| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Session | Specimen |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Power and driving force |
| Type | Power from force and derived speed (non-equilibrium) |
| Difficulty | Challenging +1.3 This is a Further Maths mechanics question requiring differentiation of vector functions, dot products, and energy/work calculations. Part (a) requires computing velocity and showing a dot product is always positive (straightforward calculus and algebra). Parts (b)-(e) are standard applications of kinetic energy and work-energy principles. While it involves exponential/trig functions and multiple steps, the techniques are routine for Further Maths students with no novel problem-solving required. |
| Spec | 6.02a Work done: concept and definition6.02d Mechanical energy: KE and PE concepts6.02l Power and velocity: P = Fv |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\mathbf{v} = \frac{d\mathbf{r}}{dt}\) | M1 | AO2 |
| \(\mathbf{v} = 2e^{2t}\mathbf{i} + 2\cos(2t)\mathbf{j} - 2\sin(2t)\mathbf{k}\) | A1 A1 | AO1 AO1 |
| \(\mathbf{v} \cdot \mathbf{r} = 2e^{4t} + 2\cos(2t)\sin(2t) - 2\sin(2t)\cos(2t)\) | M1 A1 | AO2 AO1 |
| \(\mathbf{v} \cdot \mathbf{r} = 2e^{4t}\) which is never 0. Hence \(\mathbf{v}\) and \(\mathbf{r}\) are never perpendicular to each other | E1 | AO2 |
| (b) \(v^2 = (2e^{2t})^2 + (2\cos(2t))^2 + (-2\sin(2t))^2\) | M1 | AO2 |
| \(v^2 = 4e^{4t} + 4\cos^2(2t) + 4\sin^2(2t) = 4e^{4t} + 4\) | A1 | AO1 |
| (c) \(KE = 0.5 \times 0.4 \times (4e^{4t} + 4) = 0.8(e^{4t} + 1)\) | B1 | AO1 |
| (d) WD = change in KE | M1 | AO1 |
| \(WD = 0.8(e^4 + 1) - 0.8(1 + 1) = 0.8(e^4 - 1) = 42.9\) (J) | A1 | AO1 |
| (e) Rate of work = \(\frac{d}{dt}\)(KE) | M1 | AO2 |
| Rate of work = \(\frac{d}{dt}(0.8(e^{4t} + 1)) = 3.2e^{4t}\) (W) | A1 | AO1 |
**(a)** $\mathbf{v} = \frac{d\mathbf{r}}{dt}$ | M1 | AO2
$\mathbf{v} = 2e^{2t}\mathbf{i} + 2\cos(2t)\mathbf{j} - 2\sin(2t)\mathbf{k}$ | A1 A1 | AO1 AO1 | correct differentiation of any one term, all correct
$\mathbf{v} \cdot \mathbf{r} = 2e^{4t} + 2\cos(2t)\sin(2t) - 2\sin(2t)\cos(2t)$ | M1 A1 | AO2 AO1 | correct dot product
$\mathbf{v} \cdot \mathbf{r} = 2e^{4t}$ which is never 0. Hence $\mathbf{v}$ and $\mathbf{r}$ are never perpendicular to each other | E1 | AO2
**(b)** $v^2 = (2e^{2t})^2 + (2\cos(2t))^2 + (-2\sin(2t))^2$ | M1 | AO2
$v^2 = 4e^{4t} + 4\cos^2(2t) + 4\sin^2(2t) = 4e^{4t} + 4$ | A1 | AO1
**(c)** $KE = 0.5 \times 0.4 \times (4e^{4t} + 4) = 0.8(e^{4t} + 1)$ | B1 | AO1
**(d)** WD = change in KE | M1 | AO1
$WD = 0.8(e^4 + 1) - 0.8(1 + 1) = 0.8(e^4 - 1) = 42.9$ (J) | A1 | AO1
**(e)** Rate of work = $\frac{d}{dt}$(KE) | M1 | AO2
Rate of work = $\frac{d}{dt}(0.8(e^{4t} + 1)) = 3.2e^{4t}$ (W) | A1 | AO1
**Total: [13]**
---Relative to a fixed origin $O$, the position vector $\mathbf{r}$ m at time $t$ s of a particle $P$, of mass 0.4 kg, is given by
$$\mathbf{r} = e^{2t}\mathbf{i} + \sin(2t)\mathbf{j} + \cos(2t)\mathbf{k}.$$
\begin{enumerate}[label=(\alph*)]
\item Show that the velocity vector $\mathbf{v}$ and the position vector $\mathbf{r}$ are never perpendicular to each other. [6]
\item Given that the speed of $P$ at time $t$ is $v$ ms$^{-1}$, show that
$$v^2 = 4e^{4t} + 4.$$ [2]
\item Find the kinetic energy of $P$ at time $t$. [1]
\item Calculate the work done by the force acting on $P$ in the interval $0 < t < 1$. [2]
\item Determine an expression for the rate at which the force acting on $P$ is working at time $t$. [2]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 Q4 [13]}}