| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with elastic string or spring support |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics problem requiring moments about a point, resolution of forces, and Hooke's law for elastic strings. While it involves multiple steps (finding tension using extension and Hooke's law, taking moments about B, resolving forces at the hinge), the setup is straightforward with clearly defined geometry and the techniques are routine for FM students. The 30° angles simplify calculations, and the problem follows a predictable structure without requiring novel insight. |
| Spec | 1.05g Exact trigonometric values: for standard angles3.04b Equilibrium: zero resultant moment and force6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(W \times a\cos 30° + 3W \times 2a\cos 30° = T \times 2a\cos 30°\) | M1 A1 | Take moments for rod about \(B\); \(\cos 30° = \sqrt{3}/2\) |
| \(T = 7W/2\) | A1 | Can earn M1 A0 A1 if e.g. \(\sin 30°\) wrongly used |
| \(T = \lambda(2a - 3a/5)/(3a/5)\); \(\lambda = (3/7)(7W/2) = 3W/2\) | M1 A1 | Find modulus \(\lambda\) using Hooke's Law |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| *EITHER*: \(X = T\cos 30° = (7\sqrt{3}/4)W\) or \(3.03W\) | B1\(\checkmark\) | Find horizontal component of force \(F\) at \(B\) |
| \(Y = 4W - T\sin 30° = 9W/4\) | B1\(\checkmark\) | Find vertical component |
| \(F = \sqrt{X^2 + Y^2} = (\sqrt{57}/2)W\) or \(3.77[5]W\) | B1 | Find magnitude of \(F\) |
| \(\tan^{-1}Y/X = \tan^{-1}3\sqrt{3}/7 = 36.6°\) or \(0.639\) radians | M1 A1 | Find direction (A0 if direction unclear) |
| *OR*: \(F_1 = (4W+T)\sin 30° = 15W/4\) | (B1\(\checkmark\)) | Find component along \(CB\) |
| \(F_2 = (4W-T)\cos 30° = (\sqrt{3}/4)W\) | (B1\(\checkmark\)) | Find normal component |
| \(F = (\sqrt{57}/2)W\) or \(3.77[5]W\) | (B1) | |
| \(\tan^{-1}F_2/F_1 = \tan^{-1}\sqrt{3}/15 = 6.6°\) or \(0.115\) radians | (M1 A1) | Direction of \(F\) (A0 if unclear) |
| *OR*: \(\pm F_1 = T - 4W\sin 30° = 3W/2\) | (B1\(\checkmark\)) | Find component parallel to string \(CA\) |
| \(\pm F_2 = 4W\cos 30° = 2\sqrt{3}W\) | (B1\(\checkmark\)) | Find normal component |
| \(F = (\sqrt{57}/2)W\) or \(3.77[5]W\) | (B1) | |
| \(\tan^{-1}F_2/F_1 = \tan^{-1}4/\sqrt{3} = 66.6°\) or \(1.16\) radians | (M1 A1) |
## Question 4(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $W \times a\cos 30° + 3W \times 2a\cos 30° = T \times 2a\cos 30°$ | M1 A1 | Take moments for rod about $B$; $\cos 30° = \sqrt{3}/2$ |
| $T = 7W/2$ | A1 | Can earn M1 A0 A1 if e.g. $\sin 30°$ wrongly used |
| $T = \lambda(2a - 3a/5)/(3a/5)$; $\lambda = (3/7)(7W/2) = 3W/2$ | M1 A1 | Find modulus $\lambda$ using Hooke's Law |
**Part total: 5 marks**
---
## Question 4(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| *EITHER*: $X = T\cos 30° = (7\sqrt{3}/4)W$ or $3.03W$ | B1$\checkmark$ | Find horizontal component of force $F$ at $B$ |
| $Y = 4W - T\sin 30° = 9W/4$ | B1$\checkmark$ | Find vertical component |
| $F = \sqrt{X^2 + Y^2} = (\sqrt{57}/2)W$ or $3.77[5]W$ | B1 | Find magnitude of $F$ |
| $\tan^{-1}Y/X = \tan^{-1}3\sqrt{3}/7 = 36.6°$ or $0.639$ radians | M1 A1 | Find direction (A0 if direction unclear) |
| *OR*: $F_1 = (4W+T)\sin 30° = 15W/4$ | (B1$\checkmark$) | Find component along $CB$ |
| $F_2 = (4W-T)\cos 30° = (\sqrt{3}/4)W$ | (B1$\checkmark$) | Find normal component |
| $F = (\sqrt{57}/2)W$ or $3.77[5]W$ | (B1) | |
| $\tan^{-1}F_2/F_1 = \tan^{-1}\sqrt{3}/15 = 6.6°$ or $0.115$ radians | (M1 A1) | Direction of $F$ (A0 if unclear) |
| *OR*: $\pm F_1 = T - 4W\sin 30° = 3W/2$ | (B1$\checkmark$) | Find component parallel to string $CA$ |
| $\pm F_2 = 4W\cos 30° = 2\sqrt{3}W$ | (B1$\checkmark$) | Find normal component |
| $F = (\sqrt{57}/2)W$ or $3.77[5]W$ | (B1) | |
| $\tan^{-1}F_2/F_1 = \tan^{-1}4/\sqrt{3} = 66.6°$ or $1.16$ radians | (M1 A1) | |
**Part total: 5 marks; Question total: 10 marks**
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\includegraphics[max width=\textwidth, alt={}, center]{eb3dccaf-d151-472d-82f3-6ba215b0b7f0-2_339_957_1482_593}
A uniform rod $B C$ of length $2 a$ and weight $W$ is hinged to a fixed point at $B$. A particle of weight $3 W$ is attached to the rod at $C$. The system is held in equilibrium by a light elastic string of natural length $\frac { 3 } { 5 } a$ in the same vertical plane as the rod. One end of the elastic string is attached to the rod at $C$ and the other end is attached to a fixed point $A$ which is at the same horizontal level as $B$. The rod and the string each make an angle of $30 ^ { \circ }$ with the horizontal (see diagram). Find\\
(i) the modulus of elasticity of the string,\\
(ii) the magnitude and direction of the force acting on the rod at $B$.
\hfill \mbox{\textit{CAIE FP2 2015 Q4 [10]}}