| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Parallel or perpendicular vectors condition |
| Difficulty | Standard +0.3 This is a straightforward mechanics question requiring (a) setting up a parallel vectors condition (ratio of components equal) leading to a quadratic equation, and (b) solving a circle inequality |r(t)| < 5. Both parts use standard techniques with clear methods, making it slightly easier than average for A-level. |
| Spec | 1.02c Simultaneous equations: two variables by elimination and substitution1.10d Vector operations: addition and scalar multiplication3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(//\) to \(\mathbf{i}-\mathbf{j}\): \(2q^2 - 3 = -(q+2)\) | M1 | |
| \(2q^2 + q - 1 = 0 \therefore (2q-1)(q+1) = 0\) | M1 A1 | |
| \(q = \frac{1}{2},\ q = -1\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(q = -1 \therefore\) vel \(= -\mathbf{i}+\mathbf{j}\) | B1 | |
| At time \(t\), pos'n vector is \((6\mathbf{i}-\mathbf{j}) + t(-\mathbf{i}+\mathbf{j}) = (6-t)\mathbf{i}+(t-1)\mathbf{j}\) | A1 | |
| \(d^2 = (6-t)^2 + (t-1)^2\) and \(d^2 < 5^2\) | M1 | |
| \(\therefore t^2 - 12t + 36 + t^2 - 2t + 1 < 25\) | M1 | |
| \(t^2 - 7t + 6 < 0 \therefore (t-1)(t-6) < 0\) | A1 | |
| \(1 < t < 6\) i.e. 5 seconds | A1 | (10) |
## Question 4:
### Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $//$ to $\mathbf{i}-\mathbf{j}$: $2q^2 - 3 = -(q+2)$ | M1 | |
| $2q^2 + q - 1 = 0 \therefore (2q-1)(q+1) = 0$ | M1 A1 | |
| $q = \frac{1}{2},\ q = -1$ | A1 | |
### Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $q = -1 \therefore$ vel $= -\mathbf{i}+\mathbf{j}$ | B1 | |
| At time $t$, pos'n vector is $(6\mathbf{i}-\mathbf{j}) + t(-\mathbf{i}+\mathbf{j}) = (6-t)\mathbf{i}+(t-1)\mathbf{j}$ | A1 | |
| $d^2 = (6-t)^2 + (t-1)^2$ and $d^2 < 5^2$ | M1 | |
| $\therefore t^2 - 12t + 36 + t^2 - 2t + 1 < 25$ | M1 | |
| $t^2 - 7t + 6 < 0 \therefore (t-1)(t-6) < 0$ | A1 | |
| $1 < t < 6$ i.e. 5 seconds | A1 | **(10)** |
---4. In this question, $\mathbf { i }$ and $\mathbf { j }$ are perpendicular horizontal unit vectors and $O$ is a fixed origin.
A pedestrian moves with constant velocity $\left[ \left( 2 q ^ { 2 } - 3 \right) \mathbf { i } + ( q + 2 ) \mathbf { j } \right] \mathrm { ms } ^ { - 1 }$.
Given that the velocity of the pedestrian is parallel to the vector $( \mathbf { i } - \mathbf { j } )$,
\begin{enumerate}[label=(\alph*)]
\item Show that one possible value of $q$ is ${ } ^ { - } 1$ and find the other possible value of $q$.
Given that $q = { } ^ { - } 1$, and that the pedestrian started walking at the point with position vector $( 6 \mathbf { i } - \mathbf { j } ) \mathrm { m }$,
\item find the length of time for which the pedestrian is less than 5 m from $O$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 Q4 [10]}}