| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2015 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Position vector at time t (constant velocity) |
| Difficulty | Easy -1.3 This is a straightforward mechanics question requiring only direct substitution (t=0 for part a), solving a simple linear equation (part b), and finding magnitude of velocity vector (part c). All steps are routine applications of basic formulas with no problem-solving insight needed, making it easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10e Position vectors: and displacement1.10h Vectors in kinematics: uniform acceleration in vector form |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\mathbf{r} = (-3\mathbf{i} + 4\mathbf{j})\) m | B1 (1) | Allow column vectors throughout |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(3.4 = 2T - 3\) or \(-12 = 4 - 5T\) | M1 A1 | M1 for clear attempt at either component equation; A1 for correct equation without \(\mathbf{i}\)'s and \(\mathbf{j}\)'s |
| \(T = 3.2\) | A1 (3) | If RHS not a single number then M0; if 3.2 and another value obtained and 3.2 not clearly chosen, A0 |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(\mathbf{r} = (-3\mathbf{i} + 4\mathbf{j}) + t(2\mathbf{i} - 5\mathbf{j})\) so \(\mathbf{v} = (2\mathbf{i} - 5\mathbf{j})\) | M1 A1 | First M1 for complete method for finding \(\mathbf{v}\); A1 for \(2\mathbf{i} - 5\mathbf{j}\) only |
| \(\text{speed} = \sqrt{2^2 + (-5)^2} = \sqrt{29} = 5.4 \text{ m s}^{-1}\) or better | M1 A1 (4) | Second M1 for attempt at magnitude \(\sqrt{2^2+(-5)^2}\); A1 for \(\sqrt{29}\) or \(5.4\) or better |
| Alt: \( | \mathbf{s} | = \sqrt{6.4^2 + (-16)^2} = 17.23...\) |
| \(\text{speed} = \frac{17.23}{3.2} = 5.4\) or better | M1 A1 (4) |
## Question 6:
### Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\mathbf{r} = (-3\mathbf{i} + 4\mathbf{j})$ m | B1 (1) | Allow column vectors throughout |
### Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $3.4 = 2T - 3$ or $-12 = 4 - 5T$ | M1 A1 | M1 for clear attempt at either component equation; A1 for correct equation without $\mathbf{i}$'s and $\mathbf{j}$'s |
| $T = 3.2$ | A1 (3) | If RHS not a single number then M0; if 3.2 and another value obtained and 3.2 not clearly chosen, A0 |
### Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\mathbf{r} = (-3\mathbf{i} + 4\mathbf{j}) + t(2\mathbf{i} - 5\mathbf{j})$ so $\mathbf{v} = (2\mathbf{i} - 5\mathbf{j})$ | M1 A1 | First M1 for complete method for finding $\mathbf{v}$; A1 for $2\mathbf{i} - 5\mathbf{j}$ only |
| $\text{speed} = \sqrt{2^2 + (-5)^2} = \sqrt{29} = 5.4 \text{ m s}^{-1}$ or better | M1 A1 (4) | Second M1 for attempt at magnitude $\sqrt{2^2+(-5)^2}$; A1 for $\sqrt{29}$ or $5.4$ or better |
| Alt: $|\mathbf{s}| = \sqrt{6.4^2 + (-16)^2} = 17.23...$ | M1 A1 | |
| $\text{speed} = \frac{17.23}{3.2} = 5.4$ or better | M1 A1 (4) | |
---\begin{enumerate}
\item A particle $P$ is moving with constant velocity. The position vector of $P$ at time $t$ seconds $( t \geqslant 0 )$ is $\mathbf { r }$ metres, relative to a fixed origin $O$, and is given by
\end{enumerate}
$$\mathbf { r } = ( 2 t - 3 ) \mathbf { i } + ( 4 - 5 t ) \mathbf { j }$$
(a) Find the initial position vector of $P$.
The particle $P$ passes through the point with position vector $( 3.4 \mathbf { i } - 12 \mathbf { j } )$ m at time $T$ seconds.\\
(b) Find the value of $T$.\\
(c) Find the speed of $P$.
\hfill \mbox{\textit{Edexcel M1 2015 Q6 [8]}}