CAIE M2 2019 November — Question 3 6 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2019
SessionNovember
Marks6
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TopicHooke's law and elastic energy
TypeVertical elastic string: projected vertically or mid-motion analysis
DifficultyStandard +0.8 This problem requires setting up and solving simultaneous equations using both Newton's second law (F=ma with elastic force) and energy conservation (KE + EPE + GPE). Students must handle the case where the string is extended, apply Hooke's law correctly, and work with the modulus of elasticity. The multi-step nature and need to coordinate two different approaches (dynamics and energy) for two unknowns makes this moderately challenging but still within standard M2 scope.
Spec3.03d Newton's second law: 2D vectors3.03f Weight: W=mg6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle
3 A particle \(P\) of mass 0.5 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus of elasticity 12 N . The other end of the string is attached to a fixed point \(O\). The particle \(P\) is projected vertically downwards with speed \(2 \mathrm {~m} \mathrm {~s} ^ { - 1 }\) from the point 0.5 m vertically below \(O\). For an instant when the acceleration of \(P\) is \(4 \mathrm {~m} \mathrm {~s} ^ { - 2 }\) downwards, find the extension of the string and the speed of \(P\).
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
\(0.5 \times 4 = 0.5g - T\)M1 Use Newton's Second Law vertically
\(T = \frac{12e}{0.6}\)M1 Use \(T = \frac{\lambda x}{l}\)
\(e\left(= \frac{3 \times 0.6}{12}\right) = 0.15\)A1
\(\text{EPE} = \frac{12 \times 0.15^2}{2 \times 0.6}\) and distance fallen \(= 0.6 - 0.5 + 0.15\)B1ft
\(\frac{0.5v^2}{2} = \frac{0.5 \times 2^2}{2} + 0.5g(0.6 - 0.5 + 0.15) - \frac{12 \times 0.15^2}{2 \times 0.6}\)M1 Set up a 4 term energy equation
\(v = 2.85\) m s\(^{-1}\)A1 
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $0.5 \times 4 = 0.5g - T$ | M1 | Use Newton's Second Law vertically |
| $T = \frac{12e}{0.6}$ | M1 | Use $T = \frac{\lambda x}{l}$ |
| $e\left(= \frac{3 \times 0.6}{12}\right) = 0.15$ | A1 | |
| $\text{EPE} = \frac{12 \times 0.15^2}{2 \times 0.6}$ and distance fallen $= 0.6 - 0.5 + 0.15$ | B1ft | |
| $\frac{0.5v^2}{2} = \frac{0.5 \times 2^2}{2} + 0.5g(0.6 - 0.5 + 0.15) - \frac{12 \times 0.15^2}{2 \times 0.6}$ | M1 | Set up a 4 term energy equation |
| $v = 2.85$ m s$^{-1}$ | A1 | |

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3 A particle $P$ of mass 0.5 kg is attached to one end of a light elastic string of natural length 0.6 m and modulus of elasticity 12 N . The other end of the string is attached to a fixed point $O$. The particle $P$ is projected vertically downwards with speed $2 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ from the point 0.5 m vertically below $O$. For an instant when the acceleration of $P$ is $4 \mathrm {~m} \mathrm {~s} ^ { - 2 }$ downwards, find the extension of the string and the speed of $P$.\\

\hfill \mbox{\textit{CAIE M2 2019 Q3 [6]}}
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