| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2017 |
| Session | March |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Two strings, two fixed points |
| Difficulty | Standard +0.3 This is a standard circular motion problem with connected particles requiring resolution of forces and application of F=mrω². While it involves multiple particles and 3D geometry, the setup is clearly defined with given angles, and the solution follows a systematic approach: resolve forces on P vertically and horizontally to find tension and ω, then apply the same method to Q. The multi-part structure and 7 total marks indicate moderate length, but no novel insight is required—just careful application of standard mechanics principles. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| 5(i) | Tcos60 = 1.5 + 0.4g | M1 |
| T = 11 N | A1 | |
| Tsin60 = 0.4ω2x0.5sin60 | M1 | Uses Newton's Second Law horizontally |
| Answer | Marks | Guidance |
|---|---|---|
| ω = 55 = 7.42 | A1 | |
| Total: | 4 | |
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 5(ii) | m = 0.15 (from mg = 1.5) | B1 |
| T* = 0.15 x 7.422 x 0.5sin60 | M1 | Uses Newton's Second Law horizontally |
| Answer | Marks | Guidance |
|---|---|---|
| T* = 3.57 N | A1 | |
| Total: | 3 | |
| Question | Answer | Marks |
Question 5:
--- 5(i) ---
5(i) | Tcos60 = 1.5 + 0.4g | M1 | Resolve vertically for P
T = 11 N | A1
Tsin60 = 0.4ω2x0.5sin60 | M1 | Uses Newton's Second Law horizontally
for P
ω = 55 = 7.42 | A1
Total: | 4
Question | Answer | Marks | Guidance
--- 5(ii) ---
5(ii) | m = 0.15 (from mg = 1.5) | B1 | Resolves vertically for Q
T* = 0.15 x 7.422 x 0.5sin60 | M1 | Uses Newton's Second Law horizontally
for Q
T* = 3.57 N | A1
Total: | 3
Question | Answer | Marks | Guidance\includegraphics{figure_5}
Two particles $P$ and $Q$ have masses $0.4 \text{ kg}$ and $m \text{ kg}$ respectively. $P$ is attached to a fixed point $A$ by a light inextensible string of length $0.5 \text{ m}$ which is inclined at an angle of $60°$ to the vertical. $P$ and $Q$ are joined to each other by a light inextensible vertical string. $Q$ is attached to a fixed point $B$, which is vertically below $A$, by a light inextensible string. The string $BQ$ is taut and horizontal. The particles rotate in horizontal circles about an axis through $A$ and $B$ with constant angular speed $\omega \text{ rad s}^{-1}$ (see diagram). The tension in the string joining $P$ and $Q$ is $1.5 \text{ N}$.
\begin{enumerate}[label=(\roman*)]
\item Find the tension in the string $AP$ and the value of $\omega$. [4]
\item Find $m$ and the tension in the string $BQ$. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2017 Q5 [7]}}