| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2023 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle on inner surface of sphere/bowl |
| Difficulty | Challenging +1.8 This is a substantial Further Maths mechanics problem requiring energy conservation, circular motion dynamics, and analysis of contact conditions across two connected arcs. While the individual techniques (energy equations, resolving forces in circular motion, finding when reaction becomes zero) are standard for FM students, the multi-part structure, geometric setup with two different circular arcs, and the need to verify contact conditions throughout makes this significantly harder than typical A-level questions. The 'show that' parts provide scaffolding that reduces difficulty slightly from what would be a +2.0 or higher unguided problem. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| (a)(i) Conservation of energy (\(PE = 0\) along horizontal through \(O\)): | M1 | KE and PE in dim. correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| \[v^2 = 111 \cdot 72 - 98\cos\theta \quad (= 11 \cdot 4g - 10g\cos\theta)\] | A1 | Convincing |
| Answer | Marks | Guidance |
|---|---|---|
| \[v^2 = 111 \cdot 72 - 98\cos\theta\] | (A1) | Convincing |
| (ii) N2L towards \(O\): | M1 | Dim. correct equation, \(50g\cos\theta\), \(R\) opposing |
| Answer | Marks | Guidance |
|---|---|---|
| \[R = 50g\cos\theta - \frac{50}{5}(111 \cdot 72 - 98\cos\theta)\] | m1 | Substitute for \(v^2\) (any form) |
| Answer | Marks | Guidance |
|---|---|---|
| \end{cases}\] | A1 | |
| Loses contact when \(R = 0\): | M1 | Used, FT \(R\) from (ii) or \(150g\cos\theta - 114g = 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| \[\theta = 40 \cdot 5(358\ldots)°\] | A1 | Accept 41, FT \(R\) from (ii) |
| \[\theta = 40 \cdot 5(358\ldots)° < 45°\] | A1 | FT \(R\) provided \(\theta < 45°\) |
| \(\therefore\) ring loses contact before reaching \(C\) |
| Answer | Marks | Guidance |
|---|---|---|
| Sub. \(\theta = 45°\) into expression for \(R\): | (M1) | FT \(R\) from (ii) |
| \[R = -77.7(530\ldots)\] | (A1) | FT \(R\) |
| \[R = -77.7(530\ldots) < 0\] | (A1) | FT provided \(R < 0\) |
| \(\therefore\) ring loses contact before reaching \(C\) | ||
| (iv) Rubber ring may lose contact at a greater value of \(\theta\) or Rubber ring may remain in contact until \(C\) | E1 | |
| (b) N2L towards \(D\): | M1 | Dim. correct equation |
| At \(A\), \(\cos\alpha = 0 \cdot 9\): \(R_A - 50g(0 \cdot 9) = 50a\) | A1 | Any correct equation including \(R' - 50g\cos\alpha = 50a\), \(a = \frac{v^2}{r} > 0\) |
| Answer | Marks | Guidance |
|---|---|---|
| At \(B\), \(\cos\alpha = 1\): \(R_B - 50g = 50a\) | A1 | Convincing |
| Answer | Marks |
|---|---|
| \[R_B > R_A > 0 \text{ so must remain in contact}\] | A1 |
(a)(i) Conservation of energy ($PE = 0$ along horizontal through $O$): | M1 | KE and PE in dim. correct equation
$$mg(5 + 7(1 - \cos\alpha)) = \frac{1}{2}mv^2 + mg(5\cos\theta)$$
$5 \cdot 7g = 0 \cdot 5v^2 + 5g\cos\theta$
$55 \cdot 86 = 0 \cdot 5v^2 + 49\cos\theta$
$2793 = 25v^2 + 2450\cos\theta$
$285g = 25v^2 + 250\cos\theta$
$$v^2 = 111 \cdot 72 - 98\cos\theta \quad (= 11 \cdot 4g - 10g\cos\theta)$$ | A1 | Convincing
