| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle on outer surface of sphere |
| Difficulty | Standard +0.3 This is a standard M3 circular motion question with a particle on a sphere. Part (a) is routine centripetal force calculation. Part (b) follows the classic textbook template: energy conservation for v², resolving forces radially for R, finding acceleration components, and determining when R=0. All steps are well-practiced techniques with no novel insight required, making it slightly easier than average. |
| Spec | 6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.05b Circular motion: v=r*omega and a=v^2/r6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Resolving vertically: \(R\cos18° = mg\) | M1 | Resolving forces, no friction |
| Horizontal: \(R\sin18° = \frac{mv^2}{r}\) | M1 | Circular motion equation |
| Dividing: \(\tan18° = \frac{v^2}{rg}\) | M1 | |
| \(v^2 = 45 \times 9.8 \times \tan18°\) | ||
| \(v = 11.9 \text{ ms}^{-1}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Resolving vertically: \(R\cos18° - F\sin18° = 800g\) | M1 | |
| Resolving horizontally: \(R\sin18° + F\cos18° = \frac{800 \times 15^2}{45}\) | M1 A1 | |
| Solving simultaneously | DM1 | |
| \(F = 800 \times 5 \cos18° - 800g\sin18°\) | ||
| \(F = 1310 \text{ N}\) | A1 | |
| \(R = \frac{800g\cos18° + 1310\sin18°}{\cos^2 18°+\sin^2 18°}\) | ||
| \(R = 8210 \text{ N}\) | A1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Energy conservation: \(\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - mgL(1+\cos\theta)\) | M1 A1 | \(L\) = string length, \(\theta\) = angle from upward vertical |
| \(\frac{1}{2}m(2.8)^2 = \frac{1}{2}m(7)^2 - mgL(1+\cos\theta)\) | ||
| At slack: \(T = 0\), so \(\frac{mv^2}{L} = mg\cos\theta\) | M1 A1 | Radial equation when slack |
| \(v^2 = gL\cos\theta\) | ||
| \((2.8)^2 = gL\cos\theta\) | A1 | |
| Substituting: \(7.84 = 49 - 2gL(1+\cos\theta)\) | ||
| \(2gL(1+\cos\theta) = 41.16\) | DM1 | Solving simultaneously |
| \(L = 1.96 \text{ m}\) | A1 | |
| \(\cos\theta = \frac{7.84}{9.8 \times 1.96} = \frac{4}{9.8}\), \(\theta = 66.4°\) | A1 |
# Question 2:
## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Resolving vertically: $R\cos18° = mg$ | M1 | Resolving forces, no friction |
| Horizontal: $R\sin18° = \frac{mv^2}{r}$ | M1 | Circular motion equation |
| Dividing: $\tan18° = \frac{v^2}{rg}$ | M1 | |
| $v^2 = 45 \times 9.8 \times \tan18°$ | | |
| $v = 11.9 \text{ ms}^{-1}$ | A1 | |
## Part (a)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Resolving vertically: $R\cos18° - F\sin18° = 800g$ | M1 | |
| Resolving horizontally: $R\sin18° + F\cos18° = \frac{800 \times 15^2}{45}$ | M1 A1 | |
| Solving simultaneously | DM1 | |
| $F = 800 \times 5 \cos18° - 800g\sin18°$ | | |
| $F = 1310 \text{ N}$ | A1 | |
| $R = \frac{800g\cos18° + 1310\sin18°}{\cos^2 18°+\sin^2 18°}$ | | |
| $R = 8210 \text{ N}$ | A1 A1 | |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Energy conservation: $\frac{1}{2}mv^2 = \frac{1}{2}mu^2 - mgL(1+\cos\theta)$ | M1 A1 | $L$ = string length, $\theta$ = angle from upward vertical |
| $\frac{1}{2}m(2.8)^2 = \frac{1}{2}m(7)^2 - mgL(1+\cos\theta)$ | | |
| At slack: $T = 0$, so $\frac{mv^2}{L} = mg\cos\theta$ | M1 A1 | Radial equation when slack |
| $v^2 = gL\cos\theta$ | | |
| $(2.8)^2 = gL\cos\theta$ | A1 | |
| Substituting: $7.84 = 49 - 2gL(1+\cos\theta)$ | | |
| $2gL(1+\cos\theta) = 41.16$ | DM1 | Solving simultaneously |
| $L = 1.96 \text{ m}$ | A1 | |
| $\cos\theta = \frac{7.84}{9.8 \times 1.96} = \frac{4}{9.8}$, $\theta = 66.4°$ | A1 | |
---2
\begin{enumerate}[label=(\alph*)]
\item A moon of mass $7.5 \times 10 ^ { 22 } \mathrm {~kg}$ moves round a planet in a circular path of radius $3.8 \times 10 ^ { 8 } \mathrm {~m}$, completing one orbit in a time of $2.4 \times 10 ^ { 6 } \mathrm {~s}$. Find the force acting on the moon.
\item Fig. 2 shows a fixed solid sphere with centre O and radius 4 m . Its surface is smooth. The point A on the surface of the sphere is 3.5 m vertically above the level of O . A particle P of mass 0.2 kg is placed on the surface at A and is released from rest. In the subsequent motion, when OP makes an angle $\theta$ with the horizontal and P is still on the surface of the sphere, the speed of P is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the normal reaction acting on P is $R \mathrm {~N}$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{e0e5580a-e1f0-46f8-9304-2a96533af186-03_746_734_705_662}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Express $v ^ { 2 }$ in terms of $\theta$.
\item Show that $R = 5.88 \sin \theta - 3.43$.
\item Find the radial and tangential components of the acceleration of P when $\theta = 40 ^ { \circ }$.
\item Find the value of $\theta$ at the instant when P leaves the surface of the sphere.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 Q2 [18]}}