| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Advanced work-energy problems |
| Type | Elastic string horizontal surface projection |
| Difficulty | Challenging +1.2 This is a multi-part mechanics problem requiring energy conservation with elastic strings and analysis of normal reaction forces. While it involves several concepts (elastic PE, kinetic energy, forces at limiting equilibrium), the solution follows a standard M2 framework with clearly defined stages. The 'show that' format provides target values, reducing problem-solving demand compared to open-ended questions. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(EE = 18 \times (1.8 - 1.6)^2 / (2 \times 1.6)\) | B1 | |
| \(0.2 \times 1.5^2/2 = 18(1.8-1.6)^2/(2\times1.6) + KE_B\) | M1 | Energy equation, 3 terms |
| \(KE_B = 0\) leads to \(v_B = 0\) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T = 18 \times (1.8 - 1.6)/1.6\) | B1 | \(T = 2.25\) |
| \(T\cos\theta + R = 0.2g\) | M1 | |
| \(2.25 \times 1.6/1.8 + R = 0.2g\) | A1 | |
| \(R = 0\) | A1 [4] [7] | Needs \(g = 10\) |
## Question 3:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $EE = 18 \times (1.8 - 1.6)^2 / (2 \times 1.6)$ | B1 | |
| $0.2 \times 1.5^2/2 = 18(1.8-1.6)^2/(2\times1.6) + KE_B$ | M1 | Energy equation, 3 terms |
| $KE_B = 0$ leads to $v_B = 0$ | A1 [3] | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T = 18 \times (1.8 - 1.6)/1.6$ | B1 | $T = 2.25$ |
| $T\cos\theta + R = 0.2g$ | M1 | |
| $2.25 \times 1.6/1.8 + R = 0.2g$ | A1 | |
| $R = 0$ | A1 [4] [7] | Needs $g = 10$ |
---3 A particle $P$ of mass 0.2 kg is attached to one end of a light elastic string of natural length 1.6 m and modulus of elasticity 18 N . The other end of the string is attached to a fixed point $O$ which is 1.6 m above a smooth horizontal surface. $P$ is placed on the surface vertically below $O$ and then projected horizontally. $P$ moves with initial speed $1.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a straight line on the surface. Show that, when $O P = 1.8 \mathrm {~m}$,\\
(i) $P$ is at instantaneous rest,\\
(ii) $P$ is on the point of losing contact with the surface.
\hfill \mbox{\textit{CAIE M2 2013 Q3 [7]}}