| Exam Board | OCR MEI |
|---|---|
| Module | Paper 1 (Paper 1) |
| Year | 2021 |
| Session | November |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Distance between two positions |
| Difficulty | Standard +0.3 This is a straightforward mechanics question requiring integration of velocity to find position (part a), elimination of parameter t for Cartesian equation (part b), and using F=ma with differentiation of velocity for acceleration (part c). All steps are standard A-level techniques with no novel problem-solving required, making it slightly easier than average. |
| Spec | 1.10c Magnitude and direction: of vectors1.10h Vectors in kinematics: uniform acceleration in vector form3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{r} = \int(\mathbf{v})\, dt = \int(3t^2\mathbf{i} + 7\mathbf{j})\, dt = t^3\mathbf{i} + 7t\mathbf{j} + \mathbf{c}\) | M1* | Attempt to integrate both components; condone missing \(+\mathbf{c}\) |
| When \(t = 0\), \(\mathbf{r} = -\mathbf{i} + 2\mathbf{j} = \mathbf{c}\) | M1 dep* | Using initial conditions |
| So \(\mathbf{r} = (t^3 - 1)\mathbf{i} + (7t + 2)\mathbf{j}\) | M1 dep* | Using \(t = 2\) to find position vector or values for \(x\) and \(y\) |
| When \(t = 2\): \(\mathbf{r} = (2^3 - 1)\mathbf{i} + (7 \times 2 + 2)\mathbf{j} = 7\mathbf{i} + 16\mathbf{j}\) | A1 | Accept vector form or two clear components |
| distance \(= \sqrt{7^2 + 16^2} = \sqrt{305} = 17.5\) m | M1, A1 [6] | Using Pythagoras; FT their components |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Using \(x = t^3 - 1\), \(y = 7t + 2\) | M1 | Extracting equations for \(x\) and \(y\) from their displacement vector |
| Substitute \(t = \frac{y-2}{7}\) into equation for \(x\) | M1 dep | Attempt to eliminate \(t\). Equivalent form \(y = 7(x+1)^{\frac{1}{3}} + 2\) for M1M1A0 |
| \(x = \left(\frac{y-2}{7}\right)^3 - 1\) AG | A1 [3] | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{a} = \frac{d\mathbf{v}}{dt} = 6t\mathbf{i}\) | M1* | Must be vector acceleration |
| When \(t = 2\), \(\mathbf{a} = 12\mathbf{i}\) | M1* | Evaluating when \(t = 2\). \(a = 12\) is sufficient here |
| The force must be in that direction, so \(\mathbf{F} = 48\mathbf{i} = m\mathbf{a}\) | M1* | Newton's second law in vector form, or in \(x\)-direction only. If their a has two non-zero components, allow for dividing 48 by the magnitude of their a |
| So \(m = 4\) kg | A1 dep* [4] | cao |
## Question 13:
### Part (a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{r} = \int(\mathbf{v})\, dt = \int(3t^2\mathbf{i} + 7\mathbf{j})\, dt = t^3\mathbf{i} + 7t\mathbf{j} + \mathbf{c}$ | M1* | Attempt to integrate both components; condone missing $+\mathbf{c}$ |
| When $t = 0$, $\mathbf{r} = -\mathbf{i} + 2\mathbf{j} = \mathbf{c}$ | M1 dep* | Using initial conditions |
| So $\mathbf{r} = (t^3 - 1)\mathbf{i} + (7t + 2)\mathbf{j}$ | M1 dep* | Using $t = 2$ to find position vector or values for $x$ and $y$ |
| When $t = 2$: $\mathbf{r} = (2^3 - 1)\mathbf{i} + (7 \times 2 + 2)\mathbf{j} = 7\mathbf{i} + 16\mathbf{j}$ | A1 | Accept vector form or two clear components |
| distance $= \sqrt{7^2 + 16^2} = \sqrt{305} = 17.5$ m | M1, A1 [6] | Using Pythagoras; FT their components |
### Part (b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Using $x = t^3 - 1$, $y = 7t + 2$ | M1 | Extracting equations for $x$ and $y$ from their displacement vector |
| Substitute $t = \frac{y-2}{7}$ into equation for $x$ | M1 dep | Attempt to eliminate $t$. Equivalent form $y = 7(x+1)^{\frac{1}{3}} + 2$ for M1M1A0 |
| $x = \left(\frac{y-2}{7}\right)^3 - 1$ **AG** | A1 [3] | cao |
### Part (c):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{a} = \frac{d\mathbf{v}}{dt} = 6t\mathbf{i}$ | M1* | Must be vector acceleration |
| When $t = 2$, $\mathbf{a} = 12\mathbf{i}$ | M1* | Evaluating when $t = 2$. $a = 12$ is sufficient here |
| The force must be in that direction, so $\mathbf{F} = 48\mathbf{i} = m\mathbf{a}$ | M1* | Newton's second law in vector form, or in $x$-direction only. If their **a** has two non-zero components, allow for dividing 48 by the magnitude of their **a** |
| So $m = 4$ kg | A1 dep* [4] | cao |
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Please share the actual mark scheme pages if you would like me to extract and format the content.13 In this question $\mathbf { i }$ and $\mathbf { j }$ are unit vectors in the $x$ - and $y$-directions respectively.\\
The velocity of a particle at time $t \mathrm {~s}$ is given by $\left( 3 t ^ { 2 } \mathbf { i } + 7 \mathbf { j } \right) \mathrm { m } \mathrm { s } ^ { - 1 }$. At time $t = 0$ the position of the particle with respect to the origin is $( - \mathbf { i } + 2 \mathbf { j } ) \mathrm { m }$.
\begin{enumerate}[label=(\alph*)]
\item Determine the distance of the particle from the origin when $t = 2$.
\item Show that the cartesian equation of the path of the particle is $x = \left( \frac { y - 2 } { 7 } \right) ^ { 3 } - 1$.
\item At time $t = 2$, the magnitude of the resultant force acting on the particle is 48 N .
Find the mass of the particle.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI Paper 1 2021 Q13 [13]}}