| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2014 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Interception: verify/find meeting point (position vector method) |
| Difficulty | Moderate -0.5 This is a standard M1 kinematics question involving constant velocity motion in 2D. Part (a) requires simple magnitude calculation, (b) uses basic trigonometry for bearings, and (c) involves equating position vectors—all routine techniques with no novel problem-solving required. Slightly easier than average due to straightforward setup and mechanical application of formulas. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Speed \(= \sqrt{3^2+(-2)^2}\) or \(\sqrt{3^2+2^2} = \sqrt{13}\) m s\(^{-1}\) | M1 | Use Pythagoras |
| A1(2) | Accept 3.6 or better. Ignore their diagram if it does not support their working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\tan\theta = \frac{2}{3}\), \(\theta = 33.7\) OR \(\tan\theta = \frac{3}{2}\), \(\theta = 56.3\) OR find another useful angle | M1 | Find a relevant angle |
| A1 | Their angle correct (seen or implied) | |
| Bearing \(= 124\) | A1(3) | Correct bearing. Accept \(124°\) or awrt \(124/124°\). Accept N 124 E or S 56 E |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{r}_B = 10\mathbf{j} + t(3\mathbf{i} - 2\mathbf{j})\) | M1 | Find the position vector of \(B\) or \(G\) at time \(t\) |
| A1 | Correct for \(B\) | |
| \(\mathbf{r}_G = 4\mathbf{i} - 2\mathbf{j} + t\left(\frac{5}{3}\mathbf{i} + 2\mathbf{j}\right)\) | A1 | Correct for \(G\) |
| \(3t = 4 + \frac{5}{3}t\) OR \(10 - 2t = -2 + 2t\) | DM1 | Compare coefficients of \(\mathbf{i}\) or of \(\mathbf{j}\) to form an equation in \(t\) |
| (i) \(t = 3\) s | A1 | Correct unambiguous conclusion |
| (ii) \(\mathbf{r} = 10\mathbf{j} + 3(3\mathbf{i} - 2\mathbf{j}) = (9\mathbf{i} + 4\mathbf{j})\) m OR \(\mathbf{r} = 4\mathbf{i} - 2\mathbf{j} + 3\left(\frac{5}{3}\mathbf{i} + 2\mathbf{j}\right) = (9\mathbf{i} + 4\mathbf{j})\) m | A1(6) | Final answer. Accept with no units. Do not ignore subsequent working |
# Question 5:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Speed $= \sqrt{3^2+(-2)^2}$ or $\sqrt{3^2+2^2} = \sqrt{13}$ m s$^{-1}$ | M1 | Use Pythagoras |
| | A1(2) | Accept 3.6 or better. Ignore their diagram if it does not support their working |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\tan\theta = \frac{2}{3}$, $\theta = 33.7$ OR $\tan\theta = \frac{3}{2}$, $\theta = 56.3$ OR find another useful angle | M1 | Find a relevant angle |
| | A1 | Their angle correct (seen or implied) |
| Bearing $= 124$ | A1(3) | Correct bearing. Accept $124°$ or awrt $124/124°$. Accept N 124 E or S 56 E |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{r}_B = 10\mathbf{j} + t(3\mathbf{i} - 2\mathbf{j})$ | M1 | Find the position vector of $B$ or $G$ at time $t$ |
| | A1 | Correct for $B$ |
| $\mathbf{r}_G = 4\mathbf{i} - 2\mathbf{j} + t\left(\frac{5}{3}\mathbf{i} + 2\mathbf{j}\right)$ | A1 | Correct for $G$ |
| $3t = 4 + \frac{5}{3}t$ OR $10 - 2t = -2 + 2t$ | DM1 | Compare coefficients of $\mathbf{i}$ or of $\mathbf{j}$ to form an equation in $t$ |
| (i) $t = 3$ s | A1 | Correct unambiguous conclusion |
| (ii) $\mathbf{r} = 10\mathbf{j} + 3(3\mathbf{i} - 2\mathbf{j}) = (9\mathbf{i} + 4\mathbf{j})$ m OR $\mathbf{r} = 4\mathbf{i} - 2\mathbf{j} + 3\left(\frac{5}{3}\mathbf{i} + 2\mathbf{j}\right) = (9\mathbf{i} + 4\mathbf{j})$ m | A1(6) | Final answer. Accept with no units. Do not ignore subsequent working |
**Total: [11]**
---\begin{enumerate}
\item \hspace{0pt} [In this question $\mathbf { i }$ and $\mathbf { j }$ are horizontal unit vectors due east and due north respectively. Position vectors are given relative to a fixed origin $O$.]
\end{enumerate}
A boy $B$ is running in a field with constant velocity ( $3 \mathbf { i } - 2 \mathbf { j }$ ) $\mathrm { m } \mathrm { s } ^ { - 1 }$. At time $t = 0 , B$ is at the point with position vector 10j m .
Find\\
(a) the speed of $B$,\\
(b) the direction in which $B$ is running, giving your answer as a bearing.
At time $t = 0$, a girl $G$ is at the point with position vector $( 4 \mathbf { i } - 2 \mathbf { j } ) \mathrm { m }$. The girl is running with constant velocity $\left( \frac { 5 } { 3 } \mathbf { i } + 2 \mathbf { j } \right) \mathrm { m } \mathrm { s } ^ { - 1 }$ and meets $B$ at the point $P$.\\
(c) Find\\
(i) the value of $t$ when they meet,\\
(ii) the position vector of $P$.\\
\hfill \mbox{\textit{Edexcel M1 2014 Q5 [11]}}