mathbf{r} = (-2\mathbf{i}+\mathbf{j})+t(\mathbf{i}+2\mathbf{j})\) or \(\mathbf{r} = (t-2)\mathbf{i}+(2t+1)\mathbf{j}\) | A1 cao | Correct expression in terms of \(t\), \(\mathbf{i}\) and \(\mathbf{j}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\mathbf{s} = (25\mathbf{i}+25\mathbf{j})+t(-\mathbf{i}-\mathbf{j})\) Or \(\mathbf{s} = (25-t)\mathbf{i}+(25-t)\mathbf{j}\) | B1 | Any correct expression for \(\mathbf{s}\) in terms of \(t\), \(\mathbf{i}\) and \(\mathbf{j}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Either \(\mathbf{r}-\mathbf{s}\) or \(\mathbf{s}-\mathbf{r}\) with their \(\mathbf{r}\) and \(\mathbf{s}\) substituted | M1 | (Their \(\mathbf{r}\) – their \(\mathbf{s}\)) or vice versa, unsimplified |
| \(\overrightarrow{SR} = [(2t-27)\mathbf{i}+(3t-24)\mathbf{j}]\ \text{m}\) * | A1* | Correct answer correctly obtained. Allow missing square brackets and m, but rest must be identical to given answer. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Distance \((d) = \sqrt{(2t-27)^2+(3t-24)^2}\) | M1 | Use of Pythagoras to find an expression for distance (or distance squared) |
| \((d^2) = (2t-27)^2+(3t-24)^2\) | ||
| \((d^2) = 13t^2 - 252t + 1305\) | A1 | Correct 3 term quadratic expression. N.B. If no 3 term quadratic expression is seen but a correct derivative is, award this mark. |
| \(t = \dfrac{126}{13} = 9.7\ \text{(s)}\) or better | A1 | 9.7 or better. N.B. If a fraction is given as the answer, it must be the ratio of two positive integers or a mixed fraction. |
## Question 8:
**Note:** Allow working in column vectors and penalise answers to (a) and (b) in column vector form ONCE at the first time it occurs.
---
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{v} = \dfrac{(9\mathbf{i}+23\mathbf{j})-(-2\mathbf{i}+\mathbf{j})}{11}$ | M1 | Use of displacement/time to find velocity. Allow the difference either way round. |
| Expression for $\mathbf{r}$ with correct structure | M1 | Expression for $\mathbf{r}$ with correct structure using *their* $\mathbf{v}$ and the correct initial position vector. |
| $\mathbf{r} = (-2\mathbf{i}+\mathbf{j})+t(\mathbf{i}+2\mathbf{j})$ or $\mathbf{r} = (t-2)\mathbf{i}+(2t+1)\mathbf{j}$ | A1 cao | Correct expression in terms of $t$, $\mathbf{i}$ and $\mathbf{j}$ |
**(3 marks)**
---
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\mathbf{s} = (25\mathbf{i}+25\mathbf{j})+t(-\mathbf{i}-\mathbf{j})$ **Or** $\mathbf{s} = (25-t)\mathbf{i}+(25-t)\mathbf{j}$ | B1 | Any correct expression for $\mathbf{s}$ in terms of $t$, $\mathbf{i}$ and $\mathbf{j}$ |
**(1 mark)**
---
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Either $\mathbf{r}-\mathbf{s}$ or $\mathbf{s}-\mathbf{r}$ with their $\mathbf{r}$ and $\mathbf{s}$ substituted | M1 | (Their $\mathbf{r}$ – their $\mathbf{s}$) or vice versa, unsimplified |
| $\overrightarrow{SR} = [(2t-27)\mathbf{i}+(3t-24)\mathbf{j}]\ \text{m}$ * | A1* | Correct answer correctly obtained. Allow missing square brackets and m, but rest must be identical to given answer. |
**(2 marks)**
---
**Part (d):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Distance $(d) = \sqrt{(2t-27)^2+(3t-24)^2}$ | M1 | Use of Pythagoras to find an expression for distance (or distance squared) |
| $(d^2) = (2t-27)^2+(3t-24)^2$ | | |
| $(d^2) = 13t^2 - 252t + 1305$ | A1 | Correct 3 term quadratic expression. **N.B.** If no 3 term quadratic expression is seen but a correct derivative is, award this mark. |
| $t = \dfrac{126}{13} = 9.7\ \text{(s)}$ or better | A1 | 9.7 or better. **N.B.** If a fraction is given as the answer, it must be the ratio of two positive integers or a mixed fraction. |
**(3 marks)**
**Total: 9 marks**8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f2737a11-4a15-41e9-9f87-31a705a8948b-22_792_841_246_612}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
A square floor space $A B C D$, with centre $O$, is modelled as a flat horizontal surface measuring 50 m by 50 m , as shown in Figure 5 .\\
The horizontal unit vectors $\mathbf { i }$ and $\mathbf { j }$ are in the direction of $\overrightarrow { A B }$ and $\overrightarrow { A D }$ respectively.\\
All position vectors are given relative to $O$.\\
A small robot $R$ is programmed to travel across the floor at a constant velocity.
\begin{itemize}
\item At time $t = 0 , R$ is at the point with position vector ( $- 2 \mathbf { i } + \mathbf { j }$ ) m
\item At time $t = 11 \mathrm {~s} , R$ is at the point with position vector $( 9 \mathbf { i } + 23 \mathbf { j } ) \mathrm { m }$
\item At time $t$ seconds, the position vector of $R$ is $\mathbf { r }$ metres
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $t$, i and j, an expression for $\mathbf { r }$
\end{itemize}
A second robot $S$ is at the point $C$.
\begin{itemize}
\item At time $t = 0 , S$ leaves $C$ and moves with constant velocity $( - \mathbf { i } - \mathbf { j } ) \mathrm { ms } ^ { - 1 }$
\item At time $t$ seconds, the position vector of $S$ is $\mathbf { s }$ metres
\item Write down, in terms of $t$, i and $\mathbf { j }$, an expression for $\mathbf { s }$
\item Show that
\end{itemize}
$$\overrightarrow { S R } = [ ( 2 t - 27 ) \mathbf { i } + ( 3 t - 24 ) \mathbf { j } ] \mathbf { m }$$
\item Find the time when the distance between $R$ and $S$ is a minimum.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2023 Q8 [9]}}