| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2017 |
| Session | November |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Equilibrium position with elastic string/spring |
| Difficulty | Challenging +1.8 This is a multi-part Further Maths mechanics question requiring statics analysis (moments, forces, friction) followed by geometry and elasticity calculations. While it involves several steps and careful diagram interpretation, the techniques are standard for FM2 level—resolving forces, taking moments about a hinge, and applying Hooke's law. The geometric setup requires attention but no novel insight. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.03n Equilibrium in 2D: particle under forces3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force6.02h Elastic PE: 1/2 k x^26.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T \times 2a = (5W/2) \times a\cos\theta\) | M1 | Take moments for rod about \(A\) |
| \(T = \frac{1}{2}(5W/2) \times (4/5) = W\) | A1 | to find tension \(T\) |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T\sin\theta = \mu(W + T\cos\theta)\) | M1 | Use \(F_P = \mu R_P\) at \(P\) |
| \((3/5)W = \mu(1 + 4/5)W = \mu(9/5)W\), \(\mu = 1/3\) | A1 | to find \(\mu\) |
| 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| *EITHER:* \([\pm] X = T\sin\theta = 3W/5\) or \(0.6W\); \([\pm] Y = 5W/2 - T\cos\theta\) or \(5W/2 + W - R_P\) \(= 17W/10\) or \(1.7W\) | (B1; B1) | *EITHER:* Find horizontal component \(X\) of force at \(A\); Find vertical component \(Y\) of force at \(A\) |
| *OR:* \([\pm] X = (5W/2)\sin\theta = 3W/2\) or \(1.5W\); \([\pm] Y = (5W/2)\cos\theta - T = W\) | (B1; B1) | *OR:* Find component \(X\) of force at \(A\) along \(BA\); Find component \(Y\) of force at \(A\) perpendicular to \(BA\) |
| \(R_A^2 = X^2 + Y^2 = 13W^2/4\), \(R_A = \frac{1}{2}W\sqrt{13}\) or \(1.80W\) | B1FT | Find magnitude of resultant force \(R_A\) at \(A\) (FT on \(X\), \(Y\)) |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(PB = (a + 2a\sin\theta)/\cos\theta = 5a/4 + 3a/2 = 11a/4\) or \(2.75a\); or \(x = 3a/4\) or \(0.75a\) | B1 | Find length \(PB\) or extension \(x\) of string |
| \(T = \lambda(PB - 2a)/2a\), \(\lambda = 8W/3\) or \(2.67W\) | M1 A1 | Find modulus \(\lambda\) using Hooke's Law |
| 3 |
# Question 4(i):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T \times 2a = (5W/2) \times a\cos\theta$ | **M1** | Take moments for rod about $A$ |
| $T = \frac{1}{2}(5W/2) \times (4/5) = W$ | **A1** | to find tension $T$ |
| | **2** | |
---
# Question 4(ii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T\sin\theta = \mu(W + T\cos\theta)$ | **M1** | Use $F_P = \mu R_P$ at $P$ |
| $(3/5)W = \mu(1 + 4/5)W = \mu(9/5)W$, $\mu = 1/3$ | **A1** | to find $\mu$ |
| | **2** | |
---
# Question 4(iii):
| Answer | Marks | Guidance |
|--------|-------|----------|
| *EITHER:* $[\pm] X = T\sin\theta = 3W/5$ or $0.6W$; $[\pm] Y = 5W/2 - T\cos\theta$ or $5W/2 + W - R_P$ $= 17W/10$ or $1.7W$ | **(B1**; **B1)** | *EITHER:* Find horizontal component $X$ of force at $A$; Find vertical component $Y$ of force at $A$ |
| *OR:* $[\pm] X = (5W/2)\sin\theta = 3W/2$ or $1.5W$; $[\pm] Y = (5W/2)\cos\theta - T = W$ | **(B1**; **B1)** | *OR:* Find component $X$ of force at $A$ along $BA$; Find component $Y$ of force at $A$ perpendicular to $BA$ |
| $R_A^2 = X^2 + Y^2 = 13W^2/4$, $R_A = \frac{1}{2}W\sqrt{13}$ or $1.80W$ | **B1FT** | Find magnitude of resultant force $R_A$ at $A$ (FT on $X$, $Y$) |
| | **3** | |
---
# Question 4(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $PB = (a + 2a\sin\theta)/\cos\theta = 5a/4 + 3a/2 = 11a/4$ or $2.75a$; or $x = 3a/4$ or $0.75a$ | **B1** | Find length $PB$ or extension $x$ of string |
| $T = \lambda(PB - 2a)/2a$, $\lambda = 8W/3$ or $2.67W$ | **M1 A1** | Find modulus $\lambda$ using Hooke's Law |
| | **3** | |4\\
\includegraphics[max width=\textwidth, alt={}, center]{2ab1a594-6c37-4c78-b53c-33c13bf6eb21-06_465_663_262_742}
A small ring $P$ of weight $W$ is free to slide on a rough horizontal wire, one end of which is attached to a vertical wall at $Q$. The end $A$ of a thin uniform $\operatorname { rod } A B$ of length $2 a$ and weight $\frac { 5 } { 2 } W$ is freely hinged to the wall at the point $A$ which is a distance $a$ vertically below $Q$. A light elastic string of natural length $2 a$ has one end attached to the ring $P$ and the other end attached to the rod at $B$. The string is at right angles to the rod and $A , B , P$ and $Q$ lie in a vertical plane. The system is in limiting equilibrium with $A B$ making an angle $\theta$ with the horizontal, where $\sin \theta = \frac { 3 } { 5 }$ (see diagram).\\
(i) Find the tension in the string in terms of $W$.\\
(ii) Find the coefficient of friction between the ring and the wire.\\
(iii) Find the magnitude of the resultant force on the rod at the hinge in terms of $W$.\\
(iv) Find the modulus of elasticity of the string in terms of $W$.\\
\includegraphics[max width=\textwidth, alt={}, center]{2ab1a594-6c37-4c78-b53c-33c13bf6eb21-08_862_698_260_721}
A uniform picture frame of mass $m$ is made by removing a rectangular lamina $E F G H$ in which $E F = 4 a$ and $F G = 2 a$ from a larger rectangular lamina $A B C D$ in which $A B = 6 a$ and $B C = 4 a$. The side $E F$ is parallel to the side $A B$. The point of intersection of the diagonals $A C$ and $B D$ coincides with the point of intersection of the diagonals $E G$ and $F H$. One end of a light inextensible string of length $10 a$ is attached to $A$ and the other end is attached to $B$. The frame is suspended from the mid-point $O$ of the string. A small object of mass $\frac { 11 } { 12 } m$ is fixed to the mid-point of $A B$ (see diagram).\\
\hfill \mbox{\textit{CAIE FP2 2017 Q4 [10]}}