CAIE Further Paper 3 2022 June — Question 2 5 marks

Exam BoardCAIE
ModuleFurther Paper 3 (Further Paper 3)
Year2022
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeHorizontal elastic string on rough surface
DifficultyChallenging +1.2 This is a multi-step energy problem requiring consideration of elastic potential energy, kinetic energy, and work done against friction. While it involves several components (elastic strings, friction, energy conservation), the setup is relatively standard for Further Maths mechanics with clear given parameters and a straightforward application of energy principles. The calculation is methodical rather than requiring novel insight, placing it moderately above average difficulty.
Spec3.03r Friction: concept and vector form3.03u Static equilibrium: on rough surfaces6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle

2 A particle \(P\) of mass \(m\) is attached to one end of a light elastic string of natural length \(a\) and modulus of elasticity \(\frac { 4 } { 3 } \mathrm { mg }\). The other end of the string is attached to a fixed point \(O\) on a rough horizontal surface. The particle is at rest on the surface with the string at its natural length. The coefficient of friction between \(P\) and the surface is \(\frac { 1 } { 3 }\). The particle is projected along the surface in the direction \(O P\) with a speed of \(\frac { 1 } { 2 } \sqrt { \mathrm { ga } }\). Find the greatest extension of the string during the subsequent motion.

Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{1}{2} \cdot \frac{\frac{4}{3}mg}{a} x^2\)B1 EPE term correct
\(\frac{1}{3}mgx\)B1 Work term correct
Loss in KE = gain in EPE + work done against friction: \(\frac{1}{2}mv^2 = \frac{1}{2} \times \frac{\frac{4}{3}mg}{a} x^2 + \frac{1}{3}mgx\)M1 Energy equation with 3 terms, allow sign error
\(\frac{1}{2} \times \frac{1}{4}ga = \frac{\frac{2}{3}g}{a}x^2 + \frac{1}{3}gx\); \(16x^2 + 8ax - 3a^2 = 0\); \((4x-a)(4x+3a)=0\)M1 Obtain and attempt to solve a 3-term quadratic equation
\(x = \frac{1}{4}a\)A1
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2} \cdot \frac{\frac{4}{3}mg}{a} x^2$ | B1 | EPE term correct |
| $\frac{1}{3}mgx$ | B1 | Work term correct |
| Loss in KE = gain in EPE + work done against friction: $\frac{1}{2}mv^2 = \frac{1}{2} \times \frac{\frac{4}{3}mg}{a} x^2 + \frac{1}{3}mgx$ | M1 | Energy equation with 3 terms, allow sign error |
| $\frac{1}{2} \times \frac{1}{4}ga = \frac{\frac{2}{3}g}{a}x^2 + \frac{1}{3}gx$; $16x^2 + 8ax - 3a^2 = 0$; $(4x-a)(4x+3a)=0$ | M1 | Obtain and attempt to solve a 3-term quadratic equation |
| $x = \frac{1}{4}a$ | A1 | |

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2 A particle $P$ of mass $m$ is attached to one end of a light elastic string of natural length $a$ and modulus of elasticity $\frac { 4 } { 3 } \mathrm { mg }$. The other end of the string is attached to a fixed point $O$ on a rough horizontal surface. The particle is at rest on the surface with the string at its natural length. The coefficient of friction between $P$ and the surface is $\frac { 1 } { 3 }$. The particle is projected along the surface in the direction $O P$ with a speed of $\frac { 1 } { 2 } \sqrt { \mathrm { ga } }$.

Find the greatest extension of the string during the subsequent motion.\\

\hfill \mbox{\textit{CAIE Further Paper 3 2022 Q2 [5]}}