| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2012 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Particle at midpoint of string between vertical fixed points |
| Difficulty | Challenging +1.8 This is a challenging multi-part mechanics problem requiring equilibrium analysis with two elastic strings, energy conservation with variable elastic potential energy, and verification of a specific position. The setup with a particle at the mid-point of a string attached at both ends on an inclined plane requires careful geometric reasoning, and part (ii) demands identifying that maximum speed occurs at the equilibrium position. This goes beyond standard textbook exercises and requires strong problem-solving skills across multiple mechanics concepts. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | Uses \(T = 45ev/1.5\) | |
| \(45e/1.5 = 45(1 - e)/1.5 \pm 0.6g\sin30°\) | A1 | Note either portion may be \(e\) |
| \(AP (= 0.55 + 1.5) = 2.05\) m | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | KE/EE/PE energy conservation | |
| \(45 \times 1^2/(2 \times 1.5) = 45 \times 0.55^2/(2 \times 1.5) + 45 \times 0.45^2/(2 \times 1.5) + 0.6g \times 0.45\sin30° + 0.6v^2/2\) | A1 | 3 correct EE terms |
| A1 | Correct equation | |
| \(v = 4.5\) ms\(^{-1}\) | A1 [4] |
| Answer | Marks | Guidance |
|---|---|---|
| M1 | EE/PE conservation | |
| \(45 \times 1^2/(2 \times 1.5) = 45(1.6 - 1.5)^2/(2 \times 1.5) + 45(4 - 1.6 - 1.5)^2/(2 \times 1.5) + 0.6 \times 10(2.5 - 1.6)\sin30°\) | A1 [2] | Total energy = 15 |
**(i)**
| | M1 | Uses $T = 45ev/1.5$ |
| $45e/1.5 = 45(1 - e)/1.5 \pm 0.6g\sin30°$ | A1 | Note either portion may be $e$ |
| $AP (= 0.55 + 1.5) = 2.05$ m | A1 [3] | |
**(ii)**
| | M1 | KE/EE/PE energy conservation |
| $45 \times 1^2/(2 \times 1.5) = 45 \times 0.55^2/(2 \times 1.5) + 45 \times 0.45^2/(2 \times 1.5) + 0.6g \times 0.45\sin30° + 0.6v^2/2$ | A1 | 3 correct EE terms |
| | A1 | Correct equation |
| $v = 4.5$ ms$^{-1}$ | A1 [4] | |
**(iii)**
| | M1 | EE/PE conservation |
| $45 \times 1^2/(2 \times 1.5) = 45(1.6 - 1.5)^2/(2 \times 1.5) + 45(4 - 1.6 - 1.5)^2/(2 \times 1.5) + 0.6 \times 10(2.5 - 1.6)\sin30°$ | A1 [2] | Total energy = 15 | [9] |5 A light elastic string has natural length 3 m and modulus of elasticity 45 N . A particle $P$ of mass 0.6 kg is attached to the mid-point of the string. The ends of the string are attached to fixed points $A$ and $B$ which lie on a line of greatest slope of a smooth plane inclined at $30 ^ { \circ }$ to the horizontal. The distance $A B$ is 4 m , and $A$ is higher than $B$.\\
(i) Calculate the distance $A P$ when $P$ rests on the slope in equilibrium.\\
$P$ is released from rest at the point between $A$ and $B$ where $A P = 2.5 \mathrm {~m}$.\\
(ii) Find the maximum speed of $P$.\\
(iii) Show that $P$ is at rest when $A P = 1.6 \mathrm {~m}$.
\hfill \mbox{\textit{CAIE M2 2012 Q5 [9]}}