| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2019 |
| 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 M2 circular motion problem requiring resolution of forces, application of Hooke's law for elastic strings, and use of the circular motion equation T = mrω². The setup with two strings (one inextensible, one elastic) is a classic textbook scenario, and the solution follows a straightforward method of resolving vertically and horizontally. Slightly above average difficulty due to the two-string setup and need to find natural length, but still routine for M2 students. |
| Spec | 3.03d Newton's second law: 2D vectors6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T = \dfrac{9 \times (0.8 - 0.6)}{0.6}\) | M1 | Use \(T = \dfrac{\lambda x}{l}\). Note \(OP = \dfrac{0.4}{\sin 30}\) |
| \(T = 3\text{ N}\) | A1 | |
| \(0.3a = 3 - 0.3g\sin 30\) | M1 | Use Newton's Second Law along the slope |
| \(a = 5\text{ m s}^{-1}\) | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.3g\sin 30 = \dfrac{9e}{0.6}\) | M1 | Note the maximum speed is at the equilibrium position |
| \(e = 0.1\) | A1 | |
| \(\text{EPE} = \dfrac{9 \times (0.8-0.6)^2}{2 \times 0.6}\) or \(\dfrac{9 \times 0.1^2}{2 \times 0.6}\) | B1 | |
| \(\dfrac{0.3v^2}{2} = \dfrac{9\times(0.8-0.6)^2}{2\times0.6} - \dfrac{9\times0.1^2}{2\times0.6} - 0.3g\times0.1\sin30\) | M1 | Set up a 4 term energy equation |
| \(v = 0.707\text{ m s}^{-1}\) | A1 | |
| Total: 5 |
## Question 5(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T = \dfrac{9 \times (0.8 - 0.6)}{0.6}$ | M1 | Use $T = \dfrac{\lambda x}{l}$. Note $OP = \dfrac{0.4}{\sin 30}$ |
| $T = 3\text{ N}$ | A1 | |
| $0.3a = 3 - 0.3g\sin 30$ | M1 | Use Newton's Second Law along the slope |
| $a = 5\text{ m s}^{-1}$ | A1 | |
| **Total: 4** | | |
## Question 5(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.3g\sin 30 = \dfrac{9e}{0.6}$ | M1 | Note the maximum speed is at the equilibrium position |
| $e = 0.1$ | A1 | |
| $\text{EPE} = \dfrac{9 \times (0.8-0.6)^2}{2 \times 0.6}$ or $\dfrac{9 \times 0.1^2}{2 \times 0.6}$ | B1 | |
| $\dfrac{0.3v^2}{2} = \dfrac{9\times(0.8-0.6)^2}{2\times0.6} - \dfrac{9\times0.1^2}{2\times0.6} - 0.3g\times0.1\sin30$ | M1 | Set up a 4 term energy equation |
| $v = 0.707\text{ m s}^{-1}$ | A1 | |
| **Total: 5** | | |5 A particle $P$ of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus of elasticity 9 N . The other end of the string is attached to a fixed point $O$ on a smooth plane inclined at $30 ^ { \circ }$ to the horizontal. $O A$ is a line of greatest slope of the plane with $A$ below the level of $O$ and $O A = 0.8 \mathrm {~m}$. The particle $P$ is released from rest at $A$.\\
(i) Find the initial acceleration of $P$.\\
(ii) Find the greatest speed of $P$.\\
$6 \quad A$ and $B$ are two fixed points on a vertical axis with $A 0.6 \mathrm {~m}$ above $B$. A particle $P$ of mass 0.3 kg is attached to $A$ by a light inextensible string of length 0.5 m . The particle $P$ is attached to $B$ by a light elastic string with modulus of elasticity 46 N . The particle $P$ moves with constant angular speed $8 \mathrm { rad } \mathrm { s } ^ { - 1 }$ in a horizontal circle with centre at the mid-point of $A B$.\\
(i) Find the speed of $P$.\\
(ii) Calculate the tension in the string $B P$ and hence find the natural length of this string.\\
\includegraphics[max width=\textwidth, alt={}, center]{42de91da-d65e-40e7-8de5-f40eda927850-10_540_574_260_781}\\
$A B C$ is the cross-section through the centre of mass of a uniform prism which rests with $A B$ on a rough horizontal surface. $A B = 0.4 \mathrm {~m}$ and $C$ is 0.9 m above the surface (see diagram). The prism is on the point of toppling about its edge through $B$.\\
(i) Show that angle $B A C = 48.4 ^ { \circ }$, correct to 3 significant figures.\\
A force of magnitude 18 N acting in the plane of the cross-section and perpendicular to $A C$ is now applied to the prism at $C$. The prism is on the point of rotating about its edge through $A$.\\
(ii) Calculate the weight of the prism.\\
(iii) Given also that the prism is on the point of slipping, calculate the coefficient of friction between the prism and the surface.\\
If you use the following lined page to complete the answer(s) to any question(s), the question number(s) must be clearly shown.\\
\hfill \mbox{\textit{CAIE M2 2019 Q5 [9]}}