CAIE M2 2019 November — Question 1 3 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2019
SessionNovember
Marks3
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeHorizontal elastic string on smooth surface
DifficultyStandard +0.3 This is a standard energy conservation problem with elastic strings requiring students to equate initial kinetic energy to elastic potential energy at maximum extension. The calculation is straightforward with given values, involving one main equation (½mv² = ½λx²/l) and basic algebra to solve for x, making it slightly easier than average.
Spec6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle

1 A particle of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus of elasticity 9 N . The other end of the string is attached to a fixed point \(O\) on a smooth horizontal surface. The particle is projected horizontally from \(O\) with speed \(4 \mathrm {~m} \mathrm {~s} ^ { - 1 }\). Find the greatest distance of the particle from \(O\).

Question 1:
AnswerMarks Guidance
AnswerMarks Guidance
\(\frac{0.3 \times 4^2}{2} = \frac{9e^2}{2 \times 0.6}\)M1 Set up an energy equation. Note the final velocity is zero.
\(e = 0.566\) or \(\frac{2\sqrt{2}}{5}\)A1
Distance \(= 1.17\) mA1
## Question 1:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{0.3 \times 4^2}{2} = \frac{9e^2}{2 \times 0.6}$ | M1 | Set up an energy equation. Note the final velocity is zero. |
| $e = 0.566$ or $\frac{2\sqrt{2}}{5}$ | A1 | |
| Distance $= 1.17$ m | A1 | |

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1 A particle of mass 0.3 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus of elasticity 9 N . The other end of the string is attached to a fixed point $O$ on a smooth horizontal surface. The particle is projected horizontally from $O$ with speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Find the greatest distance of the particle from $O$.\\

\hfill \mbox{\textit{CAIE M2 2019 Q1 [3]}}