| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Bearing and speed from velocity vector |
| Difficulty | Moderate -0.3 This is a straightforward M1 vectors question requiring standard techniques: calculating magnitude for speed, using arctan for bearing, and setting components equal to zero or proportional for parallel motion. All parts follow routine procedures with no conceptual challenges, making it slightly easier than average but not trivial due to the multi-part nature and need for careful algebraic manipulation. |
| Spec | 1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication |
| Answer | Marks |
|---|---|
| \(t = 0\) gives \(\mathbf{v} = \mathbf{i} - 3\mathbf{j}\) | B1 |
| speed \(= \sqrt{1^2 + (-3)^2}\) | M1 |
| \(= \sqrt{10} = 3.2\) or better | A1 |
| (3) |
| Answer | Marks |
|---|---|
| \(t = 2\) gives \(\mathbf{v} = (-3\mathbf{i} + 3\mathbf{j})\) | M1 |
| Bearing is \(315°\) | A1 |
| (2) |
| Answer | Marks |
|---|---|
| \(1 - 2t = 0 \Rightarrow t = 0.5\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(-(3t - 3) = -3(1 - 2t)\) | M1 A1 |
| Solving for \(t\) | DM1 |
| \(t = 2/3, 0.67\) or better | A1 |
| (6) | |
| [11] |
## Part (a)
$t = 0$ gives $\mathbf{v} = \mathbf{i} - 3\mathbf{j}$ | B1
speed $= \sqrt{1^2 + (-3)^2}$ | M1
$= \sqrt{10} = 3.2$ or better | A1
| (3)
## Part (b)
$t = 2$ gives $\mathbf{v} = (-3\mathbf{i} + 3\mathbf{j})$ | M1
Bearing is $315°$ | A1
| (2)
## Part (c)(i)
$1 - 2t = 0 \Rightarrow t = 0.5$ | M1 A1
## Part (c)(ii)
$-(3t - 3) = -3(1 - 2t)$ | M1 A1
Solving for $t$ | DM1
$t = 2/3, 0.67$ or better | A1
| (6)
| [11]
**Notes for Question 7:**
**Q7(a)**
- B1 for $\mathbf{i} - 3\mathbf{j}$.
- M1 for $\sqrt{\text{(sum of squares of cpt.s)}}$
- A1 for $\sqrt{10}, 3.2$ or better
**Q7(b)**
- M1 for clear attempt to sub $t = 2$ into given expression.
- A1 for 315.
**(c)(i)**
- First M1 for $1 - 2t = 0$.
- First A1 for $t = 0.5$.
- N.B. If they offer two solutions, by equating both the $\mathbf{i}$ and $\mathbf{j}$ components to zero, give M0.
**Q7(c)(ii)**
- Second M1 for $\frac{1-2t}{3t-3} = \pm(\frac{-1}{-3})$ o.e. (Must be an equation in $t$ only)
- First A1 for a correct equation (the + sign)
- Third M1, dependent on first M1, for solving for $t$.
- Second A1 for $2/3, 0.67$ or better.
---[In this question, the horizontal unit vectors $\mathbf{i}$ and $\mathbf{j}$ are directed due east and due north respectively.]
The velocity, $\mathbf{v} \text{ m s}^{-1}$, of a particle $P$ at time $t$ seconds is given by
$$\mathbf{v} = (1 - 2t)\mathbf{i} + (3t - 3)\mathbf{j}$$
\begin{enumerate}[label=(\alph*)]
\item Find the speed of $P$ when $t = 0$ [3]
\item Find the bearing on which $P$ is moving when $t = 2$ [2]
\item Find the value of $t$ when $P$ is moving
\begin{enumerate}[label=(\roman*)]
\item parallel to $\mathbf{j}$,
\item parallel to $(-\mathbf{i} - 3\mathbf{j})$. [6]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2013 Q7 [11]}}