| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2022 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: speed at specific point |
| Difficulty | Challenging +1.2 This is a standard Further Maths vertical circle problem requiring energy conservation and circular motion equations (T = mv²/r ± mg cos θ). Part (a) involves two simultaneous equations with straightforward algebra, while part (b) is a routine projectile motion calculation. The setup is typical textbook material with no novel insights required, though the multi-step nature and Further Maths context place it slightly above average difficulty. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mga\cos\theta\), \(\left[kag = \frac{1}{3}ag + 2ag\cos\theta\right]\) | B1 | Energy equation |
| \(T - mg\cos\theta = \frac{m}{a}v^2\), so \(\frac{11}{6}mg - mg\cos\theta = \frac{m}{a}\cdot kag\), \(\frac{11}{6} - \cos\theta = k\) | B1 | N2L at B |
| Solve simultaneously | M1 | |
| \(k = \frac{4}{3}\), \(\cos\theta = \frac{1}{2}\) | A1 | Both |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| Initial speed \(\uparrow = \sqrt{kag}\sin\theta\) | B1 | |
| Use \(v^2 = u^2 + 2as\): \(0 = (\sqrt{kag}\sin\theta)^2 - 2gs\) | M1 | |
| \(s = \frac{1}{2}a\) | A1 | |
| Height above lowest point \(= s + a - a\cos\theta = \frac{1}{2}a + a - \frac{1}{2}a = a\) | A1 FT |
## Question 5(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mga\cos\theta$, $\left[kag = \frac{1}{3}ag + 2ag\cos\theta\right]$ | B1 | Energy equation |
| $T - mg\cos\theta = \frac{m}{a}v^2$, so $\frac{11}{6}mg - mg\cos\theta = \frac{m}{a}\cdot kag$, $\frac{11}{6} - \cos\theta = k$ | B1 | N2L at B |
| Solve simultaneously | M1 | |
| $k = \frac{4}{3}$, $\cos\theta = \frac{1}{2}$ | A1 | Both |
## Question 5(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| Initial speed $\uparrow = \sqrt{kag}\sin\theta$ | B1 | |
| Use $v^2 = u^2 + 2as$: $0 = (\sqrt{kag}\sin\theta)^2 - 2gs$ | M1 | |
| $s = \frac{1}{2}a$ | A1 | |
| Height above lowest point $= s + a - a\cos\theta = \frac{1}{2}a + a - \frac{1}{2}a = a$ | A1 FT | |5 A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $a$. The other end of the string is attached to a fixed point $O$. The string is held taut with $O P$ horizontal. The particle $P$ is projected vertically downwards with speed $\sqrt { \frac { 1 } { 3 } \mathrm { ag } }$ and starts to move in a vertical circle. $P$ passes through the lowest point of the circle and reaches the point $Q$ where $O Q$ makes an angle $\theta$ with the downward vertical. At $Q$ the speed of $P$ is $\sqrt { \mathrm { kag } }$ and the tension in the string is $\frac { 11 } { 6 } \mathrm { mg }$.
\begin{enumerate}[label=(\alph*)]
\item Find the value of $k$ and the value of $\cos \theta$.\\
At $Q$ the particle $P$ becomes detached from the string.
\item In the subsequent motion, find the greatest height reached by $P$ above the level of the lowest point of the circle.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q5 [8]}}