| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2009 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Find velocity from position |
| Difficulty | Moderate -0.8 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—purely procedural calculus and vector manipulation. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02e Two-dimensional constant acceleration: with vectors3.03c Newton's second law: F=ma one dimension |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{v} = \mathbf{i} + (3-2t)\mathbf{j}\) | M1, A1 | Differentiating r. Allow 1 error. Could use const accn. |
| \(\mathbf{v}(4) = \mathbf{i} - 5\mathbf{j}\) | F1 | Do not award if \(\sqrt{26}\) is given as vel (accept if v given and \(v\) given as well called speed or magnitude) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{a} = -2\mathbf{j}\) | B1 | Diff v. FT their v. Award if \(-2\mathbf{j}\) seen and isw |
| Using N2L \(\mathbf{F} = 1.5 \times (-2\mathbf{j})\) | M1 | Award for \(1.5 \times (\pm\) their a or \(a)\) seen |
| so \(-3\mathbf{j}\) N | A1 | cao. Do not award if final answer is not correct. [Award M1 A1 for \(-3\mathbf{j}\) WW] |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 2+t\) and \(y = 3t - t^2\) | B1 | Must have both but may be implied |
| Substitute \(t = x-2\), so \(y = 3(x-2)-(x-2)^2\) | B1 | cao. isw. Must see the form \(y = \ldots\) |
| \([= (x-2)(5-x)]\) |
# Question 5:
## Part (i)
| $\mathbf{v} = \mathbf{i} + (3-2t)\mathbf{j}$ | M1, A1 | Differentiating **r**. Allow 1 error. Could use const accn. |
| $\mathbf{v}(4) = \mathbf{i} - 5\mathbf{j}$ | F1 | Do not award if $\sqrt{26}$ is given as vel (accept if **v** given and $v$ given as well called speed or magnitude) |
## Part (ii)
| $\mathbf{a} = -2\mathbf{j}$ | B1 | Diff **v**. FT their **v**. Award if $-2\mathbf{j}$ seen and isw |
| Using N2L $\mathbf{F} = 1.5 \times (-2\mathbf{j})$ | M1 | Award for $1.5 \times (\pm$ their **a** or $a)$ seen |
| so $-3\mathbf{j}$ N | A1 | cao. Do not award if final answer is not correct. [Award M1 A1 for $-3\mathbf{j}$ WW] |
## Part (iii)
| $x = 2+t$ and $y = 3t - t^2$ | B1 | Must have both but may be implied |
| Substitute $t = x-2$, so $y = 3(x-2)-(x-2)^2$ | B1 | cao. isw. Must see the form $y = \ldots$ |
| $[= (x-2)(5-x)]$ | | |
---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.
Section B (36 marks)\\
\hfill \mbox{\textit{OCR MEI M1 2009 Q5 [8]}}