| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Session | Specimen |
| Marks | 15 |
| 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 kinematics question using vectors with constant velocity. Parts (a) and (b) are routine applications of position = initial position + velocity × time. Parts (c) and (d) require setting up equations (bearing condition and distance formula) but involve straightforward algebraic manipulation. The multi-part structure and 15 marks suggest moderate length, but no novel insight is required—it's a textbook-style exercise slightly easier than average A-level difficulty. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10h Vectors in kinematics: uniform acceleration in vector form3.02a Kinematics language: position, displacement, velocity, acceleration |
Two cars $A$ and $B$ are moving on straight horizontal roads with constant velocities. The velocity of $A$ is $20 \text{ m s}^{-1}$ due east, and the velocity of $B$ is $(10\mathbf{i} + 10\mathbf{j}) \text{ m s}^{-1}$, where $\mathbf{i}$ and $\mathbf{j}$ are unit vectors directed due east and due north respectively. Initially $A$ is at the fixed origin $O$, and the position vector of $B$ is $300\mathbf{j}$ m relative to $O$. At time $t$ seconds, the position vectors of $A$ and $B$ are $\mathbf{r}$ metres and $\mathbf{s}$ metres respectively.
\begin{enumerate}[label=(\alph*)]
\item Find expressions for $\mathbf{r}$ and $\mathbf{s}$ in terms of $t$. [3]
\item Hence write down an expression for $\overrightarrow{AB}$ in terms of $t$. [1]
\item Find the time when the bearing of $B$ from $A$ is $045°$. [5]
\item Find the time when the cars are again 300 m apart. [6]
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q7 [15]}}