| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2016 |
| Session | November |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Composite body MI calculation |
| Difficulty | Challenging +1.2 This is a multi-part Further Maths mechanics question requiring systematic application of standard moment of inertia formulas (parallel axis theorem for discs and rods) and energy conservation. While it involves several components and careful bookkeeping, each step follows routine procedures without requiring novel insight—the parallel axis theorem applications are textbook exercises, and the energy equation is straightforward once the geometry is understood. The 'show that' format provides targets to work toward, reducing problem-solving demand. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_\text{disc} = \frac{1}{2}(\frac{1}{4}ma^2) + \frac{1}{2}m\{(3a/2)^2+(2a)^2\} = ma^2/8 + 25ma^2/8 = 13ma^2/4\) | M1 A1 | Verify MI of either disc about axis \(l\) at \(O\); A.G. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(I_{OC} = (4/3)\,ma^2\) | B1 | Find or state MI of rod \(OC\) about \(l\) |
| \(I_{AB} = \frac{1}{3}\cdot 2m\,(3a/2)^2 + 2m\,(2a)^2 = (19/2)\,ma^2\) | M1 A1 | Find MI of rod \(AB\) about \(l\) |
| \(I = 2\times 13/4 + 4/3 + 19/2 = (52/3)\,ma^2\) | A1 | Verify MI of object about \(l\); A.G. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\omega_0 = \sqrt{(2ag)/2a}\) or \(\sqrt{g/2a}\) | B1 | Find and use initial angular speed |
| \(\frac{1}{2}I\omega_0^2 = \frac{1}{2}(52/3)\,ma^2\times(g/2a) = (13/3)\,mga\) | M1 A1 | Find initial rotational KE |
| \((3mg\times 2a + mga)(1-\cos\theta)\) or \((4mg\times 7a/4)(1-\cos\theta) = 7mga(1-\cos\theta)\) | M1 A1 | Find gain in P.E. at instantaneous rest |
| \(1-\cos\theta = (13/3)\,mga/7mga = 13/21\), \(\cos\theta = 8/21\) | M1 A1 | Verify \(\cos\theta\) by equating KE and PE; A.G. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\bar{x}=5,\ \bar{y}=6\) by solving simultaneous equations | M1, A1; A1 | Find \(\bar{x},\bar{y}\) |
| \(22+x_5=5\times5,\ x_5=3\); \(25+y_5=5\times6,\ y_5=5\) | A1 A1 | Hence find missing values \(x_5, y_5\) |
| OR: \(25+y_5=5\times4.5+0.3(22+x_5)\); \(22+x_5=3(25+y_5)-5\times13\); \(y_5=0.3x_5+4.1\), \(x_5=3y_5-12\); \(x_5=3,\ y_5=5\) | (M1 A1)(A1 A1) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(r = \sqrt{(0.3\times3)}\) or \(12/\sqrt{(40\times4)} = 0.949\) | M1 A1 | A0 for \(-0.949\) or \(\pm0.949\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\sum x=25+20=45\); \(\sum x^2=165+100=265\); \(\sum y=30+17=47\); \(\sum y^2=184+69=253\); \(\sum xy=162+75=237\) | B1 | B1 needs all 5 correct |
| \(S_{xy}=237-45\times47/10=25.5\); \(S_{xx}=265-45^2/10=62.5\); \(S_{yy}=253-47^2/10=32.1\) | ||
| \(r'=S_{xy}/\sqrt{(S_{xx}S_{yy})}=25.5/\sqrt{44.79}=25.5/\sqrt{(0.408\times0.7944)}=0.569\) | M1 *A1 | M0 if based on \(B\) only |
| \(H_0: \rho=0\), \(H_1: \rho\neq0\) | B1 | State both hypotheses |
| \(r_{10,\,5\%}=0.632\) | *B1 | State or use correct tabular two-tail \(r\)-value |
| Accept \(H_0\) if \( | r' | <\) tab. value (AEF); Popln. pmcc not different from 0 |
# Question 10(a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_\text{disc} = \frac{1}{2}(\frac{1}{4}ma^2) + \frac{1}{2}m\{(3a/2)^2+(2a)^2\} = ma^2/8 + 25ma^2/8 = 13ma^2/4$ | M1 A1 | Verify MI of either disc about axis $l$ at $O$; A.G. |
---
