| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | November |
| Marks | 5 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Vectors Introduction & 2D |
| Type | Forces in equilibrium (find unknowns) |
| Difficulty | Standard +0.3 This is a straightforward application of Hooke's law and Newton's second law with symmetry simplifying the geometry. Students must find extensions, calculate tensions using T=λx/l, resolve forces vertically (which cancel by symmetry), then apply F=ma horizontally. The geometry is given clearly and the symmetry makes it more accessible than typical 2D vector problems. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 06.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T = \frac{4 \times 0.25}{0.25}\) or \(\frac{4 \times 0.5}{0.5}\) | M1 | For using \(T = \frac{\lambda x}{L}\) |
| Tension is 4N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(2 \times 4 \times 0.6 = 0.6a\) | M1, A1ft | For using Newton's second law |
| Acceleration is \(8 \text{ ms}^{-2}\) | A1 | [Total: 5] |
## Question 1:
### Part (i)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T = \frac{4 \times 0.25}{0.25}$ or $\frac{4 \times 0.5}{0.5}$ | M1 | For using $T = \frac{\lambda x}{L}$ |
| Tension is 4N | A1 | |
### Part (ii)
| Answer/Working | Mark | Guidance |
|---|---|---|
| $2 \times 4 \times 0.6 = 0.6a$ | M1, A1ft | For using Newton's second law |
| Acceleration is $8 \text{ ms}^{-2}$ | A1 | **[Total: 5]** |
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\includegraphics[max width=\textwidth, alt={}, center]{b9080e9f-2c23-43ce-b171-bd68648dc56b-2_711_398_269_877}
Each of two identical light elastic strings has natural length 0.25 m and modulus of elasticity 4 N . A particle $P$ of mass 0.6 kg is attached to one end of each of the strings. The other ends of the strings are attached to fixed points $A$ and $B$ which are 0.8 m apart on a smooth horizontal table. The particle is held at rest on the table, at a point 0.3 m from $A B$ for which $A P = B P$ (see diagram).\\
(i) Find the tension in the strings.\\
(ii) The particle is released. Find its initial acceleration.
\hfill \mbox{\textit{CAIE M2 2007 Q1 [5]}}