| Exam Board | CAIE |
|---|---|
| Module | Further Paper 3 (Further Paper 3) |
| Year | 2020 |
| Session | November |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle in hemispherical bowl |
| Difficulty | Standard +0.8 This is a standard Further Maths circular motion problem requiring resolution of forces and use of the cosine relationship for circular motion in a bowl. Part (a) is a routine 'show that' derivation. Part (b) requires setting up two equations relating x and θ for different angular speeds, then using the geometric relationship between x, r, and θ to eliminate variables—this involves more algebraic manipulation than typical but follows established methods for this topic. |
| Spec | 3.03n Equilibrium in 2D: particle under forces6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\uparrow\ N\cos\theta = mg\) | B1 | |
| \(\leftarrow\ N\sin\theta = mr\sin\theta\,\omega^2\) | B1 | |
| \(\cos\theta = \frac{mg}{N}\) so \(\cos\theta = \frac{g}{\omega^2 r}\) | B1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\cos\theta = \frac{r-x}{r} = \frac{g}{\omega^2 r}\) | B1 | Using trig of situation: must involve \(x\) |
| In new situation: \(r - 4x = r \times \frac{g}{4\omega^2 r}\) | M1 | Using new situation with \(4x\) and \(2\omega\) seen |
| \(r - x = 4(r - 4x)\) | M1 | Combining |
| \(x = \frac{1}{5}r\) | A1 |
## Question 4(a):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\uparrow\ N\cos\theta = mg$ | B1 | |
| $\leftarrow\ N\sin\theta = mr\sin\theta\,\omega^2$ | B1 | |
| $\cos\theta = \frac{mg}{N}$ so $\cos\theta = \frac{g}{\omega^2 r}$ | B1 | AG |
---
## Question 4(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\cos\theta = \frac{r-x}{r} = \frac{g}{\omega^2 r}$ | B1 | Using trig of situation: must involve $x$ |
| In new situation: $r - 4x = r \times \frac{g}{4\omega^2 r}$ | M1 | Using new situation with $4x$ and $2\omega$ seen |
| $r - x = 4(r - 4x)$ | M1 | Combining |
| $x = \frac{1}{5}r$ | A1 | |
---4 A particle $P$ of mass $m$ is moving in a horizontal circle with angular speed $\omega$ on the smooth inner surface of a hemispherical shell of radius $r$. The angle between the vertical and the normal reaction of the surface on $P$ is $\theta$.
\begin{enumerate}[label=(\alph*)]
\item Show that $\cos \theta = \frac { \mathrm { g } } { \omega ^ { 2 } \mathrm { r } }$.\\
The plane of the circular motion is at a height $x$ above the lowest point of the shell. When the angular speed is doubled, the plane of the motion is at a height $4 x$ above the lowest point of the shell.
\item Find $x$ in terms of $r$.
\end{enumerate}
\hfill \mbox{\textit{CAIE Further Paper 3 2020 Q4 [7]}}