| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2011 |
| Session | June |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Lamina on surface with string or rod support |
| Difficulty | Challenging +1.2 This is a multi-part statics problem requiring moment equilibrium, Hooke's law, and force resolution. While it involves several steps (finding centroid of triangle, taking moments, applying elasticity formula, resolving forces), each step uses standard A-level mechanics techniques without requiring novel insight. The geometry is straightforward and the 'show that' part guides students to the tension value. Slightly above average difficulty due to the combination of lamina equilibrium with elastic string properties, but well within typical M2 scope. |
| Spec | 3.04b Equilibrium: zero resultant moment and force6.02h Elastic PE: 1/2 k x^26.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(19 \times 0.6/3 + T \times 0.22 = T \times 0.6\) | M1 | Moments about A, 3 terms |
| \(T = 10\) AG | A1, A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(10 = \lambda(0.11 + 0.6 - 0.7)/0.7\) | M1 | |
| \(\lambda = 700\) | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(F^2 = 10^2 + (19 - 10)^2\) | M1 | |
| \(F = 13.5\) | A1 | |
| \(\alpha = \tan^{-1}(9/10) = 42.0°\) (with horizontal) | B1 [3] | Or \(a = \tan^{-1}(10/9) = 48°\) (with vertical) |
## Question 5:
### Part (i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $19 \times 0.6/3 + T \times 0.22 = T \times 0.6$ | M1 | Moments about A, 3 terms |
| $T = 10$ **AG** | A1, A1 [3] | |
### Part (ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $10 = \lambda(0.11 + 0.6 - 0.7)/0.7$ | M1 | |
| $\lambda = 700$ | A1 [2] | |
### Part (iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $F^2 = 10^2 + (19 - 10)^2$ | M1 | |
| $F = 13.5$ | A1 | |
| $\alpha = \tan^{-1}(9/10) = 42.0°$ (with horizontal) | B1 [3] | Or $a = \tan^{-1}(10/9) = 48°$ (with vertical) |
---5\\
\includegraphics[max width=\textwidth, alt={}, center]{18398d27-15eb-4515-8210-4f0f614d5b28-3_348_1205_251_470}\\
$A B C$ is a uniform triangular lamina of weight 19 N , with $A B = 0.22 \mathrm {~m}$ and $A C = B C = 0.61 \mathrm {~m}$. The plane of the lamina is vertical. $A$ rests on a rough horizontal surface, and $A B$ is vertical. The equilibrium of the lamina is maintained by a light elastic string of natural length 0.7 m which passes over a small smooth peg $P$ and is attached to $B$ and $C$. The portion of the string attached to $B$ is horizontal, and the portion of the string attached to $C$ is vertical (see diagram).\\
(i) Show that the tension in the string is 10 N .\\
(ii) Calculate the modulus of elasticity of the string.\\
(iii) Find the magnitude and direction of the force exerted by the surface on the lamina at $A$.
\hfill \mbox{\textit{CAIE M2 2011 Q5 [8]}}