| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle in hemispherical bowl |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem requiring resolution of forces (weight, normal reaction) and application of F=mrω². The setup is familiar (particle in bowl), the angular speed is given, and it's a single-part question requiring straightforward application of Newton's second law in circular motion. Slightly above average difficulty due to the 3D geometry and algebraic manipulation, but well within typical M3 scope. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| \(R \sin \theta = m \times 4r \sin \theta \times \frac{3g}{8r}\) | M1A1A1 | |
| \(R = \frac{3}{2}mg\) | M1A1 | |
| \(R \cos \theta = mg\) | M1A1 | |
| \(\frac{3}{2}mg \cos \theta = mg\) | M1(dep) | |
| \(\cos \theta = \frac{2}{3}\) | A1 | |
| \(OC = 4r \cos \theta = 4r \times \frac{2}{3} = \frac{8}{3}r \text{ oe}\) | M1A1 | [9] |
| Answer | Marks |
|---|---|
| \(R \sin \theta = m \times a \times \frac{3g}{8r}\) | M1A1A1 |
| \(R \cos \theta = mg\) | M1 A1 |
| \(\tan \theta = \frac{3a}{8r}\) | M1 A1 |
| \(\frac{a}{OC} = \frac{3a}{8r}\) | M1 A1 |
| \(OC = \frac{8r}{3}\) | A1 |
| Answer | Marks |
|---|---|
| M1 | for NL2 towards C. Accept use of \(v = \sqrt{\frac{3g}{8r}}\) and \(a = \frac{v^2}{r}\) as a mis-read |
| A1 | for LHS fully correct |
| A1 | for RHS fully correct |
| ALT: Work in the direction of R and obtain the same equation with \(\sin \theta\) "cancelled". Give M1A1A1 if fully correct, M0 otherwise. | |
| M1 | for resolving vertically |
| A1 | for the equation fully correct |
| M1 dep | for eliminating \(R\) between the two equations. Dependent on both above M marks |
| A1 | for \(\cos \theta = \frac{2}{3}\) |
| M1 | for attempting to use trig or Pythagoras to obtain \(OC\) |
| A1 cso | for \(OC = \frac{8}{3}r\) |
$R \sin \theta = m \times 4r \sin \theta \times \frac{3g}{8r}$ | M1A1A1 |
$R = \frac{3}{2}mg$ | M1A1 |
$R \cos \theta = mg$ | M1A1 |
$\frac{3}{2}mg \cos \theta = mg$ | M1(dep) |
$\cos \theta = \frac{2}{3}$ | A1 |
$OC = 4r \cos \theta = 4r \times \frac{2}{3} = \frac{8}{3}r \text{ oe}$ | M1A1 | [9]
**Alternative Method:**
$R \sin \theta = m \times a \times \frac{3g}{8r}$ | M1A1A1 |
$R \cos \theta = mg$ | M1 A1 |
$\tan \theta = \frac{3a}{8r}$ | M1 A1 |
$\frac{a}{OC} = \frac{3a}{8r}$ | M1 A1 |
$OC = \frac{8r}{3}$ | A1 |
**Notes for Question 1:**
M1 | for NL2 towards C. Accept use of $v = \sqrt{\frac{3g}{8r}}$ and $a = \frac{v^2}{r}$ as a mis-read |
A1 | for LHS fully correct |
A1 | for RHS fully correct |
**ALT:** Work in the direction of R and obtain the same equation with $\sin \theta$ "cancelled". Give M1A1A1 if fully correct, M0 otherwise. |
M1 | for resolving vertically |
A1 | for the equation fully correct |
M1 dep | for eliminating $R$ between the two equations. Dependent on both above M marks |
A1 | for $\cos \theta = \frac{2}{3}$ |
M1 | for attempting to use trig or Pythagoras to obtain $OC$ |
A1 cso | for $OC = \frac{8}{3}r$ |
---1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e500e20b-9060-4c69-af13-fb97b9a86dfd-02_389_524_221_712}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A hemispherical bowl of internal radius $4 r$ 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 $\sqrt { \frac { 3 g } { 8 r } }$\\
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 2014 Q1 [9]}}