| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | September |
| Marks | 8 |
| Topic | Variable acceleration (vectors) |
| Type | Find unknown constant from motion condition |
| Difficulty | Standard +0.3 This is a straightforward vector mechanics question requiring differentiation of position to find velocity and acceleration, then solving simple equations. Part (i) uses the condition that velocity has zero j-component when parallel to i. Part (ii) involves finding magnitude of acceleration and solving a quadratic. All steps are routine applications of standard techniques with no novel insight required. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02a Kinematics language: position, displacement, velocity, acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{v} = (8t - k)\mathbf{i} + (12t^2 + 4kt - 8)\mathbf{j}\) | M1 | Attempt to differentiate – all powers reduced by 1 |
| A1 | ||
| \(12 \times 2^2 + 4 \times k \times 2 - 8 = 0\) | M1 | Equate j-component to 0 for \(t = 2\) |
| \(k = -5\) | A1 | AG – sufficient working must be shown to indicate that a correct method has been applied |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{a} = 8\mathbf{i} + (24t + 4k)\mathbf{j}\) | B1ft | Differentiate their \(\mathbf{v}\) correctly |
| \(8^2 + (24t + 4k)^2 = 100\) | M1 | Setting up an equation for the magnitude = 10 or magnitude\(^2\) = 100 |
| \(24t + 4(-5) = \pm 6 \Rightarrow t = \ldots\) | M1 | Re-arrange and attempt to solve for \(t\) – leading to at least one value of \(t\) |
| \(t = \frac{13}{12}\) and \(\frac{7}{12}\) | A1 | Both correct |
### (i)
$\mathbf{v} = (8t - k)\mathbf{i} + (12t^2 + 4kt - 8)\mathbf{j}$ | M1 | Attempt to differentiate – all powers reduced by 1
| A1 |
$12 \times 2^2 + 4 \times k \times 2 - 8 = 0$ | M1 | Equate j-component to 0 for $t = 2$
$k = -5$ | A1 | AG – sufficient working must be shown to indicate that a correct method has been applied
**[4 marks total]**
### (ii)
$\mathbf{a} = 8\mathbf{i} + (24t + 4k)\mathbf{j}$ | B1ft | Differentiate their $\mathbf{v}$ correctly
$8^2 + (24t + 4k)^2 = 100$ | M1 | Setting up an equation for the magnitude = 10 or magnitude$^2$ = 100
$24t + 4(-5) = \pm 6 \Rightarrow t = \ldots$ | M1 | Re-arrange and attempt to solve for $t$ – leading to at least one value of $t$
$t = \frac{13}{12}$ and $\frac{7}{12}$ | A1 | Both correct
**[4 marks total]**
---8 At time $t$ seconds a particle $P$ has position vector $\mathbf { r }$ metres, with respect to a fixed origin $O$, where
$$\mathbf { r } = \left( 4 t ^ { 2 } - k t + 5 \right) \mathbf { i } + \left( 4 t ^ { 3 } + 2 k t ^ { 2 } - 8 t \right) \mathbf { j } , \quad t \geqslant 0 .$$
When $t = 2 , P$ is moving parallel to the vector $\mathbf { i }$.\\
(i) Show that $k = - 5$.\\
(ii) Find the values of $t$ when the magnitude of the acceleration of $P$ is $10 \mathrm {~ms} ^ { - 2 }$.
\hfill \mbox{\textit{OCR H240/03 2018 Q8 [8]}}