| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Distance between two moving objects |
| Difficulty | Moderate -0.3 This is a standard M1 relative velocity question requiring vector scaling, displacement calculation, and solving a quadratic equation from distance formula. All techniques are routine for the module, though part (c) involves several algebraic steps. Slightly easier than average due to straightforward setup and standard methods. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10b Vectors in 3D: i,j,k notation3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\text{vel. of } B = \lambda(5\mathbf{i} + 12\mathbf{j})\) | M1 A1 | |
| \(\text{mag. of vel.} = \sqrt{\lambda^2(5^2 + 12^2)} \therefore 52 = 13\lambda \text{ i.e. } \lambda = 4\) | M1 A1 | |
| \(\text{vel. of } B = 20\mathbf{i} + 48\mathbf{j}\) | A1 | |
| (b) \(\text{at } 10:15, A \text{ is at } 20\mathbf{i}, B \text{ is at } (5\mathbf{i} + 12\mathbf{j})\) | M1 | |
| \(\text{disp. vector of } B \text{ from } A = 5\mathbf{i} + 12\mathbf{j} - 20\mathbf{i} = -15\mathbf{i} + 12\mathbf{j}\) | M1 A1 | |
| (c) \(\text{disp. vector of } B \text{ from } A \text{ at time } t \text{ minutes} = \frac{1}{15}(-15\mathbf{i} + 12\mathbf{j}) \times t\) | M1 A1 | |
| \(= -\mathbf{i}t + 0.8\mathbf{j}\) | A1 | |
| \(\text{at time } t, \text{dist. between } A \text{ and } B = \sqrt{(-t)^2 + (0.8t)^2}\) | M1 | |
| \(23 = t\sqrt{1.64} \therefore t = 17.96 \text{ i.e. } t = 18 \text{ minutes (nearest minute)}\) | M1 A1 | (12 marks) |
(a) $\text{vel. of } B = \lambda(5\mathbf{i} + 12\mathbf{j})$ | M1 A1 |
$\text{mag. of vel.} = \sqrt{\lambda^2(5^2 + 12^2)} \therefore 52 = 13\lambda \text{ i.e. } \lambda = 4$ | M1 A1 |
$\text{vel. of } B = 20\mathbf{i} + 48\mathbf{j}$ | A1 |
(b) $\text{at } 10:15, A \text{ is at } 20\mathbf{i}, B \text{ is at } (5\mathbf{i} + 12\mathbf{j})$ | M1 |
$\text{disp. vector of } B \text{ from } A = 5\mathbf{i} + 12\mathbf{j} - 20\mathbf{i} = -15\mathbf{i} + 12\mathbf{j}$ | M1 A1 |
(c) $\text{disp. vector of } B \text{ from } A \text{ at time } t \text{ minutes} = \frac{1}{15}(-15\mathbf{i} + 12\mathbf{j}) \times t$ | M1 A1 |
$= -\mathbf{i}t + 0.8\mathbf{j}$ | A1 |
$\text{at time } t, \text{dist. between } A \text{ and } B = \sqrt{(-t)^2 + (0.8t)^2}$ | M1 |
$23 = t\sqrt{1.64} \therefore t = 17.96 \text{ i.e. } t = 18 \text{ minutes (nearest minute)}$ | M1 A1 | (12 marks)Two trains $A$ and $B$ leave the same station, $O$, at 10 a.m. and travel along straight horizontal tracks. $A$ travels with constant speed $80 \text{ km h}^{-1}$ due east and $B$ travels with constant speed $52 \text{ km h}^{-1}$ in the direction $(5\mathbf{i} + 12\mathbf{j})$ where $\mathbf{i}$ and $\mathbf{j}$ are unit vectors due east and due north respectively.
\begin{enumerate}[label=(\alph*)]
\item Show that the velocity of $B$ is $(20\mathbf{i} + 48\mathbf{j}) \text{ km h}^{-1}$. [3 marks]
\item Find the displacement vector of $B$ from $A$ at 10:15 a.m. [3 marks]
Given that the trains are 23 km apart $t$ minutes after 10 a.m.
\item find the value of $t$ correct to the nearest whole number. [6 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q6 [12]}}