| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2012 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod on peg or cylinder |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring multiple equilibrium conditions (forces and moments), careful geometry with the cylinder contact, and algebraic manipulation to find both a friction coefficient expression and a range of k values. It demands systematic application of statics principles with non-trivial algebra, placing it well above average difficulty but not at the extreme end for Further Maths content. |
| Spec | 6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R_C \cdot 3a = kW \cdot 4a\cos\theta + W \cdot 2a\cos\theta\) | M1 A1 | Take moments about \(A\) for rod to find \(R_C\) |
| \(R_C = (2k+1)(2W/5)\) | A1 | |
| \(F_A = R_C\sin\theta = (2k+1)(8W/25)\) | M1 A1 | Resolve horizontal forces on rod to find \(F_A\) |
| \(R_A = W + kW - R_C\cos\theta = (13k+19)(W/25)\) | M1 A1 | Resolve vertical forces on rod to find \(R_A\) |
| *OR:* Take moments about \(C\) for rod: | ||
| \(F_A \cdot 3a\sin\theta - R_A \cdot 3a\cos\theta = kWa\cos\theta - Wa\cos\theta\) | (M1 A1) | |
| \(4F_A - 3R_A = (k-1)W\) | (A1) | |
| Resolve forces parallel to rod: \(F_A\cos\theta + R_A\sin\theta = kW\sin\theta + W\sin\theta\) | (M1) | |
| \(3F_A + 4R_A = 4(k+1)W\) | (A1) | |
| Combine to find \(F_A\), \(R_A\): \(F_A = (2k+1)(8W/25)\) and \(R_A = (13k+19)(W/25)\) | (M1 A1) | |
| Use \(F_A \le \mu R_A\) to find \(\mu_{\min}\): \(\mu_{\min} = 8(2k+1)/(13k+19)\) A.G. | M1 A1 | (No use of inequality loses A1) |
| \(160k + 80 \le (\text{or} =)\ 117k + 171\) | M1 | Find bound on \(k\) (allow use of equality) |
| Bound \(= 91/43\) or \(2.12\) | A1 | |
| \(k \le 91/43\) or \(2.12\) | A1 | Correct use of inequality |
| Answer | Marks | Guidance |
|---|---|---|
| \(r^2 = (-0.5)(-1.2) = 0.6 \text{ or } 0.775^2\) | M1 A1 | |
| \(r = -0.775\) | *A1 | 3 marks |
| Answer | Marks |
|---|---|
| \(H_0: \rho = 0,\ H_1: \rho \neq 0\) | B1 |
| Answer | Marks |
|---|---|
| \(r_{5,\ 2.5\%} = 0.878\) | *B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Accept \(H_0\) if \( | r | <\) tabular value |
| Answer | Marks | Guidance |
|---|---|---|
| Coeff. does not differ from zero | A1 | 4 marks |
| Answer | Marks |
|---|---|
| \(14 + q = (-0.5)(13 + p) + 5 \times 5\) | M1 A1 |
| Answer | Marks |
|---|---|
| \(13 + p = (-1.2)(14 + q) + 5 \times 7.6\) | |
| \(or\ (13 + p)/5 = (-1.2)(14 + q)/5 + 7.6\) | A1 |
| Answer | Marks |
|---|---|
| \(p = 7,\ q = 1\) | A1, A1 |
| Answer | Marks |
|---|---|
| \(\bar{x} = 4;\ \bar{y} = 3\) | M1 A1; A1 |
| Answer | Marks | Guidance |
|---|---|---|
| \(p = 7,\ q = 1\) | A1, A1 | 5 marks |
## Question 11(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R_C \cdot 3a = kW \cdot 4a\cos\theta + W \cdot 2a\cos\theta$ | M1 A1 | Take moments about $A$ for rod to find $R_C$ |
| $R_C = (2k+1)(2W/5)$ | A1 | |
