CAIE FP2 2012 November — Question 11 EITHER

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2012
SessionNovember
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicSimple Harmonic Motion
TypeComplete motion cycle with slack phase
DifficultyChallenging +1.8 This is a challenging Further Maths mechanics problem requiring understanding of elastic SHM with a slack phase transition. Students must: (1) find equilibrium position, (2) set up SHM equation with correct amplitude and period, (3) determine when string becomes slack (extension = 0), (4) analyze projectile motion during slack phase, (5) find maximum height and calculate total distance. The multi-phase nature and need to coordinate SHM with projectile motion elevates this significantly above standard SHM questions, though the individual techniques are well-established for FM students.
Spec6.02d Mechanical energy: KE and PE concepts6.02g Hooke's law: T = k*x or T = lambda*x/l6.02i Conservation of energy: mechanical energy principle
A particle \(P\) of mass \(m\) is attached to one end of a light elastic string of modulus of elasticity \(8 m g\) and natural length \(a\). The other end of the string is attached to a fixed point \(O\). The particle is pulled vertically downwards a distance \(\frac { 1 } { 4 } a\) from its equilibrium position and released from rest. Show that the string first becomes slack after a time \(\frac { 2 \pi } { 3 } \sqrt { } \left( \frac { a } { 8 g } \right)\). Find, in terms of \(a\), the total distance travelled by \(P\) from its release until it subsequently comes to instantaneous rest for the first time.
Question 11(a):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(8mge/a = mg \Rightarrow e = a/8\)B1 Resolve vertically at equilibrium
\(m\,d^2x/dt^2 = mg - 8mg(e+x)/a\)M1 A1 Newton's Law at general point
\(d^2x/dt^2 = -(8g/a)x\), so \(\omega^2 = 8g/a\)A1 Simplify; allow stating result without derivation
OR: \(v^2 = \omega^2\{(\tfrac{1}{4}a)^2 - e^2\}\), \(3ga/8 = \omega^2(a^2/16 - a^2/64)\), \(\omega^2 = 8g/a\)(M1)(A1)(A1) Assume SHM and find \(\omega^2\) from speed \(v\) when first slack
\(\omega t = \cos^{-1}(-\tfrac{1}{2})\) or \(\tfrac{1}{2}\pi + \sin^{-1}(\tfrac{1}{2}) = 2\pi/3\)M1 A1 A1 Use \(x = \tfrac{1}{4}a\cos\omega t\) or \(\tfrac{1}{4}a\sin\omega t\)
\(t = (2\pi/3)\sqrt{(a/8g)}\) A.G.A1 Substitute \(\omega = \sqrt{(8g/a)}\)
\(v^2 = \omega^2(a^2/16 - e^2) = 3ga/8\) or \(\tfrac{1}{4}a\omega\sin 2\pi/3 = 3ga/8\)M1 A1 Find \(v^2\) when first slack
\(\tfrac{1}{2}mv^2 = \tfrac{1}{2}\cdot 8mg(e + \tfrac{1}{4}a)^2/a - mg(e + \tfrac{1}{4}a)\), \(v^2 = 9ga/8 - 3ga/4 = 3ga/8\)(M1 A1) OR: energy method
\(2gs_2 = v^2\), \(s_2 = 3a/16\)M1 A1 Find further distance \(s_2\) to rest
