| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2010 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments of inertia |
| Type | Detachment or peg impact mid-motion |
| Difficulty | Challenging +1.8 This is a challenging Further Maths mechanics problem requiring: (1) calculation of moment of inertia using parallel axis theorem for multiple composite bodies (disc, ring, three rods), (2) energy conservation with rotational KE, and (3) analysis after detachment requiring recalculation of moment of inertia and further energy conservation. The multi-stage nature, composite body calculations, and post-detachment analysis place this well above average difficulty, though the techniques are standard for Further Maths students. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(I_{disc} = \frac{1}{2}\cdot 4ma^2 = [2ma^2\) or \(18ma^2]\) | B1 | Find MI of disc about \(O\) [or \(A\)] |
| \(I_{ring} = m(2a)^2 = [4ma^2\) or \(8ma^2]\) | B1 | Find MI of ring about \(O\) [or \(A\)] |
| \(I_{rod} = (4/3)ma^2\; [\) or \(22ma^2/3]\) | B1 | Find MI of \(AO\) about \(O\) [or \(BO\) about \(A\)] |
| \(I_{wheel} = 10ma^2 + 8m(2a)^2 = 42ma^2\) A.G. | M1, A1 | Find MI of wheel about \(A\) |
| \(\frac{1}{2}I_{wheel}\omega^2 = 8mg \times 2a\sin 30°\); \(\omega^2 = 8mga^2/21ma^2\) | M1 A1 | Find angular speed \(\omega\) using energy |
| \(\omega = \sqrt{8g/21a}\) (A.E.F.) or \(1.95/\sqrt{a}\) | A1 | |
| \(I_{new} = 8ma^2 + 4m(2a)^2 = 24ma^2\) | M1 A1 | Find new MI about \(A\) |
| \(\frac{1}{2}I_{new}\omega^2 = M_{new}g \times 2a\sin\theta\) | M1 | Find required angle \(\theta\) using energy |
| \(M_{new} = m + 3m = 4m\) | A1 | Find and use new mass |
| \((32/7)mga = 8mga\sin\theta\) (A.E.F.) | A1 | Substitute for \(I_{new},\; M_{new},\; \omega^2\) |
| \(\theta = \sin^{-1}(4/7) = 0.608\;\text{rad}\) or \(34.8°\) | A1 | Solve for \(\theta\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Integrate to find \(F(t)\) for \(t \geq 2\): \(F(t) = c - (t-1)^{-2}\) | M1 | \(c\) needed |
| Use \(F(2) = 0\) to find \(c\): \(F(t) = 1 - (t-1)^{-2}\) | A1 | |
| Find \(p = P(T > 5)\): \(p = 1 - F(5) = 1 - (1 - 4^{-2}) = \frac{1}{16}\) | B1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| State or imply distribution: \(P(N > n) = p(1-p)^{n-1}\) *or* geometric distribution with parameter \(p\) | M1 | |
| Find \(P(N > E(N))\): \((1-p)^{1/p} = (15/16)^{16} = 0.356\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Relate distribution function \(G(y)\) of \(Y\) to \(T\): \(G(y) = P(Y < y) = P(1/(T-1) < y)\) | M1 | |
| Rearrange: \(= P(T > 1 + 1/y)\) | A1 | |
| Relate to distribution function \(F\): \(= 1 - F(1 + 1/y)\) | M1 | |
| Substitute expression for \(F\): \(= 1 - \{1 - (1 + 1/y - 1)^{-2}\}\) | A1 | |
| Simplify: \(= y^2\) | A1 | |
| Differentiate to find probability density function: \(g(y) = 2y\) | M1 A1 | |
| Give complete statement of \(g(y)\): \(g(y) = 2y \ (0 \leq y \leq 1)\), \(0\) otherwise | A1 |
## Question 11:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $I_{disc} = \frac{1}{2}\cdot 4ma^2 = [2ma^2$ or $18ma^2]$ | B1 | Find MI of disc about $O$ [or $A$] |
| $I_{ring} = m(2a)^2 = [4ma^2$ or $8ma^2]$ | B1 | Find MI of ring about $O$ [or $A$] |
