| Exam Board | CAIE |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2009 |
| Session | June |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Simple Harmonic Motion |
| Type | Prove SHM and find period: given force or equation of motion directly |
| Difficulty | Challenging +1.2 This is a standard Further Maths mechanics question combining elastic strings, Newton's laws, and SHM. While it requires multiple steps (forces on both particles, showing SHM form, finding equilibrium, applying friction condition, solving for time), each step follows a well-established procedure taught in FM2. The SHM identification and period calculation are routine once the equation is derived. The friction calculation adds complexity but is straightforward application of limiting friction. This is harder than average A-level due to being Further Maths content with multiple integrated concepts, but represents a typical FM2 exam question rather than requiring exceptional insight. |
| Spec | 4.10f Simple harmonic motion: x'' = -omega^2 x |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T = 4mgx/a\) | B1 | Find tension \(T\) |
| \(m\,\mathrm{d}^2x/\mathrm{d}t^2 = mg - T\) | M1 | Apply Newton's law of motion to \(B\) |
| \(\mathrm{d}^2x/\mathrm{d}t^2 = -(g/a)(4x-a)\) | A1 | Combine. A.G. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\mathrm{d}^2y/\mathrm{d}t^2 = -(4g/a)\,y\) | M1 A1 | Substitute e.g. \(y = x - \tfrac{1}{4}a\) and rearrange |
| \(x_c = \tfrac{1}{4}a\) | A1 | State centre of motion, or derive from \(y=0\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T = \tfrac{1}{3}mg,\; x_s = a/12\) | M1 A1 | Find \(x\) when \(A\) starts to slip using \(F = \mu R\) |
| \(y = y_{max}\cos\omega t\) *or* \(y_{max}\sin\omega t\) | M1 | Valid use of SHM equation to find \(t_s\) to slipping |
| \(y_s = y_{max}\cos\omega t_s\) | M1 | EITHER: Valid use of cosine form |
| \(t_s = t_1 - t_2,\; y_{max} = y_{max}\sin\omega t_1\) | M1 | OR: Valid use of sine form |
| \(y_s = y_{max}\sin\omega t_2\) | (M1) | |
| \(y_s/y_{max} = (x_c - x_s)/x_c\) | M1 | Substitute for \(y_s\), \(y_{max}\) |
| \(= (a/6)/(a/4) = \tfrac{2}{3}\) | A1 | |
| \(t_s = (\cos^{-1}\tfrac{2}{3})/\omega\) | ||
| *or* \((\tfrac{1}{2}\pi - \sin^{-1}\tfrac{2}{3})/\omega\) | M1 | |
| \(t_s = 0.421\sqrt{a/g}\) | A1 | Substitute \(\omega = 2\sqrt{g/a}\) and evaluate |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(H_0: \mu_E = \mu_W,\; H_1: \mu_E \neq \mu_W\) | B1 | State hypotheses |
| Two populations have Normal distributions and common variance | B1 | State assumption [A.E.F.] |
| \(\sigma^2 = (5\times0.0231 + 4\times0.0195)/9\) | M1 | Estimate common variance |
| \(= 0.0215\) *or* \(0.1473^2\) | A1 | |
| \(t_{9,\,0.975} = 2.26[2]\) | B1 | Use correct tabular value of \(t\) |
| \(\lvert\bar{X}_E - \bar{X}_W\rvert \geq t\sigma\sqrt{1/6+1/5}\) | M1 | Formulate rejection region (with any \(t\); allow \(>\)) |
| \(= 0.201\) | A1 | |
| \(0.253 > 0.201\sqrt{\phantom{x}}\) | Compare actual sample means with region | |
| *or* \(2.85 > 2.26[2]\sqrt{\phantom{x}}\) | M1 A1 | *or* compare calculated \(t\) with tabular \(t\) |
| Mean acidity levels do differ | A1 | Consistent conclusion (A.E.F.; dep values above) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\bar{X}_E - \bar{X}_W - a \geq t\sigma\sqrt{1/6+1/5}\) | M1 | State condition on \(a\) (with any \(t\); allow \(>\) or \(=\)) |
| \(t_{9,\,0.95} = 1.83[3]\) | A1 | Use correct tabular value of \(t\) |
| \(a_{max} = 0.253 - 0.163 = 0.09\) (2 dp) | M1 A1 | Substitute to find largest value of \(a\) |
# Question 11 (EITHER - Mechanics, 14 marks):
## Part 1 (3 marks)
| Answer | Mark | Guidance |
