| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Hooke's law and elastic energy |
| Type | Elastic string equilibrium and statics |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring coordinate geometry to find string extension, moment equilibrium about the hinge, force resolution, and elastic energy principles. The multi-step nature, need to work with given trigonometric values across multiple parts, and requirement to synthesize several mechanics concepts (Hooke's law, moments, force components) makes it significantly harder than average, though the individual techniques are standard for Further Maths. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.04a Calculate moments: about a point3.04b 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 |
|---|---|---|
| 4(i) | T × 3a sin θ = W × 1⋅5a cos 2θ | M1 A1 |
| T = 7W/30 or 0⋅233W | A1 | Find tension T |
| Total: | 3 | |
| Question | Answer | Marks |
| Answer | Marks | Guidance |
|---|---|---|
| 4(ii) | OB = (5a – 3a cos 2θ) / cos θ = 26a/5 or 5⋅2a | M1 A1 |
| T = λ (OB – 4a)/4a | M1 | Find modulus λ using Hooke’s Law |
| = λ (6a/5)/4a = 3λ /10, λ = 7W/9 or 0⋅778W | A1 | |
| Total: | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| 4(iii) | X = T cos θ [= 14W/75 or 0⋅187W] | M1 |
| Answer | Marks | Guidance |
|---|---|---|
| φ = tan–1 (Y/X) = tan–1 (171/28) | M1 | Find vertical component Y of force at A |
| Answer | Marks | Guidance |
|---|---|---|
| = 80⋅7° or 1⋅41 radians | B1 | |
| Total: | 3 | |
| Question | Answer | Marks |
Question 4:
--- 4(i) ---
4(i) | T × 3a sin θ = W × 1⋅5a cos 2θ | M1 A1 | Take moments for rod about A
T = 7W/30 or 0⋅233W | A1 | Find tension T
Total: | 3
Question | Answer | Marks | Guidance
--- 4(ii) ---
4(ii) | OB = (5a – 3a cos 2θ) / cos θ = 26a/5 or 5⋅2a | M1 A1 | Find length OB of string
T = λ (OB – 4a)/4a | M1 | Find modulus λ using Hooke’s Law
= λ (6a/5)/4a = 3λ /10, λ = 7W/9 or 0⋅778W | A1
Total: | 4
--- 4(iii) ---
4(iii) | X = T cos θ [= 14W/75 or 0⋅187W] | M1 | Find horizontal component X of force at A
Y = T sin θ + W [= 57W/50 or 1⋅14W]
φ = tan–1 (Y/X) = tan–1 (171/28) | M1 | Find vertical component Y of force at A
Find angle φ which force at A makes with horizontal
= 80⋅7° or 1⋅41 radians | B1
Total: | 3
Question | Answer | Marks | Guidance\includegraphics{figure_4}
A uniform rod $AB$ of length $3a$ and weight $W$ is freely hinged to a fixed point at the end $A$. The end $B$ is below the level of $A$ and is attached to one end of a light elastic string of natural length $4a$. The other end of the string is attached to a point $O$ on a vertical wall. The horizontal distance between $A$ and the wall is $5a$. The string and the rod make angles $\theta$ and $2\theta$ respectively with the horizontal (see diagram). The system is in equilibrium with the rod and the string in the same vertical plane. It is given that $\sin \theta = \frac{3}{5}$ and you may use the fact that $\cos 2\theta = \frac{7}{25}$.
\begin{enumerate}[label=(\roman*)]
\item Find the tension in the string in terms of $W$. [3]
\item Find the modulus of elasticity of the string in terms of $W$. [4]
\item Find the angle that the force acting on the rod at $A$ makes with the horizontal. [3]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2017 Q4 [10]}}