| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2018 |
| Session | June |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Variable acceleration (1D) |
| Type | Vector motion with components |
| Difficulty | Standard +0.3 This is a straightforward Further Maths mechanics question requiring differentiation of vector functions. Part (a) involves differentiating trigonometric functions and solving v=0, part (b) is immediate recall (p=mv), and part (c) requires one more differentiation (F=ma). All techniques are standard with no novel insight required, though the multi-component vectors and solving the vector equation for rest adds minor complexity beyond basic A-level. |
| Spec | 3.02f Non-uniform acceleration: using differentiation and integration3.03d Newton's second law: 2D vectors6.03a Linear momentum: p = mv |
| Answer | Marks | Guidance |
|---|---|---|
| 4(a) | \(v = \frac{d}{dt}x\) | M1 |
| \(v = 3\cos t \mathbf{i} + 8\sin 2t \mathbf{j} + 5\cos t \mathbf{k}\) | A1 | all correct |
| For \(v = 0\), \(\cos t = 0\) | M1 | equating either component to 0 |
| \(t = \pi/2, (3\pi/2, \ldots)\) | A1 | |
| and \(\sin 2t = 0\) | M1 | equating other component to 0 |
| \(2t = 0, \pi, (2\pi, \ldots)\) | ||
| \(t = 0, \pi/2, (\pi, \ldots)\) | A1 | |
| Hence smallest value of \(t\) when \(v = 0\) is \(\pi/2\) | A1 | cao |
| 4(b) | Mom. vector \(= 3(3\cos t \mathbf{i} + 8\sin 2t \mathbf{j} + 5\cos t \mathbf{k})\) | B1 |
| 4(c) | \(F = ma\) | M1 |
| \(a = -3\sin t \mathbf{i} + 16\cos 2t \mathbf{j} - 5\sin t \mathbf{k}\) | M1 | \(v\) differentiated |
| \(F = 3(-3\sin t \mathbf{i} + 16\cos 2t \mathbf{j} - 5\sin t \mathbf{k})\) | A1 | ft \(v\) |
| \(F = -9\sin t \mathbf{i} + 48\cos 2t \mathbf{j} - 15\sin t \mathbf{k}\) | isw |
4(a) | $v = \frac{d}{dt}x$ | M1 | used |
| $v = 3\cos t \mathbf{i} + 8\sin 2t \mathbf{j} + 5\cos t \mathbf{k}$ | A1 | all correct |
| For $v = 0$, $\cos t = 0$ | M1 | equating either component to 0 |
| $t = \pi/2, (3\pi/2, \ldots)$ | A1 | |
| and $\sin 2t = 0$ | M1 | equating other component to 0 |
| $2t = 0, \pi, (2\pi, \ldots)$ | | |
| $t = 0, \pi/2, (\pi, \ldots)$ | A1 | |
| Hence smallest value of $t$ when $v = 0$ is $\pi/2$ | A1 | cao |
4(b) | Mom. vector $= 3(3\cos t \mathbf{i} + 8\sin 2t \mathbf{j} + 5\cos t \mathbf{k})$ | B1 | ft $v$ isw |
4(c) | $F = ma$ | M1 | used |
| $a = -3\sin t \mathbf{i} + 16\cos 2t \mathbf{j} - 5\sin t \mathbf{k}$ | M1 | $v$ differentiated |
| $F = 3(-3\sin t \mathbf{i} + 16\cos 2t \mathbf{j} - 5\sin t \mathbf{k})$ | A1 | ft $v$ |
| $F = -9\sin t \mathbf{i} + 48\cos 2t \mathbf{j} - 15\sin t \mathbf{k}$ | | isw |The position vector $\mathbf{x}$ metres at time $t$ seconds of an object of mass 3 kg may be modelled by
$$\mathbf{x} = 3\sin t \mathbf{i} - 4\cos 2t \mathbf{j} + 5\sin t \mathbf{k}.$$
\begin{enumerate}[label=(\alph*)]
\item Find an expression for the velocity vector $\mathbf{v}\text{ ms}^{-1}$ at time $t$ seconds and determine the least value of $t$ when the object is instantaneously at rest. [7]
\item Write down the momentum vector at time $t$ seconds. [1]
\item Find, in vector form, an expression for the force acting on the object at time $t$ seconds. [3]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2018 Q4 [11]}}