\(\tfrac{1}{4}a + e + s_2 = 9a/16\) or \(0 \cdot 562[5]a\)M1 A1 Find total distance
Question 11(b):
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(k + 3k = 1\), \(k = \tfrac{1}{4}\)M1 A1 Find \(k\) by equating area under graph to 1
\(\tfrac{1}{2}kx = x/8\) and \(k = \tfrac{1}{4}\) A.G.B1 \(f(x)\) for \(0 < x \leq 2\) and \(2 < x \leq 5\)
Part (i): \(F(x) = x^2/16\) \((0 \leq x \leq 2)\)M1 A1 Integrate to find \(F(x)\)
\(\tfrac{1}{4}x - \tfrac{1}{4}\) \((2 < x \leq 5)\)M1 A1
\(G(y) = P(Y < y) = P(X^2 < y) = P(X < y^{1/2}) = F(y^{1/2})\)M1 A1 Relate dist. fn. \(G(y)\) of \(Y\) to \(X\)
\(= y/16\) and \(\tfrac{1}{4}y^{1/2} - \tfrac{1}{4}\)M1 A1
\(g(y) = 1/16\) or \(0 \cdot 0625\) \((0 \leq y \leq 4)\)M1 A1 Differentiate to find \(g(y)\); both results required for M1
\(\tfrac{1}{8}\sqrt{y}\) \((4 < y \leq 25)\); \([0\) otherwise\(]\)M1 A1
Part (ii): \(E(Y) = (1/16)\int y\,dy + (1/8)\int y^{1/2}\,dy\)M1 *EITHER*: Find \(E(Y)\) using \(\int y\,g(y)\,dy\)
\(= [y^2/32]_0^4 + [y^{3/2}/12]_4^{25} = \tfrac{1}{2} + 117/12 = 10 \cdot 25\) A.G.A1 A1
\(E(Y) = (1/8)\int x^3\,dx + \tfrac{1}{4}\int x^2\,dx = [x^4/32]_0^2 + [x^3/12]_2^5 = \tfrac{1}{2} + 117/12 = 10 \cdot 25\) A.G.(M1)(A1)(A1) *OR*: Find \(E(Y)\) using \(\int x^2 f(x)\,dx\)
Part (iii): \(F(m_x) = \tfrac{1}{4}m_x - \tfrac{1}{4} = \tfrac{1}{2}\), \(m_x = 3\)M1 A1 *EITHER*: Find median \(m_x\) of \(X\)
\(F(m_y) = \tfrac{1}{4}m_y^{1/2} - \tfrac{1}{4} = \tfrac{1}{2}\), \(m_y = 9\)M1 A1 median \(m_y\) of \(Y\)
\(P(Y < m_x^2) = P(X^2 < m_x^2) = P(X < m_x)\)(M1 A1) *OR*: Show \(m_y = m_x^2\) 
# Question 11(a):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $8mge/a = mg \Rightarrow e = a/8$ | B1 | Resolve vertically at equilibrium |
| $m\,d^2x/dt^2 = mg - 8mg(e+x)/a$ | M1 A1 | Newton's Law at general point |
| $d^2x/dt^2 = -(8g/a)x$, so $\omega^2 = 8g/a$ | A1 | Simplify; allow stating result without derivation |
| **OR:** $v^2 = \omega^2\{(\tfrac{1}{4}a)^2 - e^2\}$, $3ga/8 = \omega^2(a^2/16 - a^2/64)$, $\omega^2 = 8g/a$ | (M1)(A1)(A1) | Assume SHM and find $\omega^2$ from speed $v$ when first slack |
| $\omega t = \cos^{-1}(-\tfrac{1}{2})$ or $\tfrac{1}{2}\pi + \sin^{-1}(\tfrac{1}{2}) = 2\pi/3$ | M1 A1 A1 | Use $x = \tfrac{1}{4}a\cos\omega t$ or $\tfrac{1}{4}a\sin\omega t$ |
| $t = (2\pi/3)\sqrt{(a/8g)}$ **A.G.** | A1 | Substitute $\omega = \sqrt{(8g/a)}$ |
| $v^2 = \omega^2(a^2/16 - e^2) = 3ga/8$ or $\tfrac{1}{4}a\omega\sin 2\pi/3 = 3ga/8$ | M1 A1 | Find $v^2$ when first slack |
| $\tfrac{1}{2}mv^2 = \tfrac{1}{2}\cdot 8mg(e + \tfrac{1}{4}a)^2/a - mg(e + \tfrac{1}{4}a)$, $v^2 = 9ga/8 - 3ga/4 = 3ga/8$ | (M1 A1) | OR: energy method |
| $2gs_2 = v^2$, $s_2 = 3a/16$ | M1 A1 | Find further distance $s_2$ to rest |
| $\tfrac{1}{4}a + e + s_2 = 9a/16$ or $0 \cdot 562[5]a$ | M1 A1 | Find total distance |

