Challenging +1.2 This is a standard energy conservation problem in elastic strings requiring students to find equilibrium extension, then apply conservation of energy with elastic PE, gravitational PE, and KE. While it involves multiple energy terms and algebraic manipulation, it follows a well-established method taught in Further Mechanics with no novel insight required. The 6 marks reflect routine multi-step working rather than conceptual difficulty.
A light elastic string has natural length \(a\) and modulus of elasticity \(12mg\). One end of the string is attached to a fixed point \(O\). The other end of the string is attached to a particle of mass \(m\). The particle hangs in equilibrium vertically below \(O\). The particle is pulled vertically down and released from rest with the extension of the string equal to \(e\), where \(e > \frac{1}{4}a\). In the subsequent motion the particle has speed \(\sqrt{2ga}\) when it has ascended a distance \(\frac{1}{4}a\).
Find \(e\) in terms of \(a\). [6]
Question 3:
3 | 1 12mge2 1 12mg a 2 2mg( )
Loss in EPE = × − × × e− = 6e−a
2 a 2 a 3 3 | B1 | Either term correct.
1 mga
Gain in KE = mv2 and Gain in GPE =
2 3 | B1
Gain in KE + Gain in GPE = Loss in EPE | M1 | KE, GPE and at least one EPE term.
1 mv2 + mga = 2mg( 6e−a )
2 3 3 | A1 | All terms correct.
Simplify to a linear equation in e. | M1
1
e= a
2 | A1
6
Question | Answer | Marks | Guidance
A light elastic string has natural length $a$ and modulus of elasticity $12mg$. One end of the string is attached to a fixed point $O$. The other end of the string is attached to a particle of mass $m$. The particle hangs in equilibrium vertically below $O$. The particle is pulled vertically down and released from rest with the extension of the string equal to $e$, where $e > \frac{1}{4}a$. In the subsequent motion the particle has speed $\sqrt{2ga}$ when it has ascended a distance $\frac{1}{4}a$.
Find $e$ in terms of $a$. [6]
\hfill \mbox{\textit{CAIE Further Paper 3 2021 Q3 [6]}}