| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Relative velocity: find resultant velocity (magnitude and/or direction) |
| Difficulty | Moderate -0.8 This is a straightforward M1 relative velocity question requiring only standard vector operations: magnitude calculation using Pythagoras, vector subtraction, and finding an angle using arctangent. All three parts are routine textbook exercises with no problem-solving insight needed, making it easier than average but not trivial since it requires correct application of multiple techniques. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks | Guidance |
|---|---|---|
| (a) speed of \(B = \sqrt{[15^2 + (8)^2]} = 17 \text{ ms}^{-1}\) | M1 A1 | |
| (b) vel. of \(B\) rel. to \(A = (15\hat{i} - 8\hat{j}) - (8\hat{i} - 3\hat{j})\) \(= (7\hat{i} - 5\hat{j}) \text{ ms}^{-1}\) | M1 A1 | |
| (c) req'd angle \(= \tan^{-1}\frac{5}{7} = 35.5°\) to 1 dp | M1 A1 | (6) |
**(a)** speed of $B = \sqrt{[15^2 + (8)^2]} = 17 \text{ ms}^{-1}$ | M1 A1 |
**(b)** vel. of $B$ rel. to $A = (15\hat{i} - 8\hat{j}) - (8\hat{i} - 3\hat{j})$ $= (7\hat{i} - 5\hat{j}) \text{ ms}^{-1}$ | M1 A1 |
**(c)** req'd angle $= \tan^{-1}\frac{5}{7} = 35.5°$ to 1 dp | M1 A1 | (6)
---Particle $A$ has velocity $(8\mathbf{i} - 3\mathbf{j})$ ms$^{-1}$ and particle $B$ has velocity $(15\mathbf{i} - 8\mathbf{j})$ ms$^{-1}$ where $\mathbf{i}$ and $\mathbf{j}$ are perpendicular, horizontal unit vectors.
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $B$. [2 marks]
\item Find the velocity of $B$ relative to $A$. [2 marks]
\item Find the acute angle between the relative velocity found in part (b) and the vector $\mathbf{i}$, giving your answer in degrees correct to 1 decimal place. [2 marks]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q2 [6]}}