| Exam Board | WJEC |
|---|---|
| Module | Further Unit 3 (Further Unit 3) |
| Year | 2018 |
| Session | June |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring energy conservation, circular motion dynamics, and analysis of critical conditions for complete circles. Part (a) involves standard circular motion and energy methods but with a non-standard starting angle. Part (b) requires finding when the string goes slack (critical condition at top), which is a classic but non-trivial application. Part (c) extends to rigid rod motion, requiring understanding that rods can have negative tension. The multi-part structure, need for careful energy bookkeeping from 60° initial position, and the proof elements place this above average difficulty, though it follows recognizable patterns for Further Maths students. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| 5(a)(i) | Conservation of energy | M1 |
| \(0.5mu^2 = 0.5mv^2 - mgl(\cos\theta - \cos 60°)\) | A1 | PE correct |
| A1 | KE correct | |
| \(v^2 = u^2 + 2lg\cos\theta - lg\) | A1 | si |
| N2L towards centre | M1 | dim correct \(T\) and component wt opposing |
| \(T - mg\cos\theta = \frac{mv^2}{l}\) | A1 | cao |
| \(T = mg\cos\theta + \frac{m(u^2 + 2lg\cos\theta - lg)}{l}\) | m1 | |
| \(T = \frac{mu^2}{l} + 3mg\cos\theta - mg\) | A1 | |
| 5(a)(ii) | For complete circles, when \(\theta=180°\), \(T>0\) | M1 |
| \(\frac{mu^2}{l} > 4mg\) | ||
| \(u^2 > 4lg\) | A1 | |
| 5(b) | Circular motion ceases when \(T=0\), \(u^2=3lg\) | M1 |
| \(T = 3mg + 3mg\cos\theta - mg = 0\) | ||
| \(\cos\theta = -\frac{2}{3}\), \(\theta = 131.81°\) | A1 | cao |
| When circular motion ceases, the particle \(P\) is subject to gravity and behaves as a projectile (with initial velocity upwards and tangential to the circular path). | E1 | |
| 5(c) | For complete circles, when \(\theta=180°\), \(v^2>0\) | M1 |
| \(u^2 - 2lg - lg > 0\) | ||
| \(u^2 > 3lg\) | A1 |
5(a)(i) | **Conservation of energy** | M1 | PE and KE equation |
| $0.5mu^2 = 0.5mv^2 - mgl(\cos\theta - \cos 60°)$ | A1 | PE correct |
| | A1 | KE correct |
| $v^2 = u^2 + 2lg\cos\theta - lg$ | A1 | si |
| **N2L towards centre** | M1 | dim correct $T$ and component wt opposing |
| $T - mg\cos\theta = \frac{mv^2}{l}$ | A1 | cao |
| $T = mg\cos\theta + \frac{m(u^2 + 2lg\cos\theta - lg)}{l}$ | m1 | |
| $T = \frac{mu^2}{l} + 3mg\cos\theta - mg$ | A1 | |
5(a)(ii) | For complete circles, when $\theta=180°$, $T>0$ | M1 | |
| $\frac{mu^2}{l} > 4mg$ | | |
| $u^2 > 4lg$ | A1 | |
5(b) | Circular motion ceases when $T=0$, $u^2=3lg$ | M1 | |
| $T = 3mg + 3mg\cos\theta - mg = 0$ | | |
| $\cos\theta = -\frac{2}{3}$, $\theta = 131.81°$ | A1 | cao |
| When circular motion ceases, the particle $P$ is subject to gravity and behaves as a projectile (with initial velocity upwards and tangential to the circular path). | E1 | |
5(c) | For complete circles, when $\theta=180°$, $v^2>0$ | M1 | |
| $u^2 - 2lg - lg > 0$ | | |
| $u^2 > 3lg$ | A1 | |A particle $P$, of mass $m$ kg, is attached to one end of a light inextensible string of length $l$ m. The other end of the string is attached to a fixed point $O$. Initially, $P$ is held at rest with the string just taut and making an angle of 60° with the downward vertical. It is then given a velocity $u\text{ ms}^{-1}$ perpendicular to the string in a downward direction.
\begin{enumerate}[label=(\alph*)]
\item
\begin{enumerate}[label=(\roman*)]
\item When the string makes an angle $\theta$ with the downward vertical, the velocity of the particle is $v$ and the tension in the string is $T$. Find an expression for $T$ in terms of $m$, $l$, $v^2$ and $\theta$.
\item Given that $P$ describes complete circles in the subsequent motion, show that $u^2 > 4lg$. [10]
\end{enumerate}
\item Given that now $u^2 = 3lg$, find the position of the string when circular motion ceases. Briefly describe the motion of $P$ after circular motion has ceased. [3]
\item The string is replaced by a light rigid rod. Given that $P$ describes complete circles in the subsequent motion, show that $u^2 > klg$, where $k$ is to be determined. [2]
\end{enumerate}
\hfill \mbox{\textit{WJEC Further Unit 3 2018 Q5 [15]}}