| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Conical pendulum – particle on horizontal surface |
| Difficulty | Standard +0.3 This is a standard circular motion problem requiring resolution of forces and application of F=mv²/r. Part (i) involves straightforward force resolution with given speed; part (ii) requires setting up simultaneous equations with T=R condition. The geometry is clearly defined and the methods are routine for M2 level, making it slightly easier than average but still requiring careful multi-step working. |
| Spec | 6.05a Angular velocity: definitions6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| 5(i) | 0.1×1.52/0.4 = Tcosθ | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| T = 0.703 | A1 | |
| R = 0.1g – Tsinθ | M1 | Resolve vertically for P |
| R = 0.578 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 5(ii) | T + Tsinθ = 0.1g | M1 |
| T = 0.625 | A1 | |
| 0.1ω 2×0.4 = 0.625cosθ | M1 | Use Newton's Second Law horizontally |
| ω = 3.54 rad s–1 | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 5:
--- 5(i) ---
5(i) | 0.1×1.52/0.4 = Tcosθ | M1 | Note r = 0.4, cosθ = 0.8, sinθ = 0.6
Use Newton's Second Law horizontally
T = 0.703 | A1
R = 0.1g – Tsinθ | M1 | Resolve vertically for P
R = 0.578 | A1
4
Question | Answer | Marks | Guidance
--- 5(ii) ---
5(ii) | T + Tsinθ = 0.1g | M1 | Resolve vertically for P
T = 0.625 | A1
0.1ω 2×0.4 = 0.625cosθ | M1 | Use Newton's Second Law horizontally
ω = 3.54 rad s–1 | A1
4
Question | Answer | Marks | GuidanceA particle $P$ of mass $0.1\text{ kg}$ is attached to one end of a light inextensible string of length $0.5\text{ m}$. The other end of the string is attached to a fixed point $A$. The particle $P$ moves in a circle which has its centre $O$ on a smooth horizontal surface $0.3\text{ m}$ below $A$. The tension in the string has magnitude $T\text{ N}$ and the magnitude of the force exerted on $P$ by the surface is $R\text{ N}$.
\begin{enumerate}[label=(\roman*)]
\item Given that the speed of $P$ is $1.5\text{ m s}^{-1}$, calculate $T$ and $R$. [4]
\item Given instead that $T = R$, calculate the angular speed of $P$. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2018 Q5 [8]}}