| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2001 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Velocity from two position vectors |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring basic vector arithmetic (displacement divided by time for velocity), simple angle calculation using tan^(-1), and distance calculation using Pythagoras. All steps are routine applications of standard formulas with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part nature. |
| Spec | 1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At \(t=0\): \(\underline{r}_P = 2\underline{i} - \underline{j}\); At \(t=2\): \(\underline{r}_P = 6\underline{i} + \underline{j}\) | ||
| Velocity of P constant \(\Rightarrow \underline{v}_P = \frac{(6\underline{i}+\underline{j})-(2\underline{i}-\underline{j})}{2}\) | M1A1 | |
| \(\underline{v}_P = 2\underline{i} + \underline{j}\) (one slip in \(\underline{i}\) or \(\underline{j}\) only) | A1 f.t. | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\arctan\frac{1}{2}\) (or \(\arctan 2\) allowed for M1) | M1 | |
| \(26.6°\) only | A1 | (2) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\overrightarrow{OC} = 2\underline{i} - \underline{j} + 5(2\underline{i}+\underline{j})\) or \(6\underline{i}+\underline{j}+3(2\underline{i}+\underline{j})\) | M1 | |
| \(\overrightarrow{OC} = 12\underline{i} + 4\underline{j}\) | A1 f.t. | |
| \( | \overrightarrow{OC} | = \sqrt{12^2 + 4^2}\) |
| \(OC = 12.6\) only or equivalent f.t. answer | A1 f.t. | (4) Given to 1 decimal place; also depends on M1+M1 |
## Question 4:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| At $t=0$: $\underline{r}_P = 2\underline{i} - \underline{j}$; At $t=2$: $\underline{r}_P = 6\underline{i} + \underline{j}$ | | |
| Velocity of P constant $\Rightarrow \underline{v}_P = \frac{(6\underline{i}+\underline{j})-(2\underline{i}-\underline{j})}{2}$ | M1A1 | |
| $\underline{v}_P = 2\underline{i} + \underline{j}$ (one slip in $\underline{i}$ or $\underline{j}$ only) | A1 f.t. | (3) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\arctan\frac{1}{2}$ (or $\arctan 2$ allowed for M1) | M1 | |
| $26.6°$ only | A1 | (2) |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\overrightarrow{OC} = 2\underline{i} - \underline{j} + 5(2\underline{i}+\underline{j})$ or $6\underline{i}+\underline{j}+3(2\underline{i}+\underline{j})$ | M1 | |
| $\overrightarrow{OC} = 12\underline{i} + 4\underline{j}$ | A1 f.t. | |
| $|\overrightarrow{OC}| = \sqrt{12^2 + 4^2}$ | M1 | |
| $OC = 12.6$ only or equivalent f.t. answer | A1 f.t. | (4) Given to 1 decimal place; also depends on M1+M1 |
---4. A particle $P$ moves in a straight line with constant velocity. Initially $P$ is at the point $A$ with position vector $( 2 \mathbf { i } - \mathbf { j } ) \mathrm { m }$ relative to a fixed origin $O$, and 2 s later it is at the point $B$ with position vector $( 6 \mathbf { i } + \mathbf { j } ) \mathrm { m }$.
\begin{enumerate}[label=(\alph*)]
\item Find the velocity of $P$.
\item Find, in degrees to one decimal place, the size of the angle between the direction of motion of $P$ and the vector $\mathbf { i }$.\\
(2 marks)\\
Three seconds after it passes $B$ the particle $P$ reaches the point $C$.
\item Find, in m to one decimal place, the distance $O C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2001 Q4 [9]}}