| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: kinematics extension |
| Difficulty | Moderate -0.5 This is a straightforward application of Newton's second law (F=ma) in vector form with standard kinematics. Part (i) requires simple division of a vector by a scalar, part (ii) uses basic trigonometry (arctan), and part (iii) applies the SUVAT equation s=ut+½at² with vectors. All steps are routine with no problem-solving insight required, making it slightly easier than average. |
| Spec | 1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10h Vectors in kinematics: uniform acceleration in vector form3.03c Newton's second law: F=ma one dimension |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\begin{pmatrix}6\\9\end{pmatrix} = 1.5\mathbf{a}\) giving \(\mathbf{a} = \begin{pmatrix}4\\6\end{pmatrix}\) so \(\begin{pmatrix}4\\6\end{pmatrix}\) m s\(^{-2}\) | M1 | Use of N2L with an attempt to find a. Condone spurious notation |
| A1 | Must be a vector in proper form. Penalise only once in paper |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Angle is \(\arctan\left(\frac{6}{4}\right)\) | M1 | Use of arctan with their \(\frac{6}{4}\) or \(\frac{4}{6}\) or equiv. May use F. FT their a provided both components are positive and non-zero |
| \(= 56.309...\) so \(56.3°\) (3 s.f.) | F1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Using \(\mathbf{s} = t\mathbf{u} + 0.5t^2\mathbf{a}\) | M1 | Appropriate single uvast (or equivalent sequence). If integration used twice condone omission of \(\mathbf{r}(0)\) but not \(\mathbf{v}(0)\) |
| \(\mathbf{s} = 2\begin{pmatrix}-2\\3\end{pmatrix} + 0.5 \times 4\begin{pmatrix}4\\6\end{pmatrix}\) | A1 | FT their a only |
| so \(\begin{pmatrix}4\\18\end{pmatrix}\) m | A1 | cao. isw for magnitude subsequently found. Vector must be in proper form (penalise only once in paper) |
## Question 6:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\begin{pmatrix}6\\9\end{pmatrix} = 1.5\mathbf{a}$ giving $\mathbf{a} = \begin{pmatrix}4\\6\end{pmatrix}$ so $\begin{pmatrix}4\\6\end{pmatrix}$ m s$^{-2}$ | M1 | Use of N2L with an attempt to find **a**. Condone spurious notation |
| | A1 | Must be a vector in proper form. Penalise only once in paper |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Angle is $\arctan\left(\frac{6}{4}\right)$ | M1 | Use of arctan with their $\frac{6}{4}$ or $\frac{4}{6}$ or equiv. May use **F**. FT their **a** provided both components are positive and non-zero |
| $= 56.309...$ so $56.3°$ (3 s.f.) | F1 | |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Using $\mathbf{s} = t\mathbf{u} + 0.5t^2\mathbf{a}$ | M1 | Appropriate single **uvast** (or equivalent sequence). If integration used twice condone omission of $\mathbf{r}(0)$ but not $\mathbf{v}(0)$ |
| $\mathbf{s} = 2\begin{pmatrix}-2\\3\end{pmatrix} + 0.5 \times 4\begin{pmatrix}4\\6\end{pmatrix}$ | A1 | FT their **a** only |
| so $\begin{pmatrix}4\\18\end{pmatrix}$ m | A1 | cao. isw for magnitude subsequently found. Vector must be in proper form (penalise only once in paper) |6 The force acting on a particle of mass 1.5 kg is given by the vector $\binom { 6 } { 9 } \mathrm {~N}$.\\
(i) Give the acceleration of the particle as a vector.\\
(ii) Calculate the angle that the acceleration makes with the direction $\binom { 1 } { 0 }$.\\
(iii) At a certain point of its motion, the particle has a velocity of $\binom { - 2 } { 3 } \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Calculate the displacement of the particle over the next two seconds.
\hfill \mbox{\textit{OCR MEI M1 Q6 [7]}}