| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Impulse and momentum (advanced) |
| Difficulty | Challenging +1.2 This is a standard conical pendulum problem with two strings requiring resolution of forces in vertical and horizontal directions, followed by circular motion calculations. While it involves multiple steps (resolving forces, finding angular velocity, calculating KE) and requires careful trigonometry with the given angles, the solution path is methodical and follows well-established mechanics techniques. The problem is harder than average A-level due to the two-string setup and Further Maths context, but doesn't require novel insight—just systematic application of F=ma in circular motion. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| \(\sin \theta = 0 \cdot 6\); \(\cos \theta = 0 \cdot 8\) | ||
| Resolving vertically, | M1 | All forces, dim. correct |
| Answer | Marks | Guidance |
|---|---|---|
| \(T = 55 \cdot 125\) (N) | A1, A1, A1[4] | -1 each error; cao |
| Answer | Marks | Guidance |
|---|---|---|
| Using N2L towards \(C\), | M1 | All forces, dim. correct |
| Answer | Marks | Guidance |
|---|---|---|
| \((55 \cdot 125)(0 \cdot 6) + (39 \cdot 2)\left(\frac{\sqrt{3}}{2}\right) = (2 \cdot 5) \omega^2(0 \cdot 9)\) | A1 | Correct equation |
| m1 | \(a = \left\{\begin{array}{l}\frac{v^2}{r} \\ \omega^2 r\end{array}\right.\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\omega = 5 \cdot 45(78463 \ldots)\) (rad s\(^{-1}\)) | B1, A1[5] | \(r = 1 \cdot 5 \sin \theta\); \(r = 1 \cdot 5 \times 0 \cdot 6 = 0 \cdot 9\); cao |
| Answer | Marks | Guidance |
|---|---|---|
| \(v = 4 \cdot 91206 \ldots\) | M1 | FT \(\omega\) and \(r = 1 \cdot 5\); FT \(v\) |
| Answer | Marks | Guidance |
|---|---|---|
| KE = 30·16(0438 \ldots)\( (J) | m1, A1[3] | FT \)v$; cao |
## Part (a)
| | $\sin \theta = 0 \cdot 6$; $\cos \theta = 0 \cdot 8$
Resolving vertically, | M1 | All forces, dim. correct
$T \cos \theta = (39 \cdot 2)\cos 60 + 2 \cdot 5g$
$T(0 \cdot 8) = (39 \cdot 2)(0 \cdot 5) + (2 \cdot 5)(9 \cdot 8)$
$T = 55 \cdot 125$ (N) | A1, A1, A1[4] | -1 each error; cao
## Part (b)
Using N2L towards $C$, | M1 | All forces, dim. correct
$T \sin \theta + (39 \cdot 2) \sin 60 = 2 \cdot 5a$
$(55 \cdot 125)(0 \cdot 6) + (39 \cdot 2)\left(\frac{\sqrt{3}}{2}\right) = (2 \cdot 5) \omega^2(0 \cdot 9)$ | A1 | Correct equation
| m1 | $a = \left\{\begin{array}{l}\frac{v^2}{r} \\ \omega^2 r\end{array}\right.$
$\omega^2 = 29 \cdot 78808 \ldots$
$\omega = 5 \cdot 45(78463 \ldots)$ (rad s$^{-1}$) | B1, A1[5] | $r = 1 \cdot 5 \sin \theta$; $r = 1 \cdot 5 \times 0 \cdot 6 = 0 \cdot 9$; cao
## Part (c)
$v = \omega r$
$v = 5 \cdot 45 \ldots \times 0 \cdot 9$
$v = 4 \cdot 91206 \ldots$ | M1 | FT $\omega$ and $r = 1 \cdot 5$; FT $v$
KE = $\frac{1}{2}(2 \cdot 5)(4 \cdot 91206 \ldots)^2$
KE = 30·16(0438 \ldots)$ (J) | m1, A1[3] | FT $v$; cao
**Total for Question 7: 12**The diagram below shows a particle $P$, of mass 2.5 kg, attached by means of two light inextensible strings fixed at points $A$ and $B$. Point $A$ is vertically above point $B$. $BP$ makes an angle of $60°$ with the upward vertical and $AP$ is inclined at an angle $\theta$ to the downward vertical where $\cos\theta = 0.8$. The particle $P$ describes a horizontal circle with constant angular speed $\omega$ radians per second about centre $C$ with both strings taut.
\includegraphics{figure_7}
The tension in the string $BP$ is 39.2 N.
\begin{enumerate}[label=(\alph*)]
\item Calculate the tension in the string $AP$. [4]
\item Given that the length of the string $AP$ is 1.5 m, find the value of $\omega$. [5]
\item Calculate the kinetic energy of $P$. [3]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2022 Q7 [12]}}