Pre-U Pre-U 9795/2 2018 June — Question 8 5 marks

Exam BoardPre-U
ModulePre-U 9795/2 (Pre-U Further Mathematics Paper 2)
Year2018
SessionJune
Marks5
TopicHooke's law and elastic energy
TypeVertical elastic string: released from rest at natural length or above (string initially slack)
DifficultyStandard +0.8 This is a two-part elastic energy problem requiring (i) straightforward equilibrium using Hooke's law, then (ii) energy conservation with careful consideration of when the string becomes slack. Part (ii) requires students to recognize that elastic potential energy only applies while stretched, making it moderately challenging but still a standard Further Maths mechanics question.
Spec6.02g Hooke's law: T = k*x or T = lambda*x/l6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle
8 A light elastic string of natural length 0.2 m and modulus of elasticity 8 N has one end fixed to a point \(P\) on a horizontal ceiling. A particle of mass 0.4 kg is attached to the other end of the string.
  1. Find the extension of the string when the particle hangs in equilibrium vertically below \(P\).
  2. The particle is held at rest, with the string stretched, at a point \(x \mathrm {~m}\) vertically below \(P\) and is then released. Find the smallest value of \(x\) for which the particle will reach the ceiling.
Question 8(ii)
- \(\frac{\lambda}{2l}(x-0.2)^2\) M1 using formula for EPE
- \(\frac{\lambda}{2l}(x-0.2)^2 = mgx\) M1 equating EPE and GPE
- \(20(x-0.2)^2 = 4x\) A1 correct equation
- \(25x^2 - 15x + 1 = 0\) oe A1 correct simplified quadratic
- \(x = \frac{3+\sqrt{5}}{10} = 0.524\) (or \(x \geqslant 0.524\)) A1 \([0.523, 0.524]\) or \(\frac{1}{10}(3+\sqrt{5})\) oe
Alternative 1 (from slack position):
- \(\frac{\lambda e^2}{2l}\) M1 using formula for EPE
- \(\frac{\lambda e^2}{2l} = mg(e+0.2)\) M1 equating EPE and GPE
- \(20e^2 = 4(e+0.2)\) A1 correct equation
- \(25e^2 - 5e - 1 = 0\) oe A1 correct simplified quadratic
- \(x = 0.2 + \frac{1\pm\sqrt{5}}{10} = 0.524\) (or \(x \geqslant 0.524\)) A1 \([0.523, 0.524]\) or \(\frac{1}{10}(3+\sqrt{5})\) oe
Alternative 2 (from eqm position):
- \(\frac{\lambda}{2l}(h+0.1)^2\) M1 using formula for EPE
- \(\frac{1}{2}\frac{\lambda}{l}(h+0.1)^2 = mg(h+0.3)\) M1 equating EPE and GPE
- \(20(h+0.1)^2 = 4(h+0.3)\) A1 correct equation
- \(20h^2 - 1 = 0\) oe A1 correct simplified quadratic
- \(x = 0.3 + \frac{1}{\sqrt{20}} = 0.524\) (or \(x \geqslant 0.524\)) A1 \([0.523, 0.524]\) or \(\frac{1}{10}(3+\sqrt{5})\) oe
Alternative 3 (using SHM equation):
- \(0.4g - \frac{8(x-0.2)}{0.2} = 0.4\ddot{x}\) leading to M1 deriving SHM formula in recognisable form
- \(\ddot{x} + 100x = 30\); \(0.4\times10\times0.2 = \frac{1}{2}\times0.4\times v^2 \Rightarrow v = 2\) M1 Dep on previous M. Energy consideration leading to \(v\) as string goes slack
- \(x = A\cos10t + B\sin10t + 0.3\) or \(x = R\cos(10t+\phi) + 0.3\) A1 full correct solution with \(\omega = 10\) and 2 arbitrary constants
- \(R\cos\phi = -0.1\) and \(10R\cos\phi = 2\) and attempt to solve simultaneously M1 using \(t=0\), \(x=0.2\) and \(v=2\) in correct equations to find \(A\) and \(B\) or \(R\) and \(\psi\)
- \(R = \sqrt{A^2+B^2} = \frac{\sqrt{5}}{10} = 0.224\) so \(x = 0.524\) A1 cao
Alternative 4 (using SHM energy equation):
- \(0.4g - \frac{8(x-0.2)}{0.2} = 0.4\ddot{x}\) leading to M1 deriving SHM formula in recognisable form
- \(\ddot{x}+100x = 30\); \(0.4\times10\times0.2 = \frac{1}{2}\times0.4\times v^2 \Rightarrow v=2\) M1 Dep on previous M. Energy consideration leading to \(v\) as string goes slack
- \(\omega = 10 \Rightarrow v^2 + 100(x-0.3)^2 = 100R^2\) A1 value of \(\omega\) soi and fully correct energy equation with 3 terms and an arbitrary constant
- When \(x = 0.2\), \(v = 2\) so \(100R^2 = 5\) M1 using \(x=0.2\) and \(v=2\) in correct equation to find \(R\)
- \(\therefore x = \frac{\sqrt{5}}{10} + 0.3 = 0.524\) A1 cao
**Question 8(ii)**

