| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2012 |
| 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 -0.8 This is a standard M1 mechanics question testing basic vector kinematics with constant velocity. Parts (a)-(c) involve routine calculations (bearing from velocity components, position vector formula r = r₀ + vt, distance at given time). Part (d) requires recognizing that 'due north' means equal i-components, leading to a simple equation. All techniques are textbook exercises with no novel problem-solving required. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\arctan\dfrac{7.5}{12} = 32°\) | M1 A1 | M1 for \(\arctan\left(\dfrac{\pm7.5}{\pm12}\right)\); A1 for correct value, usually \(32°\) or \(58°\) |
| Bearing is \(302°\) | A1 | Allow more accurate answers |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{s} = 40\mathbf{i} - 6\mathbf{j} + t(-12\mathbf{i} + 7.5\mathbf{j})\) | M1 A1 | M1 for clear attempt at \((40\mathbf{i}-6\mathbf{j})+t(-12\mathbf{i}+7.5\mathbf{j})\); A1 for any correct expression |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(t = 3\): \(\mathbf{s} = 4\mathbf{i} + 16.5\mathbf{j}\) | M1 | B1 for \(4\mathbf{i}+16.5\mathbf{j}\) seen or implied (can be unsimplified) |
| \(\mathbf{s} - \mathbf{b} = -3\mathbf{i} + 4\mathbf{j}\) | M1 | For subtraction \(\mathbf{s}-\mathbf{b}\) or \(\mathbf{b}-\mathbf{s}\) |
| \(SB = \sqrt{(-3)^2 + 4^2} = 5 \text{ (km)}\) | DM1 A1 | Dependent on second M1; for finding magnitude of \(\mathbf{s}-\mathbf{b}\) or \(\mathbf{b}-\mathbf{s}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Equating \(\mathbf{i}\) components: \(40 - 12t = 7\) or \(-33 + 12t = 0\) | M1 | M1 for equating \(\mathbf{i}\)-component of (b) to 7, or \(\mathbf{i}\)-component of \(\mathbf{s}-\mathbf{b}\) or \(\mathbf{b}-\mathbf{s}\) to zero |
| \(t = 2\dfrac{3}{4}\) | A1 | cao |
| When \(t = 2\dfrac{3}{4}\): \(\mathbf{s} = (7\mathbf{i}) + 14\dfrac{5}{8}\mathbf{j}\) | M1 | Independent M1 for finding \(\mathbf{j}\)-component at their \(t = 2.75\) |
| \(SB = 2\dfrac{1}{8}\) km, \(2.125\), \(2.13\) | A1 | cao |
| OR: When \(t = 2\dfrac{3}{4}\), \(7.5t - 18.5 = 2.125\), \(2.13\) | M1 A1 |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\arctan\dfrac{7.5}{12} = 32°$ | M1 A1 | M1 for $\arctan\left(\dfrac{\pm7.5}{\pm12}\right)$; A1 for correct value, usually $32°$ or $58°$ |
| Bearing is $302°$ | A1 | Allow more accurate answers |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{s} = 40\mathbf{i} - 6\mathbf{j} + t(-12\mathbf{i} + 7.5\mathbf{j})$ | M1 A1 | M1 for clear attempt at $(40\mathbf{i}-6\mathbf{j})+t(-12\mathbf{i}+7.5\mathbf{j})$; A1 for any correct expression |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $t = 3$: $\mathbf{s} = 4\mathbf{i} + 16.5\mathbf{j}$ | M1 | B1 for $4\mathbf{i}+16.5\mathbf{j}$ seen or implied (can be unsimplified) |
| $\mathbf{s} - \mathbf{b} = -3\mathbf{i} + 4\mathbf{j}$ | M1 | For subtraction $\mathbf{s}-\mathbf{b}$ or $\mathbf{b}-\mathbf{s}$ |
| $SB = \sqrt{(-3)^2 + 4^2} = 5 \text{ (km)}$ | DM1 A1 | Dependent on second M1; for finding magnitude of $\mathbf{s}-\mathbf{b}$ or $\mathbf{b}-\mathbf{s}$ |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Equating $\mathbf{i}$ components: $40 - 12t = 7$ or $-33 + 12t = 0$ | M1 | M1 for equating $\mathbf{i}$-component of (b) to 7, or $\mathbf{i}$-component of $\mathbf{s}-\mathbf{b}$ or $\mathbf{b}-\mathbf{s}$ to zero |
| $t = 2\dfrac{3}{4}$ | A1 | cao |
| When $t = 2\dfrac{3}{4}$: $\mathbf{s} = (7\mathbf{i}) + 14\dfrac{5}{8}\mathbf{j}$ | M1 | Independent M1 for finding $\mathbf{j}$-component at their $t = 2.75$ |
| $SB = 2\dfrac{1}{8}$ km, $2.125$, $2.13$ | A1 | cao |
| OR: When $t = 2\dfrac{3}{4}$, $7.5t - 18.5 = 2.125$, $2.13$ | M1 A1 | |
---6. [In this question $\mathbf { i }$ and $\mathbf { j }$ are horizontal unit vectors due east and due north respectively and position vectors are given with respect to a fixed origin.]
A ship $S$ is moving with constant velocity $( - 12 \mathbf { i } + 7.5 \mathbf { j } ) \mathrm { km } \mathrm { h } ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Find the direction in which $S$ is moving, giving your answer as a bearing.
At time $t$ hours after noon, the position vector of $S$ is $\mathbf { s } \mathrm { km }$. When $t = 0 , \mathbf { s } = 40 \mathbf { i } - 6 \mathbf { j }$.
\item Write down $\mathbf { s }$ in terms of $t$.
A fixed beacon $B$ is at the point with position vector $( 7 \mathbf { i } + 12.5 \mathbf { j } ) \mathrm { km }$.
\item Find the distance of $S$ from $B$ when $t = 3$
\item Find the distance of $S$ from $B$ when $S$ is due north of $B$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2012 Q6 [13]}}