| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2010 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle in hemispherical bowl |
| Difficulty | Standard +0.3 This is a standard circular motion question covering horizontal circles in a bowl and vertical circular motion with energy conservation. Parts (i)-(iii) are routine applications of F=ma in circular motion and energy methods. Parts (iv)-(v) require combining Newton's second law with energy conservation and resolving acceleration, but follow predictable patterns for M3 level. The geometry is straightforward and all techniques are standard textbook exercises. |
| Spec | 6.05a Angular velocity: definitions6.05c Horizontal circles: conical pendulum, banked tracks6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R\cos\theta = mg\) [\(\theta\) is angle between OB and vertical] | M1 | Resolving vertically |
| \(R \times 0.8 = 0.4 \times 9.8\) | A1 | |
| Normal reaction is \(4.9\) N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R\sin\theta = m\frac{v^2}{r}\) | M1 | For acceleration \(\frac{v^2}{r}\) or \(r\omega^2\) |
| \(4.9 \times 0.6 = 0.4 \times \frac{v^2}{1.5}\) | A1 | or \(4.9 \times 0.6 = 0.4 \times 1.5\omega^2\) |
| \(v^2 = 11.025\) | ||
| Speed is \(3.32\ \text{ms}^{-1}\) (3 sf) | A1 | ft is \(1.5\sqrt{R}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| By conservation of energy: \(\frac{1}{2}mu^2 = mg \times 2.5\) | M1, A1 | Equation involving KE and PE |
| \(u^2 = 5g \quad (u=7)\) | ||
| \(R - mg = m \times \frac{u^2}{2.5}\) | M1 | Vertical equation of motion (must have three terms) |
| \(R - mg = 2mg\) | ||
| \(R = 3mg\) | E1 | Correctly shown; or \(R = 11.76\) and \(3 \times 0.4 \times 9.8 = 11.76\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}mv^2 = mg \times 2.5\cos\theta\) | M1, A1 | *Mark (iv) and (v) as one part.* Equation involving KE, PE and angle (\(\theta\) is angle with vertical); \([\frac{1}{2}mv^2 = mgh\) can earn M1A1, but only if \(\cos\theta = h/2.5\) appears somewhere] |
| \(v^2 = 5g\cos\theta\) | ||
| \(R - mg\cos\theta = m \times \frac{v^2}{2.5}\) | M1 | Equation of motion towards O (must have three terms, and the weight must be resolved) |
| When \(R = 2mg\ (= 7.84)\): | ||
| \(2mg - mg\cos\theta = \frac{mv^2}{2.5}\) | ||
| \(2mg - \frac{mv^2}{5} = \frac{mv^2}{2.5}\) | M1 | Obtaining an equation for \(v\) |
| \(7.84 - 0.08v^2 = 0.16v^2\) | M1 | Obtaining an equation for \(\theta\); *These two marks are each dependent on M1M1 above* |
| \(v^2 = \frac{98}{3}\) | ||
| Speed is \(5.72\ \text{ms}^{-1}\) (3 sf) | A1 | |
| \(\cos\theta = \frac{v^2}{5g} = \frac{2}{3} \quad (\theta = 48.2°\) or \(0.841\) rad\()\) | ||
| Tangential acceleration is \(g\sin\theta\) | M1 | [\(g\sin\theta\) in isolation only earns M1 if the angle \(\theta\) is clearly indicated] |
| Tangential acceleration is \(7.30\ \text{ms}^{-2}\) (3 sf) | A1 |
# Question 2:
## Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R\cos\theta = mg$ [$\theta$ is angle between OB and vertical] | M1 | Resolving vertically |
| $R \times 0.8 = 0.4 \times 9.8$ | A1 | |
| Normal reaction is $4.9$ N | A1 | |
## Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R\sin\theta = m\frac{v^2}{r}$ | M1 | For acceleration $\frac{v^2}{r}$ or $r\omega^2$ |
| $4.9 \times 0.6 = 0.4 \times \frac{v^2}{1.5}$ | A1 | or $4.9 \times 0.6 = 0.4 \times 1.5\omega^2$ |
| $v^2 = 11.025$ | | |
| Speed is $3.32\ \text{ms}^{-1}$ (3 sf) | A1 | ft is $1.5\sqrt{R}$ |
## Part (iii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| By conservation of energy: $\frac{1}{2}mu^2 = mg \times 2.5$ | M1, A1 | Equation involving KE and PE |
| $u^2 = 5g \quad (u=7)$ | | |
| $R - mg = m \times \frac{u^2}{2.5}$ | M1 | Vertical equation of motion (must have three terms) |
| $R - mg = 2mg$ | | |
| $R = 3mg$ | E1 | Correctly shown; or $R = 11.76$ and $3 \times 0.4 \times 9.8 = 11.76$ |
## Parts (iv) and (v)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 = mg \times 2.5\cos\theta$ | M1, A1 | *Mark (iv) and (v) as one part.* Equation involving KE, PE and angle ($\theta$ is angle with vertical); $[\frac{1}{2}mv^2 = mgh$ can earn M1A1, but only if $\cos\theta = h/2.5$ appears somewhere] |
| $v^2 = 5g\cos\theta$ | | |
| $R - mg\cos\theta = m \times \frac{v^2}{2.5}$ | M1 | Equation of motion towards O (must have three terms, and the weight must be resolved) |
| When $R = 2mg\ (= 7.84)$: | | |
| $2mg - mg\cos\theta = \frac{mv^2}{2.5}$ | | |
| $2mg - \frac{mv^2}{5} = \frac{mv^2}{2.5}$ | M1 | Obtaining an equation for $v$ |
| $7.84 - 0.08v^2 = 0.16v^2$ | M1 | Obtaining an equation for $\theta$; *These two marks are each dependent on M1M1 above* |
| $v^2 = \frac{98}{3}$ | | |
| Speed is $5.72\ \text{ms}^{-1}$ (3 sf) | A1 | |
| $\cos\theta = \frac{v^2}{5g} = \frac{2}{3} \quad (\theta = 48.2°$ or $0.841$ rad$)$ | | |
| Tangential acceleration is $g\sin\theta$ | M1 | [$g\sin\theta$ in isolation only earns M1 if the angle $\theta$ is clearly indicated] |
| Tangential acceleration is $7.30\ \text{ms}^{-2}$ (3 sf) | A1 | |
---2 A hollow hemisphere has internal radius 2.5 m and is fixed with its rim horizontal and uppermost. The centre of the hemisphere is O . A small ball B of mass 0.4 kg moves in contact with the smooth inside surface of the hemisphere.
At first, B is moving at constant speed in a horizontal circle with radius 1.5 m , as shown in Fig. 2.1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c93aed95-f655-45cb-805f-7114a15acccf-3_392_661_529_742}
\captionsetup{labelformat=empty}
\caption{Fig. 2.1}
\end{center}
\end{figure}
(i) Find the normal reaction of the hemisphere on $B$.\\
(ii) Find the speed of $\mathbf { B }$.
The ball B is now released from rest on the inside surface at a point on the same horizontal level as O . It then moves in part of a vertical circle with centre O and radius 2.5 m , as shown in Fig. 2.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c93aed95-f655-45cb-805f-7114a15acccf-3_378_663_1427_740}
\captionsetup{labelformat=empty}
\caption{Fig. 2.2}
\end{center}
\end{figure}
(iii) Show that, when $B$ is at its lowest point, the normal reaction is three times the weight of $B$.
For an instant when the normal reaction is twice the weight of $\mathbf { B }$, find\\
(iv) the speed of $\mathbf { B }$,\\
(v) the tangential component of the acceleration of $\mathbf { B }$.
\hfill \mbox{\textit{OCR MEI M3 2010 Q2 [18]}}