| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2022 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle in hemispherical bowl |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem requiring resolution of forces (normal reaction and weight) and application of F=mrω². The setup is straightforward with given angular speed, and solving involves basic trigonometry and substitution. While it requires understanding of the conical pendulum model, it's a routine textbook-style question with no novel insight needed. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R\sin\theta = m \times 6r\sin\theta \times \frac{g}{4r}\) | M1A1A1 | Attempt NL2 along \(CP\) with correct number of terms and forces resolved. First A1: either side correct. Second A1: fully correct equation. Note: if \(\sin\theta\) cancelled from both sides, \(R = m \times 6r \times \frac{g}{4r}\) scores M1A1A1. If \(r\) used instead of radius: \(R\sin\theta = m \times r \times \frac{g}{4r}\) scores M1A1A0. |
| \(R = \frac{3}{2}mg\) | ||
| \(R\cos\theta = mg\) | M1A1 | Resolve vertically. A1: correct equation. |
| \(\frac{3}{2}mg\cos\theta = mg\) | DM1 | Eliminate \(R\) between the two equations. Depends on both M marks above. |
| \(\cos\theta = \frac{2}{3}\) | A1 | Correct value for \(\cos\theta\) seen or implied. |
| \(OC = 6r\cos\theta = 6r \times \frac{2}{3} = 4r\) | M1A1 [9] | M1: attempt to obtain \(OC\) (allow sin/cos confusion). A1: \(OC = 4r\). Note: if \(\theta\) is angle with horizontal, \(\sin\theta\) and \(\cos\theta\) reversed throughout. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R\sin\theta = mL\omega^2\) | M1A1A1 | |
| \(R\cos\theta = mg\) | M1A1 | |
| \(\tan\theta = \frac{L\omega^2}{g} = \frac{L}{4r}\) | DM1A1 | |
| \(\tan\theta = \frac{L}{OC} \Rightarrow OC = 4r\) | M1A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(mg\sin\theta = m \times (6r\sin\theta) \times \frac{g}{4r}\cos\theta\) | M1A1A1 M1A1 DM1 | scores M1A1A1 M1A1 DM1 |
| \(\cos\theta = \frac{2}{3}\) | A1 | leads straight to answer |
# Question 2:
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R\sin\theta = m \times 6r\sin\theta \times \frac{g}{4r}$ | M1A1A1 | Attempt NL2 along $CP$ with correct number of terms and forces resolved. First A1: either side correct. Second A1: fully correct equation. Note: if $\sin\theta$ cancelled from both sides, $R = m \times 6r \times \frac{g}{4r}$ scores M1A1A1. If $r$ used instead of radius: $R\sin\theta = m \times r \times \frac{g}{4r}$ scores M1A1A0. |
| $R = \frac{3}{2}mg$ | | |
| $R\cos\theta = mg$ | M1A1 | Resolve vertically. A1: correct equation. |
| $\frac{3}{2}mg\cos\theta = mg$ | DM1 | Eliminate $R$ between the two equations. Depends on both M marks above. |
| $\cos\theta = \frac{2}{3}$ | A1 | Correct value for $\cos\theta$ seen or implied. |
| $OC = 6r\cos\theta = 6r \times \frac{2}{3} = 4r$ | M1A1 **[9]** | M1: attempt to obtain $OC$ (allow sin/cos confusion). A1: $OC = 4r$. Note: if $\theta$ is angle with horizontal, $\sin\theta$ and $\cos\theta$ reversed throughout. |
**ALT 1:**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R\sin\theta = mL\omega^2$ | M1A1A1 | |
| $R\cos\theta = mg$ | M1A1 | |
| $\tan\theta = \frac{L\omega^2}{g} = \frac{L}{4r}$ | DM1A1 | |
| $\tan\theta = \frac{L}{OC} \Rightarrow OC = 4r$ | M1A1 | |
**ALT 2 (resolving tangentially, $R$ never seen):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $mg\sin\theta = m \times (6r\sin\theta) \times \frac{g}{4r}\cos\theta$ | M1A1A1 M1A1 DM1 | scores M1A1A1 M1A1 DM1 |
| $\cos\theta = \frac{2}{3}$ | A1 | leads straight to answer |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2e837bb9-4ada-4f0f-8b21-2730611335f2-04_390_515_246_772}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A hemispherical bowl of internal radius $6 r$ is fixed with its circular rim horizontal. The centre of the circular rim is $O$ and the point $A$ on the surface of the bowl is vertically below $O$. A particle $P$ moves in a horizontal circle, with centre $C$, on the smooth inner surface of the bowl. The particle moves with constant angular speed $\sqrt { \frac { g } { 4 r } }$. The point $C$ lies on $O A$, as shown in Figure 1.
Find, in terms of $r$, the distance $O C$
\hfill \mbox{\textit{Edexcel M3 2022 Q2 [9]}}