| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2022 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find position by integrating velocity |
| Difficulty | Standard +0.3 This is a straightforward mechanics question requiring differentiation to find acceleration (finding k from magnitude), integration of velocity to find position, and solving a simple geometric condition. All steps are standard techniques with no novel insight required, making it slightly easier than average. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{a} = \dfrac{d\mathbf{v}}{dt} = 2kt\mathbf{i} + 6\mathbf{j}\) | M1 | Differentiating the v vector |
| When \(t=2\): \(\mathbf{a} = 2\times 2k\mathbf{i} + 6\mathbf{j}\) | M1 | Substituting \(t=2\) into their a vector |
| \( | \mathbf{a} | = \sqrt{(4k)^2 + 6^2} = 10\) |
| \(16k^2 + 36 = 100\), so \(k = 2\) | A1 | Must choose positive value if two values seen |
| [4] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{r} = \int \mathbf{v}\, dt = \dfrac{kt^3}{3}\mathbf{i} + 3t^2\mathbf{j} + \mathbf{c}\) | M1 | Integrating with their \(k\) or general \(k\); Allow for vector or both components separately integrated |
| Particle at origin when \(t=0\) so \(\mathbf{c} = \mathbf{0}\); So \(\mathbf{r} = \dfrac{kt^3}{3}\mathbf{i} + 3t^2\mathbf{j} = \left[\dfrac{2t^3}{3}\mathbf{i} + 3t^2\mathbf{j}\right]\) | A1 | Condone missing \(+\mathbf{c}\) or \(+c\); FT their \(k\) if positive or general \(k\) used; Must be in vector form |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Northeast when \(\mathbf{i}\) component \(= \mathbf{j}\) component: \(\dfrac{2t^3}{3} = 3t^2\) | M1 | FT their r |
| \(t = 4.5\) s \([t=0\) rejected as particle is at origin\(]\) | A1 | www |
| [2] |
# Question 9:
## Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{a} = \dfrac{d\mathbf{v}}{dt} = 2kt\mathbf{i} + 6\mathbf{j}$ | M1 | Differentiating the **v** vector |
| When $t=2$: $\mathbf{a} = 2\times 2k\mathbf{i} + 6\mathbf{j}$ | M1 | Substituting $t=2$ into their **a** vector |
| $|\mathbf{a}| = \sqrt{(4k)^2 + 6^2} = 10$ | M1 | Equate magnitude of their **a** vector to 10 |
| $16k^2 + 36 = 100$, so $k = 2$ | A1 | Must choose positive value if two values seen |
| **[4]** | | |
## Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{r} = \int \mathbf{v}\, dt = \dfrac{kt^3}{3}\mathbf{i} + 3t^2\mathbf{j} + \mathbf{c}$ | M1 | Integrating with their $k$ or general $k$; Allow for vector or both components separately integrated |
| Particle at origin when $t=0$ so $\mathbf{c} = \mathbf{0}$; So $\mathbf{r} = \dfrac{kt^3}{3}\mathbf{i} + 3t^2\mathbf{j} = \left[\dfrac{2t^3}{3}\mathbf{i} + 3t^2\mathbf{j}\right]$ | A1 | Condone missing $+\mathbf{c}$ or $+c$; FT their $k$ if positive or general $k$ used; Must be in vector form |
| **[2]** | | |
## Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Northeast when $\mathbf{i}$ component $= \mathbf{j}$ component: $\dfrac{2t^3}{3} = 3t^2$ | M1 | FT their **r** |
| $t = 4.5$ s $[t=0$ rejected as particle is at origin$]$ | A1 | www |
| **[2]** | | |9 In this question, the vectors $\mathbf { i }$ and $\mathbf { j }$ are directed east and north respectively.\\
The velocity $\mathbf { v } \mathrm { ms } ^ { - 1 }$ of a particle at time $t$ s is given by $\mathbf { v } = k t ^ { 2 } \mathbf { i } + 6 t$, where $k$ is a positive constant. The magnitude of the acceleration when $t = 2$ is $10 \mathrm {~ms} ^ { - 2 }$.
\begin{enumerate}[label=(\alph*)]
\item Calculate the value of $k$.
The particle is at the origin when $t = 0$.
\item Determine an expression for the position vector of the particle at time $t$.
\item Determine the time when the particle is directly north-east of the origin.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2022 Q9 [8]}}