| 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 a separate equilibrium problem. While it involves several steps and careful geometric reasoning with the rod-string-ring system, the techniques are standard for FM2 statics: resolving forces, taking moments about a hinge, and applying limiting equilibrium conditions. The calculations are methodical rather than requiring novel insight, though the multi-component setup and need to track multiple unknowns across parts elevates it above routine questions. |
| 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:* \([\pm]\ X = T\sin\theta = 3W/5\ \text{or}\ 0.6W\); \([\pm]\ Y = 5W/2 - T\cos\theta\ \text{or}\ 5W/2 + W - R_P\) | (B1 | *EITHER:* Find horizontal component \(X\) of force at \(A\); find vertical component \(Y\) of force at \(A\) |
| \(= 17W/10\ \text{or}\ 1.7W\) | B1) | |
| *OR:* \([\pm]\ X = (5W/2)\sin\theta = 3W/2\ \text{or}\ 1.5W\) | (B1 | *OR:* Find component \(X\) of force at \(A\) along \(BA\) |
| \([\pm]\ Y = (5W/2)\cos\theta - T = W\) | B1) | Find component \(Y\) of force at \(A\) perp. to \(BA\) |
| \(R_A^2 = X^2 + Y^2 = 13W^2/4,\ R_A = \frac{1}{2}W\sqrt{13}\ \text{or}\ 1.80W\) | B1FT | Find magnitude of resultant force \(R_A\) at \(A\) (FT on \(X\), \(Y\)) |
| 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(PB = (a + 2a\sin\theta)/\cos\theta = 5a/4 + 3a/2 = 11a/4\ \text{or}\ 2.75a\); \(\text{or}\ x = 3a/4\ \text{or}\ 0.75a\) | B1 | Find length \(PB\) or extension \(x\) of string |
| \(T = \lambda(PB - 2a)/2a,\ \lambda = 8W/3\ \text{or}\ 2.67W\) | 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:* $[\pm]\ X = T\sin\theta = 3W/5\ \text{or}\ 0.6W$; $[\pm]\ Y = 5W/2 - T\cos\theta\ \text{or}\ 5W/2 + W - R_P$ | **(B1** | *EITHER:* Find horizontal component $X$ of force at $A$; find vertical component $Y$ of force at $A$ |
| $= 17W/10\ \text{or}\ 1.7W$ | **B1)** | |
| *OR:* $[\pm]\ X = (5W/2)\sin\theta = 3W/2\ \text{or}\ 1.5W$ | **(B1** | *OR:* Find component $X$ of force at $A$ along $BA$ |
| $[\pm]\ Y = (5W/2)\cos\theta - T = W$ | **B1)** | Find component $Y$ of force at $A$ perp. to $BA$ |
| $R_A^2 = X^2 + Y^2 = 13W^2/4,\ R_A = \frac{1}{2}W\sqrt{13}\ \text{or}\ 1.80W$ | **B1FT** | Find magnitude of resultant force $R_A$ at $A$ (FT on $X$, $Y$) |
| | **3** | |
---
# Question 4(iv):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $PB = (a + 2a\sin\theta)/\cos\theta = 5a/4 + 3a/2 = 11a/4\ \text{or}\ 2.75a$; $\text{or}\ x = 3a/4\ \text{or}\ 0.75a$ | **B1** | Find length $PB$ or extension $x$ of string |
| $T = \lambda(PB - 2a)/2a,\ \lambda = 8W/3\ \text{or}\ 2.67W$ | **M1 A1** | Find modulus $\lambda$ using Hooke's Law |
| | **3** | |4\\
\includegraphics[max width=\textwidth, alt={}, center]{1651d08b-b20f-4f2e-9f47-0a1a5d0a839a-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]{1651d08b-b20f-4f2e-9f47-0a1a5d0a839a-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]}}