Question 10 EITHER - A-Level Maths

CAIE FP2 2013 June — Question 10 EITHER

Exam Board	CAIE
Module	FP2 (Further Pure Mathematics 2)
Year	2013
Session	June
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Hooke's law and elastic energy
Type	Vertical elastic string: general symbolic/proof questions
Difficulty	Challenging +1.8 This is a challenging Further Maths mechanics problem requiring integration of elastic strings with SHM. It demands multiple sophisticated techniques: energy methods with variable elastic force, finding equilibrium/amplitude for SHM, and calculating period for motion in an elastic medium. The multi-part structure with 'show that' proofs and the final time calculation involving both free fall and SHM phases places it well above average difficulty, though the path is relatively standard for FP2 students who have studied this topic.
Spec	6.02h Elastic PE: 1/2 k x^2 6.02i Conservation of energy: mechanical energy principle 6.05f Vertical circle: motion including free fall

A light elastic string has modulus of elasticity $\frac { 3 } { 2 } m g$ and natural length $a$. A particle of mass $m$ is attached to one end of the string. The other end of the string is attached to a fixed point $A$. The particle is released from rest at $A$. Show that when the particle has fallen a distance $k a$ from $A$, where $k > 1$, its kinetic energy is $$\frac { 1 } { 4 } m g a \left( 10 k - 3 - 3 k ^ { 2 } \right) .$$ Show that the particle first comes to instantaneous rest at the point $B$ which is at a distance $3 a$ vertically below $A$. Show that the time taken by the particle to travel from $A$ to $B$ is $$\sqrt { } \left( \frac { 2 a } { g } \right) + \frac { 2 \pi } { 3 } \sqrt { } \left( \frac { 2 a } { 3 g } \right)$$

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Question 10(a):

Finding KE after falling ka from PE:

Answer	Marks	Guidance
$mg \times ka - \frac{1}{2}(3mg/2)(ka-a)^2/a$	M1=[A1	Energy equation with EPE term

$= mga(k - 3k^2/4 + 3k/2 - \frac{3}{4})$

Answer	Marks	Guidance
$= \frac{1}{4}mga(10k - 3 - 3k^2)$ A.G.	A1	3 marks total

Finding equation for k (or ka) at B using KE = 0:

Answer	Marks
$3k^2 - 10k + 3 = 0$ (A.E.F.)	M1

Finding both roots and selecting root > 1:

Answer	Marks	Guidance
Roots 3 and $\frac{1}{3}$, so $k = 3$ A.G.	M1 A1	3 marks total

Finding time $t_1$ to fall $a$ from $A$ (under no tension):

Answer	Marks
$a = \frac{1}{2}gt_1^2$, $t_1 = \sqrt{(2a/g)}$	B1

Finding ext. $e$ or distance fallen at equilibrium point O:

Answer	Marks
$mg = 3mge/2a$, $e = 2a/3$ or $AO = 5a/3$	B1

State or find SHM equation at x below (or above) O:

Answer	Marks
$m\frac{d^2x}{dt^2} = mg - 3mg(x+e)/2a$	M1
$\frac{d^2x}{dt^2} = -3gx/2a$	A1

State or use correct amplitude $x_0$ and $\omega^2$:

Answer	Marks
$x_0 = 4a/3$ and $\omega^2 = 3g/2a$	B1

Finding time $t_2$ to fall from $a$ to $3a$ below A:

Answer	Marks
$[-]\ e = x_0\cos\omega t_2$	M1
$t_2 = \sqrt{(2a/3g)}\cos^{-1}(-\frac{1}{2})$ (A.E.F.)	A1

Finding total time $t_1 + t_2$:

Answer	Marks	Guidance
$\sqrt{(2a/g)} + (2\pi/3)\sqrt{(2a/3g)}$ A.G.	A1	8 marks, 14 total

Question 10(b):

Finding 4 summations required:

$\Sigma x = 12 + p$, $\Sigma x^2 = 38 + p^2$

$\Sigma y = 23$, $[\Sigma y^2 = 125]$

Answer	Marks
$\Sigma xy = 63 + 2p$	M1

EITHER: Substitute $b_1 = 1$ in formula for gradient:

$63 + 2p - (12+p)\times 23/5$

$= 38 + p^2 - (12+p)^2/5$

Answer	Marks	Guidance
or $39 - 13p = 4p^2 - 24p + 46$	M1 A1 A1	A1 for each side

OR: Substitute in normal equations:

Answer	Marks	Guidance
$5k + (12+p) = 23$ and $(12+p)k + (38+p^2) = 63 + 2p$	(M1 A1 A1)	A1 for each equation

Obtain and solve quadratic for p (A.E.F):

$0.8p^2 - 2.2p + 1.4 = 0$

Answer	Marks	Guidance
or $4p^2 - 11p + 7 = 0$	M1 A1
$p = 1,\ 1.75$ (or $7/4$)	A1, A1	8 marks total

Question 10(b)(i):

Finding equation for correlation coefficient with $p = 1$:

EITHER:

