| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Interception: verify/find meeting point (position vector method) |
| Difficulty | Moderate -0.3 This is a standard M1 kinematics question with vectors involving straightforward application of position vector formulas (r = râ‚€ + vt), verification of a point lying on a path, and finding minimum distance using calculus or completing the square. All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10h Vectors in kinematics: uniform acceleration in vector form3.02e Two-dimensional constant acceleration: with vectors |
| VIXV SIHIANI III IM IONOO | VIAV SIHI NI JYHAM ION OO | VI4V SIHI NI JLIYM ION OO |
| Answer | Marks |
|---|---|
| \((11\mathbf{i}+11\mathbf{j})+t(3\mathbf{i}-\mathbf{j})\) | M1A1 (2) |
| Answer | Marks | Guidance |
|---|---|---|
| When \(t=6\), \(\mathbf{r}_A = (29\mathbf{i}+5\mathbf{j})\) | M1 | |
| \(\mathbf{r}_B = (7\mathbf{i}+16\mathbf{j})+t(4\mathbf{i}-2\mathbf{j}) = (29\mathbf{i}+5\mathbf{j})\) | M1 | |
| Solve both \(4t+7=29\) and \(16-2t=5\) explicitly to give \(t=5.5\) for both equations | DM1, A1* (4) | Division by vectors is DM0; need to see 5.5 occurring twice |
| Answer | Marks | Guidance |
|---|---|---|
| \(\overrightarrow{AB} = (7\mathbf{i}+16\mathbf{j})+t(4\mathbf{i}-2\mathbf{j})-\{(11\mathbf{i}+11\mathbf{j})+t(3\mathbf{i}-\mathbf{j})\}\) | M1 | |
| \(\overrightarrow{AB} = [(t-4)\mathbf{i}+(5-t)\mathbf{j}]\) m — GIVEN ANSWER | A1* (2) | M0 if starting with \(\mathbf{r}_A=\mathbf{r}_B\); A1* for correctly establishing *exactly* the given expression, writing out in full the difference before simplifying |
| Answer | Marks | Guidance |
|---|---|---|
| \(AB^2 = (t-4)^2+(5-t)^2\) | M1 | Correct expression seen or implied |
| \(= 2(t-4.5)^2+0.5\) | A1 | Correct quadratic in completed square form |
| Complete method using above to find minimum | M1 | |
| Minimum \(AB = \sqrt{0.5} = 0.71\) m (or better) | A1 (4) |
| Answer | Marks | Guidance |
|---|---|---|
| \(AB^2 = (t-4)^2+(5-t)^2\) | M1 | |
| \(4t-18\) or \(2(t-4)-2(5-t)\) | A1 | Correct derivative (can be implied by \(t=4.5\)) |
| Complete method using derivative to find minimum | M1 | |
| Minimum \(AB = \sqrt{0.5} = 0.71\) m (or better) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(2t^2-18t+(41-d^2)=0\) \((d=AB)\) | A1 | Correct equation |
| Complete method using \(b^2-4ac=0\): \((-18)^2-4\times2(41-d^2)=0\) | M1 | |
| Minimum \(AB = \sqrt{0.5} = 0.71\) m (or better) | A1 (4) |
## Question 6:
### Part 6(a):
$(11\mathbf{i}+11\mathbf{j})+t(3\mathbf{i}-\mathbf{j})$ | M1A1 (2) |
### Part 6(b):
When $t=6$, $\mathbf{r}_A = (29\mathbf{i}+5\mathbf{j})$ | M1 |
$\mathbf{r}_B = (7\mathbf{i}+16\mathbf{j})+t(4\mathbf{i}-2\mathbf{j}) = (29\mathbf{i}+5\mathbf{j})$ | M1 |
Solve **both** $4t+7=29$ and $16-2t=5$ **explicitly** to give $t=5.5$ for both equations | DM1, A1* (4) | Division by vectors is DM0; need to see 5.5 occurring **twice**
### Part 6(c):
$\overrightarrow{AB} = (7\mathbf{i}+16\mathbf{j})+t(4\mathbf{i}-2\mathbf{j})-\{(11\mathbf{i}+11\mathbf{j})+t(3\mathbf{i}-\mathbf{j})\}$ | M1 |
$\overrightarrow{AB} = [(t-4)\mathbf{i}+(5-t)\mathbf{j}]$ m — GIVEN ANSWER | A1* (2) | M0 if starting with $\mathbf{r}_A=\mathbf{r}_B$; A1* for correctly establishing *exactly* the given expression, writing out **in full** the difference before simplifying
### Part 6(d):
$AB^2 = (t-4)^2+(5-t)^2$ | M1 | Correct expression seen or implied
$= 2(t-4.5)^2+0.5$ | A1 | Correct quadratic in completed square form
Complete method using above to find minimum | M1 |
Minimum $AB = \sqrt{0.5} = 0.71$ m (or better) | A1 (4) |
**OR:**
$AB^2 = (t-4)^2+(5-t)^2$ | M1 |
$4t-18$ **or** $2(t-4)-2(5-t)$ | A1 | Correct derivative (can be implied by $t=4.5$)
Complete method using derivative to find minimum | M1 |
Minimum $AB = \sqrt{0.5} = 0.71$ m (or better) | A1 |
**OR:**
$2t^2-18t+(41-d^2)=0$ $(d=AB)$ | A1 | Correct equation
Complete method using $b^2-4ac=0$: $(-18)^2-4\times2(41-d^2)=0$ | M1 |
Minimum $AB = \sqrt{0.5} = 0.71$ m (or better) | A1 (4) |
---6. Two girls, Agatha and Brionie, are roller skating inside a large empty building. The girls are modelled as particles.
At time $t = 0$, Agatha is at the point with position vector $( 11 \mathbf { i } + 11 \mathbf { j } ) \mathrm { m }$ and Brionie is at the point with position vector $( 7 \mathbf { i } + 16 \mathbf { j } ) \mathrm { m }$. The position vectors are given relative to the door, $O$, and $\mathbf { i }$ and $\mathbf { j }$ are horizontal perpendicular unit vectors.
Agatha skates with constant velocity ( $3 \mathbf { i } - \mathbf { j }$ ) $\mathrm { m } \mathrm { s } ^ { - 1 }$\\
Brionie skates with constant velocity ( $4 \mathbf { i } - 2 \mathbf { j }$ ) $\mathrm { m } \mathrm { s } ^ { - 1 }$
\begin{enumerate}[label=(\alph*)]
\item Find the position vector of Agatha at time $t$ seconds.
At time $t = 6$ seconds, Agatha passes through the point $P$.
\item Show that Brionie also passes through $P$ and find the value of $t$ when this occurs.
At time $t$ seconds, Agatha is at the point $A$ and Brionie is at the point $B$.
\item Show that $\overrightarrow { A B } = [ ( t - 4 ) \mathbf { i } + ( 5 - t ) \mathbf { j } ] \mathrm { m }$
\item Find the distance between the two girls when they are closest together.
\includegraphics[max width=\textwidth, alt={}, center]{ca445c1e-078c-4a57-94df-de90f30f8efd-13_2255_50_314_34}\\
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
VIXV SIHIANI III IM IONOO & VIAV SIHI NI JYHAM ION OO & VI4V SIHI NI JLIYM ION OO \\
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\end{tabular}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2021 Q6 [12]}}