| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2018 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod hinged to wall with elastic string or spring support |
| Difficulty | Challenging +1.8 This is a multi-part statics problem requiring moment equilibrium, force resolution, and Hooke's law. While it involves several steps and geometric reasoning with the angle condition, the techniques are standard for Further Maths mechanics: taking moments about the hinge, resolving forces, and applying elasticity formulas. The given tan θ simplifies calculations significantly. More challenging than typical A-level mechanics due to the geometric setup and being Further Maths content, but follows predictable solution patterns. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.02h Elastic PE: 1/2 k x^26.02j Conservation with elastics: springs and strings |
| Answer | Marks | Guidance |
|---|---|---|
| 4(i) | T × 3a sin 2θ = W × 2a cos θ + ½ W × 4a cos θ | M1A1 |
| 6a T sin θ cos θ = 4a W × a cos θ | M1 | Verify tension T |
| T = 2W / 3 sin θ = 2W / 3(8/17) = 17W / 12 AG | A1 | using sin θ = 8/17, cos θ = 15/17 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(ii) | EITHER: X = T cos θ = (5/4) W or 1.25W | B1 |
| Y = W + ½ W – T sin θ = (5/6) W or 0.833W | B1 | Find vertical component Y of force at A |
| F = √(X2 + Y2) = (5√13 / 12) W or 1.50W | B1 | Find magnitude of F |
| Answer | Marks | Guidance |
|---|---|---|
| tan–1 Y/X = tan–1 2/3 = 33.7° or 0.588 radians | M1A1 | Find direction of F |
| Answer | Marks | Guidance |
|---|---|---|
| = (305/12 × 17) W or 1.45W | (B1 | Find component R parallel to AB of force at A |
| Answer | Marks | Guidance |
|---|---|---|
| = (5/34) W or 0.147W | B1 | Find component R normal to AB of force at A |
| Answer | Marks | Guidance |
|---|---|---|
| P N | B1 | Find magnitude of F |
| Answer | Marks | Guidance |
|---|---|---|
| = 5.6° or 0.098 radians | M1A1) | Find direction of F |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 4(iii) | T = λ (CD – 2a)/2a | M1 |
| CD = 3a so λ = 2T = 17W / 6 or 2.83W | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Question | Answer | Marks |
Question 4:
--- 4(i) ---
4(i) | T × 3a sin 2θ = W × 2a cos θ + ½ W × 4a cos θ | M1A1 | Take moments for rod about A
6a T sin θ cos θ = 4a W × a cos θ | M1 | Verify tension T
T = 2W / 3 sin θ = 2W / 3(8/17) = 17W / 12 AG | A1 | using sin θ = 8/17, cos θ = 15/17
4
--- 4(ii) ---
4(ii) | EITHER: X = T cos θ = (5/4) W or 1.25W | B1 | Find horizontal component X of force at A
Y = W + ½ W – T sin θ = (5/6) W or 0.833W | B1 | Find vertical component Y of force at A
F = √(X2 + Y2) = (5√13 / 12) W or 1.50W | B1 | Find magnitude of F
Upward force at angle to horizontal of
tan–1 Y/X = tan–1 2/3 = 33.7° or 0.588 radians | M1A1 | Find direction of F
(AEF; A0 if direction unclear)
OR: R = 1.5 W sin θ + T cos 2θ
P
= (305/12 × 17) W or 1.45W | (B1 | Find component R parallel to AB of force at A
P
R = 1.5 W cos θ – T cos 2θ
N
= (5/34) W or 0.147W | B1 | Find component R normal to AB of force at A
N
F = √( R 2 + R 2) = (5√13 / 12) W or 1.50W
P N | B1 | Find magnitude of F
Upward force at angle to AB of tan–1 R /R = tan–1 6/61
N P
= 5.6° or 0.098 radians | M1A1) | Find direction of F
(AEF; A0 if direction unclear)
5
Question | Answer | Marks | Guidance
--- 4(iii) ---
4(iii) | T = λ (CD – 2a)/2a | M1 | Find modulus λ using Hooke’s Law
CD = 3a so λ = 2T = 17W / 6 or 2.83W | A1
2
Question | Answer | Marks | GuidanceA uniform rod $AB$ of length $4a$ and weight $W$ is smoothly hinged to a vertical wall at the end $A$. The rod is held at an angle $\theta$ above the horizontal by a light elastic string. One end of the string is attached to the point $C$ on the rod, where $AC = 3a$. The other end of the string is attached to a point $D$ on the wall, with $D$ vertically above $A$ and such that angle $ACD = 2\theta$. A particle of weight $\frac{1}{4}W$ is attached to the rod at $B$. It is given that $\tan \theta = \frac{5}{12}$.
\begin{enumerate}[label=(\roman*)]
\item Show that the tension in the string is $\frac{17}{12}W$. [4]
\item Find the magnitude and direction of the reaction at the hinge. [5]
\item Given that the natural length of the string is $2a$, find its modulus of elasticity. [2]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2018 Q4 [11]}}