| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Closest approach of two objects |
| Difficulty | Moderate -0.8 This is a standard M1 mechanics question testing basic vector kinematics with straightforward calculations: finding speed from velocity components (Pythagoras), writing position vectors using r = r₀ + vt, finding relative position vectors, and solving simultaneous equations to find when vectors are equal. All steps are routine textbook exercises requiring no problem-solving insight, making it easier than average A-level material. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \( | \mathbf{v} | = \sqrt{1.2^2 + (-0.9)^2} = 1.5 \text{ m s}^{-1}\) |
| (2) | ||
| (b) \((\mathbf{r}_H) = 100\mathbf{j} + t(1.2\mathbf{i} - 0.9\mathbf{j}) \text{ m}\) | M1 A1 | |
| (2) | ||
| (c) \((\mathbf{r}_K) = 9\mathbf{i} + 46\mathbf{j} + t(0.75\mathbf{i} + 1.8\mathbf{j}) \text{ m}\) | M1 A1 | |
| (d) \(\overline{HK} = \mathbf{r}_K - \mathbf{r}_H = (9 - 0.45t)\mathbf{i} + (2.7t - 54)\mathbf{j} \text{ m}\) | M1 A1 | |
| Printed Answer | ||
| Meet when \(\overline{HK} = \mathbf{0}\) | ||
| \((9 - 0.45t) = 0\) and \((2.7t - 54) = 0\) | M1 A1 | |
| \(t = 20\) from both equations | A1 | |
| \(\mathbf{r}_K = \mathbf{r}_H = (24\mathbf{i} + 82\mathbf{j}) \text{ m}\) | DM1 A1 cso | |
| (5) | ||
| Total: [13] |
**(a)** $|\mathbf{v}| = \sqrt{1.2^2 + (-0.9)^2} = 1.5 \text{ m s}^{-1}$ | M1 A1 |
| **(2)** |
**(b)** $(\mathbf{r}_H) = 100\mathbf{j} + t(1.2\mathbf{i} - 0.9\mathbf{j}) \text{ m}$ | M1 A1 |
| **(2)** |
**(c)** $(\mathbf{r}_K) = 9\mathbf{i} + 46\mathbf{j} + t(0.75\mathbf{i} + 1.8\mathbf{j}) \text{ m}$ | M1 A1 |
**(d)** $\overline{HK} = \mathbf{r}_K - \mathbf{r}_H = (9 - 0.45t)\mathbf{i} + (2.7t - 54)\mathbf{j} \text{ m}$ | M1 A1 |
Printed Answer | |
Meet when $\overline{HK} = \mathbf{0}$ | |
$(9 - 0.45t) = 0$ and $(2.7t - 54) = 0$ | M1 A1 |
$t = 20$ from both equations | A1 |
$\mathbf{r}_K = \mathbf{r}_H = (24\mathbf{i} + 82\mathbf{j}) \text{ m}$ | DM1 A1 cso |
| **(5)** |
| **Total: [13]** |[In this question $\mathbf{i}$ and $\mathbf{j}$ are horizontal unit vectors due east and due north respectively.]
A hiker $H$ is walking with constant velocity $(1.2\mathbf{i} - 0.9\mathbf{j})$ m s$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $H$. [2]
\end{enumerate}
\includegraphics{figure_3}
A horizontal field $OABC$ is rectangular with $OA$ due east and $OC$ due north, as shown in Figure 3. At twelve noon hiker $H$ is at the point $Y$ with position vector $100\mathbf{j}$ m, relative to the fixed origin $O$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Write down the position vector of $H$ at time $t$ seconds after noon. [2]
\end{enumerate}
At noon, another hiker $K$ is at the point with position vector $(9\mathbf{i} + 46\mathbf{j})$ m. Hiker $K$ is moving with constant velocity $(0.75\mathbf{i} + 1.8\mathbf{j})$ m s$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{2}
\item Show that, at time $t$ seconds after noon,
$$\overrightarrow{HK} = [(9 - 0.45t)\mathbf{i} + (2.7t - 54)\mathbf{j}] \text{ metres.}$$ [4]
\end{enumerate}
Hence,
\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item show that the two hikers meet and find the position vector of the point where they meet. [5]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2009 Q8 [13]}}