| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2021 |
| Session | October |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Position vector at time t (constant velocity) |
| Difficulty | Moderate -0.8 This is a straightforward M1 mechanics question requiring basic vector operations: substituting t=2 to find a specific position vector, calculating an angle using dot product (or component ratios), and solving a quadratic equation from |r|=2.5. All techniques are standard with no problem-solving insight needed, making it easier than average but not trivial due to the algebraic manipulation required. |
| Spec | 1.10b Vectors in 3D: i,j,k notation1.10c Magnitude and direction: of vectors1.10f Distance between points: using position vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\mathbf{r} = -\mathbf{i} - 3\mathbf{j}\) | B1 | cao |
| \(\tan\theta = \pm\dfrac{1}{3}\) or \(\pm\dfrac{3}{1}\) | M1 | For any trig ratio of a relevant angle from their \(\mathbf{r}\) (trig ratio could be implied by a relevant angle; cosine could come from scalar product of their \(\mathbf{r}\) with \(\mathbf{j}\)) |
| \(162°\) or \(198°\) nearest degree | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sqrt{(t-3)^2 + (1-2t)^2} = 2.5\) | M1 | oe |
| \(4t^2 - 8t + 3 = 0\ \ (5t^2 - 10t + 3.75 = 0)\) | DM1A1 | Dependent on first M1, for simplifying to 3-term quadratic or to a form from completing the square. A1 correct quadratic |
| \(t = \dfrac{1}{2}\) or \(\dfrac{3}{2}\) isw | M(A)1 A1 | M(A)1 for \(t = 0.5\); A1 for \(t = 1.5\) |
# Question 4:
## Part 4(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\mathbf{r} = -\mathbf{i} - 3\mathbf{j}$ | B1 | cao |
| $\tan\theta = \pm\dfrac{1}{3}$ or $\pm\dfrac{3}{1}$ | M1 | For any trig ratio of a relevant angle from their $\mathbf{r}$ (trig ratio could be implied by a relevant angle; cosine could come from scalar product of their $\mathbf{r}$ with $\mathbf{j}$) |
| $162°$ or $198°$ nearest degree | A1 | cao |
## Part 4(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sqrt{(t-3)^2 + (1-2t)^2} = 2.5$ | M1 | oe |
| $4t^2 - 8t + 3 = 0\ \ (5t^2 - 10t + 3.75 = 0)$ | DM1A1 | Dependent on first M1, for simplifying to 3-term quadratic or to a form from completing the square. A1 correct quadratic |
| $t = \dfrac{1}{2}$ or $\dfrac{3}{2}$ isw | M(A)1 A1 | M(A)1 for $t = 0.5$; A1 for $t = 1.5$ |
---\begin{enumerate}
\item The position vector, $\mathbf { r }$ metres, of a particle $P$ at time $t$ seconds, relative to a fixed origin $O$, is given by
\end{enumerate}
$$\mathbf { r } = ( t - 3 ) \mathbf { i } + ( 1 - 2 t ) \mathbf { j }$$
(a) Find, to the nearest degree, the size of the angle between $\mathbf { r }$ and the vector $\mathbf { j }$, when $t = 2$\\
(b) Find the values of $t$ for which the distance of $P$ from $O$ is 2.5 m .
\hfill \mbox{\textit{Edexcel M1 2021 Q4 [8]}}