Answer	Marks
$r = (65 - 13\times 23/5)\ /\ \sqrt{\{(39 - 13^2/5)(125 - 23^2/5)\}}$	M1 A1

$= 5.2\ /\ \sqrt{(5.2\times 19.2)}$ or $1.04\ /\ \sqrt{(1.04\times 3.84)}$

OR:

$r^2 = 1\times(65 - 13\times 23/5)/(125 - 23^2/5)$

or $1\times(39 - 13^2/5)/(125 - 23^2/5)$

Answer	Marks
$= 5.2/19.2$ or $1.04/3.84$	M1 A1

Evaluate r:

Answer	Marks	Guidance
$r = \sqrt{39/12}$ or $0.52$	A1	3 marks total

Question 10(b)(ii):

EITHER: Recall or find gradient $b_2$ of line:

Answer	Marks
$b_2 = 5.2/19.2$ or $r^2\ [= 0.2708]$	M1

Find regression line of x on y:

$x - 13/5 = b_2(y - 23/5)$

$x = (13/48)y + 65/48$

Answer	Marks
or $0.271y + 1.35$	M1, A1

OR: Use normal equations for $x = a_2 + b_2 y$:

Answer	Marks
$5a_2 + 23b_2 = 13$ and $23a_2 + 125b_2 = 65$	(M1), (M1)

Solve for $a_2$, $b_2$:

$b_2 = 13/48$ or $0.271$

Answer	Marks	Guidance
And $a_2 = 65/48$ or $1.35$	(A1)	3 marks, 14 total

# Question 10(a):

**Finding KE after falling ka from PE:**

$mg \times ka - \frac{1}{2}(3mg/2)(ka-a)^2/a$ | M1=[A1 | Energy equation with EPE term

$= mga(k - 3k^2/4 + 3k/2 - \frac{3}{4})$

$= \frac{1}{4}mga(10k - 3 - 3k^2)$ **A.G.** | A1 | 3 marks total

---

**Finding equation for k (or ka) at B using KE = 0:**

$3k^2 - 10k + 3 = 0$ (A.E.F.) | M1 |

---

**Finding both roots and selecting root > 1:**

Roots 3 and $\frac{1}{3}$, so $k = 3$ **A.G.** | M1 A1 | 3 marks total

---

**Finding time $t_1$ to fall $a$ from $A$ (under no tension):**

$a = \frac{1}{2}gt_1^2$, $t_1 = \sqrt{(2a/g)}$ | B1 |

---

**Finding ext. $e$ or distance fallen at equilibrium point O:**

$mg = 3mge/2a$, $e = 2a/3$ or $AO = 5a/3$ | B1 |

---

**State or find SHM equation at x below (or above) O:**

$m\frac{d^2x}{dt^2} = mg - 3mg(x+e)/2a$ | M1 |

$\frac{d^2x}{dt^2} = -3gx/2a$ | A1 |

---

**State or use correct amplitude $x_0$ and $\omega^2$:**

$x_0 = 4a/3$ and $\omega^2 = 3g/2a$ | B1 |

---

**Finding time $t_2$ to fall from $a$ to $3a$ below A:**

$[-]\ e = x_0\cos\omega t_2$ | M1 |

$t_2 = \sqrt{(2a/3g)}\cos^{-1}(-\frac{1}{2})$ (A.E.F.) | A1 |

---

**Finding total time $t_1 + t_2$:**

$\sqrt{(2a/g)} + (2\pi/3)\sqrt{(2a/3g)}$ **A.G.** | A1 | 8 marks, 14 total

---

# Question 10(b):

**Finding 4 summations required:**

$\Sigma x = 12 + p$, $\Sigma x^2 = 38 + p^2$

$\Sigma y = 23$, $[\Sigma y^2 = 125]$

$\Sigma xy = 63 + 2p$ | M1 |

---

**EITHER: Substitute $b_1 = 1$ in formula for gradient:**

$63 + 2p - (12+p)\times 23/5$

$= 38 + p^2 - (12+p)^2/5$

or $39 - 13p = 4p^2 - 24p + 46$ | M1 A1 A1 | A1 for each side

---

**OR: Substitute in normal equations:**

$5k + (12+p) = 23$ and $(12+p)k + (38+p^2) = 63 + 2p$ | (M1 A1 A1) | A1 for each equation

