| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2016 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | SUVAT in 2D & Gravity |
| Type | Particle motion: 2D constant acceleration |
| Difficulty | Moderate -0.8 This is a straightforward mechanics question requiring differentiation of position vectors to find velocity and acceleration, then basic algebraic manipulation to eliminate the parameter. All steps are routine applications of standard SUVAT vector techniques with no problem-solving insight needed, making it easier than average. |
| Spec | 3.02a Kinematics language: position, displacement, velocity, acceleration3.02e Two-dimensional constant acceleration: with vectors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(u = \begin{pmatrix} 2 \\ 6 \end{pmatrix}\) | B1 | |
| \(a = \begin{pmatrix} 0 \\ -8 \end{pmatrix}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(v = u + at\) | ||
| \(t = 2 \Rightarrow v = \begin{pmatrix} 2 \\ 6 \end{pmatrix} + \begin{pmatrix} 0 \\ -8 \end{pmatrix} \times 2\) | M1 | Or equivalent. FT for their \(u\) and \(a\) |
| \(= \begin{pmatrix} 2 \\ -10 \end{pmatrix}\) | A1 | Continue the FT for this mark |
| Speed \(= \sqrt{2^2 + (-10)^2} = 10.2\) m s\(^{-1}\) (to 3 sf) | B1 | FT from their \(v\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = ut \Rightarrow x = 2t \Rightarrow t = \frac{x}{2}\) | M1 | This mark may also be obtained for substituting \(x\) for \(2t\) in the expression for \(y\). |
| \(y = 6t - 4t^2\) | B1 | |
| \(y = 6 \times \frac{x}{2} - 4 \times \left(\frac{x}{2}\right)^2 = 3x - x^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(x = 2t\) | ||
| Substitute for \(x\) in given answer | M1 | |
| \(y = 3x - x^2 \Rightarrow y = 6t - 4t^2\) | A1 | |
| This is the given expression for \(y\) | B1 |
**(i)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $u = \begin{pmatrix} 2 \\ 6 \end{pmatrix}$ | B1 | |
| $a = \begin{pmatrix} 0 \\ -8 \end{pmatrix}$ | B1 | |
**(ii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $v = u + at$ | | |
| $t = 2 \Rightarrow v = \begin{pmatrix} 2 \\ 6 \end{pmatrix} + \begin{pmatrix} 0 \\ -8 \end{pmatrix} \times 2$ | M1 | Or equivalent. FT for their $u$ and $a$ |
| $= \begin{pmatrix} 2 \\ -10 \end{pmatrix}$ | A1 | Continue the FT for this mark |
| Speed $= \sqrt{2^2 + (-10)^2} = 10.2$ m s$^{-1}$ (to 3 sf) | B1 | FT from their $v$ |
**(iii)**
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = ut \Rightarrow x = 2t \Rightarrow t = \frac{x}{2}$ | M1 | This mark may also be obtained for substituting $x$ for $2t$ in the expression for $y$. |
| $y = 6t - 4t^2$ | B1 | |
| $y = 6 \times \frac{x}{2} - 4 \times \left(\frac{x}{2}\right)^2 = 3x - x^2$ | A1 | |
---
# Question 4(iii) - Alternative Method
| Answer | Mark | Guidance |
|--------|------|----------|
| $x = 2t$ | | |
| Substitute for $x$ in given answer | M1 | |
| $y = 3x - x^2 \Rightarrow y = 6t - 4t^2$ | A1 | |
| This is the given expression for $y$ | B1 | |
---4 A particle is initially at the origin, moving with velocity $\mathbf { u }$. Its acceleration $\mathbf { a }$ is constant.
At time $t$ its displacement from the origin is $\mathbf { r } = \binom { x } { y }$, where $\binom { x } { y } = \binom { 2 } { 6 } t - \binom { 0 } { 4 } t ^ { 2 }$.\\
(i) Write down $\mathbf { u }$ and $\mathbf { a }$ as column vectors.\\
(ii) Find the speed of the particle when $t = 2$.\\
(iii) Show that the equation of the path of the particle is $y = 3 x - x ^ { 2 }$.
\hfill \mbox{\textit{OCR MEI M1 2016 Q4 [8]}}