|--------|------|----------|
| $T = 4mgx/a$ | B1 | Find tension $T$ |
| $m\,\mathrm{d}^2x/\mathrm{d}t^2 = mg - T$ | M1 | Apply Newton's law of motion to $B$ |
| $\mathrm{d}^2x/\mathrm{d}t^2 = -(g/a)(4x-a)$ | A1 | Combine. **A.G.** |
## Part 2 (3 marks)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\mathrm{d}^2y/\mathrm{d}t^2 = -(4g/a)\,y$ | M1 A1 | Substitute e.g. $y = x - \tfrac{1}{4}a$ and rearrange |
| $x_c = \tfrac{1}{4}a$ | A1 | State centre of motion, or derive from $y=0$ |
## Part 3 (8 marks)
| Answer | Mark | Guidance |
|--------|------|----------|
| $T = \tfrac{1}{3}mg,\; x_s = a/12$ | M1 A1 | Find $x$ when $A$ starts to slip using $F = \mu R$ |
| $y = y_{max}\cos\omega t$ *or* $y_{max}\sin\omega t$ | M1 | Valid use of SHM equation to find $t_s$ to slipping |
| $y_s = y_{max}\cos\omega t_s$ | M1 | EITHER: Valid use of cosine form |
| $t_s = t_1 - t_2,\; y_{max} = y_{max}\sin\omega t_1$ | M1 | OR: Valid use of sine form |
| $y_s = y_{max}\sin\omega t_2$ | (M1) | |
| $y_s/y_{max} = (x_c - x_s)/x_c$ | M1 | Substitute for $y_s$, $y_{max}$ |
| $= (a/6)/(a/4) = \tfrac{2}{3}$ | A1 | |
| $t_s = (\cos^{-1}\tfrac{2}{3})/\omega$ | | |
| *or* $(\tfrac{1}{2}\pi - \sin^{-1}\tfrac{2}{3})/\omega$ | M1 | |
| $t_s = 0.421\sqrt{a/g}$ | A1 | Substitute $\omega = 2\sqrt{g/a}$ and evaluate |
---
# Question 11 (OR - Statistics, 14 marks):
## Part 1 (9 marks)
| Answer | Mark | Guidance |
|--------|------|----------|
| $H_0: \mu_E = \mu_W,\; H_1: \mu_E \neq \mu_W$ | B1 | State hypotheses |
| Two populations have Normal distributions and common variance | B1 | State assumption [A.E.F.] |
| $\sigma^2 = (5\times0.0231 + 4\times0.0195)/9$ | M1 | Estimate common variance |
| $= 0.0215$ *or* $0.1473^2$ | A1 | |
| $t_{9,\,0.975} = 2.26[2]$ | B1 | Use correct tabular value of $t$ |
| $\lvert\bar{X}_E - \bar{X}_W\rvert \geq t\sigma\sqrt{1/6+1/5}$ | M1 | Formulate rejection region (with any $t$; allow $>$) |
| $= 0.201$ | A1 | |
| $0.253 > 0.201\sqrt{\phantom{x}}$ | | Compare actual sample means with region |
| *or* $2.85 > 2.26[2]\sqrt{\phantom{x}}$ | M1 A1 | *or* compare calculated $t$ with tabular $t$ |
| Mean acidity levels **do** differ | A1 | Consistent conclusion (A.E.F.; dep values above) |
## Part 2 (4 marks)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{X}_E - \bar{X}_W - a \geq t\sigma\sqrt{1/6+1/5}$ | M1 | State condition on $a$ (with any $t$; allow $>$ or $=$) |
| $t_{9,\,0.95} = 1.83[3]$ | A1 | Use correct tabular value of $t$ |
| $a_{max} = 0.253 - 0.163 = 0.09$ (2 dp) | M1 A1 | Substitute to find largest value of $a$ |\begin{center}
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\end{center}
Two particles $A$ and $B$, of equal mass $m$, are connected by a light elastic string of natural length $a$ and modulus of elasticity $4 m g$. Particle $A$ rests on a rough horizontal table at a distance $a$ from the edge of the table. The string passes over a small smooth pulley $P$ fixed at the edge of the table. At time $t = 0 , B$ is released from rest at $P$ and falls vertically. At time $t , B$ has fallen a distance $x$, without $A$ slipping (see diagram). Show that
$$\ddot { x } = - \frac { g } { a } ( 4 x - a ) .$$
Deduce that, while $A$ does not slip, $B$ moves in simple harmonic motion and identify the centre of the motion.
Given that the coefficient of friction between $A$ and the table is $\frac { 1 } { 3 }$, find the value of $x$ when $A$ starts to slip, and the corresponding value of $t$, expressing this answer in the form $k \sqrt { } \left( \frac { a } { g } \right)$. Give the value of $k$ correct to 3 decimal places.
\hfill \mbox{\textit{CAIE FP2 2009 Q11 EITHER}}