| Exam Board | OCR |
|---|---|
| Module | H240/03 (Pure Mathematics and Mechanics) |
| Year | 2018 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: kinematics extension |
| Difficulty | Moderate -0.8 This is a straightforward application of Newton's second law in vector form with standard kinematics. Students need to sum forces (including weight), apply F=ma to find acceleration, then use SUVAT equations - all routine procedures with no problem-solving insight required. Easier than average due to its mechanical, step-by-step nature. |
| Spec | 1.10h Vectors in kinematics: uniform acceleration in vector form3.02e Two-dimensional constant acceleration: with vectors3.03d Newton's second law: 2D vectors |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{g} = \begin{pmatrix} 0 \\ -9.8 \end{pmatrix}\) | B1 | |
| \(\begin{pmatrix} 15 \\ -8 \end{pmatrix} + \begin{pmatrix} -7 \\ -2 \end{pmatrix} + 5\begin{pmatrix} 0 \\ -9.8 \end{pmatrix} = 5\mathbf{a}\) | M1 | Use of \(\mathbf{F} = m\mathbf{a}\) with correct \(m\) and two terms of \(\mathbf{F}\) correct |
| \(\mathbf{a} = \begin{pmatrix} 1.6 \\ -11.8 \end{pmatrix}\) or \(\begin{pmatrix} 1.6 \\ -g-2 \end{pmatrix}\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| \(\mathbf{s} = \frac{1}{2}\begin{pmatrix} 1.6 \\ -11.8 \end{pmatrix}(10)^2\) | M1 | Use of \(\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2\) with \(t = 10\) |
| \(\mathbf{s} = \begin{pmatrix} 80 \\ -590 \end{pmatrix}\) | A1ft | \(50\mathbf{a}\) |
| Position vector is \(\begin{pmatrix} 82 \\ -545 \end{pmatrix}\) | A1 [3] |
## Question 8:
### Part (i):
$\mathbf{g} = \begin{pmatrix} 0 \\ -9.8 \end{pmatrix}$ | B1 | |
$\begin{pmatrix} 15 \\ -8 \end{pmatrix} + \begin{pmatrix} -7 \\ -2 \end{pmatrix} + 5\begin{pmatrix} 0 \\ -9.8 \end{pmatrix} = 5\mathbf{a}$ | M1 | Use of $\mathbf{F} = m\mathbf{a}$ with correct $m$ and two terms of $\mathbf{F}$ correct |
$\mathbf{a} = \begin{pmatrix} 1.6 \\ -11.8 \end{pmatrix}$ or $\begin{pmatrix} 1.6 \\ -g-2 \end{pmatrix}$ | A1 [3] | |
### Part (ii):
$\mathbf{s} = \frac{1}{2}\begin{pmatrix} 1.6 \\ -11.8 \end{pmatrix}(10)^2$ | M1 | Use of $\mathbf{s} = \mathbf{u}t + \frac{1}{2}\mathbf{a}t^2$ with $t = 10$ |
$\mathbf{s} = \begin{pmatrix} 80 \\ -590 \end{pmatrix}$ | A1ft | $50\mathbf{a}$ |
Position vector is $\begin{pmatrix} 82 \\ -545 \end{pmatrix}$ | A1 [3] | |
---8 In this question $\binom { 1 } { 0 }$ and $\binom { 0 } { 1 }$ denote unit vectors which are horizontal and vertically upwards respectively.\\
A particle of mass 5 kg , initially at rest at the point with position vector $\binom { 2 } { 45 } \mathrm {~m}$, is acted on by gravity and also by two forces $\binom { 15 } { - 8 } \mathrm {~N}$ and $\binom { - 7 } { - 2 } \mathrm {~N}$.\\
(i) Find the acceleration vector of the particle.\\
(ii) Find the position vector of the particle after 10 seconds.
\hfill \mbox{\textit{OCR H240/03 2018 Q8 [6]}}