| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find velocity from position |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question requiring standard differentiation of position vectors to find velocity and acceleration, followed by applying F=ma and eliminating the parameter. All techniques are routine for M1 level with no problem-solving insight needed, making it slightly easier than average but not trivial due to the multi-part nature and vector component work. |
| Spec | 1.07s Parametric and implicit differentiation1.10h Vectors in kinematics: uniform acceleration in vector form3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\begin{pmatrix}-1\\14\\-8\end{pmatrix}+\begin{pmatrix}3\\-9\\10\end{pmatrix}+\mathbf{F}=4\begin{pmatrix}-1\\2\\4\end{pmatrix}\) | M1 | N2L. Allow sign errors. Do not condone \(\mathbf{F}=mg\mathbf{a}\). Allow one given force omitted |
| M1 | Attempt to add \(\begin{pmatrix}-1\\14\\-8\end{pmatrix}\) and \(\begin{pmatrix}3\\-9\\10\end{pmatrix}\) | |
| \(\mathbf{F}=\begin{pmatrix}-6\\3\\14\end{pmatrix}\) | A1 | Two components correct |
| A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\mathbf{v}=\begin{pmatrix}-3\\3\\6\end{pmatrix}+3\begin{pmatrix}-1\\2\\4\end{pmatrix}=\begin{pmatrix}-6\\9\\18\end{pmatrix}\) so \(\begin{pmatrix}-6\\9\\18\end{pmatrix}\) m s\(^{-1}\) | M1 | \(\mathbf{v}=\mathbf{u}+t\mathbf{a}\) with given \(\mathbf{u}\) and \(\mathbf{a}\). If integration used, require arbitrary constant |
| A1 | cao isw | |
| speed is \(\sqrt{(-6)^2+9^2+18^2}=21\) m s\(^{-1}\) | M1 | Allow \(-6^2\) even if interpreted as \(-36\). Only FT their \(\mathbf{v}\) |
| F1 | FT their \(\mathbf{v}\) only. [Award M1 F1 for 21 seen WWW] |
## Question 5:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\begin{pmatrix}-1\\14\\-8\end{pmatrix}+\begin{pmatrix}3\\-9\\10\end{pmatrix}+\mathbf{F}=4\begin{pmatrix}-1\\2\\4\end{pmatrix}$ | M1 | N2L. Allow sign errors. Do not condone $\mathbf{F}=mg\mathbf{a}$. Allow one given force omitted |
| | M1 | Attempt to add $\begin{pmatrix}-1\\14\\-8\end{pmatrix}$ and $\begin{pmatrix}3\\-9\\10\end{pmatrix}$ |
| $\mathbf{F}=\begin{pmatrix}-6\\3\\14\end{pmatrix}$ | A1 | Two components correct |
| | A1 | cao |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\mathbf{v}=\begin{pmatrix}-3\\3\\6\end{pmatrix}+3\begin{pmatrix}-1\\2\\4\end{pmatrix}=\begin{pmatrix}-6\\9\\18\end{pmatrix}$ so $\begin{pmatrix}-6\\9\\18\end{pmatrix}$ m s$^{-1}$ | M1 | $\mathbf{v}=\mathbf{u}+t\mathbf{a}$ with given $\mathbf{u}$ and $\mathbf{a}$. If integration used, require arbitrary constant |
| | A1 | cao isw |
| speed is $\sqrt{(-6)^2+9^2+18^2}=21$ m s$^{-1}$ | M1 | Allow $-6^2$ even if interpreted as $-36$. Only FT their $\mathbf{v}$ |
| | F1 | FT their $\mathbf{v}$ only. [Award M1 F1 for 21 seen WWW] |5 The position vector of a toy boat of mass 1.5 kg is modelled as $\mathbf { r } = ( 2 + t ) \mathbf { i } + \left( 3 t - t ^ { 2 } \right) \mathbf { j }$ where lengths are in metres, $t$ is the time in seconds, $\mathbf { i }$ and $\mathbf { j }$ are horizontal, perpendicular unit vectors and the origin is O .\\
(i) Find the velocity of the boat when $t = 4$.\\
(ii) Find the acceleration of the boat and the horizontal force acting on the boat.\\
(iii) Find the cartesian equation of the path of the boat referred to $x$ - and $y$-axes in the directions of $\mathbf { i }$ and $\mathbf { j }$, respectively, with origin O . You are not required to simplify your answer.
\hfill \mbox{\textit{OCR MEI M1 Q5 [8]}}