| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Cartesian equation of path |
| Difficulty | Standard +0.3 This is a straightforward mechanics question involving position vectors and parametric equations. Part (i) requires simple substitution, part (ii) is routine elimination of parameter, and part (iii) involves finding velocity and setting components equal—all standard M1 techniques with no novel problem-solving required. Slightly above average due to the multi-part nature and the angle condition in part (iii), but still well within typical textbook exercises. |
| Spec | 1.07a Derivative as gradient: of tangent to curve1.07r Chain rule: dy/dx = dy/du * du/dx and connected rates1.10a Vectors in 2D: i,j notation and column vectors1.10e Position vectors: and displacement |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = 2 \Rightarrow t = 4\) | B1 | cao |
| \(t = 4 \Rightarrow y = 16 - 1 = 15\) | F1 | FT their \(t\) and \(y\). Accept \(15\mathbf{j}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(x = \frac{1}{2}t\) and \(y = t^2 - 1\); eliminating \(t\) gives \(y = (2x)^2 - 1 = 4x^2 - 1\) | M1 | Attempt at elimination of expressions for \(x\) and \(y\) in terms of \(t\) |
| \(y = 4x^2 - 1\) | E1 | Accept seeing \((2x)^2 - 1 = 4x^2 - 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| We require \(\frac{dy}{dx} = 1\) | M1 | This may be implied |
| so \(8x = 1\) | B1 | Differentiating correctly to obtain \(8x\) |
| \(x = \frac{1}{8}\) and the point is \(\left(\frac{1}{8}, -\frac{15}{16}\right)\) | A1 | |
| or Differentiate to find v; equate i and j components | M1, M1 | Equating the i and j cpts of their v |
| so \(t = \frac{1}{4}\) and the point is \(\left(\frac{1}{8}, -\frac{15}{16}\right)\) | A1 |
## Question 2:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = 2 \Rightarrow t = 4$ | B1 | cao |
| $t = 4 \Rightarrow y = 16 - 1 = 15$ | F1 | FT their $t$ and $y$. Accept $15\mathbf{j}$ |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $x = \frac{1}{2}t$ and $y = t^2 - 1$; eliminating $t$ gives $y = (2x)^2 - 1 = 4x^2 - 1$ | M1 | Attempt at elimination of expressions for $x$ and $y$ in terms of $t$ |
| $y = 4x^2 - 1$ | E1 | Accept seeing $(2x)^2 - 1 = 4x^2 - 1$ |
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| We require $\frac{dy}{dx} = 1$ | M1 | This may be implied |
| so $8x = 1$ | B1 | Differentiating correctly to obtain $8x$ |
| $x = \frac{1}{8}$ and the point is $\left(\frac{1}{8}, -\frac{15}{16}\right)$ | A1 | |
| **or** Differentiate to find **v**; equate **i** and **j** components | M1, M1 | Equating the **i** and **j** cpts of their **v** |
| so $t = \frac{1}{4}$ and the point is $\left(\frac{1}{8}, -\frac{15}{16}\right)$ | A1 | |
---2 The position vector of a particle at time $t$ is given by
$$\mathbf { r } = \frac { 1 } { 2 } t \mathbf { i } + \left( t ^ { 2 } - 1 \right) \mathbf { j } .$$
referred to an origin $O$ where $\mathbf { i }$ and $\mathbf { j }$ are the standard unit vectors in the directions of the cartesian axes Ox and Oy respectively.\\
(i) Write down the value of $t$ for which the $x$-coordinate of the position of the particle is 2 . Find the $y$-coordinate at this time.\\
(ii) Show that the cartesian equation of the path of the particle is $y = 4 x ^ { 2 } - 1$.\\
(iii) Find the coordinates of the point where the particle is moving at $45 ^ { \circ }$ to both Ox and Oy .
\hfill \mbox{\textit{OCR MEI M1 Q2 [7]}}