| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Coupled circular motions |
| Difficulty | Challenging +1.2 This is a multi-part circular motion problem requiring force analysis in rotating frames with friction and coupled particles. While it involves several steps (resolving forces, applying F=mrω², solving simultaneous equations), the techniques are standard M2 content with clear physical setup. The coupled particle aspect adds moderate complexity beyond basic circular motion, but the problem structure guides students through systematic application of Newton's laws. |
| Spec | 3.03r Friction: concept and vector form3.03t Coefficient of friction: F <= mu*R model6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.3\omega^2 \times 0.5 = T + 0.36 \times 0.3g\) | M1 | Newton's Second Law, 3 terms |
| \(0.2\omega^2 \times 0.5 = T - 0.36 \times 0.2g\) | A1 | Both correct |
| \(0.1\omega^2 \times 0.5 = 0.36 \times 0.5g\) | M1 | |
| \(\omega = 6\) | A1 | |
| \(T = 0.3 \times 6^2 \times 0.5 - 0.36 \times 0.3 \times 10\) | M1 | |
| \(T = 4.32\) | A1 [6] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.2\omega^2 r = 0.3\omega^2(1-r)\) | M1 | \(0.3\omega^2 R = 0.2\omega^2(1-R)\) |
| \(r = 0.6\) | A1 | \(R = 0.4\) |
| \(r_P = 0.6\) m and \(r_Q = 0.4\) m | A1ft [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.48 = 0.2v_P^2/0.6\) or \(0.48 = 0.3v_Q^2/0.4\) | M1 | Newton's Second Law radially |
| \(v_P = 1.2\) | A1 | |
| \(v_Q = 0.8\) | A1 [3] |
## Question 7:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.3\omega^2 \times 0.5 = T + 0.36 \times 0.3g$ | M1 | Newton's Second Law, 3 terms |
| $0.2\omega^2 \times 0.5 = T - 0.36 \times 0.2g$ | A1 | Both correct |
| $0.1\omega^2 \times 0.5 = 0.36 \times 0.5g$ | M1 | |
| $\omega = 6$ | A1 | |
| $T = 0.3 \times 6^2 \times 0.5 - 0.36 \times 0.3 \times 10$ | M1 | |
| $T = 4.32$ | A1 [6] | |
### Part (ii)(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.2\omega^2 r = 0.3\omega^2(1-r)$ | M1 | $0.3\omega^2 R = 0.2\omega^2(1-R)$ |
| $r = 0.6$ | A1 | $R = 0.4$ |
| $r_P = 0.6$ m and $r_Q = 0.4$ m | A1ft [3] | |
### Part (ii)(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.48 = 0.2v_P^2/0.6$ or $0.48 = 0.3v_Q^2/0.4$ | M1 | Newton's Second Law radially |
| $v_P = 1.2$ | A1 | |
| $v_Q = 0.8$ | A1 [3] | |7\\
\includegraphics[max width=\textwidth, alt={}, center]{18398d27-15eb-4515-8210-4f0f614d5b28-4_713_933_258_605}
A narrow groove is cut along a diameter in the surface of a horizontal disc with centre $O$. Particles $P$ and $Q$, of masses 0.2 kg and 0.3 kg respectively, lie in the groove, and the coefficient of friction between each of the particles and the groove is $\mu$. The particles are attached to opposite ends of a light inextensible string of length 1 m . The disc rotates with angular velocity $\omega \mathrm { rad } \mathrm { s } ^ { - 1 }$ about a vertical axis passing through $O$ and the particles move in horizontal circles (see diagram).\\
(i) Given that $\mu = 0.36$ and that both $P$ and $Q$ move in the same horizontal circle of radius 0.5 m , calculate the greatest possible value of $\omega$ and the corresponding tension in the string.\\
(ii) Given instead that $\mu = 0$ and that the tension in the string is 0.48 N , calculate
\begin{enumerate}[label=(\alph*)]
\item the radius of the circle in which $P$ moves and the radius of the circle in which $Q$ moves,
\item the speeds of the particles.
\end{enumerate}
\hfill \mbox{\textit{CAIE M2 2011 Q7 [12]}}