| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2023 |
| 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 involving vector kinematics. Part (a) requires basic position vector formula and distance calculation. Part (b)(i) involves differentiating trigonometric functions and showing constant speed using Pythagorean identity—a standard technique. Part (b)(ii) requires finding when velocity vectors are perpendicular using dot product equals zero, then solving a simple trigonometric equation. All techniques are routine for Further Maths students with no novel problem-solving required. |
| Spec | 1.10e Position vectors: and displacement3.02g Two-dimensional variable acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| \end{cases}\] | B1 | |
| At \(t = 5\): \(r_A = 21i + 16j + 12k\) | M1 | Use of \(t = 5\) |
| \( | r_A | = \sqrt{21^2 + 16^2 + 12^2}\) |
| \( | r_A | = \sqrt{841} = 29\) (m) |
| (b)(i) \(v = \frac{dr}{dt}\) | M1 | |
| \[v_B = \frac{3}{2}\cos\left(\frac{t}{2}\right)i + \frac{3}{2}\sin\left(\frac{t}{2}\right)j\] | A1 | |
| \[ | v_B | = \sqrt{\left(\frac{3}{2}\cos\left(\frac{t}{2}\right)\right)^2 + \left(\frac{3}{2}\sin\left(\frac{t}{2}\right)\right)^2}\] |
| \[ | v_B | = \sqrt{\frac{9}{4}} = \frac{3}{2} = 1 \cdot 5 \text{ (ms}^{-1}\text{)}\] |
| (ii) Dot product, \(v_A \cdot v_B = 0\) | M1 | |
| \[(3i - j + 4k) \cdot \left(\frac{3}{2}\cos\left(\frac{t}{2}\right)i + \frac{3}{2}\sin\left(\frac{t}{2}\right)j\right) = 0\] | m1 |
| Answer | Marks | Guidance |
|---|---|---|
| \[t = 2 \cdot 4(9809 \ldots) \text{ (s)}\] | A1 | cao |
(a) $$r_A = \begin{cases}
6i + 21j - 8k + (3i - j + 4k)t \\
\begin{pmatrix} 6 + 3t \\ 21 - t \\ 4t - 8 \end{pmatrix} \\
(6 + 3t)i + (21 - t)j + (4t - 8)k
\end{cases}$$ | B1 |
At $t = 5$: $r_A = 21i + 16j + 12k$ | M1 | Use of $t = 5$
$|r_A| = \sqrt{21^2 + 16^2 + 12^2}$ | m1 | Attempt to find $|r_A|$
$|r_A| = \sqrt{841} = 29$ (m) | A1 | cao
(b)(i) $v = \frac{dr}{dt}$ | M1 |
$$v_B = \frac{3}{2}\cos\left(\frac{t}{2}\right)i + \frac{3}{2}\sin\left(\frac{t}{2}\right)j$$ | A1 |
$$|v_B| = \sqrt{\left(\frac{3}{2}\cos\left(\frac{t}{2}\right)\right)^2 + \left(\frac{3}{2}\sin\left(\frac{t}{2}\right)\right)^2}$$ | m1 | FT their $v_B$ throughout
$$|v_B| = \sqrt{\frac{9}{4}} = \frac{3}{2} = 1 \cdot 5 \text{ (ms}^{-1}\text{)}$$ | A1 | cao (which is constant)
(ii) Dot product, $v_A \cdot v_B = 0$ | M1 |
$$(3i - j + 4k) \cdot \left(\frac{3}{2}\cos\left(\frac{t}{2}\right)i + \frac{3}{2}\sin\left(\frac{t}{2}\right)j\right) = 0$$ | m1 |
$$\frac{9}{2}\cos\left(\frac{t}{2}\right) - \frac{3}{2}\sin\left(\frac{t}{2}\right) = 0$$
$$\tan\left(\frac{t}{2}\right) = 3$$
$$t = 2 \cdot 4(9809 \ldots) \text{ (s)}$$ | A1 | cao
---At time $t = 0$ seconds, a particle $A$ has position vector $(6\mathbf{i} + 2\mathbf{j} - 8\mathbf{k})$ m relative to a fixed origin $O$ and is moving with constant velocity $(3\mathbf{i} - \mathbf{j} + 4\mathbf{k})$ ms$^{-1}$.
\begin{enumerate}[label=(\alph*)]
\item Write down the position vector of particle $A$ at time $t$ seconds and hence find the distance $OA$ when $t = 5$. [4]
\item The position vector, $\mathbf{r}_B$ metres, of another particle $B$ at time $t$ seconds is given by
$$\mathbf{r}_B = 3\sin\left(\frac{t}{2}\right)\mathbf{i} - 3\cos\left(\frac{t}{2}\right)\mathbf{j} + 5\mathbf{k}.$$
\begin{enumerate}[label=(\roman*)]
\item Show that $B$ is moving with constant speed.
\item Determine the smallest value of $t$ such that particles $A$ and $B$ are moving perpendicular to each other. [7]
\end{enumerate}
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2023 Q2 [11]}}