CAIE M2 2005 June — Question 1 4 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2005
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicHooke's law and elastic energy
TypeParticle at midpoint of string between vertical fixed points
DifficultyStandard +0.3 This is a straightforward elastic string equilibrium problem requiring application of Hooke's law to two string segments and resolving forces vertically. While it involves multiple steps (finding extensions, applying Hooke's law twice, and resolving forces), the method is standard and well-practiced in M2. The setup is clear with given numerical values, making it slightly easier than average A-level mechanics questions.
Spec3.03b Newton's first law: equilibrium6.02h Elastic PE: 1/2 k x^2

1 \includegraphics[max width=\textwidth, alt={}, center]{6fe2c5e0-0496-4fb4-95d2-354b90607b5b-2_643_218_264_959} A particle \(P\) of mass \(m \mathrm {~kg}\) is attached to the mid-point of a light elastic string of natural length 0.8 m and modulus of elasticity 8 N . One end of the string is attached to a fixed point \(A\) and the other end is attached to a fixed point \(B\) which is 2 m vertically below \(A\). When the particle is in equilibrium the distance \(A P\) is 1.1 m (see diagram). Find the value of \(m\).

Question 1:
AnswerMarks Guidance
Answer/WorkingMark Guidance
\(T_A = 8 \times 0.7 \div 0.4\) or \(T_B = 8 \times 0.5 \div 0.4\)B1, M1 For resolving forces on P vertically (3 terms needed)
\(8 \times 0.7 \div 0.4 = 8 \times 0.5 \div 0.4 + 10m\)A1 Correct unsimplified equation
\(m = 0.4\)A1 [4 marks]
## Question 1:

| Answer/Working | Mark | Guidance |
|---|---|---|
| $T_A = 8 \times 0.7 \div 0.4$ or $T_B = 8 \times 0.5 \div 0.4$ | B1, M1 | For resolving forces on P vertically (3 terms needed) |
| $8 \times 0.7 \div 0.4 = 8 \times 0.5 \div 0.4 + 10m$ | A1 | Correct unsimplified equation |
| $m = 0.4$ | A1 | **[4 marks]** |

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\includegraphics[max width=\textwidth, alt={}, center]{6fe2c5e0-0496-4fb4-95d2-354b90607b5b-2_643_218_264_959}

A particle $P$ of mass $m \mathrm {~kg}$ is attached to the mid-point of a light elastic string of natural length 0.8 m and modulus of elasticity 8 N . One end of the string is attached to a fixed point $A$ and the other end is attached to a fixed point $B$ which is 2 m vertically below $A$. When the particle is in equilibrium the distance $A P$ is 1.1 m (see diagram). Find the value of $m$.

\hfill \mbox{\textit{CAIE M2 2005 Q1 [4]}}