| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Forces, equilibrium and resultants |
| Type | Forces in vector form: equilibrium (find unknowns) |
| Difficulty | Standard +0.3 This is a straightforward Further Maths mechanics question requiring standard techniques: equilibrium of forces (vector addition), verifying parallel vectors (scalar multiple check), work done calculation (F·d), and work-energy principle application. All steps are routine with no novel problem-solving required, making it slightly easier than average even for Further Maths. |
| Spec | 3.03n Equilibrium in 2D: particle under forces6.02b Calculate work: constant force, resolved component6.02i Conservation of energy: mechanical energy principle |
| Answer | Marks |
|---|---|
| \(\mathbf{F}_3 = -15\mathbf{i} + \mathbf{j} + 9\mathbf{k}\) (N) | M1, A1[2] |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 6\mathbf{i} + 4\mathbf{j} - 8\mathbf{k}\) | M1, A1 | or BA; oe, cao |
| \(\mathbf{F}_1 = \frac{2}{3}\mathbf{AB}\) or \(\mathbf{AB} = \frac{3}{2}\mathbf{F}_1\) (\(\therefore\) parallel) | A1 | Convincing |
| Answer | Marks | Guidance |
|---|---|---|
| \(= 174\) (J) | M1, A1[3] | Used. FT AB; FT their AB |
| Answer | Marks | Guidance |
|---|---|---|
| Work done = change in KE | M1 | FT their '174' |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = 26 \cdot 38(18 \ldots)\) (ms\(^{-1}\)) | A1 | \(v = \sqrt{696} = 2\sqrt{174}\); FT their '174' |
## Part (a)
$(9\mathbf{i} + 6\mathbf{j} - 12\mathbf{k}) + (6\mathbf{i} - 7\mathbf{j} + 3\mathbf{k}) + \mathbf{F}_3 = 0$
$\mathbf{F}_3 = -15\mathbf{i} + \mathbf{j} + 9\mathbf{k}$ (N) | M1, A1[2] |
## Part (b)
### (i)
AB = $\mathbf{r}_B - \mathbf{r}_A = (8\mathbf{i} - 5\mathbf{j} - \mathbf{k}) - (2\mathbf{i} - 9\mathbf{j} + 7\mathbf{k})$
$= 6\mathbf{i} + 4\mathbf{j} - 8\mathbf{k}$ | M1, A1 | or BA; oe, cao
$\mathbf{F}_1 = \frac{2}{3}\mathbf{AB}$ or $\mathbf{AB} = \frac{3}{2}\mathbf{F}_1$ ($\therefore$ parallel) | A1 | Convincing
### (ii)
Work done by $\mathbf{F}_1 = \mathbf{F}_1 \cdot \mathbf{AB}$
$= (9\mathbf{i} + 6\mathbf{j} - 12\mathbf{k}) \cdot (6\mathbf{i} + 4\mathbf{j} - 8\mathbf{k})$
$= (9)(6) + (6)(4) + (-12)(-8)$
$= 174$ (J) | M1, A1[3] | Used. FT AB; FT their AB
### (iii)
Work done = change in KE | M1 | FT their '174'
$174 = \frac{1}{2}(0 \cdot 5)v^2 - 0$
$v = 26 \cdot 38(18 \ldots)$ (ms$^{-1}$) | A1 | $v = \sqrt{696} = 2\sqrt{174}$; FT their '174'
**[7]**
**Total for Question 4: 9**
---A particle $P$ of mass 0.5 kg is in equilibrium under the action of three forces $\mathbf{F}_1$, $\mathbf{F}_2$ and $\mathbf{F}_3$.
$$\mathbf{F}_1 = (9\mathbf{i} + 6\mathbf{j} - 12\mathbf{k})\text{N} \quad \text{and} \quad \mathbf{F}_2 = (6\mathbf{i} - 7\mathbf{j} + 3\mathbf{k})\text{N}.$$
\begin{enumerate}[label=(\alph*)]
\item Find the force $\mathbf{F}_3$. [2]
\item Forces $\mathbf{F}_2$ and $\mathbf{F}_3$ are removed so that $P$ moves in a straight line $AB$ under the action of the single force $\mathbf{F}_1$. The points $A$ and $B$ have position vectors $(2\mathbf{i} - 9\mathbf{j} + 7\mathbf{k})$ m and $(8\mathbf{i} - 5\mathbf{j} - \mathbf{k})$ m respectively. The particle $P$ is initially at rest at $A$.
\begin{enumerate}[label=(\roman*)]
\item Verify that $\mathbf{F}_1$ acts parallel to the vector $\overrightarrow{AB}$.
\item Find the work done by the force $\mathbf{F}_1$ as the particle moves from $A$ to $B$.
\item By using the work-energy principle, find the speed of $P$ as it reaches $B$. [7]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2022 Q4 [9]}}