| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 17 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Interception: verify/find meeting point (position vector method) |
| Difficulty | Standard +0.3 This is a standard M1 kinematics question involving vectors and relative motion. Parts (a) and (b) require routine application of speed-direction relationships and position vector equations. Parts (c) and (d) involve showing given results through algebraic manipulation of vectors. While multi-step with 17 marks total, each part uses well-practiced techniques without requiring novel insight—slightly easier than average due to the scaffolded structure and 'show that' format providing the answer. |
| 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 multiplication1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(SO = 10\mathbf{i} - 24\mathbf{j}\), having magnitude 26 km | B1 B1 | |
| \(v_S = \frac{12}{26}(10\mathbf{i} - 24\mathbf{j}) = (20\mathbf{i} - 48\mathbf{j})\) km h\(^{-1}\) → \(v_T = -50\mathbf{j}\) km h\(^{-1}\) | M1 A1 A1 | |
| (b) \(r_S = -10\mathbf{i} + 24\mathbf{j} + \frac{t}{26}(20\mathbf{i} - 48\mathbf{j}) = (\frac{t}{3} - 10)\mathbf{i} + (24 - \frac{4}{5}t)\mathbf{j}\) | M1 A1 A1 | |
| \(r_T = 25\mathbf{j} + \frac{t}{60}(-50\mathbf{j}) = (25 - \frac{5}{6}t)\mathbf{j}\) | M1 A1 | |
| (c) \(ST = (10 - \frac{1}{3}t)\mathbf{i} + (1 - \frac{1}{30}t)\mathbf{j}\) → \(\tan \theta = (10 - \frac{1}{3}t) \div (1 - \frac{1}{30}t) = 10\) | M1 A1 M1 A1 | |
| Bearing \(= 084.3°\) | A1 | |
| (d) When \(t = 30\), \(r_S = r_T = 0\), so trains collide at O | M1 A1 | Total: 17 marks |
**(a)** $SO = 10\mathbf{i} - 24\mathbf{j}$, having magnitude 26 km | B1 B1 |
$v_S = \frac{12}{26}(10\mathbf{i} - 24\mathbf{j}) = (20\mathbf{i} - 48\mathbf{j})$ km h$^{-1}$ → $v_T = -50\mathbf{j}$ km h$^{-1}$ | M1 A1 A1 |
**(b)** $r_S = -10\mathbf{i} + 24\mathbf{j} + \frac{t}{26}(20\mathbf{i} - 48\mathbf{j}) = (\frac{t}{3} - 10)\mathbf{i} + (24 - \frac{4}{5}t)\mathbf{j}$ | M1 A1 A1 |
$r_T = 25\mathbf{j} + \frac{t}{60}(-50\mathbf{j}) = (25 - \frac{5}{6}t)\mathbf{j}$ | M1 A1 |
**(c)** $ST = (10 - \frac{1}{3}t)\mathbf{i} + (1 - \frac{1}{30}t)\mathbf{j}$ → $\tan \theta = (10 - \frac{1}{3}t) \div (1 - \frac{1}{30}t) = 10$ | M1 A1 M1 A1 |
Bearing $= 084.3°$ | A1 |
**(d)** When $t = 30$, $r_S = r_T = 0$, so trains collide at O | M1 A1 | **Total: 17 marks**Two trains $S$ and $T$ are moving with constant speeds on straight tracks which intersect at the point $O$. At 9.00 a.m. $S$ has position vector $(-10\mathbf{i} + 24\mathbf{j})$ km and $T$ has position vector $25\mathbf{j}$ km relative to $O$, where $\mathbf{i}$ and $\mathbf{j}$ are unit vectors in the directions due east and due north respectively. $S$ is moving with speed 52 km h$^{-1}$ and $T$ is moving with speed 50 km h$^{-1}$, both towards $O$.
\begin{enumerate}[label=(\alph*)]
\item Show that the velocity vector of $S$ is $(20\mathbf{i} - 48\mathbf{j})$ km h$^{-1}$ and find the velocity vector of $T$. \hfill [5 marks]
\item Find expressions for the position vectors of $S$ and $T$ at time $t$ minutes after 9.00 a.m. \hfill [5 marks]
\item Show that the bearing of $T$ from $S$ remains constant during the motion, and find this bearing. \hfill [5 marks]
\item Show that if the trains continue at the given speeds they will collide. \hfill [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q7 [17]}}