---

# Question 11(b):

| Answer/Working | Mark | Guidance |
|---|---|---|
| $k + 3k = 1$, $k = \tfrac{1}{4}$ | M1 A1 | Find $k$ by equating area under graph to 1 |
| $\tfrac{1}{2}kx = x/8$ and $k = \tfrac{1}{4}$ **A.G.** | B1 | $f(x)$ for $0 < x \leq 2$ and $2 < x \leq 5$ |
| **Part (i):** $F(x) = x^2/16$ $(0 \leq x \leq 2)$ | M1 A1 | Integrate to find $F(x)$ |
| $\tfrac{1}{4}x - \tfrac{1}{4}$ $(2 < x \leq 5)$ | M1 A1 | |
| $G(y) = P(Y < y) = P(X^2 < y) = P(X < y^{1/2}) = F(y^{1/2})$ | M1 A1 | Relate dist. fn. $G(y)$ of $Y$ to $X$ |
| $= y/16$ and $\tfrac{1}{4}y^{1/2} - \tfrac{1}{4}$ | M1 A1 | |
| $g(y) = 1/16$ or $0 \cdot 0625$ $(0 \leq y \leq 4)$ | M1 A1 | Differentiate to find $g(y)$; both results required for M1 |
| $\tfrac{1}{8}\sqrt{y}$ $(4 < y \leq 25)$; $[0$ otherwise$]$ | M1 A1 | |
| **Part (ii):** $E(Y) = (1/16)\int y\,dy + (1/8)\int y^{1/2}\,dy$ | M1 | *EITHER*: Find $E(Y)$ using $\int y\,g(y)\,dy$ |
| $= [y^2/32]_0^4 + [y^{3/2}/12]_4^{25} = \tfrac{1}{2} + 117/12 = 10 \cdot 25$ **A.G.** | A1 A1 | |
| $E(Y) = (1/8)\int x^3\,dx + \tfrac{1}{4}\int x^2\,dx = [x^4/32]_0^2 + [x^3/12]_2^5 = \tfrac{1}{2} + 117/12 = 10 \cdot 25$ **A.G.** | (M1)(A1)(A1) | *OR*: Find $E(Y)$ using $\int x^2 f(x)\,dx$ |
| **Part (iii):** $F(m_x) = \tfrac{1}{4}m_x - \tfrac{1}{4} = \tfrac{1}{2}$, $m_x = 3$ | M1 A1 | *EITHER*: Find median $m_x$ of $X$ |
| $F(m_y) = \tfrac{1}{4}m_y^{1/2} - \tfrac{1}{4} = \tfrac{1}{2}$, $m_y = 9$ | M1 A1 | median $m_y$ of $Y$ |
| $P(Y < m_x^2) = P(X^2 < m_x^2) = P(X < m_x)$ | (M1 A1) | *OR*: Show $m_y = m_x^2$ |
A particle $P$ of mass $m$ is attached to one end of a light elastic string of modulus of elasticity $8 m g$ and natural length $a$. The other end of the string is attached to a fixed point $O$. The particle is pulled vertically downwards a distance $\frac { 1 } { 4 } a$ from its equilibrium position and released from rest. Show that the string first becomes slack after a time $\frac { 2 \pi } { 3 } \sqrt { } \left( \frac { a } { 8 g } \right)$.

Find, in terms of $a$, the total distance travelled by $P$ from its release until it subsequently comes to instantaneous rest for the first time.

\hfill \mbox{\textit{CAIE FP2 2012 Q11 EITHER}}
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