| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2014 |
| Session | November |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod or block on rough surface in limiting equilibrium (no wall) |
| Difficulty | Challenging +1.2 This is a multi-step statics problem requiring resolution of forces, moments about a point, friction at limiting equilibrium, and Hooke's law. While it involves several components (rod, wall, floor, spring) and requires careful geometric setup with the given angle, it follows standard mechanics procedures: taking moments to find reactions, using equilibrium conditions, and applying the spring formula. The 'show that' part provides a target to verify, reducing problem-solving demand. For Further Maths students, this represents a moderately challenging but routine application of statics principles. |
| Spec | 6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks |
|---|---|
| 4 (i) | 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Page 6 | Mark Scheme | Syllabus |
| Cambridge International A Level – October/November 2014 | 9231 | 22 |
| (ii) | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| 4 20 5 | 7 | |
| 4 | [11] | |
| Page 7 | Mark Scheme | Syllabus |
| Cambridge International A Level – October/November 2014 | 9231 | 22 |
| Answer | Marks |
|---|---|
| (iii) | Find extn. of either string by equating equil. tensions: |
| Answer | Marks |
|---|---|
| 10 | 4 |
| Answer | Marks |
|---|---|
| 2 | [12] |
Question 4:
--- 4 (i) ---
4 (i) | 1
Relate FA and RA using F = µR: FA = RA B1
3
1
Resolve horizontally: RB = FA [= RA] B1
3
Resolve vertically (may not be needed here): S = mg – RA B1
1 3
EITHER: Take moments about C: RB l sin α + FA l sin α
4 4
1 3
+ mg l cos α = RA l cos α M1 A1
4 4
3
Combine, using tanα = , to find RA: RA + 3RA + 4mg = 12RA
4
1
RA = mg A.G. M1 A1
2
3
OR: Take moments about A: RB l sin α + S l cos α
4
1
= mg l cos α (M1 A1)
2
3
Combine, using tanα = , to find RA: RA + 3(mg – RA) = 2mg
4
1
RA = mg A.G. (M1 A1)
2
1
OR: Take moments about B: FA l sin α + mg l cos α
2
1
= RA l cos α + S l cos α (M1 A1)
4
Page 6 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2014 | 9231 | 22
(ii) | 3
Combine, using tanα = , to find RA: RA + 2mg = 4RA + mg – RA
4
1
RA = mg A.G. (M1 A1)
2
3
OR: Take moments about D: RA l cos α
4
1
= RB l sin α + mg l cos α (M1 A1)
4
3
Combine, using tanα = , to find RA: 3RA = RA + mg
4
1
RA = mg A.G. (M1 A1)
2
Use Hooke’s Law to relate extn. e and nat. length L:
1 2mge 1
S = mg = , e = L B1
2 L 4
3 9l
Find length of CD: CD = l sin α = B1
4 20
1 9l 3l
Combine to find L: L – L = , L = M1 A1
4 20 5 | 7
4 | [11]
Page 7 | Mark Scheme | Syllabus | Paper
Cambridge International A Level – October/November 2014 | 9231 | 22
5 (i)
(ii)
(iii) | Find extn. of either string by equating equil. tensions:
6mge mg(3a−e )
A = A
3a 2a
6mg(3a−e ) mge
or B = B M1 A1
3a 2a
3a 12a
eA = or eB = A1
5 5
Find AO: A.G. AO = 3a + eA or 6a – eB = 3⋅6a B1
d2x mg(3a−e −x)
Apply Newton’s law at general point, e.g.: m = A
dt2 2a
6mg(e +x)
(lose A1 for each incorrect term) – A
3a
d2y mg(3a−e + y)
or m = − A
dt2 2a
6mg(e − y)
+ A M1 A2
3a
d2 x 5gx
= −
Simplify to give standard SHM eqn, e.g.: A1
dt 2 2a
S.R.: B1 if no derivation (max 3/6)
5g 2a
Find period T using SHM with ω = : T = 2π (A.E.F.) M1 A1
2a 5g
5g
Find max speed using ωA with A = 0⋅2 a: v max = × 0⋅2a
2a
ag
= or √a (A.E.F.) M1 A1
10 | 4
6
2 | [12]\includegraphics{figure_4}
A uniform rod $AB$, of length $l$ and mass $m$, rests in equilibrium with its lower end $A$ on a rough horizontal floor and the end $B$ against a smooth vertical wall. The rod is inclined to the horizontal at an angle $\alpha$, where $\tan \alpha = \frac{4}{3}$, and is in a vertical plane perpendicular to the wall. The rod is supported by a light spring $CD$ which is in compression in a vertical line with its lower end $D$ fixed on the floor. The upper end $C$ is attached to the rod at a distance $\frac{4l}{5}$ from $B$ (see diagram). The coefficient of friction at $A$ between the rod and the floor is $\frac{1}{2}$ and the system is in limiting equilibrium.
\begin{enumerate}[label=(\roman*)]
\item Show that the normal reaction of the floor at $A$ has magnitude $\frac{1}{2}mg$ and find the force in the spring. [7]
\item Given that the modulus of elasticity of the spring is $2mg$, find the natural length of the spring. [4]
\end{enumerate}
\hfill \mbox{\textit{CAIE FP2 2014 Q4 [11]}}