| $I_{rod} = (4/3)ma^2\; [$ or $22ma^2/3]$ | B1 | Find MI of $AO$ about $O$ [or $BO$ about $A$] |
| $I_{wheel} = 10ma^2 + 8m(2a)^2 = 42ma^2$ **A.G.** | M1, A1 | Find MI of wheel about $A$ |
| $\frac{1}{2}I_{wheel}\omega^2 = 8mg \times 2a\sin 30°$; $\omega^2 = 8mga^2/21ma^2$ | M1 A1 | Find angular speed $\omega$ using energy |
| $\omega = \sqrt{8g/21a}$ (A.E.F.) or $1.95/\sqrt{a}$ | A1 | |
| $I_{new} = 8ma^2 + 4m(2a)^2 = 24ma^2$ | M1 A1 | Find new MI about $A$ |
| $\frac{1}{2}I_{new}\omega^2 = M_{new}g \times 2a\sin\theta$ | M1 | Find required angle $\theta$ using energy |
| $M_{new} = m + 3m = 4m$ | A1 | Find and use new mass |
| $(32/7)mga = 8mga\sin\theta$ (A.E.F.) | A1 | Substitute for $I_{new},\; M_{new},\; \omega^2$ |
| $\theta = \sin^{-1}(4/7) = 0.608\;\text{rad}$ or $34.8°$ | A1 | Solve for $\theta$ |
**Total: 14 marks**
## Question 11 OR:
### Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Integrate to find $F(t)$ for $t \geq 2$: $F(t) = c - (t-1)^{-2}$ | M1 | $c$ needed |
| Use $F(2) = 0$ to find $c$: $F(t) = 1 - (t-1)^{-2}$ | A1 | |
| Find $p = P(T > 5)$: $p = 1 - F(5) = 1 - (1 - 4^{-2}) = \frac{1}{16}$ | B1 | |
**Part mark: 3**
---
### Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| State or imply distribution: $P(N > n) = p(1-p)^{n-1}$ *or* geometric distribution with parameter $p$ | M1 | |
| Find $P(N > E(N))$: $(1-p)^{1/p} = (15/16)^{16} = 0.356$ | M1 A1 | |
**Part mark: 3**
---
### Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Relate distribution function $G(y)$ of $Y$ to $T$: $G(y) = P(Y < y) = P(1/(T-1) < y)$ | M1 | |
| Rearrange: $= P(T > 1 + 1/y)$ | A1 | |
| Relate to distribution function $F$: $= 1 - F(1 + 1/y)$ | M1 | |
| Substitute expression for $F$: $= 1 - \{1 - (1 + 1/y - 1)^{-2}\}$ | A1 | |
| Simplify: $= y^2$ | A1 | |
| Differentiate to find probability density function: $g(y) = 2y$ | M1 A1 | |
| Give complete statement of $g(y)$: $g(y) = 2y \ (0 \leq y \leq 1)$, $0$ otherwise | A1 | |
**Part mark: 8**
**Total: [14]**\begin{center}
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\end{center}
A uniform disc, of mass $4 m$ and radius $a$, and a uniform ring, of mass $m$ and radius $2 a$, each have centre $O$. A wheel is made by fixing three uniform rods, $O A , O B$ and $O C$, each of mass $m$ and length $2 a$, to the disc and the ring, as shown in the diagram. Show that the moment of inertia of the wheel about an axis through $A$, perpendicular to the plane of the wheel, is $42 m a ^ { 2 }$.
The axis through $A$ is horizontal, and the wheel can rotate freely about this axis. The wheel is released from rest with $O$ above the level of $A$ and $A O$ making an angle of $30 ^ { \circ }$ with the horizontal. Find the angular speed of the wheel when $A O$ is horizontal.
When $A O$ is horizontal the disc becomes detached from the wheel. Find the angle that $A O$ makes with the horizontal when the wheel first comes to instantaneous rest.
\hfill \mbox{\textit{CAIE FP2 2010 Q11 EITHER}}