| Exam Board | OCR MEI |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2005 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (vectors) |
| Type | Cartesian equation of path |
| Difficulty | Moderate -0.3 This is a straightforward mechanics question involving position vectors and velocity. Part (i) requires simple substitution, part (ii) is routine elimination of parameter, and part (iii) involves differentiating to find velocity and setting components equal (for 45° angle). All techniques are standard M1 material with no novel problem-solving required, making it slightly 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 |
|---|---|---|
| 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 \(t\) |
| E1 | Accept seeing \((2x)^2 - 1 = 4x^2 - 1\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Either: 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 components of their v |
| so \(t = \frac{1}{4}\) and the point is \(\left(\frac{1}{8}, -\frac{15}{16}\right)\) | A1 |
## Question 5:
### 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 $t$ |
| | E1 | Accept seeing $(2x)^2 - 1 = 4x^2 - 1$ |
### Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| **Either:** 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** components of **their v** |
| so $t = \frac{1}{4}$ and the point is $\left(\frac{1}{8}, -\frac{15}{16}\right)$ | A1 | |
---5 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 $\mathbf { O }$ where $\mathbf { i }$ and $\mathbf { j }$ are the standard unit vectors in the directions of the cartesian axes $\mathrm { O } x$ 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 $\mathrm { O } x$ and $\mathrm { O } y$.
Section B (36 marks)\\
\hfill \mbox{\textit{OCR MEI M1 2005 Q5 [7]}}