| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | March |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Conical pendulum (horizontal circle) |
| Difficulty | Standard +0.3 This is a standard conical pendulum problem on a sphere requiring resolution of forces, circular motion equations (F=mrω²), and solving simultaneous equations. Part (i) is guided, parts (ii-iii) follow routine procedures for Further Maths M2. Slightly above average due to the geometry and multi-part nature, but uses well-practiced techniques without requiring novel insight. |
| Spec | 3.03b Newton's first law: equilibrium6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R\cos60 + T\cos30 = 0.2g\) | M1 | Resolving vertically |
| \(R + T\sqrt{3} = 4\) | A1 (AG) | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T\sin30 - R\sin60 = 0.2\omega^2 \times 0.6\sin60\) | M1 | N2L horizontally with accn \(= \omega^2 r\) |
| \((T - R\sqrt{3} = 0.12\omega^2\sqrt{3})\) | A1 | Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(R\cos60\sin30 + R\sin60\cos30 = 2\sin30 - 0.2 \times 2^2 \times 0.6\sin60\cos30\) | M1 | Substitutes \(\omega = 2\) and eliminates \(T\) from (i) and (ii) |
| \(R = 0.64\ \text{N}\) | A1 | Total: 2 |
| OR | ||
| \(R + 3R = 4 - 0.12 \times 2^2 \times 3\) | M1 | |
| \(R = 0.64\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T\cos30 = 2\) | M1 | |
| \(T = 2.31\) | A1 | When \(R = 0\), \(T = 4\sqrt{3}/3\) or \(4/\sqrt{3}\) |
| \(2.31\sin30 = 0.2\omega^2 \times 0.6\sin60\) AND \(v = \omega \times 0.6\sin60\) | M1 | |
| \(v = 1.73\ \text{m s}^{-1}\) | A1 | Total: 4 |
| OR | ||
| \(2.31\sin30 = 0.2v^2/(0.6\sin60)\) | M1 | |
| \(v = 1.73\ \text{m s}^{-1}\) | A1 | Final pair of marks |
## Question 7:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R\cos60 + T\cos30 = 0.2g$ | M1 | Resolving vertically |
| $R + T\sqrt{3} = 4$ | A1 (AG) | **Total: 2** | $g = 10$ must be used |
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T\sin30 - R\sin60 = 0.2\omega^2 \times 0.6\sin60$ | M1 | N2L horizontally with accn $= \omega^2 r$ |
| $(T - R\sqrt{3} = 0.12\omega^2\sqrt{3})$ | A1 | **Total: 2** | Accept with trig ratios |
### Part (iii)(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $R\cos60\sin30 + R\sin60\cos30 = 2\sin30 - 0.2 \times 2^2 \times 0.6\sin60\cos30$ | M1 | Substitutes $\omega = 2$ and eliminates $T$ from (i) and (ii) |
| $R = 0.64\ \text{N}$ | A1 | **Total: 2** | Accept answers between 0.639 and 0.641 inclusive |
| **OR** | | |
| $R + 3R = 4 - 0.12 \times 2^2 \times 3$ | M1 | |
| $R = 0.64$ | A1 | |
### Part (iii)(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T\cos30 = 2$ | M1 | |
| $T = 2.31$ | A1 | When $R = 0$, $T = 4\sqrt{3}/3$ or $4/\sqrt{3}$ |
| $2.31\sin30 = 0.2\omega^2 \times 0.6\sin60$ AND $v = \omega \times 0.6\sin60$ | M1 | |
| $v = 1.73\ \text{m s}^{-1}$ | A1 | **Total: 4** |
| **OR** | | |
| $2.31\sin30 = 0.2v^2/(0.6\sin60)$ | M1 | |
| $v = 1.73\ \text{m s}^{-1}$ | A1 | Final pair of marks |7\\
\includegraphics[max width=\textwidth, alt={}, center]{334b4bdf-6d9c-4208-9032-572eb7c5f9ee-3_451_432_1434_852}
One end of a light inextensible string is attached to the highest point $A$ of a solid fixed sphere with centre $O$ and radius 0.6 m . The other end of the string is attached to a particle $P$ of mass 0.2 kg which rests in contact with the smooth surface of the sphere. The angle $A O P = 60 ^ { \circ }$ (see diagram). The sphere exerts a contact force of magnitude $R \mathrm {~N}$ on $P$ and the tension in the string is $T \mathrm {~N}$.\\
(i) By resolving vertically, show that $R + ( \sqrt { } 3 ) T = 4$.\\
$P$ is now set in motion, and moves with angular speed $\omega \mathrm { rad } \mathrm { s } ^ { - 1 }$ in a horizontal circle on the surface of the sphere.\\
(ii) Find an equation involving $R , T$ and $\omega$.\\
(iii) Hence
\begin{enumerate}[label=(\alph*)]
\item calculate $R$ when $\omega = 2$,
\item find the greatest possible value of $T$ and the corresponding speed of $P$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2016 Q7 [10]}}