**Alternative solution** ($PE = 0$ along horizontal through $B$):
$$mg(7(1 - \cos\alpha)) = \frac{1}{2}mv^2 - mg(5(1 - \cos\theta))$$
$0 \cdot 7g = 0 \cdot 5v^2 - 5g + 5g\cos\theta$
$6 \cdot 86 = 0 \cdot 5v^2 - 49 + 49\cos\theta$
$343 = 25v^2 - 2450 + 2450\cos\theta$
$$v^2 = 111 \cdot 72 - 98\cos\theta$$ | (A1) | Convincing
(ii) N2L towards $O$: | M1 | Dim. correct equation, $50g\cos\theta$, $R$ opposing
$$50g\cos\theta - R = \frac{50v^2}{5}$$
$$R = 50g\cos\theta - \frac{50v^2}{5}$$
$$R = 50g\cos\theta - \frac{50}{5}(111 \cdot 72 - 98\cos\theta)$$ | m1 | Substitute for $v^2$ (any form)
$$R = \begin{cases}
1470\cos\theta - 1117 \cdot 2 \\
150g\cos\theta - 114g
\end{cases}$$ | A1 |
**Loses contact when $R = 0$**: | M1 | Used, FT $R$ from (ii) or $150g\cos\theta - 114g = 0$
$$1470\cos\theta - 1117 \cdot 2 = 0$$
$$\cos\theta = \frac{114}{150} \quad (= \frac{1117 \cdot 2}{1470} = \frac{19}{23})$$
$$\theta = 40 \cdot 5(358\ldots)°$$ | A1 | Accept 41, FT $R$ from (ii)
$$\theta = 40 \cdot 5(358\ldots)° < 45°$$ | A1 | FT $R$ provided $\theta < 45°$
$\therefore$ ring loses contact before reaching $C$ |
**Alternative Solution** (iii):
Sub. $\theta = 45°$ into expression for $R$: | (M1) | FT $R$ from (ii)
$$R = -77.7(530\ldots)$$ | (A1) | FT $R$
$$R = -77.7(530\ldots) < 0$$ | (A1) | FT provided $R < 0$
$\therefore$ ring loses contact before reaching $C$ |
(iv) Rubber ring may lose contact at a greater value of $\theta$ or Rubber ring may remain in contact until $C$ | E1 |
(b) N2L towards $D$: | M1 | Dim. correct equation
At $A$, $\cos\alpha = 0 \cdot 9$: $R_A - 50g(0 \cdot 9) = 50a$ | A1 | Any correct equation including $R' - 50g\cos\alpha = 50a$, $a = \frac{v^2}{r} > 0$
$$R_A = 50a + 45g > 45g = 441$$
At $B$, $\cos\alpha = 1$: $R_B - 50g = 50a$ | A1 | Convincing
$$R_B = 50a + 50g > 50g = 490$$
$$R_B > R_A > 0 \text{ so must remain in contact}$$ | A1 |
---The diagram shows a slide, $ABC$, at a water park. The shape of the slide may be modelled by two circular arcs, $AB$ and $BC$, in the same vertical plane. Arc $AB$ has radius $7$ m and subtends an angle $\alpha$ at its centre $D$, where $\cos \alpha = \frac{9}{10}$. Arc $BC$ has radius $5$ m and subtends an angle of $45°$ at its centre, $O$. The straight line $DBO$ is vertical.
\includegraphics{figure_6}
Users of the slide are required to sit in a rubber ring and are released from rest at point $A$. A girl decides to use the slide. The combined mass of the girl and the rubber ring is $50$ kg.
\begin{enumerate}[label=(\alph*)]
\item When the rubber ring is at a point $P$ on the circular arc $BC$, its speed is $v$ ms$^{-1}$ and $OP$ makes an angle $\theta$ with the upward vertical.
\begin{enumerate}[label=(\roman*)]
\item Show that $v^2 = 111.72 - 98\cos\theta$. [4]
\item Find, in terms of $\theta$, the reaction between the rubber ring and the slide at $P$. [4]
\item Show that, according to this model, the rubber ring loses contact with the slide before reaching $C$. [3]
\item In reality, there will be resistive forces opposing the motion of the rubber ring. Explain how this fact will affect your answer to (iii). [1]
\end{enumerate}
\item Show that the rubber ring will remain in contact with the slide along the arc $AB$. [3]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2023 Q6 [15]}}