| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2013 |
| Session | January |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Constant acceleration vector problems |
| Difficulty | Moderate -0.8 This is a straightforward kinematics question with constant acceleration in 2D using vector notation. It requires direct application of standard SUVAT-style integration formulas (v = u + at, r = r₀ + ut + ½at²) with no problem-solving insight needed. The vector arithmetic is elementary, and finding distance from displacement is a routine Pythagoras calculation. Easier than average A-level mechanics. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{v} = \mathbf{u} + \mathbf{a}t\) | M1 | May be implied by either of the next two answers but not the final answer. Evidence of use of vectors necessary |
| Velocity \(\mathbf{v} = \begin{pmatrix}2\\0\end{pmatrix} + t\begin{pmatrix}-1\\1\end{pmatrix} \left(= \begin{pmatrix}2-t\\t\end{pmatrix}\right)\) | A1 | |
| When \(t=8\), \(\mathbf{v} = \begin{pmatrix}-6\\8\end{pmatrix}\) | A1 | May be implied by the final answer |
| Speed \(= \sqrt{(-6)^2 + 8^2} = 10\text{ m s}^{-1}\) | A1 | Cao but condone no units; Give SC2 for 10 without working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) | M1 | Use of correct equation with substitution. Condone omission of \(\mathbf{r}_0\). Or equivalent equation |
| \(\mathbf{r} = \begin{pmatrix}0\\-2\end{pmatrix} + \begin{pmatrix}2\\0\end{pmatrix}\times8 + \frac{1}{2}\times\begin{pmatrix}-1\\1\end{pmatrix}\times8^2\) | A1 | Condone omission of \(\mathbf{r}_0\). Follow through for their value of \(\mathbf{v}\) |
| \(\mathbf{r} = \begin{pmatrix}-16\\30\end{pmatrix}\) | A1 | Cao but may be implied by a correct final answer |
| Distance \(= 34\text{ m}\) | A1 | Allow for \(35.77...\) from \(\mathbf{r}=\begin{pmatrix}-16\\32\end{pmatrix}\) and \(37.57...\) from \(\mathbf{r}=\begin{pmatrix}-16\\34\end{pmatrix}\) |
## Question 2:
**Part (i)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{v} = \mathbf{u} + \mathbf{a}t$ | M1 | May be implied by either of the next two answers but not the final answer. Evidence of use of vectors necessary |
| Velocity $\mathbf{v} = \begin{pmatrix}2\\0\end{pmatrix} + t\begin{pmatrix}-1\\1\end{pmatrix} \left(= \begin{pmatrix}2-t\\t\end{pmatrix}\right)$ | A1 | |
| When $t=8$, $\mathbf{v} = \begin{pmatrix}-6\\8\end{pmatrix}$ | A1 | May be implied by the final answer |
| Speed $= \sqrt{(-6)^2 + 8^2} = 10\text{ m s}^{-1}$ | A1 | Cao but condone no units; Give SC2 for 10 without working |
**Part (ii)**
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{r} = \mathbf{r}_0 + \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$ | M1 | Use of correct equation with substitution. Condone omission of $\mathbf{r}_0$. Or equivalent equation |
| $\mathbf{r} = \begin{pmatrix}0\\-2\end{pmatrix} + \begin{pmatrix}2\\0\end{pmatrix}\times8 + \frac{1}{2}\times\begin{pmatrix}-1\\1\end{pmatrix}\times8^2$ | A1 | Condone omission of $\mathbf{r}_0$. Follow through for their value of $\mathbf{v}$ |
| $\mathbf{r} = \begin{pmatrix}-16\\30\end{pmatrix}$ | A1 | Cao but may be implied by a correct final answer |
| Distance $= 34\text{ m}$ | A1 | Allow for $35.77...$ from $\mathbf{r}=\begin{pmatrix}-16\\32\end{pmatrix}$ and $37.57...$ from $\mathbf{r}=\begin{pmatrix}-16\\34\end{pmatrix}$ |
---2 In this question, the unit vectors $\binom { 1 } { 0 }$ and $\binom { 0 } { 1 }$ are in the directions east and north.\\
Distance is measured in metres and time, $t$, in seconds.\\
A radio-controlled toy car moves on a flat horizontal surface. A child is standing at the origin and controlling the car.\\
When $t = 0$, the displacement of the car from the origin is $\binom { 0 } { - 2 } \mathrm {~m}$, and the car has velocity $\binom { 2 } { 0 } \mathrm {~ms} ^ { - 1 }$. The acceleration of the car is constant and is $\binom { - 1 } { 1 } \mathrm {~ms} ^ { - 2 }$.\\
(i) Find the velocity of the car at time $t$ and its speed when $t = 8$.\\
(ii) Find the distance of the car from the child when $t = 8$.
\hfill \mbox{\textit{OCR MEI M1 2013 Q2 [8]}}