# Question 10(a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $I_{OC} = (4/3)\,ma^2$ | B1 | Find or state MI of rod $OC$ about $l$ |
| $I_{AB} = \frac{1}{3}\cdot 2m\,(3a/2)^2 + 2m\,(2a)^2 = (19/2)\,ma^2$ | M1 A1 | Find MI of rod $AB$ about $l$ |
| $I = 2\times 13/4 + 4/3 + 19/2 = (52/3)\,ma^2$ | A1 | Verify MI of object about $l$; A.G. |
---
# Question 10(a)(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\omega_0 = \sqrt{(2ag)/2a}$ or $\sqrt{g/2a}$ | B1 | Find and use initial angular speed |
| $\frac{1}{2}I\omega_0^2 = \frac{1}{2}(52/3)\,ma^2\times(g/2a) = (13/3)\,mga$ | M1 A1 | Find initial rotational KE |
| $(3mg\times 2a + mga)(1-\cos\theta)$ or $(4mg\times 7a/4)(1-\cos\theta) = 7mga(1-\cos\theta)$ | M1 A1 | Find gain in P.E. at instantaneous rest |
| $1-\cos\theta = (13/3)\,mga/7mga = 13/21$, $\cos\theta = 8/21$ | M1 A1 | Verify $\cos\theta$ by equating KE and PE; A.G. |
---
# Question 10(b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\bar{x}=5,\ \bar{y}=6$ by solving simultaneous equations | M1, A1; A1 | Find $\bar{x},\bar{y}$ |
| $22+x_5=5\times5,\ x_5=3$; $25+y_5=5\times6,\ y_5=5$ | A1 A1 | Hence find missing values $x_5, y_5$ |
| OR: $25+y_5=5\times4.5+0.3(22+x_5)$; $22+x_5=3(25+y_5)-5\times13$; $y_5=0.3x_5+4.1$, $x_5=3y_5-12$; $x_5=3,\ y_5=5$ | (M1 A1)(A1 A1) | |
---
# Question 10(b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $r = \sqrt{(0.3\times3)}$ or $12/\sqrt{(40\times4)} = 0.949$ | M1 A1 | A0 for $-0.949$ or $\pm0.949$ |
---
# Question 10(b)(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\sum x=25+20=45$; $\sum x^2=165+100=265$; $\sum y=30+17=47$; $\sum y^2=184+69=253$; $\sum xy=162+75=237$ | B1 | B1 needs all 5 correct |
| $S_{xy}=237-45\times47/10=25.5$; $S_{xx}=265-45^2/10=62.5$; $S_{yy}=253-47^2/10=32.1$ | | |
| $r'=S_{xy}/\sqrt{(S_{xx}S_{yy})}=25.5/\sqrt{44.79}=25.5/\sqrt{(0.408\times0.7944)}=0.569$ | M1 *A1 | M0 if based on $B$ only |
| $H_0: \rho=0$, $H_1: \rho\neq0$ | B1 | State both hypotheses |
| $r_{10,\,5\%}=0.632$ | *B1 | State or use correct tabular two-tail $r$-value |
| Accept $H_0$ if $|r'|<$ tab. value (AEF); Popln. pmcc not different from 0 | M1 DA1 | |\begin{center}
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\end{center}
A thin uniform rod $A B$ has mass $2 m$ and length $3 a$. Two identical uniform discs each have mass $\frac { 1 } { 2 } m$ and radius $a$. The centre of one of the discs is rigidly attached to the end $A$ of the rod and the centre of the other disc is rigidly attached to the end $B$ of the rod. The plane of each disc is perpendicular to the rod $A B$. A second thin uniform rod $O C$ has mass $m$ and length $2 a$. The end $C$ of this rod is rigidly attached to the mid-point of $A B$, with $O C$ perpendicular to $A B$ (see diagram). The object consisting of the two discs and two rods is free to rotate about a horizontal axis $l$, through $O$, which is perpendicular to both rods.\\
(i) Show that the moment of inertia of one of the discs about $l$ is $\frac { 13 } { 4 } m a ^ { 2 }$.\\
(ii) Show that the moment of inertia of the object about $l$ is $\frac { 52 } { 3 } m a ^ { 2 }$.
When the object is suspended from $O$ and is hanging in equilibrium, the point $C$ is given a speed of $\sqrt { } ( 2 a g )$ in the direction parallel to $A B$. In the subsequent motion, the angle through which $O C$ has turned before the object comes to instantaneous rest is $\theta$.\\
(iii) Show that $\cos \theta = \frac { 8 } { 21 }$.
\hfill \mbox{\textit{CAIE FP2 2016 Q10 EITHER}}