| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | November |
| Marks | 9 |
| 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 two strings requiring resolution of forces and application of F=mrω². The geometry is given (60° angles), and parts follow a clear progression using standard techniques. Slightly above average difficulty due to the two-string setup and multiple parts, but all steps are routine for M2 level. |
| Spec | 3.03b Newton's first law: equilibrium3.03d Newton's second law: 2D vectors6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| (i) | \([\frac{1}{2} \times 12(7^2 - 3^2)]\) | M1 |
| Increase is 240 J | A1 | 2 marks total |
| (ii) | M1 | |
| \(12g \times AB\sin10° = 240\) | A1ft | |
| Distance is 11.5 m | A1 | 3 marks total |
| SR for candidates who avoid 'hence' (max 2/3): For using Newton's Second Law and \(v^2 = u^2 + 2as\) \([12g\sin 10° = 12a\) then \(7^2 = 3^2 + 2(g\sin10° \times AB)]\) M1; \(11.5\) m A1 | ||
| (iii) | M1 | |
| \(F \times 11.5\cos10° = 240\) or \(F\cos10° - 12g\sin10° = 0\) | A1ft | |
| Magnitude is 21.2 N | A1 | 3 marks total |
(i) | $[\frac{1}{2} \times 12(7^2 - 3^2)]$ | M1 | For using $KE = \frac{1}{2}m(v_B^2 - v_A^2)$ |
| Increase is 240 J | A1 | 2 marks total |
(ii) | | M1 | For using $mgh = KE$ gain |
| $12g \times AB\sin10° = 240$ | A1ft | |
| Distance is 11.5 m | A1 | 3 marks total |
| | | SR for candidates who avoid 'hence' (max 2/3): For using Newton's Second Law and $v^2 = u^2 + 2as$ $[12g\sin 10° = 12a$ then $7^2 = 3^2 + 2(g\sin10° \times AB)]$ M1; $11.5$ m A1 |
(iii) | | M1 | For using $F(AB)\cos10° = PE$ gain or for using Newton's 2nd law with $a = 0$ |
| $F \times 11.5\cos10° = 240$ or $F\cos10° - 12g\sin10° = 0$ | A1ft | |
| Magnitude is 21.2 N | A1 | 3 marks total |
---5 A small ball $B$ of mass 0.2 kg is attached to fixed points $P$ and $Q$ by two light inextensible strings of equal length. $P$ is vertically above $Q$, the strings are taut and each is inclined at $60 ^ { \circ }$ to the vertical. $B$ moves with constant speed in a horizontal circle of radius 0.6 m .\\
(i) Given that the tension in the string $P B$ is 7 N , calculate
\begin{enumerate}[label=(\alph*)]
\item the tension in string $Q B$,
\item the speed of $B$.\\
(ii) Given instead that $B$ is moving with angular speed $7 \mathrm { rad } \mathrm { s } ^ { - 1 }$, calculate the tension in the string $Q B$.
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2012 Q5 [9]}}