| $F_A = R_C\sin\theta = (2k+1)(8W/25)$ | M1 A1 | Resolve horizontal forces on rod to find $F_A$ |
| $R_A = W + kW - R_C\cos\theta = (13k+19)(W/25)$ | M1 A1 | Resolve vertical forces on rod to find $R_A$ |
| *OR:* Take moments about $C$ for rod: | | |
| $F_A \cdot 3a\sin\theta - R_A \cdot 3a\cos\theta = kWa\cos\theta - Wa\cos\theta$ | (M1 A1) | |
| $4F_A - 3R_A = (k-1)W$ | (A1) | |
| Resolve forces parallel to rod: $F_A\cos\theta + R_A\sin\theta = kW\sin\theta + W\sin\theta$ | (M1) | |
| $3F_A + 4R_A = 4(k+1)W$ | (A1) | |
| Combine to find $F_A$, $R_A$: $F_A = (2k+1)(8W/25)$ and $R_A = (13k+19)(W/25)$ | (M1 A1) | |
| Use $F_A \le \mu R_A$ to find $\mu_{\min}$: $\mu_{\min} = 8(2k+1)/(13k+19)$ **A.G.** | M1 A1 | (No use of inequality loses A1) |
| $160k + 80 \le (\text{or} =)\ 117k + 171$ | M1 | Find bound on $k$ (allow use of equality) |
| Bound $= 91/43$ or $2.12$ | A1 | |
| $k \le 91/43$ or $2.12$ | A1 | Correct use of inequality |
**Total: 12 marks**
## Question 11(b):
**Find correlation coefficient $r$:**
$r^2 = (-0.5)(-1.2) = 0.6 \text{ or } 0.775^2$ | M1 A1 |
$r = -0.775$ | *A1 | 3 marks
---
**State both hypotheses (B0 for $r$...):**
$H_0: \rho = 0,\ H_1: \rho \neq 0$ | B1 |
---
**State or use correct tabular one-tail $r$ value:**
$r_{5,\ 2.5\%} = 0.878$ | *B1 |
---
**Valid method for reaching conclusion:**
Accept $H_0$ if $|r| <$ tabular value | M1 |
---
**Correct conclusion (AEF, dep *A1, *B1):**
Coeff. does not differ from zero | A1 | 4 marks
---
**EITHER: Find two indep. eqns for $p, q$:**
$14 + q = (-0.5)(13 + p) + 5 \times 5$ | M1 A1 |
$or\ (14 + q)/5 = (-0.5)(13 + p)/5 + 5$
$13 + p = (-1.2)(14 + q) + 5 \times 7.6$ | |
$or\ (13 + p)/5 = (-1.2)(14 + q)/5 + 7.6$ | A1 |
---
**Combine to find $p, q$:**
$p = 7,\ q = 1$ | A1, A1 |
---
**OR: Combine regression lines to find $\bar{x},\ \bar{y}$:**
$\bar{x} = 4;\ \bar{y} = 3$ | M1 A1; A1 |
**Use sample data to find $p, q$:**
$p = 7,\ q = 1$ | A1, A1 | 5 marks
---
**Total: [12]**
\begin{center}
\includegraphics[max width=\textwidth, alt={}]{71a3b842-9d31-4c25-b894-ca6d1f47d84b-5_474_796_479_676}
\end{center}
The diagram shows a uniform rod $A B$, of length $4 a$ and weight $W$, resting in equilibrium with its end $A$ on rough horizontal ground. The rod rests at $C$ on the surface of a smooth cylinder whose axis is horizontal. The cylinder rests on the ground and is fixed to it. The rod is in a vertical plane perpendicular to the axis of the cylinder and is inclined at an angle $\theta$ to the horizontal, where $\cos \theta = \frac { 3 } { 5 }$. A particle of weight $k W$ is attached to the rod at $B$. Given that $A C = 3 a$, show that the least possible value of the coefficient of friction $\mu$ between the rod and the ground is $\frac { 8 ( 2 k + 1 ) } { 13 k + 19 }$.
Given that $\mu = \frac { 9 } { 10 }$, find the set of values of $k$ for which equilibrium is possible.
\hfill \mbox{\textit{CAIE FP2 2012 Q11 EITHER}}