- $\frac{\lambda}{2l}(x-0.2)^2$ **M1** using formula for EPE

- $\frac{\lambda}{2l}(x-0.2)^2 = mgx$ **M1** equating EPE and GPE

- $20(x-0.2)^2 = 4x$ **A1** correct equation

- $25x^2 - 15x + 1 = 0$ oe **A1** correct simplified quadratic

- $x = \frac{3+\sqrt{5}}{10} = 0.524$ (or $x \geqslant 0.524$) **A1** $[0.523, 0.524]$ or $\frac{1}{10}(3+\sqrt{5})$ oe

**Alternative 1 (from slack position):**

- $\frac{\lambda e^2}{2l}$ **M1** using formula for EPE

- $\frac{\lambda e^2}{2l} = mg(e+0.2)$ **M1** equating EPE and GPE

- $20e^2 = 4(e+0.2)$ **A1** correct equation

- $25e^2 - 5e - 1 = 0$ oe **A1** correct simplified quadratic

- $x = 0.2 + \frac{1\pm\sqrt{5}}{10} = 0.524$ (or $x \geqslant 0.524$) **A1** $[0.523, 0.524]$ or $\frac{1}{10}(3+\sqrt{5})$ oe

**Alternative 2 (from eqm position):**

- $\frac{\lambda}{2l}(h+0.1)^2$ **M1** using formula for EPE

- $\frac{1}{2}\frac{\lambda}{l}(h+0.1)^2 = mg(h+0.3)$ **M1** equating EPE and GPE

- $20(h+0.1)^2 = 4(h+0.3)$ **A1** correct equation

- $20h^2 - 1 = 0$ oe **A1** correct simplified quadratic

- $x = 0.3 + \frac{1}{\sqrt{20}} = 0.524$ (or $x \geqslant 0.524$) **A1** $[0.523, 0.524]$ or $\frac{1}{10}(3+\sqrt{5})$ oe

**Alternative 3 (using SHM equation):**

- $0.4g - \frac{8(x-0.2)}{0.2} = 0.4\ddot{x}$ leading to **M1** deriving SHM formula in recognisable form

- $\ddot{x} + 100x = 30$; $0.4\times10\times0.2 = \frac{1}{2}\times0.4\times v^2 \Rightarrow v = 2$ **M1** Dep on previous M. Energy consideration leading to $v$ as string goes slack

- $x = A\cos10t + B\sin10t + 0.3$ or $x = R\cos(10t+\phi) + 0.3$ **A1** full correct solution with $\omega = 10$ and 2 arbitrary constants

- $R\cos\phi = -0.1$ and $10R\cos\phi = 2$ and attempt to solve simultaneously **M1** using $t=0$, $x=0.2$ and $v=2$ in correct equations to find $A$ and $B$ or $R$ and $\psi$

- $R = \sqrt{A^2+B^2} = \frac{\sqrt{5}}{10} = 0.224$ so $x = 0.524$ **A1** cao

**Alternative 4 (using SHM energy equation):**

- $0.4g - \frac{8(x-0.2)}{0.2} = 0.4\ddot{x}$ leading to **M1** deriving SHM formula in recognisable form

- $\ddot{x}+100x = 30$; $0.4\times10\times0.2 = \frac{1}{2}\times0.4\times v^2 \Rightarrow v=2$ **M1** Dep on previous M. Energy consideration leading to $v$ as string goes slack

- $\omega = 10 \Rightarrow v^2 + 100(x-0.3)^2 = 100R^2$ **A1** value of $\omega$ soi and fully correct energy equation with 3 terms and an arbitrary constant

- When $x = 0.2$, $v = 2$ so $100R^2 = 5$ **M1** using $x=0.2$ and $v=2$ in correct equation to find $R$

- $\therefore x = \frac{\sqrt{5}}{10} + 0.3 = 0.524$ **A1** cao
8 A light elastic string of natural length 0.2 m and modulus of elasticity 8 N has one end fixed to a point $P$ on a horizontal ceiling. A particle of mass 0.4 kg is attached to the other end of the string.\\
(i) Find the extension of the string when the particle hangs in equilibrium vertically below $P$.\\
(ii) The particle is held at rest, with the string stretched, at a point $x \mathrm {~m}$ vertically below $P$ and is then released. Find the smallest value of $x$ for which the particle will reach the ceiling.

\hfill \mbox{\textit{Pre-U Pre-U 9795/2 2018 Q8 [5]}}
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