---

**Obtain and solve quadratic for p (A.E.F):**

$0.8p^2 - 2.2p + 1.4 = 0$

or $4p^2 - 11p + 7 = 0$ | M1 A1 |

$p = 1,\ 1.75$ (or $7/4$) | A1, A1 | 8 marks total

---

# Question 10(b)(i):

**Finding equation for correlation coefficient with $p = 1$:**

**EITHER:**

$r = (65 - 13\times 23/5)\ /\ \sqrt{\{(39 - 13^2/5)(125 - 23^2/5)\}}$ | M1 A1 |

$= 5.2\ /\ \sqrt{(5.2\times 19.2)}$ or $1.04\ /\ \sqrt{(1.04\times 3.84)}$

**OR:**

$r^2 = 1\times(65 - 13\times 23/5)/(125 - 23^2/5)$

or $1\times(39 - 13^2/5)/(125 - 23^2/5)$

$= 5.2/19.2$ or $1.04/3.84$ | M1 A1 |

**Evaluate r:**

$r = \sqrt{39/12}$ or $0.52$ | A1 | 3 marks total

---

# Question 10(b)(ii):

**EITHER: Recall or find gradient $b_2$ of line:**

$b_2 = 5.2/19.2$ or $r^2\ [= 0.2708]$ | M1 |

**Find regression line of x on y:**

$x - 13/5 = b_2(y - 23/5)$

$x = (13/48)y + 65/48$

or $0.271y + 1.35$ | M1, A1 |

---

**OR: Use normal equations for $x = a_2 + b_2 y$:**

$5a_2 + 23b_2 = 13$ and $23a_2 + 125b_2 = 65$ | (M1), (M1) |

**Solve for $a_2$, $b_2$:**

$b_2 = 13/48$ or $0.271$

And $a_2 = 65/48$ or $1.35$ | (A1) | 3 marks, 14 total

Show LaTeX source

A light elastic string has modulus of elasticity $\frac { 3 } { 2 } m g$ and natural length $a$. A particle of mass $m$ is attached to one end of the string. The other end of the string is attached to a fixed point $A$. The particle is released from rest at $A$. Show that when the particle has fallen a distance $k a$ from $A$, where $k > 1$, its kinetic energy is

$$\frac { 1 } { 4 } m g a \left( 10 k - 3 - 3 k ^ { 2 } \right) .$$

Show that the particle first comes to instantaneous rest at the point $B$ which is at a distance $3 a$ vertically below $A$.

Show that the time taken by the particle to travel from $A$ to $B$ is

$$\sqrt { } \left( \frac { 2 a } { g } \right) + \frac { 2 \pi } { 3 } \sqrt { } \left( \frac { 2 a } { 3 g } \right)$$

\hfill \mbox{\textit{CAIE FP2 2013 Q10 EITHER}}

This paper (11 questions)

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Q1 8 Q2 11 Q3 12 Q4 12 Q5 4 Q6 7 Q7 9 Q8 9 Q9 14 Q10 EITHER Q10 OR

CAIE FP2 2013 June — Question 10 EITHER

Question 10(a):

Finding KE after falling ka from PE:

\(= mga(k - 3k^2/4 + 3k/2 - \frac{3}{4})\)

Finding equation for k (or ka) at B using KE = 0:

Finding both roots and selecting root > 1:

Finding time \(t_1\) to fall \(a\) from \(A\) (under no tension):

Finding ext. \(e\) or distance fallen at equilibrium point O:

State or find SHM equation at x below (or above) O:

State or use correct amplitude \(x_0\) and \(\omega^2\):

Finding time \(t_2\) to fall from \(a\) to \(3a\) below A:

Finding total time \(t_1 + t_2\):

Question 10(b):

Finding 4 summations required:

\(\Sigma x = 12 + p\), \(\Sigma x^2 = 38 + p^2\)

\(\Sigma y = 23\), \([\Sigma y^2 = 125]\)

EITHER: Substitute \(b_1 = 1\) in formula for gradient:

\(63 + 2p - (12+p)\times 23/5\)

\(= 38 + p^2 - (12+p)^2/5\)

OR: Substitute in normal equations:

Obtain and solve quadratic for p (A.E.F):

\(0.8p^2 - 2.2p + 1.4 = 0\)

Question 10(b)(i):

Finding equation for correlation coefficient with \(p = 1\):

EITHER:

\(= 5.2\ /\ \sqrt{(5.2\times 19.2)}\) or \(1.04\ /\ \sqrt{(1.04\times 3.84)}\)

OR:

\(r^2 = 1\times(65 - 13\times 23/5)/(125 - 23^2/5)\)

or \(1\times(39 - 13^2/5)/(125 - 23^2/5)\)

Evaluate r:

Question 10(b)(ii):

EITHER: Recall or find gradient \(b_2\) of line:

Find regression line of x on y:

\(x - 13/5 = b_2(y - 23/5)\)

\(x = (13/48)y + 65/48\)

OR: Use normal equations for \(x = a_2 + b_2 y\):

Solve for \(a_2\), \(b_2\):

\(b_2 = 13/48\) or \(0.271\)

This paper (11 questions)

Answer	Marks
\(m\frac{d^2x}{dt^2} = mg - 3mg(x+e)/2a\)	M1
\(\frac{d^2x}{dt^2} = -3gx/2a\)	A1

Answer	Marks
\([-]\ e = x_0\cos\omega t_2\)	M1
\(t_2 = \sqrt{(2a/3g)}\cos^{-1}(-\frac{1}{2})\) (A.E.F.)	A1

Answer	Marks	Guidance
or \(4p^2 - 11p + 7 = 0\)	M1 A1
\(p = 1,\ 1.75\) (or \(7/4\))	A1, A1	8 marks total