| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Position vector at time t (constant velocity) |
| Difficulty | Standard +0.3 This is a standard M1 mechanics question on constant velocity motion with vectors. Part (a) is routine calculation of velocity from displacement and time. Parts (b)-(d) require setting up position vectors and using directional constraints (due east, north-east, due south), which are standard textbook techniques. The multi-part structure and geometric reasoning elevate it slightly above trivial, but it remains easier than average A-level questions overall. |
| Spec | 1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| \((4\mathbf{i}+6\mathbf{j}) = (-2\mathbf{i}+9\mathbf{j}) + 0.6\mathbf{v}\) | M1 | Correct structure with \(t = 0.6\) |
| \((10\mathbf{i}-5\mathbf{j})\) \((\text{km h}^{-1})\) | A1* | Given answer correctly obtained; N.B. 1 more line of intermediate working needed and must state answer in \(\mathbf{i}\)-\(\mathbf{j}\) form. Allow verification. (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{r} = (-2\mathbf{i}+9\mathbf{j}) + t(10\mathbf{i}-5\mathbf{j})\) (km) | M1A1 | Correct structure; cao (2 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = 1.8\): \(\mathbf{r} = (-2\mathbf{i}+9\mathbf{j})+1.8(10\mathbf{i}-5\mathbf{j})\) | M1 | Correct unsimplified substitution of \(t=1.8\) into their \(\mathbf{r}\), OR use \(t=1.2\) with \((4\mathbf{i}+6\mathbf{j})\) as start point |
| \(\mathbf{r} = 16\mathbf{i}\) | A1 | cao |
| \(t = 2\): \(\mathbf{r} = (-2\mathbf{i}+9\mathbf{j})+2(10\mathbf{i}-5\mathbf{j})\) OR \(\mathbf{r} = 16\mathbf{i}+0.2(10\mathbf{i}-5\mathbf{j})\) | M1 | Correct unsimplified substitution of \(t=2\) into their \(\mathbf{r}\), OR use \(t=1.4\) with \((4\mathbf{i}+6\mathbf{j})\) as start point, OR use \(t=0.2\) and their first answer as start point |
| \(\mathbf{r}_L = 19\mathbf{i}\) (km) | A1 | cao (4 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| \(-2+10t = 19\) | M1 | Equating \(\mathbf{i}\) component of their \(\mathbf{r}\) to \(\mathbf{i}\) component of their \(\mathbf{r}_L\) |
| \(t = 2.1\) | A1 | cao |
| \(\mathbf{r} = (-2\mathbf{i}+9\mathbf{j})+2.1(10\mathbf{i}-5\mathbf{j})\) | DM1 | Dependent on previous M1, for substituting their value of \(t\) into their \(\mathbf{r}\) |
| \(\mathbf{r} = (19\mathbf{i}-1.5\mathbf{j})\) (km) | A1 | cao (4 marks) |
# Question 8:
## Part 8(a):
$(4\mathbf{i}+6\mathbf{j}) = (-2\mathbf{i}+9\mathbf{j}) + 0.6\mathbf{v}$ | M1 | Correct structure with $t = 0.6$
$(10\mathbf{i}-5\mathbf{j})$ $(\text{km h}^{-1})$ | A1* | **Given answer correctly obtained**; N.B. 1 more line of intermediate working needed and must state answer in $\mathbf{i}$-$\mathbf{j}$ form. Allow verification. **(2 marks)**
## Part 8(b):
$\mathbf{r} = (-2\mathbf{i}+9\mathbf{j}) + t(10\mathbf{i}-5\mathbf{j})$ (km) | M1A1 | Correct structure; cao **(2 marks)**
## Part 8(c):
$t = 1.8$: $\mathbf{r} = (-2\mathbf{i}+9\mathbf{j})+1.8(10\mathbf{i}-5\mathbf{j})$ | M1 | Correct unsimplified substitution of $t=1.8$ into their $\mathbf{r}$, **OR** use $t=1.2$ with $(4\mathbf{i}+6\mathbf{j})$ as start point
$\mathbf{r} = 16\mathbf{i}$ | A1 | cao
$t = 2$: $\mathbf{r} = (-2\mathbf{i}+9\mathbf{j})+2(10\mathbf{i}-5\mathbf{j})$ **OR** $\mathbf{r} = 16\mathbf{i}+0.2(10\mathbf{i}-5\mathbf{j})$ | M1 | Correct unsimplified substitution of $t=2$ into their $\mathbf{r}$, **OR** use $t=1.4$ with $(4\mathbf{i}+6\mathbf{j})$ as start point, **OR** use $t=0.2$ and their first answer as start point
$\mathbf{r}_L = 19\mathbf{i}$ (km) | A1 | cao **(4 marks)**
## Part 8(d):
$-2+10t = 19$ | M1 | Equating $\mathbf{i}$ component of their $\mathbf{r}$ to $\mathbf{i}$ component of their $\mathbf{r}_L$
$t = 2.1$ | A1 | cao
$\mathbf{r} = (-2\mathbf{i}+9\mathbf{j})+2.1(10\mathbf{i}-5\mathbf{j})$ | DM1 | Dependent on previous M1, for substituting their value of $t$ into their $\mathbf{r}$
$\mathbf{r} = (19\mathbf{i}-1.5\mathbf{j})$ (km) | A1 | cao **(4 marks)**
**Total: 12 marks**8. [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.]
At 7 am a ship leaves a port and moves with constant velocity. The position vector of the port is $( - 2 \mathbf { i } + 9 \mathbf { j } ) \mathrm { km }$.
At 7.36 am the ship is at the point with position vector $( 4 \mathbf { i } + 6 \mathbf { j } ) \mathrm { km }$.
\begin{enumerate}[label=(\alph*)]
\item Show that the velocity of the ship is $( 10 \mathbf { i } - 5 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }$
\item Find the position vector of the ship $t$ hours after leaving port.
At 8.48 am a passenger on the ship notices that a lighthouse is due east of the ship. At 9 am the same passenger notices that the lighthouse is now north east of the ship.
\item Find the position vector of the lighthouse.
\item Find the position vector of the ship when it is due south of the lighthouse.\\
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{151d9232-5a78-4bc1-a57e-6c9cae80e473-32_2258_53_308_1980}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2021 Q8 [12]}}