| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2018 |
| Session | November |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Vertical elastic string: projected from equilibrium or other point |
| Difficulty | Standard +0.3 This is a standard A-level mechanics question on elastic strings with energy conservation. Part (i) requires applying conservation of energy with elastic potential energy (EPE = λx²/2l), gravitational PE, and kinetic energy—a routine M2 technique. Part (ii) requires recognizing that maximum speed occurs when net force is zero (at equilibrium position). While multi-step, these are textbook applications of well-practiced methods with no novel insight required, making it slightly easier than average. |
| Spec | 6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{15(0.2+x)^2}{2\times0.6}\) or \(\frac{15\times0.2^2}{2\times0.6}\) | B1 | |
| M1 | Set up a 4 term energy equation | |
| \(\frac{15(0.2+x)^2}{2\times0.6} = \frac{15\times0.2^2}{2\times0.6} + 0.7gx + \frac{(0.7\times2^2)}{2}\) | A1 | \(0.5 + 5x + 12.5x^2 = 0.5 + 7x + 1.4\) |
| \(x = 0.424\) m | A1 | |
| 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(0.7g = 15e/0.6\) | M1 | Use \(T = \lambda x/L\) |
| \(e = 0.28\) m | A1 | |
| \(\frac{0.7\times2^2}{2} + 0.7g(0.28-0.2) + \frac{15\times0.2^2}{2\times0.6} = \frac{0.7\times v^2}{2} + \frac{15\times0.28^2}{2\times0.6}\) | M1 | Set up a 5 term energy equation |
| \(v = 2.06\ \text{m s}^{-1}\) | A1 | |
| 4 |
## Question 5:
### Part (i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{15(0.2+x)^2}{2\times0.6}$ **or** $\frac{15\times0.2^2}{2\times0.6}$ | B1 | |
| | M1 | Set up a 4 term energy equation |
| $\frac{15(0.2+x)^2}{2\times0.6} = \frac{15\times0.2^2}{2\times0.6} + 0.7gx + \frac{(0.7\times2^2)}{2}$ | A1 | $0.5 + 5x + 12.5x^2 = 0.5 + 7x + 1.4$ |
| $x = 0.424$ m | A1 | |
| | **4** | |
### Part (ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.7g = 15e/0.6$ | M1 | Use $T = \lambda x/L$ |
| $e = 0.28$ m | A1 | |
| $\frac{0.7\times2^2}{2} + 0.7g(0.28-0.2) + \frac{15\times0.2^2}{2\times0.6} = \frac{0.7\times v^2}{2} + \frac{15\times0.28^2}{2\times0.6}$ | M1 | Set up a 5 term energy equation |
| $v = 2.06\ \text{m s}^{-1}$ | A1 | |
| | **4** | |
---5 A particle $P$ of mass 0.7 kg is attached to a fixed point $O$ by a light elastic string of natural length 0.6 m and modulus of elasticity 15 N . The particle $P$ is projected vertically downwards from the point $A , 0.8 \mathrm {~m}$ vertically below $O$. The initial speed of $P$ is $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Find the distance below $A$ of the point at which $P$ comes to instantaneous rest.\\
(ii) Find the greatest speed of $P$ in the motion.\\
\includegraphics[max width=\textwidth, alt={}, center]{f922bf53-94a0-4ccc-8c38-959d2f795629-10_478_652_260_751}
The diagram shows a uniform lamina $A B C D E F G H$. The lamina consists of a quarter-circle $O A B$ of radius $r \mathrm {~m}$, a rectangle $D E F G$ and two isosceles right-angled triangles $C O D$ and $G O H$. The rectangle has $D G = E F = r \mathrm {~m}$ and $D E = F G = x \mathrm {~m}$.\\
(i) Given that the centre of mass of the lamina is at $O$, express $x$ in terms of $r$.\\
(ii) Given instead that the rectangle $D E F G$ is a square with edges of length $r \mathrm {~m}$, state with a reason whether the centre of mass of the lamina lies within the square or the quarter-circle.\\
\includegraphics[max width=\textwidth, alt={}, center]{f922bf53-94a0-4ccc-8c38-959d2f795629-12_384_693_258_726}
A rough horizontal rod $A B$ of length 0.45 m rotates with constant angular velocity $6 \mathrm { rad } \mathrm { s } ^ { - 1 }$ about a vertical axis through $A$. A small ring $R$ of mass 0.2 kg can slide on the rod. A particle $P$ of mass 0.1 kg is attached to the mid-point of a light inextensible string of length 0.6 m . One end of the string is attached to $R$ and the other end of the string is attached to $B$, with angle $R P B = 60 ^ { \circ }$ (see diagram). $R$ and $P$ move in horizontal circles as the system rotates. $R$ is in limiting equilibrium.\\
(i) Show that the tension in the portion $P R$ of the string is 1.66 N , correct to 3 significant figures.\\
(ii) Find the coefficient of friction between the ring and the rod.\\
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 2018 Q5 [8]}}