| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2006 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Position from velocity and initial conditions |
| Difficulty | Moderate -0.8 This is a straightforward M1 mechanics question requiring basic differentiation of position vectors and elimination of parameters. Part (i) involves setting y=0 and solving a quadratic, part (ii) is direct differentiation with simple interpretation, and part (iii) is standard parameter elimination. All techniques are routine with no problem-solving insight required, making it easier than average. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.10h Vectors in kinematics: uniform acceleration in vector form3.02f Non-uniform acceleration: using differentiation and integration |
| Answer | Marks | Guidance |
|---|---|---|
| Need j component \(= 0\) so \(18t^2 - 1 = 0\) | M1 | Need not solve |
| \(\Rightarrow t^2 = \frac{1}{18}\). Only one root as \(t > 0\) | E1 | Must establish only one of the two roots is valid |
| or Establish sign change in j component | B1 | |
| Establish only one root | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{v} = 3\mathbf{i} + 36t\mathbf{j}\) | M1, A1 | Differentiate. Allow i or j omitted |
| Need i component \(= 0\) and this never happens | E1 | Clear explanation. Accept 'i component always there' or equiv |
| Answer | Marks | Guidance |
|---|---|---|
| \(x = 3t\) and \(y = 18t^2 - 1\) | B1 | Award for these two expressions seen |
| Eliminate \(t\) to give \(y = 18\left(\frac{x}{3}\right)^2 - 1\) | M1 | \(t\) properly eliminated. Accept any form and brackets missing |
| so \(y = 2x^2 - 1\) | A1 | cao |
# Question 4:
## Part (i)
| Need **j** component $= 0$ so $18t^2 - 1 = 0$ | M1 | Need not solve |
| $\Rightarrow t^2 = \frac{1}{18}$. Only one root as $t > 0$ | E1 | Must establish only one of the two roots is valid |
| **or** Establish sign change in **j** component | B1 | |
| Establish only one root | B1 | |
## Part (ii)
| $\mathbf{v} = 3\mathbf{i} + 36t\mathbf{j}$ | M1, A1 | Differentiate. Allow **i** or **j** omitted |
| Need **i** component $= 0$ and this never happens | E1 | Clear explanation. Accept '**i** component always there' or equiv |
## Part (iii)
| $x = 3t$ and $y = 18t^2 - 1$ | B1 | Award for these two expressions seen |
| Eliminate $t$ to give $y = 18\left(\frac{x}{3}\right)^2 - 1$ | M1 | $t$ properly eliminated. Accept any form and brackets missing |
| so $y = 2x^2 - 1$ | A1 | cao |
---4 Fig. 4 shows the unit vectors $\mathbf { i }$ and $\mathbf { j }$ in the directions of the cartesian axes $\mathrm { O } x$ and $\mathrm { O } y$, respectively. O is the origin of the axes and of position vectors.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{4957086c-fd1c-4cdc-bbdb-1959b3b21b2d-3_383_383_424_840}
\captionsetup{labelformat=empty}
\caption{Fig. 4}
\end{center}
\end{figure}
The position vector of a particle is given by $\mathbf { r } = 3 t \mathbf { i } + \left( 18 t ^ { 2 } - 1 \right) \mathbf { j }$ for $t \geqslant 0$, where $t$ is time.\\
(i) Show that the path of the particle cuts the $x$-axis just once.\\
(ii) Find an expression for the velocity of the particle at time $t$.
Deduce that the particle never travels in the j direction.\\
(iii) Find the cartesian equation of the path of the particle, simplifying your answer.
\hfill \mbox{\textit{OCR MEI M1 2006 Q4 [8]}}